objectives and methods of analysis and design

Module 1

Objectives and Methods of Analysis and Design,

and Properties of Concrete and Steel

Version 2 CE IIT, Kharagpur

Lesson 1

Objectives and Methods of Analysis and Design


Instructional Objectives: At the end of this lesson, the student should be able to: • state the four objectives of the design of reinforced concrete structures, • name the three methods of design of concrete structure and identify the best

method of design, • state the basis of the analysis of the structure, • express the design loads in terms of characteristic loads in limit state and

working stress methods, • define the characteristic load, • name the different loads, forces and effects to be considered in the design, • state the basis of determining the combination of different loads acting on the

structure 1.1.1 Introduction

Reinforced concrete, as a composite material, has occupied a special place in the modern construction of different types of structures due to its several advantages. Italian architect Ponti once remarked that concrete liberated us from the rectangle. Due to its flexibility in form and superiority in performance, it has replaced, to a large extent, the earlier materials like stone, timber and steel. Further, architect's scope and imaginations have widened to a great extent due to its mouldability and monolithicity. Thus, it has helped the architects and engineers to build several attractive shell forms and other curved structures. However, its role in several straight line structural forms like multistoried frames, bridges, foundations etc. is enormous. The design of these modern reinforced concrete structures may appear to be highly complex. However, most of these structures are the assembly of several basic structural elements such as beams, columns, slabs, walls and foundations (Anim. 1.1.1). Accordingly, the designer has to learn the design of these basic reinforced concrete elements. The joints and connections are then carefully developed.

Anim. 1.1.1

Design of reinforced concrete structures started in the beginning of last century following purely empirical approach. Thereafter came the so called rigorous elastic theory where the levels of stresses in concrete and steel are limited so that stress-deformations are taken to be linear. However, the limit state method, though semi-empirical approach, has been found to be the best for the design of reinforced concrete structures (see sec. 1.1.3.1 also). The constraints and applicabilities of both the methods will be discussed later.


1.1.2 Objectives of the Design of Reinforced Concrete Structures

Every structure has got its form, function and aesthetics. Normally, we consider that the architects will take care of them and the structural engineers will be solely responsible for the strength and safety of the structure. However, the roles of architects and structural engineers are very much interactive and a unified approach of both will only result in an "Integrated" structure, where every material of the total structure takes part effectively for form, function, aesthetics, strength as well as safety and durability. This is possible when architects have some basic understanding of structural design and the structural engineers also have the basic knowledge of architectural requirements.

Both the engineer and the architect should realize that the skeletal structure without architecture is barren and mere architecture without the structural strength and safety is disastrous. Safety, here, includes consideration of reserve strength, limited deformation and durability. However, some basic knowledge of architectural and structural requirements would facilitate to appreciate the possibilities and limitations of exploiting the reinforced concrete material for the design of innovative structures.

Before proceeding to the design, one should know the objectives of the

design of concrete structures. The objectives of the design are as follows: 1.1.2.1 The structures so designed should have an acceptable probability of performing satisfactorily during their intended life.

This objective does not include a guarantee that every structure must perform satisfactorily during its intended life. There are uncertainties in the design process both in the estimation of the loads likely to be applied on the structure and in the strength of the material. Moreover, full guarantee would only involve more cost. Thus, there is an acceptable probability of performance of structures as given in standard codes of practices of different countries.

1.1.2.2 The designed structure should sustain all loads and deform within limits for construction and use.

Anim. 1.1.2 Anim. 1.1.3 Anim. 1.1.4

Adequate strengths and limited deformations are the two requirements of the designed structure. The structure should have sufficient strength and the deformations must be within prescribed limits due to all loads during construction and use as seen in Anim. 1.1.2. Animation 1.1.3 shows the structure having insufficient strength of concrete which fails in bending compression with the increase of load, though the deformation of the structure is not alarming. On the


other hand, Anim. 1.1.4 shows another situation where the structure, having sufficient strength, deforms excessively. Both are undesirable during normal construction and use.

However, sometimes structures are heavily loaded beyond control. The

structural engineer is not responsible to ensure the strength and deformation within limit under such situation. The staircases in residential buildings during festival like marriage etc., roof of the structures during flood in the adjoining area or for buildings near some stadium during cricket or football matches are some of the examples when structures get overloaded. Though, the structural designer is not responsible for the strength and deformations under these situations, he, however, has to ensure that the failure of the structures should give sufficient time for the occupants to vacate. The structures, thus, should give sufficient warning to the occupants and must not fail suddenly. 1.1.2.3 The designed structures should be durable.

The materials of reinforced concrete structures get affected by the environmental conditions. Thus, structures having sufficient strength and permissible deformations may have lower strength and exhibit excessive deformations in the long run. The designed structures, therefore, must be checked for durability. Separate checks for durability are needed for the steel reinforcement and concrete. This will avoid problems of frequent repairing of the structure. 1.1.2.4 The designed structures should adequately resist to the effects of misuse and fire.

Structures may be misused to prepare fire works, store fire works, gas and other highly inflammable and/or explosive chemicals. Fire may also take place as accidents or as secondary effects during earthquake by overturning kerosene stoves or lantern, electrical short circuiting etc. Properly designed structures should allow sufficient time and safe route for the persons inside to vacate the structures before they actually collapse. How to fulfill the objectives? All the above objectives can be fulfilled by understanding the strength and deformation characteristics of the materials used in the design as also their deterioration under hostile exposure. Out of the two basic materials concrete and steel, the steel is produced in industries. Further, it is available in form of standard bars and rods of specific diameters. However, sample testing and checking are important to ensure the quality of these steel bars or rods. The concrete, on the other hand, is prepared from several materials (cement, sand,


coarse aggregate, water and admixtures, if any). Therefore, it is important to know the characteristic properties of each of the materials used to prepare concrete. These materials and the concrete after its preparation are also to be tested and checked to ensure the quality. The necessary information regarding the properties and characteristic strength of these materials are available in the standard codes of practices of different countries. It is necessary to follow these clearly defined standards for materials, production, workmanship and maintenance, and the performance of structures in service. 1.1.3 Method of Design

Three methods of design are accepted in cl. 18.2 of IS 456:2000 (Indian Standard Plain and Reinforced Concrete - Code of Practice, published by the Bureau of Indian Standards, New Delhi). They are as follows: 1.1.3.1 Limit state method

The term “Limit states” is of continental origin where there are three limit states - serviceability / crack opening / collapse. For reasons not very clear, in English literature limit state of collapse is termed as limit state.

As mentioned in sec. 1.1.1, the semi-empirical limit state method of design has been found to be the best for the design of reinforced concrete members. More details of this method are explained in Module 3 (Lesson 4). However, because of its superiority to other two methods (see sections 2.3.2 and 2.3.3 of Lesson 3), IS 456:2000 has been thoroughly updated in its fourth revision in 2000 taking into consideration the rapid development in the field of concrete technology and incorporating important aspects like durability etc. This standard has put greater emphasis to limit state method of design by presenting it in a full section (section 5), while the working stress method has been given in Annex B of the same standard. Accordingly, structures or structural elements shall normally be designed by limit state method. 1.1.3.2 Working stress method

This method of design, considered as the method of earlier times, has several limitations. However, in situations where limit state method cannot be conveniently applied, working stress method can be employed as an alternative. It is expected that in the near future the working stress method will be completely replaced by the limit state method. Presently, this method is put in Annex B of IS 456:2000.


1.1.3.3 Method based on experimental approach

The designer may perform experimental investigations on models or full size structures or elements and accordingly design the structures or elements. However, the four objectives of the structural design (sec. 1.1.2) must be satisfied when designed by employing this approach. Moreover, the engineer-in-charge has to approve the experimental details and the analysis connected therewith. Though the choice of the method of design is still left to the designer as per cl. 18.2 of IS 456:2000, the superiority of the limit state method is evident from the emphasis given to this method by presenting it in a full section (Section 5), while accommodating the working stress method in Annex B of IS 456:2000, from its earlier place of section 6 in IS 456:1978. It is expected that a gradual change over to the limit state method of design will take place in the near future after overcoming the inconveniences of adopting this method in some situations. 1.1.4 Analysis of Structures

Structures when subjected to external loads (actions) have internal reactions in the form of bending moment, shear force, axial thrust and torsion in individual members. As a result, the structures develop internal stresses and undergo deformations. Essentially, we analyse a structure elastically replacing each member by a line (with EI values) and then design the section using concepts of limit state of collapse. Figure 1.1.1 explains the internal and external reactions of a simply supported beam under external loads. The external loads to be applied on the structures are the design loads and the analyses of structures are based on linear elastic theory (vide cl. 22 of IS 456:2000). 1.1.5 Design Loads

The design loads are determined separately for the two methods of design as mentioned below after determining the combination of different loads. 1.1.5.1 In the limit state method, the design load is the characteristic load with appropriate partial safety factor (vide sec. 2.3.2.3 for partial safety factors).


1.1.5.2 In the working stress method, the design load is the characteristic load only. What is meant by characteristic load? Characteristic load (cl. 36.2 of IS 456:2000) is that load which has a ninety-five per cent probability of not being exceeded during the life of the structure.


The various loads acting on structures consist of dead loads, live loads, wind or earthquake loads etc. These are discussed in sec. 1.1.6. However, the researches made so far fail to estimate the actual loads on the structure. Accordingly, the loads are predicted based on statistical approach, where it is assumed that the variation of the loads acting on structures follows the normal distribution (Fig. 1.1.2). Characteristic load should be more than the average/mean load. Accordingly,

Characteristic load = Average/mean load + K (standard deviation for load) The value of K is assumed such that the actual load does not exceed the characteristic load during the life of the structure in 95 per cent of the cases. 1.1.6 Loads and Forces

The following are the different types of loads and forces acting on the structure. As mentioned in sec. 1.1.5, their values have been assumed based on earlier data and experiences. It is worth mentioning that their assumed values as stipulated in IS 875 have been used successfully. 1.1.6.1 Dead loads

Anims. 1.1.5 a and


b

These are the self weight of the structure to be designed (see Anim. 1.1.5a). Needless to mention that the dimensions of the cross section are to be assumed initially which enable to estimate the dead loads from the known unit weights of the materials of the structure. The accuracy of the estimation thus depends on the assumed values of the initial dimensions of the cross section. The values of unit weights of the materials are specified in Part 1 of IS 875. 1.1.6.2 Imposed loads

They are also known as live loads (Anim. 1.1.5a) and consist of all loads other than the dead loads of the structure. The values of the imposed loads depend on the functional requirement of the structure. Residential buildings will have comparatively lower values of the imposed loads than those of school or office buildings. The standard values are stipulated in Part 2 of IS 875. 1.1.6.3 Wind loads

These loads (Anim. 1.1.5a) depend on the velocity of the wind at the location of the structure, permeability of the structure, height of the structure etc. They may be horizontal or inclined forces depending on the angle of inclination of the roof for pitched roof structures. They can even be suction type of forces depending on the angle of inclination of the roof or geometry of the buildings (Anim. 1.1.5b). Wind loads are specified in Part 3 of IS 875. 1.1.6.4 Snow loads

These are important loads for structures located in areas having snow fall, which gets accumulated in different parts of the structure depending on projections, height, slope etc. of the structure (Anim. 1.1.6). The standard values of snow loads are specified in Part 4 of IS 875.

Anim. 1.1.6

1.1.6.5 Earthquake forces

Anim. 1.1.7

Earthquake generates waves which move from the origin of its location (epicenter) with velocities depending on the intensity and magnitude of the earthquake. The impact of earthquake on structures depends on the stiffness of the structure, stiffness of the soil media, height and location of the structure etc. (Anim. 1.1.7). Accordingly, the country has been divided into several zones depending on the magnitude of the earthquake. The earthquake forces are


prescribed in IS 1893. Designers have adopted equivalent static load approach or spectral method. 1.1.6.6 Shrinkage, creep and temperature effects

Anim. 1.1.8, 9 and 10

Shrinkage, creep and temperature (high or low) may produce stresses and

cause deformations like other loads and forces (Anim. 1.1.8, 9 and 10). Hence, these are also considered as loads which are time dependent. The safety and serviceability of structures are to be checked following the stipulations of cls. 6.2.4, 5 and 6 of IS 456:2000 and Part 5 of IS 875. 1.1.6.7 Other forces and effects

It is difficult to prepare an exhaustive list of loads, forces and effects

coming onto the structures and affecting the safety and serviceability of them. However, IS 456:2000 stipulates the following forces and effects to be taken into account in case they are liable to affect materially the safety and serviceability of the structures. The relevant codes as mentioned therein are also indicated below:

• Foundation movement (IS 1904) (Fig. 1.1.3) • Elastic axial shortening


• Soil and fluid pressures (vide IS 875 - Part 5) • Vibration • Fatigue • Impact (vide IS 875 - Part 5) • Erection loads (Please refer to IS 875 - Part 2) (Fig. 1.1.4) • Stress concentration effect due to point of application of load and the

like. 1.1.6.8 Combination of loads

Design of structures would have become highly expensive in order to maintain their serviceability and safety if all types of forces would have acted on all structures at all times. Accordingly, the concept of characteristic loads has been accepted to ensure that in at least 95 per cent of the cases, the characteristic loads considered will be higher than the actual loads on the structure. However, the characteristic loads are to be calculated on the basis of average/mean load of some logical combinations of all the loads mentioned in sec. 1.1.6.1 to 7. These logical combinations are based on (i) the natural phenomena like wind and earthquake do not occur simultaneously, (ii) live loads on roof should not be present when wind loads are considered; to name a few. IS 875 Part 5 stipulates the combination of loads to be considered in the design of structures. 1.1.7 Practice Questions and Problems with Answers Q.1: Show two reasons why concrete is superior to stone, timber and steel? A.1: (i) Stone, timber and steel cannot be fitted to any mould, but concrete

during its green stage can fit to any mould. (ii) Structures made of stone, timber and steel have several joints, but

different elements of concrete structures can be cast monolithically. Q.2: Define integrated structure. A.2: Integrated structure is one where each part of the structure satisfies both

the structural requirements of strength and serviceability and architectural requirements of aesthetics and functionality.

Q.3: State four objectives of the design of reinforced concrete structure. A.3: Properly designed reinforced concrete structures should:

(i) have acceptable probability of performing satisfactorily during their intended life,


(ii) sustain all loads with limited deformations during construction and

use,

(iii) be durable,

(iv) adequately resist the effects of misuse and fire. Q.4: How to fulfil the four objectives of the design of reinforced concrete structures? A.4: The four objectives can be fulfilled by:

(i) understanding the strength and deformation characteristics of concrete and steel,

(ii) following the clearly defined standards for materials, production,

workmanship and maintenance, and use of structures in service,

(iii) adopting measures needed for durability. Q.5: What are the three methods of design of reinforced concrete structural elements? A.5: The three methods are:

(i) limit state method, (ii) working stress method, (iii) method based on experimental approach

Q.6: Which of the three methods is the best? A.6: Limit state method is the best of the three methods when clearly applicable. Q.7: What is the basis of the analysis of structures to be designed? A.7: The basis of the analysis is the employment of linear elastic theory. Q.8: How to estimate the design loads in (i) limit state method, and (ii) working

stress method? A.8: (i) In limit state method,


Design loads = Characteristic loads multiplied by the partial safety factor for loads

(ii) In working stress method, Design loads = Characteristic loads Q.9: Define characteristic load. A.9: Characteristic load is that load which has a ninety-five per cent probability

of not being exceeded during the life of the structure. Q.10: What are the main (i) loads, (ii) forces and (iii) effects to be considered

while designing the structures? A.10: (i) The main loads are:

(a) Dead loads (b) Imposed loads or live loads (c) Wind loads (d) Snow loads (e) Erection loads

(ii) The main force is:

(a) Earthquake force

(iii) The main effects are: (a) Shrinkage, creep and temperature effects (b) Foundation movements (c) Elastic axial shortening (d) Soil and fluid pressures (e) Vibration (f) Fatigue (g) Impact (h) Stress concentration effects due to application of point loads

Q.11: What are the basis of combining different loads for the design? A.11: Natural phenomenon and common sense are the basis of selecting the

combination different loads acting on the structure while designing.


1.1.8 References

1. Reinforced Concrete Limit State Design, 6th Edition, by Ashok K. Jain, Nem Chand & Bros, Roorkee, 2002.

2. Limit State Design of Reinforced Concrete, 2nd Edition, by P.C.Varghese, Prentice-Hall of India Pvt. Ltd., New Delhi, 2002.

3. Advanced Reinforced Concrete Design, by P.C.Varghese, Prentice-Hall of India Pvt. Ltd., New Delhi, 2001.

4. Reinforced Concrete Design, 2nd Edition, by S.Unnikrishna Pillai and Devdas Menon, Tata McGraw-Hill Publishing Company Limited, New Delhi, 2003.

5. Limit State Design of Reinforced Concrete Structures, by P.Dayaratnam, Oxford & I.B.H. Publishing Company Pvt. Ltd., New Delhi, 2004.

6. Reinforced Concrete Design, 1st Revised Edition, by S.N.Sinha, Tata McGraw-Hill Publishing Company. New Delhi, 1990.

7. Reinforced Concrete, 6th Edition, by S.K.Mallick and A.P.Gupta, Oxford & IBH Publishing Co. Pvt. Ltd. New Delhi, 1996.

8. Behaviour, Analysis & Design of Reinforced Concrete Structural Elements, by I.C.Syal and R.K.Ummat, A.H.Wheeler & Co. Ltd., Allahabad, 1989.

9. Reinforced Concrete Structures, 3rd Edition, by I.C.Syal and A.K.Goel, A.H.Wheeler & Co. Ltd., Allahabad, 1992.

10. Textbook of R.C.C, by G.S.Birdie and J.S.Birdie, Wiley Eastern Limited, New Delhi, 1993.

11. Design of Concrete Structures, 13th Edition, by Arthur H. Nilson, David Darwin and Charles W. Dolan, Tata McGraw-Hill Publishing Company Limited, New Delhi, 2004.

12. Concrete Technology, by A.M.Neville and J.J.Brooks, ELBS with Longman, 1994.

13. Properties of Concrete, 4th Edition, 1st Indian reprint, by A.M.Neville, Longman, 2000.

14. Reinforced Concrete Designer’s Handbook, 10th Edition, by C.E.Reynolds and J.C.Steedman, E & FN SPON, London, 1997.

15. Indian Standard Plain and Reinforced Concrete – Code of Practice (4th Revision), IS 456: 2000, BIS, New Delhi.

16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 1.1.9 Test 1 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Tick the correct answer: (4 x 7 = 28 marks) (i) Properly designed concrete structures should


(a) sustain all loads likely to come during next 50 years (b) sustain all loads and deformations without collapse or any damage (c) sustain all loads with limited deformations during construction and use (d) sustain characteristic loads and deformations during the next 50 years A.TQ.1: (i): (c) (ii) In the limit state method, the design load is (a) the characteristic load (b) the ultimate load (c) the characteristic load divided by the partial safety factor for loads (d) the characteristic load multiplied by the partial safety for loads A.TQ.1: (ii): (d) (iii) In the limit state method, the basis of the analysis of structure is (a) linear elastic theory (b) non-linear theory (c) plastic method of analysis (d) involving fracture mechanics A.TQ.1: (iii): (a) (iv) The characteristic load is (a) the load at first crack (b) that load which has a probability of ninety-five per cent of not being exceeded

during the life of the structure (c) the ultimate collapse load (d) the ultimate collapse load multiplied by the partial safety factor for loads


A.TQ.1: (iv): (b) TQ.2: Name the main (i) loads, (ii) forces and (iii) effects to be considered while

designing the structures. (5 + 1 + 8 = 14 marks)

(5 marks for five loads, 1 mark for one force and 8 marks for eight effects) A.TQ.2: See A.10 of sec. 1.1.7. TQ.3: State the basis of determining the combination of different loads acting on

the structure. (8 marks)

A.TQ.3: See A.11 of sec. 1.1.7. 1.1.10 Summary of this Lesson This lesson explains the four objectives of the design of reinforced concrete structures. It also mentions the three methods of design and states about the superiority of the limit state method. The basis of the analysis of structure is discussed. The terminologies like design loads, characteristic loads etc. are defined and explained. The different loads, forces and effects to be considered have been listed and discussed. Finally, the basis of the combination of different loads to be considered is also explained.


Module 1

Objectives and Methods of Analysis and Design,

and Properties of Concrete and Steel


Lesson 2

Properties of Concrete and Steel


Instructional Objectives: At the end of this lesson, the student should be able to: • know the properties of concrete in respect of strength, deformation and

durability • know the properties of steel used as reinforcement in concrete structures • understand the importance of quality control, inspection and testing of

concrete and steel in several steps from its basic preparation to the removal of formwork after the construction

• recommend the acceptance of good concrete based on sample test of

specimens, core tests, load test and non-destructive tests 1.2.1 Introduction

It is essential that the designer has to acquire a fair knowledge of the

materials to be used in the design of reinforced concrete structure. This lesson summarises the characteristic properties of concrete and steel, the two basic materials used for the design. This summary, though not exhaustive, provides the minimum information needed for the design. 1.2.2 Properties of Concrete

Plain concrete is prepared by mixing cement, sand (also known as fine

aggregate), gravel (also known as coarse aggregate) and water with specific proportions. Mineral admixtures may also be added to improve certain properties of concrete. Thus, the properties of concrete regarding its strength and deformations depend on the individual properties of cement, sand, gravel, water and admixtures. Clauses 5 and 6 of IS 456:2000 stipulate the standards and requirements of the individual material and concrete, respectively. Plain concrete after preparation and placement needs curing to attain strength. However, plain concrete is very good in compression but weak in tension. That is why steel is used as reinforcing material to make the composite sustainable in tension also. Plain concrete, thus when reinforced with steel bars in appropriate locations is known as reinforced concrete.

The strength and deformation characteristics of concrete thus depend on

the grade and type of cement, aggregates, admixtures, environmental conditions and curing. The increase of strength with its age during curing is considered to be marginal after 28 days. Blended cements (like fly ash cement) have slower rate of strength gain than ordinary Portland cement as recognized by code,


Depending on several factors during its preparation, placement and curing, concrete has a wide range of compressive strength and the material is graded on the basis of its compressive strength on 28th day also known as "characteristic strength" as defined below while discussing various strength and deformation properties. (a) Characteristic strength property

Characteristic strength is defined as the strength below which not more than five per cent of the test results are expected to fall. Concrete is graded on the basis of its characteristic compressive strength of 150 mm size cube at 28 days and expressed in N/mm2. The grades are designated by one letter M (for mix) and a number from 10 to 80 indicating the characteristic compressive strength (fck) in N/mm2. As per IS 456 (Table 2), concrete has three groups as (i) ordinary concrete (M 10 to M 20), (ii) standard concrete (M 25 to M 55) and (iii) high strength concrete (M 60 to M 80). The size of specimen for determining characteristic strength may be different in different countries. (b) Other strengths of concrete

In addition to its good compressive strength, concrete has flexural and

splitting tensile strengths too. The flexural and splitting tensile strengths are obtained as described in IS 516 and IS 5816, respectively. However, the following expression gives an estimation of flexural strength (fcr) of concrete from its characteristic compressive strength (cl. 6.2.2)

2N/mmin 70 kcrc f.f =

(1.1) (c) Elastic deformation of concrete


Figure 1.2.1 shows a typical stress-strain curve of concrete in compression, where Ec = initial tangent modulus at the origin, also known as short term static

modulus Es = secant modulus at A Et = tangent modulus at A εe = elastic strain at A εi = inelastic strain at A

It is seen that the initial tangent modulus is much higher than Et (tangent modulus at A). Near the failure, the actual strain consists of both εe and εi (elastic and inelastic respectively) components of strain. The initial tangent modulus Ec in N/mm2 is estimated from

kcc fE 5000= (1.2)

where fck = characteristic compressive strength of concrete at 28 days The initial tangent modulus Ec is also known as short term static modulus of elasticity of concrete in N/mm2 and is used to calculate the elastic deflections. (d) Shrinkage of concrete Shrinkage is the time dependent deformation, generally compressive in nature. The constituents of concrete, size of the member and environmental conditions are the factors on which the total shrinkage of concrete depends. However, the total shrinkage of concrete is most influenced by the total amount of water present in the concrete at the time of mixing for a given humidity and temperature. The cement content, however, influences the total shrinkage of concrete to a lesser extent. The approximate value of the total shrinkage strain for design is taken as 0.0003 in the absence of test data (cl. 6.2.4.1).


(e) Creep of concrete

Creep is another time dependent deformation of concrete by which it

continues to deform, usually under compressive stress. The creep strains recover partly when the stresses are released. Figure 1.2.2 shows the creep recovery in two parts. The elastic recovery is immediate and the creep recovery is slow in nature. Thus, the long term deflection will be added to the short term deflection to get the total deflection of the structure. Accordingly, the long term modulus Ece or the effective modulus of concrete will be needed to include the effect of creep due to permanent loads. The relationship between Ece and Ec is obtained as follows: εc = fc/Ec (1.3) where εc = short term strain at the age of loading at a stress value of fc εcr = ultimate creep strain

θ = creep coefficient = c

rc

εε

(cl. 6.2.5.1 of IS 456)

(1.4) The values of θ on 7th, 28th and 365th day of loading are 2.2, 1.6 and 1.1 respectively.


Then the total strain = εc + εcr = ce

c

Ef

(1.5) where Ece = effective modulus of concrete From the above Eq. (1.5), we have

θ

EεεEε

εεfE c

rcc

cc

rcc

cce +

=+

=+

=1

(1.6) The effective modulus of Ece of concrete is used only in the calculation of creep deflection. It is seen that the value of creep coefficient θ (Eq. 1.4) is reducing with the age of concrete at loading. It may also be noted that the ultimate creep strain εcr does not include short term strain εc. The creep of concrete is influenced by

• Properties of concrete • Water/cement ratio • Humidity and temperature of curing • Humidity during the period of use • Age of concrete at first loading • Magnitude of stress and its duration • Surface-volume ratio of the member

(f) Thermal expansion of concrete

The knowledge of thermal expansion of concrete is very important as it is prepared and remains in service at a wide range of temperature in different countries having very hot or cold climates. Moreover, concrete will be having its effect of high temperature during fire. The coefficient of thermal expansion depends on the nature of cement, aggregate, cement content, relative humidity and size of the section. IS 456 stipulates (cl. 6.2.6) the values of coefficient of thermal expansion for concrete / oC for different types of aggregate. 1.2.3 Workability and Durability of Concrete

Workability and durability of concrete are important properties to be considered. The relevant issues are discussed in the following:

(a) Concrete mix proportioning


The selected mix proportions of cement, aggregates (fine and coarse) and

water ensure: • the workability of fresh concrete, • required strength, durability and surface finish when concrete is hardened.

Recently more than forty per cent of concrete poured world over would contain admixtures. (b) Workability

It is the property which determines the ease and homogeneity with which concrete can be mixed, placed, compacted and finished. A workable concrete will not have any segregation or bleeding. Segregation causes large voids and hence concrete becomes less durable. Bleeding results in several small pores on the surface due to excess water coming up. Bleeding also makes concrete less durable. The degree of workability of concrete is classified from very low to very high with the corresponding value of slump in mm (cl. 7 of IS 456). (c) Durability of concrete

A durable concrete performs satisfactorily in the working environment during its anticipated exposure conditions during service. The durable concrete should have low permeability with adequate cement content, sufficient low free water/cement ratio and ensured complete compaction of concrete by adequate curing. For more information, please refer to cl. 8 of IS 456. (d) Design mix and nominal mix concrete

In design mix, the proportions of cement, aggregates (sand and gravel), water and mineral admixtures, if any, are actually designed, while in nominal mix, the proportions are nominally adopted. The design mix concrete is preferred to the nominal mix as the former results in the grade of concrete having the specified workability and characteristic strength (vide cl. 9 of IS 456). (e) Batching

Mass and volume are the two types of batching for measuring cement, sand, coarse aggregates, admixtures and water. Coarse aggregates may be gravel, grade stone chips or other man made aggregates. The quantities of cement, sand, coarse aggregates and solid admixtures shall be measured by mass. Liquid admixtures and water are measured either by volume or by mass (cl. 10 of IS 456). 1.2.4 Properties of Steel


As mentioned earlier in sec. 1.2.2, steel is used as the reinforcing material in concrete to make it good in tension. Steel as such is good in tension as well as in compression. Unlike concrete, steel reinforcement rods are produced in steel plants. Moreover, the reinforcing bars or rods are commercially available in some specific diameters. Normally, steel bars up to 12 mm in diameter are designated as bars which can be coiled for transportation. Bars more than 12 mm in diameter are termed as rods and they are transported in standard lengths. Like concrete, steel also has several types or grades. The four types of steel used in concrete structures as specified in cl. 5.6 of IS 456 are given below:

(i) Mild steel and medium tensile steel bars conforming to IS 432 (Part 1) (ii) High yield strength deformed (HYSD) steel bars conforming to IS 1786

(iii) Hard-drawn steel wire fabric conforming to IS 1566

(iv) Structural steel conforming to Grade A of IS 2062.

Mild steel bars had been progressively replaced by HYSD bars and subsequently TMT bars are promoted in our country. The implications of adopting different kinds of blended cement and reinforcing steel should be examined before adopting. Stress-strain curves for reinforcement


Figures 1.2.3 and 1.2.4 show the representative stress-strain curves for steel having definite yield point and not having definite yield point, respectively. The characteristic yield strength fy of steel is assumed as the minimum yield stress or 0.2 per cent of proof stress for steel having no definite yield point. The modulus of elasticity of steel is taken to be 200000 N/mm2. For mild steel (Fig. 1.2.3), the stress is proportional to the strain up to the yield point. Thereafter, post yield strain increases faster while the stress is assumed to remain at constant value of fy. For cold-worked bars (Fig. 1.2.4), the stress is proportional to the strain up to a stress of 0.8 fy. Thereafter, the inelastic curve is defined as given below:

Stress Inelastic strain 0.80 fy Nil 0.85 fy 0.90 fy

0.0001 0.0003

0.95 fy 0.975 fy

0.0007 0.0010

1.00 fy 0.0020


Linear interpolation is to be done for intermediate values. The two grades of cold-worked bars used as steel reinforcement are Fe 415 and Fe 500 with the values of fy as 415 N/mm2 and 500 N/mm2, respectively. Considering the material safety factor γm (vide sec. 2.3.2.3 of Lesson 3) of steel as 1.15, the design yield stress (fyd) of both mild steel and cold worked bars is computed from fyd = fy / γm (1.7) Accordingly, the representative stress-strain curve for the design is obtained by substituting fyd for fy in Figs. 1.2.3 and 1.2.4 for the two types of steel with or without the definite yield point, respectively. 1.2.5 Other Important Factors

The following are some of the important factors to be followed properly as per the stipulations in IS 456 even for the design mix concrete with materials free from impurities in order to achieve the desired strength and quality of concrete. The relevant clause numbers of IS 456 are also mentioned as ready references for each of the factors. (a) Mixing (cl. 10.3)

Concrete is mixed in a mechanical mixer at least for two minutes so as to have uniform distribution of the materials having uniform colour and consistency. (b) Formwork (cl. 11)

Properly designed formwork shall be used to maintain its rigidity during placing and compaction of concrete. It should prevent the loss of slurry from the concrete. The stripping time of formwork should be such that the concrete attains strength of at least twice the stress that the concrete may be subjected at the time of removing the formwork. As a ready reference IS 456 specifies the minimum period before striking formwork.

There is a scope for good design of formwork system so that stripping off

is efficient without undue shock to concrete and facilitating reuse of formwork. (c) Assembly of reinforcement (cl. 12)

The required reinforcement bars for the bending moment, shear force and axial thrust are to be accommodated together and proper bar bending schedules shall be prepared. The reinforcement bars should be placed over blocks, spacers, supporting bars etc. to maintain their positions so that they have the


required covers. High strength deformed steel bars should not be re-bent. The reinforcement bars should be assembled to have proper flow of concrete without obstruction or segregation during placing, compacting and vibrating. (d) Transporting, placing, compaction and curing (cl. 13)

Concrete should be transported to the formwork immediately after mixing to avoid segregation, loss of any of the ingredients, mixing of any foreign matter or loss of workability. Proper protections should be taken to prevent evaporation loss of water in hot weather and loss of heat in cold weather.

To avoid rehandling, concrete should be deposited very near to the final

position of its placing. The compaction should start before the initial setting time and should not be disturbed once the initial setting has started. While placing concrete, reinforcement bars should not be displaced and the formwork should not be moved.

The compaction of concrete using only mechanical vibrators is very

important, particularly around the reinforcement, embedded fixtures and the corners of the formwork to prevent honeycomb type of concreting. Excessive vibration leads to segregation.

Proper curing prevents loss of moisture from the concrete and maintains a

satisfactory temperature regime. In moist curing, the exposed concrete surface is kept in a damp or wet condition by ponding or covering with a layer of sacking, canvas, hessian etc. and kept constantly wet for a period of 7-14 days depending on the type of cement and weather conditions. Blended cement needs extended curing. In some situations, polyethylene sheets or similar impermeable membranes may be used to cover the concrete surface closely to prevent evaporation. (e) Sampling and strength of designed concrete mix (cl. 15)

Random samples of concrete cubes shall be cast from fresh concrete, cured and tested at 28 days as laid down in IS 516. Additional tests on beams for modulus of rupture at 3 or 7 days, or compressive strength tests at 7 days shall also be conducted. The number of samples would depend on the total quantity of concrete as given in cl. 15.2.2 and there should be three test specimens in each sample for testing at 28 days, and additional tests at 3 or 7 days. (f) Acceptance criteria (cl. 16)

Concrete should be considered satisfactory when both the mean strength of any group of four consecutive test results and any individual test result of compressive strength and flexural strength comply the limits prescribed in IS 456.


(g) Inspection and testing of structures (cl. 17)

Inspection of the construction is very important to ensure that it complies the design. Such inspection should follow a systematic procedure covering materials, records, workmanship and construction.

All the materials of concrete and reinforcement are to be tested following

the relevant standards. It is important to see that the design and detailing are capable of execution maintaining a standard with due allowance for the dimensional tolerances. The quality of the individual parts of the structure should be verified. If needed, suitable quality assurance schemes should be used. The concrete should be inspected immediately after the removal of formwork to remove any defective work before concrete has hardened.

Standard core tests (IS 516) are to be conducted at three or more points

to represent the whole concrete work in case of any doubt regarding the grade of concrete during inspection either due to poor workmanship or unsatisfactory results on cube strength obtained following the standard procedure. If the average equivalent cube strength of cores is equal to at least 85 per cent of the cube strength of that grade of concrete at that age and each of the individual cores has strength of at least 75 per cent, then only the concrete represented by the core test is considered acceptable. For unsatisfactory core test results, load tests should be conducted for the flexural members and proper analytical investigations should be made for non-flexural members.

Such load tests should be done as soon as possible after expiry of 28

days from the date of casting of the flexural members subjected to full dead load and 1.25 times the imposed load for 24 hours and then the imposed load shall be removed. The maximum deflection of the member during 24 hours under imposed load in mm should be less than 40 l2/D, where l is the effective span in m and D is the overall depth of the member in mm. For members showing more deflection, the recovery of the deflection within 24 hours of removal of the imposed load has to be noted. If the recovery is less than 75 per cent of the deflection under imposed load, the test should be repeated after a lapse of 72 hours. The structure is considered unacceptable if the recovery is less than 80 per cent.

There are further provisions of conducting non-destructive tests like

ultrasonic pulse velocity (UPV), rebound hammer, probe penetration, pull out and maturity, as options to core tests or to supplement the data obtained from a limited number of cores. However, it is important that the acceptance criteria shall be agreed upon prior to these non-destructive testing. There are reports that UPV tests conducted three days after casting after removal of side formwork give very dependable insight about the quality of concrete. 1.2.6 Concluding Remarks


The reinforced concrete consisting of plain concrete and steel

reinforcement opened a new vista fulfilling the imaginations of architect with a unified approach of the architect and structural engineer. This has been made possible due to mouldability and monolithicity of concrete in addition to its strength in both tension and compression when reinforced with steel. However, concrete is produced by mixing cement, sand, gravel, water and mineral admixtures, if needed. Therefore, the final strength of concrete depends not only on the individual properties of its constituent materials, but also on the proportions of the material and the manner in which it is prepared, transported, placed, compacted and cured. Moreover, durability of the concrete is also largely influenced by all the steps of its preparation.

Steel reinforcement though produced in steel plants and made available in

form of bars and rods of specific diameter also influences the final strength of reinforced concrete by its quality and durability due to environmental effects.

Concrete cover provides the protective environment to embedded steel

from rusting that would need presence of both oxygen and moisture. Not only the extent of cover but the quality of cover is important for this reason.

Accordingly, inspection of concrete work, sample testing of specimens,

core tests, load tests and non-destructive tests are very important to maintain the quality, strength and durability of reinforced concrete structures. Moreover, it is equally important to remove small defects or make good of it after removing the formwork before it has thoroughly hardened.

Thus, starting from the selection of each constitutive material to the

satisfactory construction of the structure, the designer's responsibility will only produce the desired concrete structure which will satisfy the functional requirements as well as will have its aesthetic values exploiting all the good properties of this highly potential material. 1.2.7 Practice Questions and Problems with Answers Q.1: What are the constituent materials of plain concrete? A.1: The constituent materials of plain concrete are cement, sand (fine

aggregate), gravel (coarse aggregate), water and mineral admixtures in some special cases.

Q.2: Define characteristic strength fck of concrete.


A.2: Characteristic strength of concrete is defined as the compressive strength of 150 mm size cube at 28 days and expressed in N/mm2 below which not more than five per cent of the test results are expected to fall.

Q.3: How and when the characteristic compressive strength fck is determined? A.3: Characteristic compressive strength is determined by conducting

compressive strength tests on specified number of 150 mm concrete cubes at 28 days after casting. It is expressed in N/mm2.

Q.4: What do the symbols M and 20 mean for grade M 20 concrete? A.4: The symbol M refers to mix and the number 20 indicates that the

characteristic strength fck of grade M 20 is 20 N/mm2. Q.5: Express the relation between flexural strength (fcr) and characteristic

compression strength fck of concrete. A.5: The generally accepted relation is: fcr = 0.7 ckf where fcr and fck are in

N/mm2. Q.6: Draw stress-strain curve of concrete and show the following: (a) Initial tangent modulus Ec, (b) Secant modulus Es at any point A on the

stress-strain curve, (c) Tangent modulus Et at A and (d) elastic and inelastic strain components of the total strain at A.

A.6: Please refer to Fig. 1.2.1

Q.7: Express the short term static modulus Ec in terms of the characteristic compressive strength fck of concrete.

A.7: The suggested expression is : Ec = 5000 ckf where Ec and fck are in

N/mm2. Q.8: State the approximate value of total shrinkage strain of concrete to be

taken for the design purpose and mention the relevant clause no. of IS code.

A.8: As per cl. 6.2.4.1 of IS 456:2000, the approximate value of total shrinkage

strain of concrete is to be taken as 0.0003. Q.9: Define creep coefficient θ of concrete and express the relation between the

effective modulus Ece, short term static modulus Ec and creep coefficient θ of concrete.


A.9: Creep coefficient θ is the ratio of ultimate creep strain εcr and short term

strain at the age of loading (θ = εcr/εc).

The required relation is Ece = θ

cE

+1. {The derivation of Eq. 1.6 is given in

sec. 1.2.2 part (e)}. Q.10: Define workability of concrete. A.10: Workability of concrete is the property which determines the ease and

homogeneity with which concrete can be mixed, placed, compacted and finished.

Q.11: Differentiate between design mix and nominal mix concrete. A.11: In design mix, the proportions of cement, aggregates (sand and gravel),

water and mineral admixtures, if any are actually determined by actual design to have a desired strength. In nominal mix, however, these proportions are nominally adopted.

Q.12: What are the different types of batching in mixing the constituent materials

of concrete and name the type of batching to be adopted for different materials?

A.12: Mass and volume are the two types of batching. The quantities of cement,

aggregates (sand and gravel) and solid admixtures shall be measured by mass batching. Liquid admixtures and water are measured either by mass or volume batching.

Q.13: Differentiate between steel bars and rods. A.13: Bars are steel bars of diameter up to 12 mm which are coiled during

transportation. Rods are steel bars of diameter greater than 12 mm and cannot be coiled. They are transported in standard lengths.

Q.14: Name the types of steel and their relevant IS standards to be used as

reinforcement in concrete. A.14: Please refer to sec. 1.2.4(i), (ii), (iii) and (iv). Q.15: Draw stress-strain curve of steel bars with or without definite yield point

and indicate the yield stress fy of them.


A.15: Please refer to Figs. 1.2.3 and 1.2.4. For steel bars of Fig. 1.2.3, fy = 250 N/mm2 and for steel bars of Fig. 1.2.4, fy = 415 and 500 N/mm2 for the two different grades.

Q.16: What are the criteria of properly mixed concrete and how to achieve

them?

A.16: Properly mixed concrete will have uniform distribution of materials having uniform colour and consistency.

These are achieved by mixing the constituent materials in a mechanical

mixer at least for two minutes or such time till those qualities are achieved. Q.17: What should be the expected strength of concrete structure at the time of

removal of formwork? A.17: The concrete at the time of removing the formwork should have strength

of at least twice the stress that it may be subjected to at the time of removal of formwork.

Q.18: Name the sample tests to be performed for checking the strength of

concrete. A.18: The main test to be performed is 150 mm cube strength at 28 days made

of fresh concrete and cured. Additional tests should also be conducted on 150 mm cubes at 7 days and beam tests to determine modulus of rupture at 3 or 7 days. There should be at least 3 or more samples of such specimens to represent the entire concrete work. Each sample should have at least three specimens for conducting each of the above-mentioned tests.

Q.19: Mention the specific acceptance criteria of the sample tests of cubes and

beams. A.19: Concrete should be considered satisfactory when both the mean strength

determined from any group of four consecutive test results and any individual test result of compressive and flexural strength tests comply the prescribed limits of cl. 16 of IS 456.

Q.20: When is it essential to conduct standard core test? A.20: Standard core tests are needed if the inspection of concrete work raises

doubt regarding the grade of concrete either due to poor workmanship or unsatisfactory cube strength results performed following standard procedure.

Q.21: When do you consider core test results as satisfactory?


A.21: The core test results are considered satisfactory if: (i) the average equivalent cube strength of the cores is at least 85 per cent

of the cube strength of the grade of concrete at that age, and (ii) each of the individual cores has strength of at least 75 per cent of the concrete cube strength at that age.

Q.22: What are to be done for unsatisfactory core test results? A.22: Load tests are to be conducted for the flexural members and analytical

investigations are to be performed for non-flexural members. Q.23: Prescribe the loading conditions and age of structure for conducting load

tests. A.23: Load tests are to be conducted as soon as possible after expiry of 28 days

from the date of casting. The flexural member is subjected to full dead load and 1.25 times the imposed load for 24 hours and then the imposed load has to be removed.

Q.24: When do you conclude the load tests as satisfactory? A.24: Load tests are considered satisfactory if the maximum deflection in mm of

the member during 24 hours under load is less than 40 l2/D, where l = effective span in m and D = overall depth of the member in mm.

For members showing more deflection, the recovery of the deflection within 24 hours of removal of the imposed load has to be noted. If the recovery is less than 75 per cent of the deflection under imposed load, the tests should be repeated after a lapse of 72 hours. The structure is considered unacceptable if the recovery is less than 80 per cent.

Q.25: Name the acceptable non-destructive tests to be performed on structures. A.25: The acceptable non-destructive tests are ultrasonic pulse velocity, rebound

hammer, probe penetration, pull out and maturity. 1.2.8 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 1.2.9 Test 2 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. Each question carries five marks. TQ.1: Define characteristic strength (fck) of concrete A.TQ.1: See Ans. 2 of sec. 1.2.7. TQ.2: How and when the characteristic compressive strength (fck) is

determined? A.TQ.2: See Ans. 3 of sec. 1.2.7. TQ.3: What do the symbols M and 20 mean for grade M 20 concrete?


A.TQ.3: See Ans. 4 of sec. 1.2.7.

TQ.4: Express the short term static modulus (Ec) in terms of the characteristic compressive strength (fck) of concrete.

A.TQ.4: See Ans. 7 of sec. 1.2.7. TQ.5: Define creep coefficient θ of concrete and express the relation between

the effective modulus (Ece), short term static modulus (Ec) and creep coefficient (θ) of concrete.

A.TQ.5: See Ans. 9 of sec. 1.2.7. TQ.6: Differentiate between design mix and nominal mix concrete. A.TQ.6: See Ans. 11 of sec. 1.2.7. TQ.7: Draw stress-strain curve of steel bars with or without definite yield point

and indicate the yield stress fy of them. A.TQ.7: See Ans. 15 of sec. 1.2.7. TQ.8: Mention the specific acceptance criteria of the sample tests of cubes and

beams. A.TQ.8: See Ans. 19 of sec. 1.2.7. TQ.9: When do you consider core test results as satisfactory? A.TQ.9: See Ans. 21 of sec. 1.2.7. TQ.10: Prescribe the loading conditions and age of structure for conducting load

tests. A.TQ.10: See Ans. 23 of sec. 1.2.7. 1.2.10 Summary of this Lesson Properties of concrete and steel are essential to be thoroughly known and understood by the designer. These properties in respect of strength, deformation and durability are summarised in this lesson. The importance of quality control, inspection and testing are emphasised starting from the basic preparation of concrete to the removal of formwork after the construction. The recommendations of Indian Standards are discussed regarding the acceptance of


good concrete based on sample tests of specimens, core tests and other non-destructive tests.


Module 2

Philosophies of Design by Limit State Method


Lesson 3

Philosophies of Design by Limit State Method


Instructional Objectives: At the end of this lesson, the student should be able to: • categorically state the four common steps to be followed in any method of

design, • define the limit states, • identify and differentiate the different limit states, • state if the structures are to be designed following all the limit states, • explain the concept of separate partial safety factors for loads and material

strengths depending on the limit state being considered, • justify the "size effect" of concrete in its strength, • name the theory for the analysis of structures to be designed by limit states, • name the method of analysis of statically indeterminate beams and frames,

and slabs spanning in two direction at right angles, • justify the need to redistribute the moments in statically indeterminate beams

and frames, • identify four reasons to justify the design of structures or parts of the structure

by limit state method. 2.3.1 Introduction

In any method of design, the following are the common steps to be followed: (i) To assess the dead loads and other external loads and forces likely to

be applied on the structure, (ii) To determine the design loads from different combinations of loads,

(iii) To estimate structural responses (bending moment, shear force, axial

thrust etc.) due to the design loads,

(iv) To determine the cross-sectional areas of concrete sections and amounts of reinforcement needed.


Many of the above steps have lot of uncertainties. Estimation of loads and

evaluation of material properties are to name a few. Hence, some suitable factors of safety should be taken into consideration depending on the degrees of such uncertainties.

Limit state method is one of the three methods of design as per IS

456:2000. The code has put more emphasis on this method by presenting it in a full section (Section 5), while accommodating the working stress method in Annex B of the code (IS 456). Considering rapid development in concrete technology and simultaneous development in handling problems of uncertainties, the limit state method is a superior method where certain aspects of reality can be explained in a better manner. 2.3.2 Limit State Method 2.3.2.1 What are limit states?

Limit states are the acceptable limits for the safety and serviceability requirements of the structure before failure occurs. The design of structures by this method will thus ensure that they will not reach limit states and will not become unfit for the use for which they are intended. It is worth mentioning that structures will not just fail or collapse by violating (exceeding) the limit states. Failure, therefore, implies that clearly defined limit states of structural usefulness has been exceeded.

Limit state of collapse was found / detailed in several countries in

continent fifty years ago. In 1960 Soviet Code recognized three limit states: (i) deformation, (ii) cracking and (iii) collapse. 2.3.2.2 How many limit states are there?


There are two main limit states: (i) limit state of collapse and (ii) limit state of serviceability (see Fig. 2.3.1).

(i) Limit state of collapse deals with the strength and stability of structures subjected to the maximum design loads out of the possible combinations of several types of loads. Therefore, this limit state ensures that neither any part nor the whole structure should collapse or become unstable under any combination of expected overloads. (ii) Limit state of serviceability deals with deflection and cracking of structures under service loads, durability under working environment during their anticipated exposure conditions during service, stability of structures as a whole, fire resistance etc.

All relevant limit states have to be considered in the design to ensure adequate degree of safety and serviceability. The structure shall be designed on the basis of the most critical limit state and shall be checked for other limit states (see Fig. 2.3.2).


2.3.2.3 Partial safety factors

The characteristic values of loads as discussed in sec. 1.1.5 are based on statistical data. It is assumed that in ninety-five per cent cases the characteristic loads will not be exceeded during the life of the structures (Fig. 2.3.3). However, structures are subjected to overloading also. Hence, structures should be designed with loads obtained by multiplying the characteristic loads with suitable factors of safety depending on the nature of loads or their combinations, and the limit state being considered. These factors of safety for loads are termed as partial safety factors (γf) for loads. Thus, the design loads are calculated as (Design load Fd) = (Characteristic load F) (Partial safety factor for load γf) (2.1) Respective values of γf for loads in the two limit states as given in Table 18 of IS 456 for different combinations of loads are furnished in Table 2.1.


Table 2.1 Values of partial safety factor γf for loads

Limit state of collapse Limit state of serviceability

(for short term effects only)

Load combinations

DL IL WL DL IL WL DL + IL 1.5 1.0 1.0 1.0 -

DL + WL 1.5 or 0.91)

- 1.5 1.0 - 1.0

DL + IL + WL 1.2 1.0 0.8 0.8 NOTES: 1 While considering earthquake effects, substitute EL for WL. 2 For the limit states of serviceability, the values of γf given in this table are applicable for

short term effects. While assessing the long term effects due to creep the dead load and that part of the live load likely to be permanent may only be considered.

1) This value is to be considered when stability against overturning or stress reversal is

critical.


Similarly, the characteristic strength of a material as obtained from the statistical approach is the strength of that material below which not more than five per cent of the test results are expected to fall (see Fig. 2.3.4). However, such characteristic strengths may differ from sample to sample also. Accordingly, the design strength is calculated dividing the characteristic strength further by the partial safety factor for the material (γm), where γm depends on the material and the limit state being considered. Thus,

df material theofstrength Design = mγ

f material theoffactor safety Partial

material theofstrength sticCharacteri

(2.2) Both the partial safety factors are shown schematically in Fig. 2.3.5.

Clause 36.4.2 of IS 456 states that γm for concrete and steel should be taken as 1.5 and 1.15, respectively when assessing the strength of the structures or structural members employing limit state of collapse. However, when assessing the deflection, the material properties such as modulus of elasticity should be taken as those associated with the characteristic strength of the


material. It is worth mentioning that partial safety factor for steel (1.15) is comparatively lower than that of concrete (1.5) because the steel for reinforcement is produced in steel plants and commercially available in specific diameters with expected better quality control than that of concrete. Further, in case of concrete the characteristic strength is calculated on the basis of test results on 150 mm standard cubes. But the concrete in the structure has different sizes. To take the size effect into account, it is assumed that the concrete in the structure develops a strength of 0.67 times the characteristic strength of cubes. Accordingly, in the calculation of strength employing the limit state of collapse, the characteristic strength (fck) is first multiplied with 0.67 (size effect) and then divided by 1.5 (γm for concrete) to have 0.446 fck as the maximum strength of concrete in the stress block. 2.3.3 Analysis

Analysis of structure has been briefly mentioned in sec. 1.1.4 earlier. Herein, the analysis of structure, in the two limit states (of collapse and of serviceability), is taken up. In the limit state of collapse, the strength and stability of the structure or part of the structure are ensured. The resistances to bending moment, shear force, axial thrust, torsional moment at every section shall not be less than their appropriate values at that section due to the probable most unfavourable combination of the design loads on the structure. Further, the structure or part of the structure should be assessed for rupture of one or more critical sections and buckling due to elastic or plastic instability considering the effects of sway, if it occurs or overturning.

Linear elastic theory is recommended in cl. 22 of IS 456 to analyse the

entire structural system subjected to design loads. The code further stipulates the adoption of simplified analyses for frames (cl. 22.4) and for continuous beams (cl. 22.5). For both the limit states the material strengths should be taken as the characteristic values in determining the elastic properties of members. It is worth mentioning that the statically indeterminate structures subjected to design loads will have plastic hinges at certain locations as the loads increase beyond the characteristic loads. On further increase of loads, bending moments do not increase in the locations of plastic hinges as they are already at the full capacities of bending moments. However, these plastic hinges undergo more rotations and the moments are now received by other sections which are less stressed. This phenomenon continues till the plastic hinges reach their full rotation capacities to form a mechanism of collapse of the structure. This is known as the redistribution of moments (Figs. 2.3.6 and 2.3.7). The theory and numerical problems of “Redistribution of moments” are presented elaborately in Lesson 38.


The design of structure, therefore, should also ensure that the less stressed sections can absorb further moments with a view to enabling the structure to rotate till their full capacities. This will give sufficient warming to the users before the structures collapse. Accordingly, there is a need to redistribute moments in continuous beams and frames. Clause 37.1.1 stipulates this provision and the designer has to carry out the redistribution by satisfying the stipulated conditions there.

The analysis of slabs spanning in two directions at right angles should be

performed by employing yield line theory or any other acceptable method. IS 456:2000 has illustrated alternative provisions for the simply supported and restrained slabs spanning in two directions in Annex D along with Tables 26 and 27 giving bending moment coefficients of these slabs for different possible boundary conditions. These provisions enable to determine the reinforcement


needed for bending moments in two directions and torsional reinforcement wherever needed.

2.3.4 Concluding Remarks

The limit state method is based on a stochastic process where the design parameters are determined from observations taken over a period of time. The concept of separate partial safety factors for loads and material strengths are based on statistical and probabilistic grounds. These partial safety factors for the material strengths are determined on the basis of reliability of preparations of concrete and reinforcement. The overloading of structure has been kept in mind while specifying the partial safety factors of loads.

The stress block of structures or parts of structure designed on the basis

of limit state method subjected to the designed loads or collapse loads represents the stress-strain diagram at the defined states of collapse and satisfy the requirements of strength and stability. Simultaneous checking of these structures or parts of them for the limit state of serviceability ensures the deflection and cracking to remain within their limits. Thus, this method is more rational and scientific. 2.3.5 Practice Questions and Problems with Answers Q.1: List the common steps of design of structures by any method of design. A.1: The four steps are listed in section 2.3.1 under (i) to (iv). Q.2: What are limit states? A.2: See sec. 2.3.2.1 Q.3: How many limit states are there? Should a structure be designed following

all the limit states? A.3: See sec. 2.3.2.2 Q.4: Define partial safety factors of load and material. Write the expressions to

determine the design load and design strength of the material from their respective characteristic values employing the corresponding partial safety factors.

A.4: See sec. 2.3.2.3 and Eqs.2.1 and 2.2 Q.5: What is size effect of concrete? What is its role in determining the material

strength of concrete?


A.5: The characteristic strength of concrete is determined from the results of tests conducted on cube specimens of 150 mm dimension. The dimensions of concrete structures or in members of structure are different and widely varying. This has an effect on the strength of concrete in the structure. This is known as size effect.

Due to the size effect, the characteristic strength of concrete is reduced to

2/3 of its value and then further divided by the partial safety factor of the concrete (γm = 1.5) to get the design strength of concrete (fd). Thus,

kcf..

kcf.

.)/()kc(fdf 4460

51

670

5132

===

Q.6: Which theory should be employed for the analysis of structural system to

be designed element wise, by limit state method? A.6: Linear elastic theory should be employed for the analysis of structural

system subjected to design loads. Q.7: Justify the need to do the redistribution of moments in statically

indeterminate structures. A.7: Statically indeterminate structures will have plastic hinges formed when

loads increase from the characteristic values. These locations where plastic hinges are formed will undergo rotations at constant moment when the sections of lower stresses will receive the additional moments due to further increase of loads. This process will continue till sufficient plastic hinges are formed to have a mechanism of collapse (see Figs. 2.3.6 and 2.3.7).

The structures should have such a provision to avoid sudden failure at the

failure of one critical section. The comparatively lower stressed sections, therefore, should be designed taking the redistribution of moments into account.

Q.8: What are the analytical methods for the design of simply supported and

restrained slabs? A.8: Yield line theory or any other acceptable method of analysis can be

employed for these slabs. Alternatively, the method illustrated in Annex D of IS 456:2000 can also be used for the slabs spanning in two perpendicular directions.

Q.9: Give four reasons to justify the design of structures by limit state method.


A.9: The four reasons are:

(i) Concept of separate partial safety factors of loads of different combinations in the two limit state methods.

(ii) Concept of separate partial safety factors of materials depending on

their quality control during preparation. Thus, γm for concrete is 1.5 and the same for steel is 1.15. This is more logical than one arbitrary value in the name of safety factor.

(iii) A structure designed by employing limit state method of collapse

and checked for other limit states will ensure the strength and stability requirements at the collapse under the design loads and also deflection and cracking at the limit state of serviceability. This will help to achieve the structure with acceptable probabilities that the structure will not become unfit for the use for which it is intended.

(iv) The stress block represents in a more realistic manner when the

structure is at the collapsing stage (limit state of collapse) subjected to design loads.

2.3.6 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 2.3.7 Test 3 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: List the common steps of design of structures by any method of design.

(8 marks) A.TQ.1: See Ans. 1 of sec. 2.3.5.

TQ.2: What are limit states? (6 marks) A.TQ.2: See Ans. 2 of sec. 2.3.5. TQ.3: Draw schematic figures to explain (i) the different limit states and (ii) use

of the limit states to design a structure. (6 + 6 = 12 marks)

A.TQ.3: The schematic figures are Figs. 2.3.1 and 2.3.2 and answers are in sec. 2.3.2.2 TQ.4: What is size effect of concrete? What is its role in determining the material

strength of concrete? (6 marks)

A.TQ.4: See Ans. 5 of sec. 2.3.5. TQ.5: Justify the need to do the redistribution of moments in statically

indeterminate structures. (10 marks)

A.TQ.5: See Ans. 7 of sec. 2.3.5.


TQ.6: Give four reasons to justify the design of structures by limit state method.

(8 marks) A.TQ.6: See Ans. 9 of sec. 2.3.5. 2.3.8 Summary of this Lesson This lesson defines and identifies different limit states. It explains the concept and defines the partial safety factors for loads and material strengths depending on the limit states to be considered in the design. It explains the theory for the analysis of structures to be carried out by the designer. Further, the redistribution of moments in statically indeterminate beams and frames is explained. Finally, the use of limit state method for the design of reinforced concrete structures has been justified.


Module 3

Limit State of Collapse - Flexure (Theories and

Examples)


Lesson 4

Computation of Parameters of

Governing Equations


Instructional Objectives: At the end of this lesson, the student should be able to: • identify the primary load carrying mechanisms of reinforced concrete beams

and slabs, • name three different types of reinforced concrete beam with their specific

applications, • identify the parameters influencing the effective widths of T and L-beams, • differentiate between one-way and two-way slabs, • state and explain the significance of six assumptions of the design, • draw the stress-strain diagrams across the depth of a cross-section of

rectangular beam, • write the three equations of equilibrium, • write and derive the expressions of total compression and tension forces C

and T, respectively. 3.4.1 Introduction


Reinforced concrete beams and slabs carry loads primarily by bending (Figs. 3.4.1 to 3). They are, therefore, designed on the basis of limit state of collapse in flexure. The beams are also to be checked for other limit states of shear and torsion. Slabs under normal design loadings (except in bridge decks etc.) need not be provided with shear reinforcement. However, adequate torsional reinforcement must be provided wherever needed.


This lesson explains the basic governing equations and the computation of parameters required for the design of beams and one-way slabs employing limit state of collapse in flexure. There are three types of reinforced concrete beams:

(i) Singly or doubly reinforced rectangular beams (Figs. 3.4.4 to 7) (ii) Singly or doubly reinforced T-beams (Figs. 3.4.8 to 11) (iii) Singly or doubly reinforced L-beams (Figs. 3.4.12 to 15)


During construction of reinforced concrete structures, concrete slabs and beams are cast monolithic making the beams a part of the floor deck system. While bending under positive moments near midspan, bending compression stresses at the top are taken by the rectangular section of the beams above the neutral axis and the slabs, if present in T or L-beams (Figs. 3.4.4, 5, 8, 9, 12 and 13). However, under the negative moment over the support or elsewhere, the bending compression stresses are at the bottom and the rectangular sections of rectangular, T and L-beams below the neutral axis only resist that compression (Figs. 3.4.6, 7, 10, 11, 14 and 15). Thus, in a slab-beam system the beam will be


considered as rectangular for the negative moment and T for the positive moment. While for the intermediate spans of slabs the beam under positive moment is considered as T, the end span edge beam is considered as L-beam if the slab is not projected on both the sides of the beam. It is worth mentioning that the effective width of flange of these T or L-beams is to be determined which depends on:


(a) if it is an isolated or continuous beam (b) the distance between points of zero moments in the beam (c) the width of the web (d) the thickness of the flange

Reinforced concrete slabs are classified as one-way or two-way depending on if they are spanning in one or two directions (Figs. 3.4.16 and 17). As a guideline, slabs whose ratio of longer span (ly) to the shorter span (lx) is more than two are considered as one-way slabs. One-way slabs also can be designed following the procedure of the design of beams of rectangular cross-section. Again, slabs may be isolated or continuous also.


3.4.2 Assumptions


The following are the assumptions of the design of flexural members (Figs. 3.4.18 to 20) employing limit state of collapse:

(i) Plane sections normal to the axis remain plane after bending. This assumption ensures that the cross-section of the member does not warp due to the loads applied. It further means that the strain at any point on the cross-section is directly proportional to its distance from the neutral axis. (ii) The maximum strain in concrete at the outer most compression fibre is taken as 0.0035 in bending (Figs. 3.4.19 and 20). This is a clearly defined limiting strain of concrete in bending compression beyond which the concrete will be taken as reaching the state of collapse. It is very clear that the specified limiting strain of 0.0035 does not depend on the strength of concrete. (iii) The acceptable stress-strain curve of concrete is assumed to be parabolic as shown in Fig. 1.2.1 of Lesson 2.

The maximum compressive stress-strain curve in the structure is obtained by reducing the values of the top parabolic curve (Figs. 21 of IS 456:2000) in two stages. First, dividing by 1.5 due to size effect and secondly, again dividing by 1.5 considering the partial safety factor of the material. The middle and bottom curves (Fig. 21 of IS 456:2000) represent these stages. Thus, the maximum compressive stress in bending is limited to the constant value of 0.446 fck for the strain ranging from 0.002 to 0.0035 (Figs. 3.4.19 and 20, Figs. 21 and 22 of IS 456:2000).


(iv) The tensile strength of concrete is ignored. Concrete has some tensile strength (very small but not zero). Yet, this tensile strength is ignored and the steel reinforcement is assumed to resist the tensile stress. However, the tensile strength of concrete is taken into account to check the deflection and crack widths in the limit state of serviceability. (v) The design stresses of the reinforcement are derived from the representative stress-strain curves as shown in Figs. 1.2.3 and 4 of Lesson 2 and Figs. 23A and B of IS 456:2000, for the type of steel used using the partial safety factor γm as 1.15. In the reinforced concrete structures, two types of steel are used: one with definite yield point (mild steel, Figs. 1.2.3 of Lesson 2 and Figs. 23B of IS 456:2000) and the other where the yield points are not definite (cold work deformed bars). The representative stress-strain diagram (Fig. 1.2.4 of Lesson 2 and Fig. 23A of IS 456:2000) defines the points between 0.8 fy and 1.0 fy in case of cold work deformed bars where the curve is inelastic. (vi) The maximum strain in the tension reinforcement in the section at failure shall not be less than fy/(1.15 Es) + 0.002, where fy is the characteristic strength of steel and Es = modulus of elasticity of steel (Figs. 3.4.19 and 20). This assumption ensures ductile failure in which the tensile reinforcement undergoes a certain degree of inelastic deformation before concrete fails in compression. 3.4.3 Singly Reinforce Rectangular Beams

Figure 3.4.18 shows the singly reinforced rectangular beam in flexure. The following notations are used (Figs. 3.4.19 and 20):

Ast = area of tension steel b = width of the beam C = total compressive force of concrete d = effective depth of the beam L = centre to centre distance between supports P = two constant loads acting at a distance of L/3 from the two supports

of the beam T = total tensile force of steel xu = depth of neutral axis from the top compression fibre

3.4.4 Equations of Equilibrium

The cross-sections of the beam under the applied loads as shown in Fig. 3.4.18 has three types of combinations of shear forces and bending moments: (i)


only shear force is there at the support and bending moment is zero, (ii) both bending moment (increasing gradually) and shear force (constant = P) are there between the support and the loading point and (iii) a constant moment (= PL/3) is there in the middle third zone i.e. between the two loads where the shear force is zero (Fig. 1.1.1 of Lesson 1). Since the beam is in static equilibrium, any cross-section of the beam is also in static equilibrium. Considering the cross-section in the middle zone (Fig. 3.4.18) the three equations of equilibrium are the following (Figs. 3.4.19 and 20): (i) Equilibrium of horizontal forces: Σ H = 0 gives T = C (3.1) (ii) Equilibrium of vertical shear forces: Σ V = 0 (3.2) This equation gives an identity 0 = 0 as there is no shear in the middle third zone of the beam. (iii) Equilibrium of moments: Σ M = 0, (3.3) This equation shows that the applied moment at the section is fully resisted by moment of the resisting couple T a = C a , where a is the operating lever arm between T and C (Figs. 3.4.19 and 20). 3.4.5 Computations of C and T


Figures 3.4.21a and b present the enlarged view of the compressive part of the strain and stress diagrams. The convex parabolic part of the stress block of Fig. 3.4.21b is made rectangular by dotted lines to facilitate the calculations adding another concave parabolic stress zone which is really non-existent as marked by hatch in Fig. 3.4.21b.

The different compressive forces C, C1, C2 and C3 and distances x1 to x5

and xu as marked in Fig. 3.4.21b are explained in the following:

C = Total compressive force of concrete = C1 + C2

C1 = Compressive force of concrete due to the constant stress of 0.446 fck and up to a depth of x3 from the top fibre

C2 = Compressive force of concrete due to the convex parabolic stress

block of values ranging from zero at the neutral axis to 0.446 fck at a distance of x3 from the top fibre

C3 = Compressive force of concrete due to the concave parabolic stress

block (actually non-existent) of values ranging from 0.446 fck at the neutral axis to zero at a distance of x3 from the top fibre

x1 = Distance of the line of action of C1 from the top compressive fibre

x2 = Distance of the line of action of C (= C1 + C2) from the top

compressive fibre

x3 = Distance of the fibre from the top compressive fibre, where the strain = 0.002 and stress = 0.446 fck



xu = Distance of the neutral axis from the top compressive fibre.

From the strain triangle of Fig. 3.4.21a, we have 570

74

0035000203 .

..

xxx

u

u ===− , giving

x3 = 0.43 xu (3.4) Since C1 is due to the constant stress acting from the top to a distance of x3, the distance x1 of the line of action of C1 is:


x1 = 0.5 x3 = 0.215 xu (3.5) From Fig. 3.4.21a: x5 = x3 +

43 (xu - x3) = 0.43 xu + 0.75(0.57 xu)

or x5 = 0.86 xu (3.6) The compressive force C1 due to the rectangular stress block is: C1 = b x3(0.446 fck) = 0.191 b xu fck (3.7) The compressive force C2 due to parabolic stress block is: C2 = b (xu - x3) 3

2 (0.446 fck) = 0.17 b xu fck

(3.8) Adding C1 and C2, we have C = C1 + C2 = 0.361 b xu fck = 0.36 b xu fck (say) (3.9) The non-existent compressive force C3 due to parabolic (concave) stress block is: C3 = b (xu - x3) 3

1 (0.446 fck) = 0.085 b xu fck

(3.10) Now, we can get x4 by taking moment of C2 and C3 about the top fibre as follows: C2(x4) + C3 (x5) = (C2 + C3) (x3 +

23xxu − )

which gives x4 = 0.64 xu (3.11) Similarly, x2 is obtained by taking moment of C1 and C2 about the top fibre as follows: C1(x1) + C2(x4) = C(x2) which gives x2 = 0.4153 xu


or x2 = 0.42 xu (say). (3.12) Thus, the required parameters of the stress block (Fig. 3.4.19) are C = 0.36 b xu fck (3.9) x2 = 0.42 xu (3.12) and lever arm = (d - x2) = (d - 0.42 xu) (3.13) The tensile force T is obtained by multiplying the design stress of steel with the area of steel. Thus, T = styst

y Af.A).f

( 870151

=

(3.14) 3.4.6 Practice Questions and Problems with Answers Q.1: How do the beams and slabs primarily carry the transverse loads ? A.1: The beams and slabs carry the transverse loads primarily by bending. Q.2: Name three different types of reinforced concrete beams and their specific

applications. A.2: They are:

(i) Singly reinforced and doubly reinforced rectangular beams - used in resisting negative moments in intermediate spans of continuous beams over the supports or elsewhere in slab-beam monolithic constructions, and positive moments in midspan of isolated or intermediate spans of beams with inverted slab (monolithic) constructions and lintels.

(ii) Singly reinforced and doubly reinforced T-beams - used in resisting

positive moments in isolated or intermediate spans (midspan) in slab-beam monolithic constructions and negative moments over the support for continuous spans with inverted slab (monolithic) constructions.


(iii) Singly reinforced and doubly reinforced L-beams - Same as (ii) above except that these are for end spans instead of intermediate spans.

Q.3: Name four parameters which determine the effective widths of T and L-beams. A.3: The four parameters are:

(i) isolated or continuous beams, (ii) the distance between points of zero moments in the beam, (iii) the breadth of the web, (iv) the thickness of the flange.

Q.4: Differentiate between one-way and two-way slabs. A.4: One-way slab spans in one direction and two-way slab spans in both the

directions. Slabs whose ratio of longer span (ly) to shorter span (lx) is more than 2 are called one-way. Slabs of this ratio up to 2 are called two-way slabs.

Q.5: State and explain the significance of the six assumptions of design of

flexural members employing limit state of collapse. A.5: Sec. 3.4.2 gives the full answer. Q.6: Draw a cross-section of singly reinforced rectangular beam and show the

strain and stress diagrams. A.6: Fig. 3.4.19. Q.7: Write the three equations of equilibrium needed to design the reinforced

concrete beams. A.7: Vide sec. 3.4.4 and Eqs. 3.1 to 3. Q.8: Write the final expression of the total compressive force C and tensile

force T for a rectangular reinforced concrete beam in terms of the designing parameters.

A.8: Eq. 3.9 for C and Eq. 3.14 for T.


3.4.7 References
















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 3.4.8 Test 4 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Tick the correct answer: (4 x 5 = 20 marks)


(i) Beams and slabs carry the transverse loads primarily by

(a) truss action (b) balance of shear action (c) bending (d) slab-beam interaction

A.TQ.1: (i): (c) (ii) The ratio of longer span (ly) to shorter span (lx) of a two-way slab is

(a) up to 2 (b) more than 2 (c) equal to 1 (d) more than 1

A.TQ.1: (ii): (a) (iii) An inverted T-beam is considered as a rectangular beam for the design

(a) over the intermediate support of a continuous beam where the bending moment is negative

(b) at the midspan of a continuous beam where the bending moment is

positive

(c) at the point of zero bending moment

(d) over the support of a simply supported beam A.TQ.1: (iii): (b) (iv) The maximum strain in the tension reinforcement in the section at failure

shall be

(a) more than fy /(1.15 Es) + 0.002 (b) equal to 0.0035

(c) more than fy /Es + 0.002

(d) less than fy /(1.15 Es) + 0.002

A.TQ.1: (iv): (d)


TQ.2: Draw a cross-section of singly reinforced rectangular beam and show the strain and stress diagrams. (10)

A.TQ.2: Fig. 3.4.19 TQ.3: Name four parameters which determine the effective widths of T and L-beams. (6) A.TQ.3: The four parameters are:

(i) isolated or continuous beams, (ii) the distance between points of zero moments in the beam, (iii) the breadth of the web, (iv) the thickness of the flange.

TQ.4: Derive the final expressions of the total compressive force C and tensile

force T for a rectangular reinforced concrete beam in terms of the designing parameters.

(10 + 4 = 14)

A.TQ.4: Section 3.4.5 is the full answer. 3.4.9 Summary of this Lesson Lesson 4 illustrates the primary load carrying principle in a slab-beam structural system subjected to transverse loadings. It also mentions three different types of singly and doubly reinforced beams normally used in construction. Various assumptions made in the design of these beams employing limit state of collapse are explained. The stress and strain diagrams of a singly reinforced rectangular beam are explained to write down the three equations of equilibrium. Finally, the computations of the total compressive and tensile forces are illustrated.


Module 3


Examples)


Lesson 5

Determination of Neutral Axis Depth and

Computation of Moment of Resistance


Instructional Objectives: At the end of this lesson, the student should be able to: • determine the depth of the neutral axis for a given cross-section with known

value of Ast and grades of steel and concrete, • write and derive the expressions of xu,max, pt,lim, Mu,lim/bd2 and state the

influences of grades of steel and concrete on them separately, • derive the corresponding expression of Mu when xu < xu,max, xu = xu,max and xu

> xu,max, • finally take a decision if xu > xu,max. 3.5.1 Introduction After learning the basic assumptions, the three equations of equilibrium and the computations of the total compressive and tensile forces in Lesson 4, it is now required to determine the depth of neutral axis (NA) and then to estimate the moment of resistance of the beams. These two are determined using the two equations of equilibrium (Eqs. 3.1 and 3.3). It has been explained that the depth of neutral axis has important role to estimate the moment of resistance. Accordingly, three different cases are illustrated in this lesson. 3.5.2 Computation of the Depth of Neutral Axis xu From Eqs. 3.1, 9 and 14, we have 0.87 fy Ast = 0.36 b xu fck (3.15) or

ck

styu fb.

Af.x

360870

=

(3.16) We can also write:

ck

styufdb.Af.

dx

360870

=

(3.17)


Substituting the expression of dxu from Eq. 3.17 into Eq. 3.13 of Lesson 4, we

have lever arm =

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛−

dbf.Af.

.dck

sty360870

4201

= ⎟

⎟⎠

⎞⎜⎜⎝

⎛−

dbffA

.dck

yst01511

(3.18) Ignoring multiplying factor 1.015 in Eq. 3.18, we have lever arm = ⎟

⎟⎠

⎞⎜⎜⎝

⎛−

dbffA

dck

yst1

(3.19) 3.5.3 Limiting Value of xu (= xu, max) Should there be a limiting or maximum value of xu? Equation 3.17 reveals that

dxu increases with the increase of percentage

of steel reinforcement db

Ast for fixed values of fy and fck. Thus, the depth of the

neutral axis xu will tend to reach the depth of the tensile steel. But, that should not be allowed. However, let us first find out that value of xu which will satisfy assumptions (ii) and (vi) of sec. 3.4.2 and designate that by xu, max for the present, till we confirm that xu should have a limiting value.

Version 2 C IIT, Kharagpur

Figure 3.5.1 presents the strain diagrams for the three cases: (i) when xu = xu, max ; (ii) when xu is less than xu, max and (iii) when xu is greater than xu, max . The following discussion for the three cases has the reference to Fig. 3.5.1. (i) When xu = xu, max

The compressive strain at the top concrete fibre = 0.0035 and the tensile strain at the level of steel = ⎟

⎟⎠

⎞⎜⎜⎝

⎛+ 0020

870.

Ef.

s

y . Thus, it satisfies the assumptions (ii)

and (vi) of sec. 3.4.2. (ii) When xu is less than xu, max

There are two possibilities here:


(a) If the compressive strain at the top fibre = 0.0035, the tensile strain is more than ⎟

⎟⎠

⎞⎜⎜⎝

⎛+ 0020

870.

Ef.

s

y . Thus, this possibility satisfies the two assumptions (ii)

and (vi) of sec. 3.4.2.

(b) When the steel tensile strain is ⎟⎟⎠

⎞⎜⎜⎝

⎛+ 0020

870.

Ef.

s

y , the compressive

concrete strain is less than 0.0035. Here also, both the assumptions (ii) and (vi) of sec. 3.4.2 are satisfied. (iii) When xu is more than xu, max There are two possibilities here:

(a) When the top compressive strain reaches 0.0035, the tensile steel strain is less than ⎟

⎟⎠

⎞⎜⎜⎝

⎛+ .0020

870

s

yE

f. . It violets the assumption (vi) though

assumption (ii) of sec. 3.4.2 is satisfied.

(b) When the steel tensile strain is ⎟⎟⎠

⎞⎜⎜⎝

⎛+ 0020

870.

Ef.

s

y , the compressive strain

of concrete exceeds 0.0035. Thus, this possibility violets assumption (ii) though assumption (vi) is satisfied. The above discussion clearly indicates that the depth of xu should not become more than xu, max. Therefore, the depth of the neutral axis has a limiting or maximum value = xu, max. Accordingly, if the Ast provided yields xu > xu, max, the section has to be redesigned. Since xu depends on the area of steel, we can calculate Ast, lim from Eq. 3.17. From Eq. 3.17 (using xu = xu, max and Ast = Ast, lim), we have

ck

,sty,ufdb.

Af.d

x360

870 limmax =

or )100(870

360 )100( max

mm

⎟⎟⎠

⎞⎜⎜⎝

⎛==

df.xf.

pdb

A

y

,uckt,li

li,st (3.20)

In the above equation

dx ,u max can be obtained from the strain diagram of Fig.

3.5.1 as follows:


00550

87000350max

.E

f..

d

x

s

y

,u

+=

(3.21) 3.5.4 Values of

dx ,u max and pt, lim

Equation 3.20 shows that the values of pt, lim depend on both the grades of steel and concrete, while Eq. 3.21 reveals that

dx ,u max depends on the grade of

steel alone and not on the grade of concrete at all. The respective values of pt, lim for the three grades of steel and the three grades of concrete are presented in Table 3.1. Similarly, the respective values of

dx ,u max for three grades of steel are

presented in Table 3.2. Table 3.1 Values of pt, lim

fck in N/mm2 fy = 250 N/mm2 fy = 415 N/mm2 fy = 500 N/mm2

20 1.76 0.96 0.76 25 2.20 1.19 0.94 30 2.64 1.43 1.13

Table 3.2 Values of

dx ,u max

fy in N/mm2 250 415 500

dx ,u max 0.531 = 0.53 (say) 0.479 = 0.48 (say) 0.456 = 0.46 (say)

A careful study of Tables 3.1 and 3.2 reveals the following:

(i) The pt, lim increases with lowering the grade of steel for a particular grade of concrete. The pt, lim, however, increases with increasing the grade of concrete for a specific grade of steel. (ii) The maximum depth of the neutral axis xu,max increases with lowering the grade of steel. That is more area of the section will be utilized in taking the compression with lower grade of steel. 3.5.5 Computation of Mu Equation 3.3 of Lesson 4 explains that Mu can be obtained by multiplying the tensile force T or the compressive force C with the lever arm. The expressions of C, lever arm and T are given in Eqs. 3.9, 3.13 (also 3.19) and


3.14 respectively of Lesson 4. Section 3.5.3 discusses that there are three possible cases depending on the location of xu. The corresponding expressions of Mu are given below for the three cases: (i) When xu < xu, max Figure 3.5.1 shows that when concrete reaches 0.0035, steel has started flowing showing ductility (Strain > 0020

870.

Ef.

s

y + ). So, the computation of Mu is to

be done using the tensile force of steel in this case. Therefore, using Eqs. 3.13 and 3.14 of Lesson 4, we have

Mu = T (lever arm) = 0.87 fy Ast (d - 0.42 xu) (3.22) Substituting the expressions of T and lever arm from Eqs. 3.14 of Lesson 4 and 3.19 respectively we get, ⎟

⎟⎠

⎞⎜⎜⎝

⎛−=

dbffA

dAf.Mck

yststyu 1870

(3.23) (ii) When xu = xu, max From Fig. 3.5.1, it is seen that steel just reaches the value of 0020

870.

Ef.

s

y +

and concrete also reaches its maximum value. The strain of steel can further increase but the reaching of limiting strain of concrete should be taken into consideration to determine the limiting Mu as Mu, lim here. So, we have Mu, lim = C (lever arm) Substituting the expressions of C and lever arm from Eqs. 3.9 of Lesson 4 and 3.19 respectively, we have ck

,u,u,u fdb

d

x.

d

x.M 2maxmax

lim 4201360 ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

(3.24) (iii) When xu > xu, max In this case, it is seen from Fig. 3.5.1 that when concrete reaches the strain of 0.0035, tensile strain of steel is much less than ( 0020

870.

Ef.

s

y + ) and any

further increase of strain of steel will mean failure of concrete, which is to be


avoided. On the other hand, when steel reaches 0020870

.E

f.

s

y + , the strain of

concrete far exceeds 0.0035. Hence, it is not possible. Therefore, such design is avoided and the section should be redesigned. However, in case of any existing reinforced concrete beam where xu > xu,

max, the moment of resistance Mu for such existing beam is calculated by restricting xu to xu, max only and the corresponding Mu will be as per the case when xu = xu, max. 3.5.6 Computation of Limiting Moment of Resistance Factor Equation 3.24 shows that a particular rectangular beam of given dimensions of b and d has a limiting capacity of Mu, lim for a specified grade of concrete. The limiting moment of resistance factor R,lim (= Mu,lim/bd2) can be established from Eq. 3.24 as follows: ck

,u,u,u, f

d

x.

d

x.

db

MR ⎟

⎟⎠

⎞⎜⎜⎝

⎛−== maxmax

2lim

lim 4201360

(3.25) It is seen that the limiting moment of resistance factor R,lim depends on

dx ,u max and fck. Since

dx ,u max depends on the grade of steel fy, we can say that

R,lim depends on fck and fy. Table 3.3 furnishes the values of R,lim for three grades of concrete and three grades of steel.

Table 3.3 Limiting values of 2m

m, db

MR li,u

li = factors (in N/mm2)

fck in N/mm2 fy = 250 N/mm2 fy = 415 N/mm2 fy = 500 N/mm2

20 2.98 2.76 2.66 25 3.73 3.45 3.33 30 4.47 4.14 3.99

A study of Table 3.3 reveals that the limiting moment of resistance factor R,lim increases with higher grade of concrete for a particular grade of steel. It is also seen that this factor increases with lowering the grade of steel for a particular grade of concrete. The increase of this factor due to higher grade of concrete is understandable. However, such increase of the factor with lowering the grade of steel is explained below:


Lowering the grade of steel increases the d

x ,u max (vide Table 3.2) and this

enhanced d

x ,u max increases Mu as seen from Eq. 3.24. However, one may argue

that Eq. 3.24 has two terms: d

x ,u max and ⎟⎟⎠

⎞⎜⎜⎝

⎛−

d

x ,u max42.01 .

With the increase of d

x ,u max , ⎟⎟⎠

⎞⎜⎜⎝

⎛−

dx

. ,u max4201 is decreasing. Then how

do we confirm that the product is increasing with the increase of d

x ,u max ? Actual

computation will reveal the fact. Otherwise, it can be further explained from Table 3.1 that as the grade of steel is lowered for a particular grade of concrete, the pt,

lim gets increased. Therefore, amount of steel needed to have Mu, lim with lower grade of steel is higher. Thus, higher amount of steel and higher values of

dx ,u max

show higher 2lim

db

M ,u factor with the lowering of grade of steel for a particular grade

of concrete (see Table 3.3). 3.5.7 Practice Questions and Problems with Answers Q.1: Which equation is needed to determine the depth of the neutral axis? A.1: Eq. 3.16 or 3.17. Q.2: How to find the lever arm? A.2: Eq. 3.13 of Lesson 4 or Eq. 3.19. Q.3: Should there be limiting or maximum value of up? If so, why? What is the

equation to find the maximum value of xu? How to find Ast, lim for such case?

A.3: Sec. 3.5.3 is the complete answer. Q.4: State the effects of grades of concrete and steel separately on pt, lim and

dx ,u max .

A.4: (i) pt, lim (see Eq. 3.20 and Table 3.1)

(a) pt, lim increases with lowering the grade of steel for a particular grade of concrete


(b) pt, lim increases with increasing the grade of concrete for a particular grade of steel

(ii)

dx ,u max (see Eq. 3.21 and Table 3.2)

(a)

dx ,u max increases with lowering the grade of steel

(b)

dx ,u max is independent on the grade of concrete.

Q.5: Write the corresponding expression of Mu when (i) xu < xu, max; and (ii) xu = xu, max A.5: (i) Eq. 3.23, when xu < xu, max (ii) Eq. 3.24, when xu = xu, max Q.6: What is to be done if xu > xu, max and why ? A.6: When xu > xu, max, the section has to be redesigned as this does not

ensure ductile failure of the beam. Q.7: Write the expression of limiting moment of resistance factor R,lim = 2

lim

db

M ,u .

What is its unit? A.7: ck

,u,u,u, f

dx

.d

x.

db

MR ⎟

⎟⎠

⎞⎜⎜⎝

⎛−== maxmax

2lim

lim 4201360 (Eq. 3.25). Its unit is

N/mm2. Q.8: State separately the effects of grades of concrete and steel on the limiting

moment of resistance factor R,lim = 2lim

db

M ,u .

A.8: (i) R,lim increases with increasing the grade of concrete for a particular

grade of steel (see Table 3.3). (ii) R,lim increases with lowering the grade of steel for a particular grade of

concrete (see Table 3.3).


3.5.8 References
















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 3.5.9 Test 5 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Tick the correct answer: (4 x 5 =

20) Version 2 CE IIT, Kharagpur

(i) The depth of the neutral axis is calculated from the known area of steel and it

should be

(a) less than 0.5 times the full depth of the beam (b) more than 0.5 times the effective depth of the beam

(c) less than or equal to limiting value of the neutral axis depth

(d) less than 0.43 times the effective depth of the beam

A.TQ.1: (i): (c) (ii) For a particular grade of concrete and with lowering the grade of steel, the

pt,lim

(a) increases (b) decreases

(c) sometimes increases and sometimes decreases

(d) remains constant

A.TQ.1: (ii): (a) (iii) For a particular grade of steel and with increasing the grade of concrete, the

pt,lim

(a) decreases (b) increases

(c) remains constant

(d) sometimes increases and sometimes decreases

A.TQ.1: (iii): (b) (iv) Which of the statements is correct?

(a) xu,max /d is independent of grades of concrete and steel (b) xu,max /d is independent of grade of steel but changes with grade of

steel


(c) xu,max /d changes with the grade of concrete and steel

(d) xu,max /d is independent of the grade of concrete and changes with

the grade of steel A.TQ.1: (iv): (d) TQ.2: Derive the expression of determining the depth of neutral axis and lever

arm of a singly reinforced rectangular beam with known quantity of tension steel. (10)

A.TQ.2: Section 3.5.2 is the complete answer. TQ.3: Establish the expressions of the moment of resistance of a singly

reinforced rectangular beam when (i) xu < xu,max , (ii) xu = xu,max and (iii) xu > xu,max.

(8+7=15)

A.TQ.3: Section 3.5.5 is the complete answer. TQ.4: Derive the expression of limiting moment of resistance factor and explain

how it is influenced by the grades of concrete and steel. (5)

A.TQ.4: Section 3.5.6 is the full answer. 3.5.10 Summary of this Lesson Understanding the various assumptions of the design, stress-strain diagrams of concrete and steel, computations of the total compressive and tensile forces and the equations of equilibrium in Lesson 4, this lesson illustrates the applications of the two equations of equilibrium. Accordingly, the depth of neutral axis and the moment of resistance of the beam can be computed with the expressions derived in this lesson. Different cases that arise due to different values of the depth of neutral axis are discussed to select the particular expression of the moment of resistance. Further, the expressions of determining the limiting values of percentage of steel, depth of neutral axis, moment of resistance and moment of resistance factor are established. The influences of grades of concrete and steel on them are also illustrated.


Module 3


Examples)


Lesson 6

Numerical Problems on Singly Reinforced

Rectangular Beams Version 2 CE IIT, Kharagpur

Instructional Objectives: At the end of this lesson, the student should be able to:

• identify the main two types of problems of singly reinforced rectangular

sections, • name the inputs and outputs of the two types of problems, • state the specific guidelines of assuming the breadth, depths, area of steel

reinforcement, diameter of the bars, grade of concrete and grade of steel, • determine the depth of the neutral axis for specific dimensions of beam

(breadth and depth) and amount of reinforcement, • identify the beam with known dimensions and area of steel if it is under-

reinforced or over-reinforced, • apply the principles to design a beam. 3.6.1 Types of Problems Two types of problems are possible: (i) design type and (ii) analysis type. In the design type of problems, the designer has to determine the dimensions b, d, D, Ast (Fig. 3.6.1) and other detailing of reinforcement, grades of concrete and steel from the given design moment of the beam. In the analysis type of the problems, all the above data will be known and the designer has to find out the moment of resistance of the beam. Both the types of problems are taken up for illustration in the following two lessons.


3.6.2 Design Type of Problems The designer has to make preliminary plan lay out including location of the beam, its span and spacing, estimate the imposed and other loads from the given functional requirement of the structure. The dead loads of the beam are estimated assuming the dimensions b and d initially. The bending moment, shear force and axial thrust are determined after estimating the different loads. In this illustrative problem, let us assume that the imposed and other loads are given. Therefore, the problem is such that the designer has to start with some initial dimensions and subsequently revise them, if needed. The following guidelines are helpful to assume the design parameters initially. 3.6.2.1 Selection of breadth of the beam b Normally, the breadth of the beam b is governed by: (i) proper housing of reinforcing bars and (ii) architectural considerations. It is desirable that the width of the beam should be less than or equal to the width of its supporting structure like column width, or width of the wall etc. Practical aspects should also be kept in mind. It has been found that most of the requirements are satisfied with b as 150, 200, 230, 250 and 300 mm. Again, width to overall depth ratio is normally kept between 0.5 and 0.67. 3.6.2.2 Selection of depths of the beam d and D The effective depth has the major role to play in satisfying (i) the strength requirements of bending moment and shear force, and (ii) deflection of the beam. The initial effective depth of the beam, however, is assumed to satisfy the deflection requirement depending on the span and type of the reinforcement. IS 456 stipulates the basic ratios of span to effective depth of beams for span up to 10 m as (Clause 23.2.1) Cantilever 7 Simply supported 20 Continuous 26 For spans above 10 m, the above values may be multiplied with 10/span in metres, except for cantilevers where the deflection calculations should be made. Further, these ratios are to be multiplied with the modification factor depending on reinforcement percentage and type. Figures 4 and 5 of IS 456 give the different values of modification factors. The total depth D can be determined by adding 40 to 80 mm to the effective depth. 3.6.2.3 Selection of the amount of steel reinforcement Ast The amount of steel reinforcement should provide the required tensile force T to resist the factored moment Mu of the beam. Further, it should satisfy


the minimum and maximum percentages of reinforcement requirements also. The minimum reinforcement As is provided for creep, shrinkage, thermal and other environmental requirements irrespective of the strength requirement. The minimum reinforcement As to be provided in a beam depends on the fy of steel and it follows the relation: (cl. 26.5.1.1a of IS 456)

yf

.dbsA 850

=

(3.26) The maximum tension reinforcement should not exceed 0.04 bD (cl. 26.5.1.1b of IS 456), where D is the total depth. Besides satisfying the minimum and maximum reinforcement, the amount of reinforcement of the singly reinforced beam should normally be 75 to 80% of

pt, lim. This will ensure that strain in steel will be more than )0020870

( .E

f.

s

y + as

the design stress in steel will be 0.87 fy. Moreover, in many cases, the depth required for deflection becomes more than the limiting depth required to resist Mu,

lim. Thus, it is almost obligatory to provide more depth. Providing more depth also helps in the amount of the steel which is less than that required for Mu, lim. This helps to ensure ductile failure. Such beams are designated as under-reinforced beams. 3.6.2.4 Selection of diameters of bar of tension reinforcement Reinforcement bars are available in different diameters such as 6, 8, 10, 12, 14, 16, 18, 20, 22, 25, 28, 30, 32, 36 and 40 mm. Some of these bars are less available. The selection of the diameter of bars depends on its availability, minimum stiffness to resist while persons walk over them during construction, bond requirement etc. Normally, the diameters of main tensile bars are chosen from 12, 16, 20, 22, 25 and 32 mm. 3.6.2.5 Selection of grade of concrete Besides strength and deflection, durability is a major factor to decide on the grade of concrete. Table 5 of IS 456 recommends M 20 as the minimum grade under mild environmental exposure and other grades of concrete under different environmental exposures also. 3.6.2.6 Selection of grade of steel Normally, Fe 250, 415 and 500 are in used in reinforced concrete work. Mild steel (Fe 250) is more ductile and is preferred for structures in earthquake zones or where there are possibilities of vibration, impact, blast etc.


3.6.3 Design Problem 3.1

Design a simply supported reinforced concrete rectangular beam (Fig. 3.6.2) whose centre to centre distance between supports is 8 m and supported on brick walls of 300 mm thickness. The beam is subjected to imposed loads of 7.0 kN/m. 3.6.4 Solution by Direct Computation Method The unknowns are b, d, D, Ast, grade of steel and grade of concrete. It is worth mentioning that these parameters have to satisfy different requirements and they also are interrelated. Accordingly, some of them are to be assumed which subsequently may need revision. 3.6.4.1 Grades of steel and concrete Let us assume Fe 415 and M 20 are the grades of steel and concrete respectively. As per clause 6.1.2 and Table 5 of IS 456, minimum grade of concrete is M 20 for reinforced concrete under mild exposure (durability requirement). 3.6.4.2 Effective span Leff Clause 22.2(a) of IS 456 recommends that the effective span is the lower of (i) clear span plus effective depth and (ii) centre to centre distance between two supports. Here, the clear span is 7700 mm. Thus


(i) Clear span + d = 7700 + 400 (assuming d = 400 from the specified ratio of span to effective depth as 20 and mentioned in the next section)

(ii) Centre to centre distance between two supports = 8000 mm. Hence, Leff = 8000 mm 3.6.4.3 Percentage of steel reinforcement pt The percentage of steel reinforcement to be provided is needed to determine the modification factor which is required to calculate d. As mentioned earlier in sec. 3.6.2.3, it is normally kept at 75 to 80 per cent of pt, lim. Here, pt,

lim = 0.96 (vide Table 3.1 of Lesson 5). So, percentage of steel to be provided is assumed = 0.75 (0.96) = 0.72. 3.6.4.4 Effective depth d As per clause 23.2.1 of IS 456, the basic value of span to effective depth ratio here is 20. Further, Fig. 4 of IS 456 presents the modification factor which will be multiplied with the basic span to effective depth ratio. This modification factor is determined on the value of fs where

Area of cross-section of steel required0 58Area of cross-section of steel provided

f . fs y=

= 0.58 fy (assuming that the Ast provided is the same as Ast required) = 0.58 (415) = 240.7 N/mm2. From Fig. 4 of IS 456, the required modification factor is found to be 1.1 for fs = 240.7 N/mm2 and percentage of steel = 0.72. So, the span to effective depth ratio = 22 as obtained by multiplying 20 with 1.1. Accordingly, the effective depth = 8000/22 = 363.63 mm, say 365 mm. Since this value of d is different from the d assumed at the beginning, let us check the effective span as lower of (i) 7700 + 365 and (ii) 8000 mm. Thus, the effective span remains at 8000 mm. Adding 50 mm with the effective depth of 365 mm (assuming 50 mm for cover etc.), the total depth is assumed to be 365 + 50 = 415 mm. 3.6.4.5 Breadth of the beam b Let us assume b = 250 mm to get b/D = 250/415 = 0.6024, which is acceptable as the ratio of b/D is in between 0.5 and 0.67. 3.6.4.6 Dead loads, total design loads Fd and bending moment


With the unit weight of reinforced concrete as 25 kN/m3 (cl. 19.2.1 of IS 456):

Dead load of the beam = 0.25 (0.415) (25) kN/m = 2.59 kN/m Imposed loads = 7.00 kN/m

Thus, total load = 9.59 kN/m, which gives factored load Fd as 9.59 (1.5) (partial safety factor for dead load and imposed load as 1.5) = 14.385 kN/m. We have, therefore, Mu = Factored bending moment = 14.385 (8) = 115.08 kNm. 3.6.4.7 Checking of effective depth d It is desirable to design the beam as under-reinforced so that the ductility is ensured with steel stress reaching the design value. Let us now determine the limiting effective depth when xu = xu, max and the factored moment Mu = Mu, lim = 115.08 kNm from Eq. 3.24 of Lesson 5.

cku,u,

u, fdbd

x.

dx

.M 2maxmaxlim 4201360

⎭⎬⎫

⎩⎨⎧−=

(3.24)

Table 3.2 of Lesson 5 givesd

u,x max = 0.479 for fy = 415 N/mm2. Thus:

(115.08) 106 Nmm = 0.36(0.479) [1 - 0.42(0.479)] b d2 (20) which gives d = 408.76 mm So, let us revise d = 410 mm from the earlier value of 365 mm to have the total depth = 410 + 50 = 460 mm. 3.6.4.8 Area of Steel Ast The effective depth of the beam has been revised to 408.76 mm from the limiting moment carrying capacity of the beam. Increasing that depth to 410 also has raised the Mu, lim of the beam from the design factored moment of 115.08 kNm. Therefore, the area of steel is to be calculated from the moment equation (Eq. 3.23 of Lesson 5), when steel is ensured to reach the design stress fd = 0.87 (415) = 361.05 N/mm2.

⎭⎬⎫

⎩⎨⎧

−=dbf

fAdAf.M

ck

yststyu 1870

(3.23)


Here, all but Ast are known. However, this will give a quadratic equation of Ast and one of the values, the lower one, will be provided in the beam. The above equation gives:

Nmm41025020

4151410415870Nmm)10(08115 6

⎭⎬⎫

⎩⎨⎧

−=)()(

)(A)(A)(.. st

st

= 148030.5 - 29.96715 stA 2

stA or - 4939.759 + 3840205 = 0 2

stA stA which gives = 966.5168 mmstA 2 or 3973.2422 mm2

The values of xu determined from Eq. 3.16 of Lesson 5 are 193.87 mm and 796.97 mm respectively, when Ast = 966.5168 mm2 and 3973.2422 mm2. It is seen that the value of xu with lower value of Ast is less than xu,max (= 216 mm). However, the value of xu with higher value of Ast (= 3973.2422 mm2) is more than the value of xu,max (= 0.48 d = 216 mm), which is not permissible as it exceeds the total depth of the beam (= 460 mm). In some problems, the value of xu may be less than the total depth of the beam, but it shall always be more than xu,max. The beam becomes over-reinforced. Therefore, the lower value of the area of steel is to be accepted as the tensile reinforcement out of the two values obtained from the solution of the quadratic equation involving Ast. Accepting the lower value of = 966.5168 mmstA 2, the percentage of steel becomes

0.9429(410)250

100)(966.5168= per cent

This percentage is higher than the initially assumed percentage as 0.72. By providing higher effective depth, this can be maintained as shown below. 3.6.4.9 Increase of effective depth and new Ast Increasing the effective depth to 450 mm from 410 mm, we have from Eq. 3.23 of Lesson 5,


⎭⎬⎫

⎩⎨⎧

−=)450( )250(20

4151)450()415(870)10(08115 6 )(A

A.. stst

= 162472.5 - 29.967148 stA 2

stA or - 5421.6871 + 3840205.2 = 0 2

stA stA or = 0.5 {5421.6871 ± 3746.1808} stA The lower value of now becomes 837.75315 which gives the percentage of

as stA

stA

7446.0)450(250

)100(75315.837= , which is close to earlier assumed percentage of

0.72. Therefore, let us have d = 450 mm, D = 500 m, b = 250 mm and = 837.75315 mm

stA2 for this beam.

For any design problem, this increase of depth is obligatory to satisfy the deflection and other requirements. Moreover, obtaining with increased depth employing moment equation (Eq. 3.23 of Lesson 5) as illustrated above, results in under-reinforced beam ensuring ductility.

stA

3.6.4.10 Further change of Ast due to increased dead load However, increasing the total depth of the beam to 500 mm from earlier value of 415 mm has increased the dead load and hence, the design moment Mu. This can be checked as follows:

The revised dead load = 0.25 (0.5) (25) = 3.125 kN/m Imposed loads = 7.00 kN/m Total factored load Fd = 1.5(10.125) = 15.1875 kN/m Mu = 15.1875 (8) = 121.5 kNm

The limiting moment that this beam can carry is obtained from using Mu, lim/bd2 factor as 2.76 from Table 3.3 of of Lesson 5. Thus, Mu, lim = (2.76) bd2 = (2.76) (250) (450)2 Nmm


= 139.72 kNm > (Mu = 121.5 kNm) Hence, it is under-reinforced beam. Equation 3.23 of Lesson 5 is now used to determine the for MstA u = 121.5 kNm

⎭⎬⎫

⎩⎨⎧−=

dbffA

dAf.Mck

yststyu 1870

(3.23)

or ⎭⎬⎫

⎩⎨⎧−=

)450()250(20)415(

1)450()415(870)10(5121 6 stst

AA..

= 162472.5 - 29.96715 stA 2

stA or = 0.5 {5421.6867 ± 3630.0038} = 895.84145 mmstA 2

The steel reinforcement is 450)(250

100)(895.84 = 0.7963 per cent which is 83 per cent

of pt,lim. So, we have the final parameters as b = 250 m, d = 450 mm, D = 500 mm,

= 895.84 mmstA 2. A selection of 2-20 T bars and 2-14 T bars gives the = 935 mm

stA2 (Fig. 3.6.3). Though not designed, Fig. 3.6.3 shows the holder bars

and stirrups also.


3.6.4.11 Summary of steps Table 3.4 presents the complete solution of the problem in eleven steps. Six columns of the table indicate (i) parameters assumed/determined, (ii) if they need revision, (iii) final parameters, (iv) major requirements of the parameter, (v) reference section numbers, and (vi) reference source material. Table 3.4 Steps of the illustrative problem Step

Assumed/ determined parameter(s)

(i)

If need(s) revision

(ii)

Final parameter(s)

(iii)

Major requirement of the parameter(s)

(iv)

Reference section number

(v)

Reference source Material(s)

(vi)

1 fck, fy No fck, fy Durability for fck and ductility for fy

3.6.4.1 cl.6.1.2, cl. 8 and Table 5 of IS 456

2 d Yes No 20spanc/c

=d 3.6.4.2 cl. 23.2 of IS 456

3 Leff Yes No Boundary conditions

3.6.4.2 cl.22.2 of IS 456

4 p = Ast/bd No Yes Ductility (p = 75 to 80% of pt, lim)

3.6.4.3 Table 3.1 of Lesson 5 for pt, lim

5 d Yes No Control of deflection

3.6.4.4 cl.23.2 of IS 456

6 D, b Yes for D

b Economy 3.6.4.5 D = d + (40 to 80 mm) b = (0.5 to 0.67)D

7 Fd, Mu Yes No Strength 3.6.4.6 Strength of material books

8 d Yes No Limiting depth considering Mu = Mu,lim

3.6.4.7 Eq. 3.24 of Lesson 5

9 Ast Yes No Strength 3.6.4.8 Eq. 3.23 of Lesson 5

10 d, D, Ast No d, D, Leff Under-reinforced 3.6.4.9 D = d + 50 Eq. 3.23 of Lesson 5

11 Ast No Ast Strength 3.6.4.10 Eq. 3.23 of Lesson 5


3.6.5 Use of Design Aids From the solution of the illustrative numerical problems, it is clear that b, d, D and Ast are having individual requirements and they are mutually related. Thus, any design problem has several possible sets of these four parameters. After getting one set of values, obtaining the second set, however, involves the same steps as those of the first one. The steps are simple but time consuming and hence, the designer may not have interest to compare between several sets of these parameters. The client, contractor or the architect may request for alternatives also. Thus, there is a need to get several sets of these four parameters as quickly as possible. One way is to write a computer program which also may restrict average designer not having a computer. Bureau of Indian Standard (BIS), New Delhi published SP-16, Design Aids for Reinforced Concrete to IS: 456, Special Publication No. 16, which is very convenient to get several sets of these values quickly. SP-16 provides both charts (graphs) and tables explaining their use with illustrative examples. On top left or right corner of these charts and tables, the governing parameters are provided for which that chart/table is to be used. 3.6.6 Solution by using Design Aids Charts (SP-16) The initial dimension of effective depth d of Design Problem 3.1 is modified from 400 mm to 410 mm first to satisfy the deflection and other requirements and then to 450 mm as the final dimension. While using only the charts or tables of SP-16, the final results as obtained for this problem by direct calculation method will not be available. So, we will assume the percentage of steel as 0.75 (0.96) = 0.72 initially. 3.6.6.1 Effective depth d Chart 22 of SP-16 for fy = 415 N/mm2 and fck = 20 N/mm2 gives maximum ratio of span to effective depth as 21.5 when the percentage of steel assumed = 0.75 (0.96) = 0.72. Thus, we get effective depth d = 8000/21.5 = 372.09 mm with d = 372.09 mm and effective span Leff = 8000 mm. Total depth D = 372.09 + 50 = 422.09 = 425 mm (say). 3.6.6.2 Breadth b and factored moment Mu Here also b = 250 mm is assumed and accordingly,

Dead load = 0.25 (0.425) (25) = 2.66 kN/m Imposed loads = 7.00 kN/m Total factored load, Fd = 1.5 (9.66) = 14.50 kN/m


Factored bending moment = (14.5) (8) = 116.00 kN/m 3.6.6.3 Checking of effective depth d and area of steel Ast Chart 14 of SP-16 is for fck = 20 N/mm2, fy = 415 N/mm2 and d varying from 300 to 550 mm. For this problem, Mu per metre width of the beam =

. For the percentage of reinforcement = 0.72, chart 14 gives d = 460 mm and then D = 510 mm. Area of steel reinforcement 0.72 (25) (460)/100 = 828 mm

kNm/m464

2. As in the earlier problem, the increased dead load due to the increased D to 510 mm is checked below: Revised dead load = 0.25 (0.51) (25) = 3.188 kN/m Imposed loads = 7.000 kN/m Total factored load Fd = 1.5 (10.188) = 15.282 kN/m Factored moment Mu = 15.282 (8) = 122.256 kN/m Mu per metre width of the beam = 122.256/0.25 = 489.02 kNm/m. Chart 14 of Sp-16 gives the effective depth of the beam d = 472 mm and D = 475 + 50 = 525 mm assuming d = 475 mm. Ast required = (0.72/100) (250) (475) = 855.0 mm2

Thus, we have b = 250 mm, d = 475 mm, D = 525 mm and Ast = 855 mm2

3.6.7 Solution by using Design Aids Tables (SP-16) 3.6.7.1 Effective depth d Tables 1 to 4 of SP-16 present pt for different values of Mu/bd2 covering a wide range of fy and fck. Table 2 is needed for this problem. To have more confidence while employing this method, we are starting with the effective depth d as 400 mm as in the direct computational method. The total depth D is (400 + 50) mm = 450 mm. The breadth b of the beam is taken as 250 mm. 3.6.7.2 Factored load and bending moment Dead load = 0.25 (0.45) (25) = 2.8125 kN/m


Imposed loads = 7.00 kN/m Factored load Fd = 1.5 (2.8125 + 7.00) = 14.71875 kN/m Factored bending moment Mu = 14.71875 (8) = 117.75 kNm

943752)400()400(250

)10(75117 6

2 ..db

M u ==

3.6.7.3 Use of Tables of SP-16

Table 2 of SP-16 shows that 2dbMu is restricted up to 2.76 when pt =

0.955, i.e. the limiting condition. So, increasing the effective depth by another 50 mm to have D = 500 m, the total factored moment as calculated in sec. 3.6.4.10 is 121.5 kNm,

Now, 42)450()450(250

)10(5121 6

2 ..db

M u ==

From Table 2 of SP-16, the corresponding pt becomes 0.798. Therefore, Ast = 0.01 (0.798) (250) (450) = 897.75 mm2 3.6.8 Comparison of Results of Three Methods Results of this problem by three methods: (i) direct computation method, (ii) use of charts of SP-16 and (iii) use of tables of SP-16 are summarised for the purpose of comparison. The tabular summary includes the last two values of d and Ast. Other parameters (b, fck and fy) are remaining constants in all the three methods. Table 3.5 Comparison of d and Ast by three methods

Direct computation method

Use of charts of SP-16 Use of tables of SP-16

Cycle

d (mm) Ast (mm2) d (mm) Ast (mm2) d (mm) Ast (mm2) 1 410 966.5168 460 828 400 Not

possible 2 450 895.84145

(2-20+2-14 = 935 mm2)

475 855 (2-20 + 2-12 = 854

mm2)

450 897.75 (2-20+2-14

= 935 mm2)


3.6.9 Other Alternatives using Charts and Tables of SP-16 Any alternative solution of d will involve computations of factored loads Fd and bending moment Mu. Thereafter, Eq. 3.23 of Lesson 5 has to be solved to get the value of Ast by direct computation method. On the other hand, it is very simple to get the Ast with the help of either charts or tables of SP-16 from the value of factored bending moment. Some alternatives are given below in Table 3.6 by the use of tables of SP-16. In sec. 3.6.7.3, it is observed that an effective depth of 400 mm is not acceptable. Hence, the effective depth is increased up to 450 mm at intervals of 10 mm and the corresponding Ast values are presented in Table 3.6. The width b is kept as 250 mm and M 20 and Fe 415 are used for all the alternatives.

Table 3.6 Alternative values of d, D, Fd, Mu, 2dbMu , pt and Ast

Sl. No.

d (mm)

D (mm)

Fd (kN/m)

Mu (kNm) 2db

Mu

(N/mm2

)

pt (%)

Ast (mm2)

1

2

410

420

460

470

14.825

14.9062

118.5

119.25

2.8197

2.7041

Not acceptable

0.93

Not acceptable

976.5

3

4

430

440

480

490

15.0

15.09

120.00

120.75

2.596

2.4948

0.88

0.839

946.0

922.9

5 450 500 15.1875

121.5 2.4 0.798 897.75

3.6.10 Advantages of using SP-16

The following are the advantages: (i) Alternative sets of b, d and Ast are obtained very quickly.

(ii) The results automatically exclude those possibilities where the steel reinforcement is inadmissible.

It has been mentioned that the reinforcement should be within 75 to 80 per cent of limiting reinforcement to ensure ductile failure. The values of charts and tables are given up to the limiting reinforcement. Hence, the designer should be careful to avoid the reinforcement up to the limiting amount. Moreover, these charts and tables can be used for the design of slabs also. Therefore, the values


are also taking care of the minimum reinforcement of slabs. The minimum reinforcement of beams are higher than that of slabs. Accordingly, the designer should also satisfy the requirement of minimum reinforcement for beams while using SP-16. It is further suggested to use the tables than the charts as the values of the charts may have personal error while reading from the charts. Tabular values have the advantage of numerical, which avoid personal error. Moreover, intermediate values can also be evaluated by linear interpolation. 3.6.11 Practice Questions and Problems with Answers Q.1: Mention the necessary input data and unknowns to be determined for the

two types of problems of singly reinforce beams. A.1: (i) The input data for the design type of problems are layout plan, imposed

loads, grades of steel and concrete. The unknowns to be determined are b, d, D, Ast and Leff.

(ii) The input data for the analysis type of problem are b, d, D, Ast, Leff,

grades of concrete and steel. The unknowns to be determined are Mu and service imposed loads.

Q.2: State specific guidelines to select the initial dimensions/amount/grade of

the following parameters before designing the reinforced concrete beams: (i) b, (ii) d, (iii) D, (iv) Ast, (v) diameter of reinforcing bars, (vi) grade of

concrete and (vii) grade of steel. A.2: Sections 3.6.2.1 to 6 cover the answers. Q.3: Name the three methods of solution of the design of reinforced concrete

beam problems. A.3: The three methods are: (i) Direct computation method, (ii) Use of charts of

SP-16 and (iii) Use of tables of SP-16 Q.4: Determine the imposed loads and the tensile steel Ast,lim of the singly

reinforced rectangular beam shown in Figs. 3.6.2 and 4 of L = 8.0 m simply supported, thickness of brick wall = 300 mm, width b = 300 mm, effective depth d = 550 mm, total depth D = 600 mm, grade of concrete = M 20 and grade of steel = Fe 500. Use (i) direct computation method, (ii) design chart of SP-16 and (iii) design table of SP-16.


A.4: (i) Direct computation method: The limiting moment of resistance Mu,lim is obtained from Eq. 3.24 as

follows

cku,limu,

u, fdbd

x.

dx

.M 2imllim 4201360

⎭⎬⎫

⎩⎨⎧−=

Here, xu,lim /d = 0.46 (cl. 38.1, Note of IS 456:2000)

Hence, Mu,lim = 0.36 (0.46) {1 - 0.42 (0.46)} (300) (550) (500) (20)

Nmm = 22,04,50,00,000 Nmm Tensile steel Ast,lim is obtained from Eq. 3.23 as follows:

⎭⎬⎫

⎩⎨⎧−=

dbffA

dAf.Mck

ylistlistyliu

m,m,m, 1870

(3.23) Denoting the unknown Ast,lim as A, we get: A2 - 6600 A + 6081379.31 = 0 Solving the above equation, the lower value of A is the Ast,lim which is

equal to 1107.14 mm2


(ii) Use of chart of SP-16: Using chart 17 of SP-16 for Mu,lim /b = 220.45/0.3 = 734.833 kNm/m, we

get the reinforcement percentage 100(Ast,lim )/bd = 0.67. So, Ast,lim = 0.67 (300) (550)/100 = 1105 mm2

(iii) Use of table of SP-16:

Table 2 of SP-16 for Mu,lim /bd2 = 220.45/300 (0.55) (0.55) = 2.4292

N/mm2, we get the reinforcement percentage by linear interpolation as: 0.669 + (0.007) (0.0092)/(0.02) = 0.67222. Hence, Ast,lim = 0.67222 (300) (550)/(100) = 1109.16 mm2

Comparison of results:

Method Ast,lim (mm2)

(i) 1107.14 (ii) 1105.00 (iii) 1109.16

Imposed loads: The total load W per metre can be obtained from W = 8 (Mu,lim ) / 2

effL Where, Leff is the lower of (i) 7700 + 550 or (ii) 8000 mm (cl. 22.2a of IS

456:2000) Using Leff = 8000 mm and Mu,lim = 220.45 kNm We get the total load W = 220.45/8 = 27.556 kN/m The dead load of the beam = 0.3 (0.6) (25) = 4.5 kN/m Hence, the imposed loads = 27.556 - 4.5 = 23.056 kN/m


3.6.12 References
















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 3.6.13 Test 6 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: State specific guidelines to select the initial dimensions/amount/grade of

the following parameters before designing the reinforced concrete beams:


(i) b, (ii) d, (iii) D, (iv) Ast, (v) diameter of reinforcing bars, (vi) grade of concrete and (vii) grade of steel. (6 x 5 = 30 marks)

A.TQ.1: See secs. 3.6.2.1 to 6. TQ.2: State the advantages of using SP-16 than employing direct computation

method in the design of a beam. (15 marks)

A.TQ.2: See sec. 3.6.10 (except the last para). TQ.3: Why the use of tables of SP-16 is better than the use of chart ?

(5 marks) A.TQ.3: See sec. 3.6.10 (last para only). 3.6.14 Summary of this Lesson Explaining the two types of problems and giving the necessary guidelines of the preliminary selection of the parameters, this lesson illustrates step by step method of solving design type of problems employing (i) direct computation method, (ii) use of charts of SP-16 and (iii) use of tables of SP-16. The results of a specific problem are compared. The advantages of using SP-16 in general and the superiority of using the tables of SP-16 to the charts are also discussed.


Module 3


Examples)


Lesson 7

Numerical Problems on Singly Reinforced

Rectangular Beams (Continued)


Instructional Objectives: At the end of this lesson, the student should be able to: • Apply the principles to analyse a given cross-section of a beam with specific

reinforcement to determine its moment of resistance. 3.7.1 Introduction This lesson explains the determination of moment of resistance of given singly reinforced rectangular beam sections with the help of illustrative analysis type of problem. The numerical problem is solved by (i) direct computation method, (ii) using charts of SP-16 and (iii) using tables of SP-16. Step by step solutions illustrate the procedure clearly. 3.7.2 Analysis Type of Problems It may be required to estimate the moment of resistance and hence the service load of a beam already designed earlier with specific dimensions of b, d and D and amount of steel reinforcement Ast. The grades of concrete and steel are also known. In such a situation, the designer has to find out first if the beam is under-reinforced or over-reinforced. The following are the steps to be followed for such problems. 3.7.2.1 xu, max The maximum depth of the neutral axis xu, max is determined from Table 3.2 of Lesson 5 using the known value of fy. 3.7.2.2 xu The depth of the neutral axis for the particular beam is determined from Eq. 3.16 of Lesson 5 employing the known values of fy, fck, b and Ast. 3.7.2.3 Mu and service imposed loads The moment of resistance Mu is calculated for the three different cases as follows:

(a) If xu < xu, max, the beam is under-reinforced and Mu is obtained from Eq. 3.22 of Lesson 5.

(b) If xu = xu, max, the Mu is obtained from Eq. 3.24 of Lesson 5.


(c) If xu > xu, max, the beam is over-reinforced for which xu is taken as xu,

max and then Mu is calculated from Eq. 3.24 of Lesson 5, using xu = xu,

max. With the known value of Mu, which is the factored moment, the total factored load can be obtained from the boundary condition of the beam. The total service imposed loads is then determined dividing the total factored load by partial safety factor for loads (= 1.5). The service imposed loads are then obtained by subtracting the dead load of the beam from the total service loads. 3.7.3 Analysis Problems 3.2 and 3.3 Determine the service imposed loads of two simply supported beam of same effective span of 8 m (Figs. 3.7.1 and 2) and same cross-sectional dimensions, but having two different amounts of reinforcement. Both the beams are made of M 20 and Fe 415.

3.7.4 Solution by Direct Computation Method - Problem 3.2 Given data: b = 300 mm, d = 550 mm, D = 600 mm, Ast = 1256 mm2 (4-20 T), Leff = 8 m and boundary condition = simply supported (Fig. 3.7.1). 3.7.4.1 xu, max From Table 3.2 of Lesson 5, we get xu, max = 0.479 d = 263.45 mm


3.7.4.2 xu

ck

styu fb.

Af.x

360 870

=

(3.16)

)20()300(36.0)1256()415(87.0

=

= 209.94385 mm < xu, max = (263.45 mm) Hence, the beam is under-reinforced. 3.7.4.3 Mu and service imposed loads For xu < xu, max , we have Mu = 0.87 fy Ast (d - 0.42 xu) (3.22) = 0.87 (415) (1256) {550 - 0.42(209.94385)} = 209.4272 kNm

Total factor load )8(8

)4272209(82

8 .

effLuM

dF ==

= 26.1784 kN/m

Total service load = 17.4522661.5

26.178451

==.

Fd kN/m

Dead load of the beam = 0.3 (0.6) (25) = 4.5 kN/m Hence, service imposed loads = (17.452266 - 4.5) kN/m = 12.952266 kN/m


3.7.5 Solution by Direct Computation Method - Problem 3.3

Given data: b = 300 mm, d = 550 mm, D = 600 mm, Ast = 1658 mm2 (4-20 T + 2-16 T), Leff = 8 m and boundary conditions = simply supported (Fig. 3.7.2) 3.7.5.1 xu, max From Table 3.2 of Lesson 5, we get xu, max = 0.479 d = 263.45 mm 3.7.5.2 xu

ck

styu fb.

Af.x

360870

=

(3.16)

)20()300(36.0)1658()415(87.0

=

= 277.13924 mm > xu, max = (263.45 mm) Hence, the beam is over-reinforced. 3.7.5.3 Mu and service imposed loads For xu > xu,max, we have

cku,u,

u fdbd

x.

dx

.M 2maxmax )4201(360 −= (3.24)


= 0.36 (0.479) {1 - 0.42(0.479)} (300) (550) (550) (20) Nmm = 250.01356 kNm If we use Eq. 3.22 using xu = xu,max, for Mu Mu = 0.87 fy Ast (d - 0.42 xu, max) (3.22) Then, (Mu)steel = 0.87 (415) (1658) {550 - 0.42 (263.45)} Nmm = 263.00468 kNm > (Mu)concrete (= 250.01356 kNm) The higher Mu as obtained from steel is not true because the entire amount of steel (1658 mm2) cannot yield due to over-reinforcing. Prior to that, concrete fails at 250.01356 kNm. However, we can get the same of Mu as obtained from Eq. 3.24 of Lesson 5, if we can find out how much Ast is needed to have xu = 263.45 mm. From Eq. 3.16 of Lesson 5, we can write:

y

uckxforneededst f

xfbA u 87.0

36.0)( 45.263 ==

)415(87.0

)45.263()20()300(36.0=

= 1576.1027 mm2

If we use this value for Ast in Eq. 3.22 of Lesson 5, we get )}45.263(42.0550{)1027.1576()415(87.0)( −=uM = 250.0135 (same as obtained from Eq. 3.24). From the factored moment Mu = 250.01356 kNm, we have:

Total factored load = 251695.31)8(8

)01356.250(882 ===eff

ud L

MF

kN/m

Total service load = 834463.205.1

251695.31= kN/m


Now, Dead load of the beam = 0.3 (0.6) (25) = 4.5 kN/m Hence, service imposed loads = 20.834463 - 4.5 = 16.334463 kN/m 3.7.6 Solution by Design Chart - Problems 3.2 and 3.3 For the two problems with known b, d, D, Ast, grade of concrete and grade of steel, chart 14 of SP-16 is applicable. From the effective depth d and

percentage of reinforcement )100

(db

Ap st

t = , the chart is used to find Mu per metre

width. Multiplying Mu per meter width with b, we get Mu for the beam. After that, the service imposed load is calculated using the relation

Service imposed load = 2)5.1(8

LMu - Dead load

(3.27) The results of the two problems are furnished in Table 3.7. Table 3.7 Results of Problems 3.2 and 3.3 (Chart of SP-16) Prob- lem

Ast (mm2) pt (%)

d, b (mm)

Mu (kNm/m) (Chart

14)

Mu (kNm)

25.1

8

L

Mu

(kN/m)

Dead load (kN/m)

Service imposed loads (kN/m)

Remarks

3.2 1256 0.7612

550, 300

700 210 17.5 4.5 13 Mu per m of 700 is well within the chart, hence under-reinforced.

3.3 1658 1.205818

1

550, 300

820 246 20.5 4.5 16 Maximum Mu = 820, so over-reinforced.


3.7.7 Solution by Design Tables - Problems 3.2 and 3.3 Table 2 of SP-16 presents the value of reinforcement percentage for different combinations of fy and (Mu/bd2) for M-20. Here, from the known values of p and fy, the corresponding values of Mu/bd2 are determined. These in turn give Mu of the beam. Then the service imposed load can be obtained using Eq. 3.27 as explained in the earlier section (sec. 3.7.6). The results of the two problems are presented in Table 3.8. Table 3.8 Results of Problems 3.2 and 3.3 (Table of SP-16)

Prob-lem

Ast (mm2) pt (%)

d, b (mm)

fy (N/mm2

)

2db

Mu

(N/mm2

) from

Table 2

Mu (kNm)

25.1

8

L

Mu Dead load (kN/m)

Service imposed loads (kN/m)

Remarks

3.2 1256 0.7612

550, 300

415 2.3105 209.67784

17.473155

4.5 12.973155

*

3.3 1658 1.205818

1

500, 300

415 2.76 250.47 20.8725 4.5 16.3725

**

* Linear interpolated )30.232.2()757.0765.0()757.07612.0(30.22 −

−−

+=db

Mu = 2.3105

** pt = 1.2058181 is not admissible, i.e. over-reinforced. So at pt = 0.955, Mu/bd2 =2.76. 3.7.8 Comparison of Results of Three Methods The values of Mu and service imposed loads of the under-reinforced and over-reinforced problems (Problems 3.2 and 3.3), computed by three methods, are presented in Table 3.9.


Table 3.9 Comparison of results of Problems 3.2 and 3.3

Mu (kNm) Service imposed loads (kN/m)

Proble

m Direct computation

Chart of SP-16

Table of SP-16

Direct computation

Chart of SP-16

Table of SP-16

3.2 209.4272 210 209.67784

12.9522 13 12.973155

3.3 250.01356 246 250.47 16.334463 16 16.3725 3.7.9 Practice Questions and Problems with Answers Q.1: Determine the moments of resistance Mu and service imposed loads on a

simply supported beam of effective span 10.0 m with b = 300mm, d = 500 mm, D = 550 mm and grades of concrete and steel are M20 and Fe500, respectively for the two different cases employing (a) direct computation method and (b) using charts and tables of SP-16: (i) when Ast is minimum acceptable and (ii) when Ast is maximum acceptable (Fig. 3.7.3).

A.1: Given data: b = 300 mm, d = 500 mm, D = 550 mm, Leff = 10.0 m, fck =

20 N/mm2 and fy = 500 N/mm2.

(a) Direct computation method: Case (i) When Ast is minimum acceptable (Eq. 3.26 of Lesson 6)

Minimum y

s fdbA 85.0

=

(3.26)


So, 500

)500()300(5.0=sA mm2 = 255 mm2. Providing 4-12T gives Ast = 452

mm2. Equation 3.16 of Lesson 5 gives the depth of the neutral axis xu:

03.91)20()300(36.0)452()500(87.0

36.087.0

===ck

styu fb

Afx mm

Table 3.2 of Lesson 5 gives xu, max = 0.46(500) = 230 mm. The beam is, therefore, under-reinforced (as xu < xu, max). Equation 3.22 of Lesson 5 gives the Mu as follows: Mu = 0.87 fy Ast (d - 0.42 xu) (3.22) = 0.87(500) (452) {500 - 0.42(91.03)} Nmm = 90.79 kNm

Total factored load = 26.7100

)79.90(882 ==eff

u

LM kN/m

The dead load of the beam = 0.3 (0.55) (25) = 4.125 kN/m So, the service imposed loads = {(Total factored load)/(Load factor)} - (Dead load) = 7.26/1.5 - 4.125 = 0.715 kN/m This shows that the beam can carry maximum service imposed loads, 17 per cent of its dead load only, when the acceptable minimum tensile reinforcement is 452 mm2 (4 bars of 12 mm diameter). Case (ii) when Ast is maximum acceptable: To ensure ductile failure, it is essential that the acceptable maximum tensile reinforcement should be between 75 and 80 per cent of pt, lim and not as given in clause 26.5.1.1.(b), i.e. 0.04 bD. Thus, here the maximum acceptable pt should be between 0.57 and 0.61 per cent (as pt, lim = 0.76 from Table 3.1 of Lesson 5). However, let us start with 0.76 per cent as the span is relatively large


and keeping in mind that while selecting the bar diameter, it may get reduced to some extent. So, (Ast)max = 0.76 (300) (500)/100 = 1140 mm2

Selecting 4-16 and 2-12 mm diameter bars, we get Ast = 1030 mm2 when pt becomes 0.67 per cent. So, the maximum acceptable tensile reinforcement is 1030 mm2.

43.207 )20()300(36.0)1030()500(87.0

36.087.0

===ck

styu fb

Afx mm (Eq. 3.16 of Lesson 5)

xu < xu, max (as xu, max = 230 mm; see Case (i) of this problem). Therefore, Mu is obtained from Eq. 3.22 of Lesson 5 as, Mu = 0.87 fy Ast (d - 0.42 xu) (3.22) = 0.87(500) (1030) {500 - 0.42(207.43)} Nmm = 184.99 kNm

Total factored load = 8.14100

)99.184(882 ==eff

u

LM kN/m

With dead load = 4.125 kN/m (see Case (i) of this problem), we have: Service imposed load = 14.8/1.5 - 4.125 = 5.74 kN/m This beam, therefore, is in a position to carry service imposed loads of 5.74 kN/m, about 40% higher than its own dead load. (b) Using chart and tables of SP-16:

Tables 3.10 and 3.11 present the results using charts and tables respectively of SP-16. For the benefit of the reader the different steps are given below separately for the use of chart and table respectively for the minimum acceptable reinforcement.

The steps using chart of SP-16 are given below:


Step 1: With the given fck, fy and d, choose the particular chart. Here, for fck = 20 N/mm2, fy = 500 N/mm2 and d = 500 mm, the needed chart no. is 17.

Step 2: Chart 17 shows minimum pt = 0.13% which gives Ast = 195 mmm2

Step 3: Provide bars of 4-12 mm diameter, which give pt = 0.301% Step 4: For pt = 0.301, Chart 17 shows Mu = 300 kNm per metre width, which

gives Mu = 300 (0.3) = 90 kNm for the beam. Step 5: The service imposed loads are calculated as follows:

Service imposed loads = 675.0125.4)5.1(

82 =−eff

u

LM kN/m, using

the dead load of the beam as 4.125 kN/m (see case (i) of this problem). Step 6: The capacity of the beam is to carry 0.675 kN/m which is (0.675/4.125)

100 = 16.36%. Table 3.10 Results of Q.1 using chart of SP-16 Sl. No.

Chart No. Min/Max

Minimum/Maximum pt and Ast (%, mm2)

Provided bars

(No, mm diameter)

(pt) provided

(%)

Mu(kNm/m,

kNm)

Service imposed

loads (kN/m)

Service imposed loads in

(%) 1 17

(Minimum) 0.13, 195

4-12 0.301 300, 90

0.675 16.36

2 17 (Maximum)

0.76, 1140

4-16 + 2-12

0.67 610, 183

5.635 136.61

Similar calculations are done for the maximum acceptable reinforcement. The steps are given below: Step 1: With the given fck = 20 N/mm2, fy = 500 N/mm2, Table 2 of SP-16 is

selected. Step 2: (pt)min = 0.07 from Table 2, which gives Ast = 105 mmm2

Step 3: Provide bars of 4-12 mm diameter, which give pt = 0.301% Step 4: For (pt)provided, we get (Mu/bd2) from Table 2 of SP-16 by linear

interpolation as follows:


204.1)298.0312.0(

)298.0301.0()2.125.1(20.12 =−

−−+=

dbMu

Hence, Mu = 1.204 (300) (500)2 = 90.3 kNm Step 5: Same as Step 5 while using chart of SP-16. Step 6: Same as Step 6 while using chart of SP-16. Table 3.11 Results of Q.1 using table of SP-16 Sl. No.

Table No. Min/Max

Minimum/Maximum pt and Ast (%, mm2)

Provided bars

(No, mm diameter)

(pt) provided

(%)

2db

Mu

(kNm/m, kNm)

Service imposed

loads (kN/m)

Service imposed loads in

(%)

1 2 (Minimum)

0.07, 105

4-12 0.301 1.204, 90.3

0.691 16.75

2 2 (Maximum

)

0.755, 1132

4-16 + 2-12

0.67 2.42, 181.5

5.555 134.66

3.7.10 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 3.7.11 Test 7 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes TQ.1: Determine the moment of resistance for the beams shown in Figs. 3.7.4

and 3.7.5 using M 20 and Fe 250 by direct computation and using charts and tables of SP-16.

A.TQ.1: Case A: TQ.1 A of Fig. 3.7.4

(i) Direct computation method


Ast = 1963 mm2 (4-25 φ) = 0.53 (450) = 238.5 mm (Table 3.2 of Lesson 5 gives 0.53) max,ux

66.197 )20()300(36.0)1963()250(87.0

36.087.0

===ck

styu fb

Afx (See Eq. 3.16 of Lesson 5)

So, xu < shows that it is under-reinforced section for which Mmax,ux u is obtained from Eq. 3.22 of Lesson 5

Mu = 0.87 fy Ast (d - 0.42 xu) (3.22) = 0.87(250) (1963) {450 - 0.42(197.66)} = 156.68 kNm

(ii) Chart of SP-16 Ast = 1963 mm2

45.1)450(300)100(1963)100(

===db

Ap stt

From chart 11 of SP-16, when pt = 1.45, d = 450, we get Mu per metre width = 522 kNm/m Mu = 522 (0.3) = 156.6 kNm (iii) Table of SP-16 Table 2 of SP-16 is for M-20 and Fe250. At pt = 1.451, we get

58.22 =db

Mu N/mm2

So, Mu = 2.58 (300) (450) (450) (10-6) kNm = 156.74 kNm

The three values of Mu are close to each other.


A.TQ.1: Case B: TQ.1 B of Fig. 3.7.5 (i) Direct computation method

Ast = 1963 + 981 = 2944 mm2 (6-25 φ)

44.296 )20()300(36.0)2944()250(87.0

36.087.0

===ck

styu fb

Afx mm (See Eq. 3.16 of

Lesson 5) = 0.53 (450) = 238.5 mm (Table 3.2 of Lesson 5 gives 0.53) max,ux So, xu > xu, max and the beam is over-reinforced. In such a situation, we

take xu = xu, max = 238.5 mm. The Mu will be calculated from Eq. 3.24 of Lesson 5.

ckuu

uu fdbd

xd

xMM 2max,max,

lim, )42.01(36.0 −==

(3.24) = 0.36 (0.53) {1 - 0.42 (0.53)} (300) (450)2 (20) Nmm = 180.22 kNm (ii) Chart of SP-16

18.2)450(300)100(2944

==tp . In chart 11 (for M 20 and Fe 250), maximum

admissible pt is 1.75 and for this pt when d = 450, Mu = 600 kNm/m.


So, Mu = 600 (0.3) = 180 kNm (iii) Table of SP-16 Table 2 (for M 20 and Fe 250) has the maximum pt = 1.76 and at that

value, (Mu/bd2) = 2.98. This gives

Mu = 2.98 (300) (450)2 (10-6) = 181.03 kNm. Here also the three values of Mu are close to each other.

3.7.12 Summary of this Lesson This lesson explains the use of equations derived in Lesson 5 for the analysis type of problems. The three methods (i) direct computation, (ii) use of charts of SP-16 and (iii) use of tables of SP-16 are illustrated through the step by step solutions of numerical problems. Their results are compared to show the closeness of them.


Module 4

Doubly Reinforced Beams – Theory and

Problems


Lesson 8

Doubly Reinforced Beams – Theory


Instructional Objectives: At the end of this lesson, the student should be able to: • explain the situations when doubly reinforced beams are designed, • name three cases other than doubly reinforced beams where compression

reinforcement is provided, • state the assumptions of analysis and design of doubly reinforced beams, • derive the governing equations of doubly reinforced beams, • calculate the values of fsc from (i) d'/d and (ii) calculating the strain of the

compression reinforcement, • state the minimum and maximum amounts of Asc and Ast in doubly

reinforced beams, • state the two types of numerical problems of doubly reinforced beams, • name the two methods of solving the two types of problems, and • write down the steps of the two methods for each of the two types of

problems. 4.8.1 Introduction


Concrete has very good compressive strength and almost negligible tensile strength. Hence, steel reinforcement is used on the tensile side of concrete. Thus, singly reinforced beams reinforced on the tensile face are good both in compression and tension. However, these beams have their respective limiting moments of resistance with specified width, depth and grades of concrete and steel. The amount of steel reinforcement needed is known as Ast,lim. Problem will arise, therefore, if such a section is subjected to bending moment greater than its limiting moment of resistance as a singly reinforced section.

There are two ways to solve the problem. First, we may increase the depth

of the beam, which may not be feasible in many situations. In those cases, it is possible to increase both the compressive and tensile forces of the beam by providing steel reinforcement in compression face and additional reinforcement in tension face of the beam without increasing the depth (Fig. 4.8.1). The total compressive force of such beams comprises (i) force due to concrete in compression and (ii) force due to steel in compression. The tensile force also has two components: (i) the first provided by Ast,lim which is equal to the compressive force of concrete in compression. The second part is due to the additional steel in tension - its force will be equal to the compressive force of steel in compression. Such reinforced concrete beams having steel reinforcement both on tensile and compressive faces are known as doubly reinforced beams. Doubly reinforced beams, therefore, have moment of resistance more than the singly reinforced beams of the same depth for particular grades of steel and concrete. In many practical situations, architectural or functional requirements may restrict the overall depth of the beams. However, other than in doubly reinforced beams compression steel reinforcement is provided when:

(i) some sections of a continuous beam with moving loads undergo change of sign of the bending moment which makes compression zone as tension zone or vice versa.

(ii) the ductility requirement has to be followed.

(iii) the reduction of long term deflection is needed.

It may be noted that even in so called singly reinforced beams there would be longitudinal hanger bars in compression zone for locating and fixing stirrups. 4.8.2 Assumptions

(i) The assumptions of sec. 3.4.2 of Lesson 4 are also applicable here. (ii) Provision of compression steel ensures ductile failure and hence,

the limitations of x/d ratios need not be strictly followed here.


(iii) The stress-strain relationship of steel in compression is the same as that in tension. So, the yield stress of steel in compression is 0.87 fy.

4.8.3 Basic Principle

As mentioned in sec. 4.8.1, the moment of resistance Mu of the doubly reinforced beam consists of (i) Mu,lim of singly reinforced beam and (ii) Mu2 because of equal and opposite compression and tension forces (C2 and T2) due to additional steel reinforcement on compression and tension faces of the beam (Figs. 4.8.1 and 2). Thus, the moment of resistance Mu of a doubly reinforced beam is


Mu = Mu,lim + Mu2 (4.1) The Mu,lim is as given in Eq. 3.24 of Lesson 5, i.e.,

Mu,lim = ckuu fdbd

xd

x 2max,max, )42.01()(36.0 − (4.2)

Also, Mu,lim can be written from Eq. 3.22 of Lesson 5, using xu = xu, max, i.e., Mu, lim = 0.87 Ast, lim fy (d - 0.42 xu, max)

= 0.87 pt, lim (1 - 0.42 yu fdbd

x 2max, ) (4.3)

The additional moment Mu2 can be expressed in two ways (Fig. 4.8.2): considering (i) the compressive force C2 due to compression steel and (ii) the tensile force T2 due to additional steel on tension face. In both the equations, the lever arm is (d - d'). Thus, we have )'()(2 ddffAM ccscscu −−= (4.4) )'()87.0(22 ddfAM ystu −= (4.5) where Asc = area of compression steel reinforcement fsc = stress in compression steel reinforcement fcc = compressive stress in concrete at the level of centroid of

compression steel reinforcement Ast2 = area of additional steel reinforcement Since the additional compressive force C2 is equal to the additional tensile force T2, we have Asc (fsc - fcc) = Ast2 (0.87 fy)

(4.6) Any two of the three equations (Eqs. 4.4 - 4.6) can be employed to determine Asc and Ast2. The total tensile reinforcement Ast is then obtained from:

1 2st st sA A A= + t (4.7)


where )42.0(87.0100 max,

lim,lim,1

uy

utst xdf

MdbpA−

== (4.8)

4.8.4 Determination of fsc and fcc It is seen that the values of fsc and fcc should be known before calculating Asc. The following procedure may be followed to determine the value of fsc and fcc for the design type of problems (and not for analysing a given section). For the design problem the depth of the neutral axis may be taken as xu,max as shown in Fig. 4.8.2. From Fig. 4.8.2, the strain at the level of compression steel reinforcement εsc may be written as

)'1(0035.0max,u

sc xd

−=ε (4.9)

The stress in compression steel fsc is corresponding to the strain εsc of Eq. 4.9 and is determined for (a) mild steel and (b) cold worked bars Fe 415 and 500 as given below: (a) Mild steel Fe 250 The strain at the design yield stress of 217.39 N/mm2 (fd = 0.87 fy ) is 0.0010869 (= 217.39/Es). The fsc is determined from the idealized stress-strain diagram of mild steel (Fig. 1.2.3 of Lesson 2 or Fig. 23B of IS 456) after computing the value of εsc from Eq. 4.9 as follows: (i) If the computed value of εsc ≤ 0.0010869, fsc = εsc Es = 2 (105) εsc (ii) If the computed value of εsc > 0.0010869, fsc = 217.39 N/mm2. (b) Cold worked bars Fe 415 and Fe 500 The stress-strain diagram of these bars is given in Fig. 1.2.4 of Lesson 2 and in Fig. 23A of IS 456. It shows that stress is proportional to strain up to a stress of 0.8 fy. The stress-strain curve for the design purpose is obtained by substituting fyd for fy in the figure up to 0.8 fyd. Thereafter, from 0.8 fyd to fyd, Table A of SP-16 gives the values of total strains and design stresses for Fe 415 and Fe 500. Table 4.1 presents these values as a ready reference here.


Table 4.1 Values of fsc and εsc

Fe 415

Fe 500 Stress level

Strain εsc Stress fsc(N/mm2)

Strain εsc Stress fsc(N/mm2)

0.80 fyd 0.00144 288.7 0.00174 347.8 0.85 fyd 0.00163 306.7 0.00195 369.6 0.90 fyd 0.00192 324.8 0.00226 391.3 0.95 fyd 0.00241 342.8 0.00277 413.0

0.975 fyd 0.00276 351.8 0.00312 423.9 1.0 fyd 0.00380 360.9 0.00417 434.8

Linear interpolation may be done for intermediate values. The above procedure has been much simplified for the cold worked bars by presenting the values of fsc of compression steel in doubly reinforced beams for different values of d'/d only taking the practical aspects into consideration. In most of the doubly reinforced beams, d'/d has been found to be between 0.05 and 0.2. Accordingly, values of fsc can be computed from Table 4.1 after determining the value of εsc from Eq. 4.9 for known values of d'/d as 0.05, 0.10, 0.15 and 0.2. Table F of SP-16 presents these values of fsc for four values of d'/d (0.05, 0.10, 0.15 and 0.2) of Fe 415 and Fe 500. Table 4.2 below, however, includes Fe 250 also whose fsc values are computed as laid down in sec. 4.8.4(a) (i) and (ii) along with those of Fe 415 and Fe 500. This table is very useful and easy to determine the fsc from the given value of d'/d. The table also includes strain values at yield which are explained below: (i) The strain at yield of Fe 250 =

0.0010869 (200000) 1.15250 Stress YieldDesign

==sE

Here, there is only elastic component of the strain without any inelastic strain.

(ii) The strain at yield of Fe 415 = Stress YieldDesign Strain InelasticsE

+

0.0038043 (200000) 1.15415 0.002 =+=

(iii) The strain at yield of Fe 500 = 0.0041739 (200000) 1.15500 0.002 =+


Table 4.2 Values of fsc for different values of d'/d

d'/d fy(N/mm2) 0.05 0.10 0.15 0.20

Strain at yield

250 217.4 217.4 217.4 217.4 0.0010869 415 355 353 342 329 0.0038043 500 412 412 395 370 0.0041739

4.8.5 Minimum and maximum steel 4.8.5.1 In compression There is no stipulation in IS 456 regarding the minimum compression steel in doubly reinforced beams. However, hangers and other bars provided up to 0.2% of the whole area of cross section may be necessary for creep and shrinkage of concrete. Accordingly, these bars are not considered as compression reinforcement. From the practical aspects of consideration, therefore, the minimum steel as compression reinforcement should be at least 0.4% of the area of concrete in compression or 0.2% of the whole cross-sectional area of the beam so that the doubly reinforced beam can take care of the extra loads in addition to resisting the effects of creep and shrinkage of concrete. The maximum compression steel shall not exceed 4 per cent of the whole area of cross-section of the beam as given in cl. 26.5.1.2 of IS 456. 4.8.5.2 In tension As stipulated in cl. 26.5.1.1(a) and (b) of IS 456, the minimum amount of tensile reinforcement shall be at least (0.85 bd/fy) and the maximum area of tension reinforcement shall not exceed (0.04 bD). It has been discussed in sec. 3.6.2.3 of Lesson 6 that the singly reinforced beams shall have Ast normally not exceeding 75 to 80% of Ast,lim so that xu remains less than xu,max with a view to ensuring ductile failure. However, in the case of doubly reinforced beams, the ductile failure is ensured with the presence of compression steel. Thus, the depth of the neutral axis may be taken as xu, max if the beam is over-reinforced. Accordingly, the Ast1 part of tension steel can go up to Ast, lim and the additional tension steel Ast2 is provided for the additional moment Mu - Mu, lim. The quantities of Ast1 and Ast2 together form the total Ast, which shall not exceed 0.04 bD.


4.8.6 Types of problems and steps of solution Similar to the singly reinforced beams, the doubly reinforced beams have two types of problems: (i) design type and (ii) analysis type. The different steps of solutions of these problems are taken up separately. 4.8.6.1 Design type of problems In the design type of problems, the given data are b, d, D, grades of concrete and steel. The designer has to determine Asc and Ast of the beam from the given factored moment. These problems can be solved by two ways: (i) use of the equations developed for the doubly reinforced beams, named here as direct computation method, (ii) use of charts and tables of SP-16. (a) Direct computation method Step 1: To determine Mu, lim and Ast, lim from Eqs. 4.2 and 4.8, respectively. Step 2: To determine Mu2, Asc, Ast2 and Ast from Eqs. 4.1, 4.4, 4.6 and 4.7, respectively. Step 3: To check for minimum and maximum reinforcement in compression and tension as explained in sec. 4.8.5. Step 4: To select the number and diameter of bars from known values of Asc and Ast. (b) Use of SP table Tables 45 to 56 present the pt and pc of doubly reinforced sections for d'/d = 0.05, 0.10, 0.15 and 0.2 for different fck and fy values against Mu /bd2. The values of pt and pc are obtained directly selecting the proper table with known values of Mu/bd2 and d'/d. 4.8.6.2 Analysis type of problems In the analysis type of problems, the data given are b, d, d', D, fck, fy, Asc and Ast . It is required to determine the moment of resistance Mu of such beams. These problems can be solved: (i) by direct computation method and (ii) by using tables of SP-16. (a) Direct computation method Step 1: To check if the beam is under-reinforced or over-reinforced.


First, xu,max is determined assuming it has reached limiting stage using

dxu, max coefficients as given in cl. 38.1, Note of IS 456. The strain of tensile steel

εst is computed from max

max )(

u,

u, cst x

d - x ε ε = and is checked if εst has reached the

yield strain of steel:

0020151

. (E).

f ε y

ldst at yie +=

The beam is under-reinforced or over-reinforced if εst is less than or more than the yield strain. Step 2: To determine Mu,lim from Eq. 4.2 and Ast,lim from the pt, lim given in Table 3.1 of Lesson 5. Step 3: To determine Ast2 and Asc from Eqs. 4.7 and 4.6, respectively. Step 4: To determine Mu2 and Mu from Eqs. 4.4 and 4.1, respectively. (b) Use of tables of SP-16 As mentioned earlier Tables 45 to 56 are needed for the doubly reinforced beams. First, the needed parameters d'/d, pt and pc are calculated. Thereafter, Mu/bd2 is computed in two stages: first, using d'/d and pt and then using d'/d and pc . The lower value of Mu is the moment of resistance of the beam. 4.8.7 Practice Questions and Problems with Answers Q.1: When do we go for doubly reinforced beams ? A.1: The depth of the beams may be restricted for architectural and/or

functional requirements. Doubly reinforced beams are designed if such beams of restricted depth are required to resist moment more that its Mu,

lim. Q.2: Name three situations other than doubly reinforced beams, where the

compression reinforcement is provided. A.2: Compression reinforcement is provided when:

(i) Some sections of a continuous beam with moving loads undergo change of sign of the bending moment which makes compression zone as tension zone,


(ii) the ductility requirement has to be satisfied,

(iii) the reduction of long term deflection is needed. Q.3: State the assumptions of the analysis and design of doubly reinforced beams. A.3: See sec. 4.8.2 (i), (ii) and (iii). Q.4: Derive the governing equations of a doubly reinforced beam. A.4: See sec. 4.8.3 Q.5: How do you determine fsc of mild steel and cold worked bars and fcc? A.5: See sec. 4.8.4 Q.6: State the minimum and maximum amounts of Asc and Ast in doubly

reinforced beams. A.6: See sec. 4.8.5 Q.7: State the two types of problems of doubly reinforced beams specifying the

given data and the values to be determined in the two types of problems. A.7: The two types of problems are:

(i) Design type of problems and (ii) Analysis type of problems

(i) Design type of problems:

The given data are b, d, D, fck, fy and Mu . It is required to determine Asc

and Ast. (ii) Analysis type of problems: The given data are b, d, D, fck, fy, Asc and Ast. It is required to

determine the Mu of the beam. Q.8: Name the two methods of solving the two types of problems. A.8: The two methods of solving the two types of problems are: (i) Direct computation method, and

(ii) Use of tables of SP-16.


Q.9: Write down the steps of the solution by the two methods of each of the two

types of problems. A.9: (A) For the design type of problems: (i) See sec. 4.8.6.1(a) for the steps of direct computation method, and (ii) See sec. 4.8.6.1(b) for the steps of using the tables of SP-16 (B) For the analysis type of problems: (i) See sec. 4.8.6.2 (a) for the steps of direct computation method, and (ii) See sec. 4.8.6.2 (b) for the steps of using the tables of SP-16. 4.8.8 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 4.8.9 Test 8 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Derive the governing equations of a doubly reinforced beam.

(10 marks) A.TQ.1: See sec. 4.8.3 TQ.2: State the two types of problems of doubly reinforced beams specifying the

given data and the values to be determined in the two type of problems. (8 marks)

A.TQ.2: The two types of problems are:

(i) Design type of problems and (ii) Analysis type of problems

(i) Design type of problems:

The given data are b, d, D, fck, fy and Mu . It is required to determine Asc

and Ast. (ii) Analysis type of problems: The given data are b, d, D, fck, fy, Asc and Ast. It is required to

determine the Mu of the beam. TQ.3: Write down the steps of the solution by the two methods of each of the two

types of problems. (8 marks)

A.TQ.3: (A) For the design type of problems: (i) See sec. 4.8.6.1(a) for the steps of direct computation method, and (ii) See sec. 4.8.6.1(b) for the steps of using the tables of SP-16 (B) For the analysis type of problems:


(i) See sec. 4.8.6.2 (a) for the steps of direct computation method, and (ii) See sec. 4.8.6.2 (b) for the steps of using the tables of SP-16. TQ.4: How do you determine fsc of mild steel and cold worked bars and fcc?

(8 marks) A.TQ.4: See sec. 4.8.4 TQ.5: State the assumptions of the analysis and design of doubly reinforced beams.

(8 marks) A.TQ.5: See sec. 4.8.2 (i), (ii) and (iii). TQ.6: Name three situations other than doubly reinforced beams, where the

compression reinforcement is provided. (8 marks)

A.TQ.6: Compression reinforcement is provided when:

(i) Some sections of a continuous beam with moving loads undergo change of sign of the bending moment which makes compression zone as tension zone,

(ii) the ductility requirement has to be satisfied,

(iii) the reduction of long term deflection is needed.

4.8.10 Summary of this Lesson Lesson 8 derives the governing equations of the doubly reinforced beams explaining different assumptions and situations when they are needed. The methods of determination of compressive stress in steel are illustrated. The two types of problems and the different steps of solution of them by two different methods are explained.


Module 4

Doubly Reinforced Beams – Theory and

Problems


Lesson 9

Doubly Reinforced Beams – Theory


Instructional Objectives: At the end of this lesson, the student should be able to: • design the amounts of compression and tensile reinforcement if the b, d, d',

fck, fy and Mu are given, and • determine the moment of resistance of a beam if b, d, d', fck, fy, Asc and Ast are

given. 4.9.1 Introduction

This lesson illustrates the application of the theory of doubly reinforced beams in solving the two types of problems mentioned in Lesson 8. Both the design and analysis types of problems are solved by (i) direct computation method, and (ii) using tables of SP-16. The step by step solution of the problems will help in understanding the theory of Lesson 8 and its application. 4.9.2 Numerical problems 4.9.2.1 Problem 4.1 Design a simply supported beam of effective span 8 m subjected to imposed loads of 35 kN/m. The beam dimensions and other data are: b = 300 mm, D = 700 mm, M 20 concrete, Fe 415 steel (Fig. 4.9.1). Determine fsc from d'/d as given in Table 4.2 of Lesson 8.


(a) Solution by direct computation method Dead load of the beam = 0.3 (0.7) (25) = 5.25 kN/m Imposed loads (given) = 35.00 kN/m Total loads = 5.25 + 35.00 = 40.25 kN/m

Factored bending moment = 2 1 5 40 25 8 8(1 5 482 96

8 8wl ( . ) ( . ) ( )( ). ) . = = kNm

Assuming d' = 70 mm, d = 700 - 70 = 630 mm

480max . d

x u, = gives xu, max = 0.48 (630) = 302.4 mm

Step 1: Determination of Mu, lim and Ast, lim

cku, u,

u, f b d d

x . -

dx

. M 2maxmaxlim )4201()(360=

(4.2) = 0.36(0.48) {1 – 0.42 (0.48)} (300) (630)2 (20) (10-6) kNm = 328.55 kNm

) 420(870 max

limlim

u, y

u, st, x.d - f.

M A = (6.8)

So, 26

1 mm 141809}630)480(420630{)415(870

Nmm )10(55328 . . . .

.Ast =−

=

Step 2: Determination of Mu2, Asc, Ast2 and Ast (Please refer to Eqs. 4.1, 4.4, 4.6 and 4.7 of Lesson 8.) 2 482 96 328 55 154 41u u u, lim M M - M . - . . = = = kNm Here, d'/d = 70/630 = 0.11 From Table 4.2 of Lesson 8, by linear interpolation, we get,


83505

342353353 . - - f sc == N/mm2

26

2 mm 806.517 N/mm 70) - (630 )}20( 446.08.350{

Nmm )(10 154.41 )()(

=−

==d - d' - ff

M A

ccsc

usc

22 mm694763

)415()870()9288350(517806

870)(

. .

. - . . f.

- ff A A

y

ccscscst ===

2

21 mm 8342572621783141809 . . . A A A ststst =+=+= Step 3: Check for minimum and maximum tension and compression steel. (vide sec.4.8.5 of Lesson 8) (i) In compression:

(a) Minimum 2mm 420 (700) (300) 1000.2 ==scA

(b) Maximum 2mm 8400 (700) (300) 100

4 ==scA

Thus, 420 mm2 < 806.517 mm2 < 8400 mm2 . Hence, o.k. (ii) In tension:

(a) Minimum 2mm 387.1 415

(630) (300) 0.85 850 == f

b d .Ay

st

(b) Maximum 2mm 8400 (700) (300) 100

4 ==stA

Here, 387.1 mm2 < 2572.834 mm2 < 8400 mm2 . Hence, o.k. Step 4: Selection of bar diameter and numbers. (i) for Asc: Provide 2-20 T + 2-12 T (= 628 + 226 = 854 mm2) (ii) for Ast: Provide 4-25 T + 2-20 T (= 1963 + 628 = 2591 mm2)


It may be noted that Ast is provided in two layers in order to provide adequate space for concreting around reinforcement. Also the centroid of the tensile bars is at 70 mm from bottom (Fig. 4.9.1). (b) Solution by use of table of SP-16

For this problem, 0.11. 63070 and 4.056

(630) 300)(10 482.96 2

6

2 ==== d'/db dM u

Table 50 of SP-16 gives pt and pc for 0.15. and 0.1 and 4.1 and 4 2 == d'/db dM u

The required pt and pc are determined by linear interpolation. The values are presented in Table 4.3 to get the final pt and pc of this problem. Table 4.3 Calculation of pt and pc

2b dM u

0.1 /' =dd 0.15 /' =dd 0.11 /' =dd

pt 1.337 1.360 1.342 05.0(0.01) 0.023 337.1 =+

4.0 pc 0.401

0.437 0.408

05.0(0.01) 0.036 433.0 =+

pt 1.368 1.392 1.373 05.0(0.01) 0.024 368.1 =+

4.1 pc 0.433

0.472 0.039 (0.01)0 433 0.441

0 05.

.+ =

pt Not Applicable

(NA)

NA 0.031 (0.056)1 342 1.35940 1

..

+ =

4.056 pc NA

NA 0.033 (0.056)0 408 0.426

0 1.

.+ =

So, 2m 2569.26 100

(630) (300) 1.3594 ==stA

and 2mm 805.14 100

(630) (300) 0.426 ==scA

These values are close to those obtained by direct computation method where Ast = 2572.834 mm2 and Asc = 806.517 mm2. Thus, by using table of SP-16 we


get the reinforcement very close to that of direct computation method. Hence, provide (i) for Asc: 2-20 T + 2-12 T (= 628 + 226 = 854 mm2)

(ii) for Ast: 4-25 T + 2-20 T (= 1963 + 628 = 2591 mm2) 4.9.2.2 Problem 4.2

Determine the ultimate moment capacity of the doubly reinforced beam of b = 350 mm, d' = 60 mm, d = 600 mm, Ast = 2945 mm2 (6-25 T), Asc = 1256 mm2 (4-20 T), using M 20 and Fe 415 (Fig.4.9.2). Use direct computation method only. Solution by direct computation method Step 1: To check if the beam is under-reinforced or over-reinforced. mm 288 (600) 0.48 max ==u,x

0.00379 288

288) - (600 0.0035 )(

max

max ===u,

u,cst x

d - x ε ε


Yield strain of Fe 415 = 0.002 )(10 (2) (1.15)

415 0.002 )( 15.1 5 +=+

s

y

Ef

= 0.0038 > 0.00379. Hence, the beam is over-reinforced. Step 2: To determine Mu,lim and Ast,lim (vide Eq. 4.2 of Lesson 8 and Table 3.1 of Lesson 5)

cku, u,

u fb d d

xd

xM 2maxmax

lim , ) 0.42 - 1 ( ) ( 0.36 =

= 0.36 (0.48) {1 - 0.42 (0.48)} (350) (600)2 (20) (10-6) kNm = 347.67 kNm From Table 3.1 of Lesson 5, for fck = 20 N/mm2 and fy = 415 N/mm2,

2lim, mm 2016

100(600) (350) 0.96 ==stA

Step 3: To determine Ast2 and Asc (vide Eqs.4.7 and 4.6 of Lesson 8) Ast2 = Ast - Ast, lim = 2945 - 2016 = 929 mm2

The required Asc will have the compression force equal to the tensile force as given by 929 mm2 of Ast2.

So, ) - (

) (0.87 2

ccsc

ystsc ff

fAA =

For fsc let us calculate εsc: (vide Eq. 4.9 of Lesson 8)

0.002771 288

60) - (288 0.0035 )( 0.0035

max

max ===u,

u, sc x

- d'xε

Table 4.1 of Lesson 8 gives:

)00276.000380.0(

0.002760) - (0.002771 351.8) - (360.9 351.8 −

+=scf 2N/mm 351.896 =


So, 2mm 977.956 )}20( 446.089.351{

(415) (0.87) 929 =−

=scA

Step 4: To determine Mu2, Mu and Ast (Please refer to Eqs. 4.4 and 4.1 of Lesson 8) )( ) ( 2 d - d' - ffAM ccscscu = = 977.956 {351.896 - 0.446 (20)} (600 - 60) (10-6) kNm = 181.12 kNm

Mu = Mu, lim + Mu2 = 347.67 + 181.12 = 528.79 kNm Therefore, with Ast = Ast, lim + Ast2 = 2016 + 929 = 2945 mm2 the required Asc = 977.956 mm2 (much less than the provided 1256 mm2). Hence, o.k. 4.9.3 Practice Questions and Problems with Answers Q.1: Design a doubly reinforced beam (Fig. 4.9.3) to resist Mu = 375 kNm

when b = 250 mm, d = 500 mm, d' = 75 mm, fck = 30 N/mm2 and fy = 500 N/mm2, using (i) direct computation method and (ii) using table of SP-16.


A.1: (A) Solution by direct computation method: From the given data

Mu, lim = ckuu fdbd

xd

x 2max,max, )42.01()(36.0 −

= 0.36 (0.46) {1 - 0.42 (0.46)} (250) (500)2 (30) (10-6) kNm = 250.51 kNm

Using the value of pt = 1.13 from Table 3.1 of Lesson 5 for fck = 30 N/mm2 and fy = 500 N/mm2,

2lim mm 1412.5

100(500) (250) 1.13 ==st, A

Mu2 = 375 - 250.51 = 124.49 kNm From Table 4.2 of Lesson 8, for d'/d = 75/500 = 0.15 and fy = 500 N/mm2

, we get fsc = 395 N/mm2

2

62 mm 767.56

75) - (500 )}0(3 446.0395{ )(10 124.49

)()(=

−==

d - d' - f fM

Accsc

usc

22 mm37.673

)500(870(30)} 446.0395{56.767

870)(

. -

f. - ff A

Ay

ccscscst ===

2

2lim, mm 87.208537.6735.1412 A A A ststst =+=+= Alternatively: (use of Table 4.1 of Lesson 8 to determine fsc from εsc) xu, max = 0.46 (500) = 230 mm

0.002359 230

(155) 0.0035 230

75) - (230 0.0035 ===scε

From Table 4.1


0.00226) - (0.00277

0.00226) - (0.002359 391.3) - (413.0 391.3 +=scf 2N/mm 395.512 =

26

2 mm 766.53 75) - (500 )}30( 446.0512.395{

)(10 124.49 )()(

=−

==d - d' - f f

M A

ccsc

usc

22 mm369.673

)500(870)132.382(53.766

870)(

.

f.

- f f A A

y

ccscscst ===

2

2lim , mm 869.2085369.6735.1412 A A A ststst =+=+= Check for minimum and maximum Ast and Asc

(i) Minimum 2mm 212.5 500

(500) (250) 0.85 850 == f

b d .Ay

st

(ii) Maximum 2mm 5750 (575) (250) 0.04 0.04 === b DAst

(iii) Minimum 2mm 287.5 100

(575) (250) 0.2 100

20 == b D .Ast

(iv) Maximum 2mm 5750 (575) (250) 0.04 0.04 === b DAst

Hence, the areas of reinforcement satisfy the requirements. So, provide (i) 6-20 T + 2-12 T = 1885 + 226 = 2111 mm2 for Ast

(ii) 4-16 T = 804 mm2 for Asc

(B) Solution by use of table of SP-16 From the given data, we have

6.0 (500) 250

)(10 375 2

6

2 ==b dM u


d'/d = 75/500 = 0.15

Table 56 of SP-16 gives: pt = 1.676 and pc = 0.619

So, 2mm 2095 100

(500) (250) (1.676) ==stA

and 2mm 773.75 100

(500) (250) (0.619 ==scA

These values are close to those of (A). Hence, provide 6-20 T + 2-12 T as Ast and 4-16 T as Asc.

Q.2: Determine the moment of resistance of the doubly reinforced beam (Fig. 4.9.4) with b = 300 mm, d = 600 mm, d' = 90 mm, fck = 30 N/mm2, fy = 500 N/mm2, Asc = 2236 mm2 (2-32 T + 2-20 T), and Ast = 4021 mm2 (4-32 T + 4-16 T). Use (i) direct computation method and (ii) tables of SP-16.

A.2: (i) Solution by direct computation method: xu, max = 0.46 (600) = 276 mm


0.0041086 276

276) - (600 0.0035 ==stε

εyield = 0.00417. So εst < εyield i.e. the beam is over-reinforced. For d'/d = 0.15 and fy = 500 N/mm2, Table 4.2 of Lesson 8 gives: fsc = 395

N/mm2 and with fck = 30 N/mm2 , Table 3.1 of Lesson 5 gives pt, lim = 1.13.

2lim, mm 2034

100(600) (300) 1.13 ==stA

Mu,lim = ckuu fdbd

xd

x 2max,max, )42.01()(36.0 −

= 0.36 (0.46) {1 - 0.42 (0.46)} (300) (600)2 (30) (10-6) kNm = 432.88 kNm Ast2 = 4021 - 2034 = 1987 mm2

222 mm 2236 mm 2264.94 )}0(3 446.0395{ (500) (0.87) 1987

)( (0.87)

)( >=−

== - f ffA

Accsc

ystrequiredsc

So, Ast2 of 1987 mm2 is not fully used. Let us determine Ast2 required

when Asc = 2236 mm2.

22 mm61.1961

)500()870(30)}( 446.0395{2236

870)(

.

- f.

- ff A A

y

ccscscst ===

Ast = Ast, lim + Ast2 = 2034 + 1961.61 = 3995.61 mm2 < 4021 mm2.

Hence, o.k. With Ast2 = 1961.61 mm2, Mu2 = Ast2 (0.87 fy ) (d - d') = 1961.61 (0.87) (500) (600 - 75) (10-6) kNm = 447.98268 kNm Again, when Asc = 2236 mm2 (as provided)


)( ) ( 2 d - d' - ffAM ccscscu = = 2236 {395 - 0.446 (30)} (600 - 75) (10-6) kNm = 447.9837 kNm

Mu = Mu, lim + Mu2 = 432.88 + 447.98 (Mu2 is taken the lower of the two)

= 880.86 kNm Hence, the moment of resistance of the beam is 880.86 kNm. Alternatively fsc can be determined from Table 4.1 of Lesson 8. Using the following from the above: xu, max = 276 mm Ast,lim = 2034 mm2

Mu,lim = 432.88 kNm Ast2 = 1987 mm2

To find (Asc)required

0.00236 276

90) - (276 0.0035 ==stε

Table 4.1 of Lesson 8 gives:

0.00226) - (0.00277

0.00226) - (0.00236 391.3) - (413 391.3 +=scf 2N/mm 395.55 =

22

2

mm 2236 mm 2261.68

)}03( 446.055.395{

(500) (0.87) 1987 )(

(0.87) )(

>=

−==

- f ffA

Accsc

ystrequiredsc

So, it is not o.k. Let us determine Ast2 required when Asc = 2236 mm2.


22 mm44.1964

)500()870((30)} 446.055.395{2236

870)(

. -

f. - ff A

Ay

ccscscst ===

Ast = Ast, lim + Ast2 = 2034 + 1964.44 = 3998.44 mm2 < 4021 mm2.

So, o.k. Mu2 (when Ast2 = 1964.44 mm2) = Ast2 (0.87 fy ) (d - d')

= 1964.44 (0.87) (500) (600 - 75) (10-6) kNm = 448.63 kNm For Asc = 2236 mm2,

)( ) ( 2 d - d' - ffAM ccscscu = = 2236 {(395.55 - 0.446 (30)} (600 - 75) (10-6) kNm = 2236 (382.17) (525) (10-6) kNm = 448.63 kNm Both the Mu2 values are the same. So,

Mu = Mu,lim + Mu2 = 432.88 + 448.63 = 881.51 kNm Here, the Mu = 881.51 kNm. (ii) Solution by using table of SP-16 From the given data:

2.234 )600( 300

(100) 4021 ==tp

1.242 )600( 300

(100) 2236 ==cp

d'/d = 0.15


Table 56 of SP-16 is used first considering d'/d = 0.15 and pt = 2.234,

and secondly, considering d'/d = 0.15 and pc = 1.242. The calculated values of pc and Mu/bd2 for the first and pt and Mu/bd2 for the second cases are presented below separately. Linear interpolation has been done.

(i) When d'/d = 0.15 and pt = 2.234

8.06 2.218) - (2.245

2.218) - (2.234 8.0) - (8.1 8.00 2 =+=db

M u

1.242 1.253 )027.0((0.016) 1.235) - (1.266 1.235 >=+=cp

So, this is not possible. (ii) When d'/d = 0.15 and pc = 1.242

8.022 1.235) - (1.266

1.235) - (1.242 8.0) - (8.1 8.00 2 =+=db

M u

2.234 2.224 )235.1266.1(

1.235) - (1.242 2.218) - (2.245 2.218 <=−

+=tp

So, Mu = 8.022 (300) (600)2 (10-6)= 866.376 kNm. Hence, o.k.

4.9.4 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 4.9.5 Test 9 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Design a simply supported beam of effective span 8 m subjected to

imposed loads of 35 kN/m. The beam dimensions and other data are: b = 300 mm, D = 700 mm, M 20 concrete, Fe 415 steel (Fig. 4.9.1). Determine fsc from strain εsc as given in Table 4.1 of Lesson 8.

A.TQ.1: This problem is the same as Problem 4.1 in sec. 4.9.2.1 except that here

the fsc is to be calculated using Table 4.1 instead of Table 4.2. Step 1: Here, the Step 1 will remain the same as that of Problem 4.1. Step 2: Determination of Mu2, Asc, Ast2 and Ast

. . - . - M M M u, uu 411545532896482lim2 === kNm


From strain triangle: (Fig. 4.8.2 of Lesson 8)

0.00269 4.302

70) - (302.4 0.0035 ==scε

0.00241) - (0.00269 0.00241) - (0.00276342.8) - (351.8 342.8 8)Lesson of 4.1 Table (from +=scf

2N/mm 350 =

26

2 mm 808.41 N/mm 70) - (630 )}20( 446.0350{

)(10 154.41 )()(

=−

==d - d' - ff

M A

ccsc

usc

22 mm696763

)415()870()08341(41808

870)(

. .

. . f.

- f f A A

y

ccscscst ===

2

21 mm 8362572696763141809 . . . A A A ststst =+=+= Asc = 808.41 mm2

Steps 3 & 4 will also remain the same as those of Problem 4.1. Hence, provide 2-20 T + 2-12 T (854 mm2) as Asc and 4-25 T + 2-20 T (2591 mm2) as Ast . TQ.2: Determine the ultimate moment capacity of the doubly reinforced beam of

b = 350 mm, d' = 60 mm, d = 600 mm, Ast = 2945 mm2 (6-25 T), Asc = 1256 mm2 (4-20 T), using M 20 and Fe 415 (Fig. 4.9.2). Use table of SP-16 only.

A.TQ.2: Solution by using table of SP-16

This problem is the same as that of Problem 4.2 of sec. 4.9.2.2, which has been solved by direct computation method. Here, the same is to be solved by using SP-16.

The needed parameters are:

d'/d = 60/600 = 0.1

1.402 )600(350

(100) 2945 (100)

===b d

Ap st

t


0.5981 )600( 350

(100) 1256 (100)

===b d

Ap sc

c

Here, we need to use Table 50 for fck = 20 N/mm2 and fy = 415 N/mm2. The table gives values of Mu/bd2 for (I) d'/d and pt and (ii) d'/d and pc. So, we will consider both the possibilities and determine Mu.

(i) Considering Table 50 of SP-16 when d'/d = 0.1 and pt = 1.402: Interpolating the values of Mu/bd2 at pt = 1.399 and 1.429, we get

4.21 1.399) - (1.429

1.399) - (1.402 4.2) - (4.3 4.2 1.402

2 =+=⎟⎠⎞

⎜⎝⎛

=tp

u

dbM

the corresponding ( ) 0.4692 )399.1 429.1(

1.399) - (1.402 0.466) - (0.498 0.466 1.402 =−

+==tpcp

But, pc provided is 0.5981 indicates that extra compression reinforcement has been used. So, we get when AkNm 530.46 )(10 (600) (350) (4.21) 4.21 -622 === b dM u st = 2945 mm2 and Asc = 985.32 mm2, i.e. 270.69 mm2 (= 1256 - 985.32) of compression steel is extra. (ii) Considering d'/d = 0.1 and pc = 0.5981, we get by linear interpolation

4.61 0.595) - (0.628

0.595) - (0.5981 4.6) - (4.7 4.6 0.5981

2 =+=⎟⎠⎞

⎜⎝⎛

=cp

u

dbM

the corresponding pt is:

( ) 1.5231 )595.0 628.0(

0.595) - (0.5981 1.522) - (1.533 1.522 0.5981 =−

+==cptp

The provided pt = 1.402 indicates that the tension steel is insufficient by 254.31 mm2 as shown below: Amount of additional Ast still required =


2mm 254.31 100

(600) (350) 1.402) - (1.5231=

If this additional steel is provided, then the Mu of this beam becomes: Mu = 4.61 b d2 = 4.61 (350) (600)2 (10-6) kNm = 580.86 kNm The above two results show that the moment of resistance of this beam is the lower of the two. So, Mu = 530.46 kNm. By direct computation the Mu = 528.79 kNm. The two results are in good agreement. 4.9.6 Summary of this Lesson

This lesson presents solutions of four numerical problems covering both design and analysis types. These problems are solved by two methods: (i) direct computation method and (ii) using table of SP-16. Two problems are illustrated in the lesson and the other two are given in the practice problem and test of this lesson. The solutions will help in understanding the step by step application of the theory of doubly reinforced beams given in Lesson 8.


Module 5

Flanged Beams – Theory and Numerical

Problems


Lesson 10

Flanged Beams – Theory



• identify the regions where the beam shall be designed as a flanged and where it will be rectangular in normal slab beam construction,

• define the effective and actual widths of flanged beams,

• state the requirements so that the slab part is effectively coupled with the

flanged beam, • write the expressions of effective widths of T and L-beams both for

continuous and isolated cases,

• derive the expressions of C, T and Mu for four different cases depending on the location of the neutral axis and depth of the flange.

5.10.1 Introduction


Reinforced concrete slabs used in floors, roofs and decks are mostly cast monolithic from the bottom of the beam to the top of the slab. Such rectangular beams having slab on top are different from others having either no slab (bracings of elevated tanks, lintels etc.) or having disconnected slabs as in some pre-cast systems (Figs. 5.10.1 a, b and c). Due to monolithic casting, beams and a part of the slab act together. Under the action of positive bending moment, i.e., between the supports of a continuous beam, the slab, up to a certain width greater than the width of the beam, forms the top part of the beam. Such beams having slab on top of the rectangular rib are designated as the flanged beams - either T or L type depending on whether the slab is on both sides or on one side of the beam (Figs. 5.10.2 a to e). Over the supports of a continuous beam, the bending moment is negative and the slab, therefore, is in tension while a part of the rectangular beam (rib) is in compression. The continuous beam at support is thus equivalent to a rectangular beam (Figs. 5.10.2 a, c, f and g).


The actual width of the flange is the spacing of the beam, which is the same as the distance between the middle points of the adjacent spans of the slab, as shown in Fig. 5.10.2 b. However, in a flanged beam, a part of the width less than the actual width, is effective to be considered as a part of the beam. This width of the slab is designated as the effective width of the flange. 5.10.2 Effective Width


5.10.2.1 IS code requirements The following requirements (cl. 23.1.1 of IS 456) are to be satisfied to ensure the combined action of the part of the slab and the rib (rectangular part of the beam). (a) The slab and the rectangular beam shall be cast integrally or they shall be effectively bonded in any other manner. (b) Slabs must be provided with the transverse reinforcement of at least 60 per cent of the main reinforcement at the mid span of the slab if the main reinforcement of the slab is parallel to the transverse beam (Figs. 5.10.3 a and b).

The variation of compressive stress (Fig. 5.10.4) along the actual width of the flange shows that the compressive stress is more in the flange just above the rib than the same at some distance away from it. The nature of variation is complex and, therefore, the concept of effective width has been introduced. The effective width is a convenient hypothetical width of the flange over which the compressive stress is assumed to be uniform to give the same compressive


force as it would have been in case of the actual width with the true variation of compressive stress. 5.10.2.2 IS code specifications Clause 23.1.2 of IS 456 specifies the following effective widths of T and L-beams: (a) For T-beams, the lesser of (i) bf = lo/6 + bw + 6 Df (ii) bf = Actual width of the flange (b) For isolated T-beams, the lesser of

(i) bf = wo

o b/bl

l

4 )(+

+

(ii) bf = Actual width of the flange (c) For L-beams, the lesser of (i) bf = lo/12 + bw + 3 Df (ii) bf = Actual width of the flange (d) For isolated L-beams, the lesser of

(i) bf = wo

o b/bl

l

4 )( 5.0

++

(ii) bf = Actual width of the flange where bf = effective width of the flange, lo = distance between points of zero moments in the beam, which is the

effective span for simply supported beams and 0.7 times the effective span for continuous beams and frames,

bw = beadth of the web, Df = thickness of the flange, and b = actual width of the flange.


5.10.3 Four Different Cases The neutral axis of a flanged beam may be either in the flange or in the web depending on the physical dimensions of the effective width of flange bf, effective width of web bw, thickness of flange Df and effective depth of flanged beam d (Fig. 5.10.4). The flanged beam may be considered as a rectangular beam of width bf and effective depth d if the neutral axis is in the flange as the concrete in tension is ignored. However, if the neutral axis is in the web, the compression is taken by the flange and a part of the web.

All the assumptions made in sec. 3.4.2 of Lesson 4 are also applicable for the flanged beams. As explained in Lesson 4, the compressive stress remains constant between the strains of 0.002 and 0.0035. It is important to find the depth h of the beam where the strain is 0.002 (Fig. 5.10.5 b). If it is located in the web, the whole of flange will be under the constant stress level of 0.446 fck. The


following gives the relation of Df and d to facilitate the determination of the depth h where the strain will be 0.002. From the strain diagram of Fig. 5.10.5 b:

u

u

x - hx

0035.0002.0

=

or 3 0 437u

h .x

= =

(5.1) when , we get max , uu xx =

dddxh u 0.197 and 0.205 , 0.227 73 max , == , for Fe 250, Fe 415 and Fe

500, respectively. In general, we can adopt, say h/d = 0.2 (5.2) The same relation is obtained below from the values of strains of concrete and steel of Fig. 5.10.5 b.

u

u

c

st

xd - x

=εε

or c

cst

εεε +

=ux

d (5.3)

Dividing Eq. 5.1 by Eq. 5.3

0.0035

0.0015 st +

=εd

h (5.4)

Using (0.87 ) 0.002st y sf / Eε = + in Eq. 5.4, we get h/d = 0.227, 0.205 and 0.197 for Fe 250, Fe 415 and Fe 500 respectively, and we can adopt h/d = 0.2 (as in Eq. 5.2). Thus, we get the same Eq. 5.2 from Eq. 5.4, h/d = 0.2 (5.2)


It is now clear that the three values of h are around 0.2 d for the three grades of steel. The maximum value of h may be Df, at the bottom of the flange where the strain will be 0.002, if Df /d = 0.2. This reveals that the thickness of the flange may be considered small if Df /d does not exceed 0.2 and in that case, the position of the fibre of 0.002 strain will be in the web and the entire flange will be under a constant compressive stress of 0.446 fck . On the other hand, if Df is > 0.2 d, the position of the fibre of 0.002 strain will be in the flange. In that case, a part of the slab will have the constant stress of 0.446 fck where the strain will be more than 0.002. Thus, in the balanced and over-reinforced flanged beams (when

), the ratio of Dmax , uu xx = f /d is important to determine if the rectangular stress block is for the full depth of the flange (when Df /d does not exceed 0.2) of for a part of the flange (when Df /d > 0.2). Similarly, for the under-reinforced flanged beams, the ratio of Df /xu is considered in place of Df /d. If Df /xu does not exceed 0.43 (see Eq. 5.1), the constant stress block is for the full depth of the flange. If Df /xu > 0.43, the constant stress block is for a part of the depth of the flange. Based on the above discussion, the four cases of flanged beams are as follows:


(i) Neutral axis is in the flange (xu < Df ), (Fig. 5.10.6 a to c)


(ii) Neutral axis is in the web and the section is balanced (xu = xu,max >

Df), (Figs. 5.10.7 and 8 a to e) It has two situations: (a) when Df /d does not exceed 0.2, the constant stress block is for the entire depth of the flange (Fig. 5.10.7), and (b) when Df /d > 0.2, the constant stress block is for a part of the depth of flange (Fig. 5.10.8).


(iii) Neutral axis is in the web and the section is under-reinforced (xu,max > xu > Df), (Figs. 5.10.9 and 10 a to e)


This has two situations: (a) when Df /xu does not exceed 0.43, the full depth of flange is having the constant stress (Fig. 5.10.9), and (b) when Df /xu > 0.43, the constant stress is for a part of the depth of flange (Fig. 5.10.10).

(iv) Neutral axis is in the web and the section is over-reinforced (xu >

xu,max> Df), (Figs. 5.10.7 and 8 a to e) As mentioned earlier, the value of xu is then taken as xu,max when xu> xu,max. Therefore, this case also will have two situations depending on Df /d not exceeding 0.2 or > 0.2 as in (ii) above. The governing equations of the four different cases are now taken up.

5.10.4 Governing Equations The following equations are only for the singly reinforced T-beams. Additional terms involving Mu,lim, Mu2, Asc , Ast1 and Ast2 are to be included from Eqs. 4.1 to 4.8 of sec. 4.8.3 of Lesson 8 depending on the particular case. Applications of these terms are explained through the solutions of numerical problems of doubly reinforced T-beams in Lessons 11 and 12. 5.10.4.1 Case (i): When the neutral axis is in the flange (xu < Df ), (Figs. 5.10.6 a to c) Concrete below the neutral axis is in tension and is ignored. The steel reinforcement takes the tensile force (Fig. 5.10.6). Therefore, T and L-beams are considered as rectangular beams of width bf and effective depth d. All the equations of singly and doubly reinforced rectangular beams derived in Lessons 4 to 5 and 8 respectively, are also applicable here. 5.10.4.2 Case (ii): When the neutral axis is in the web and the section is balanced (xu,max > Df ), (Figs. 5.10.7 and 8 a to e) (a) When Df /d does not exceed 0.2, (Figs. 5.10.7 a to e) As explained in sec. 5.10.3, the depth of the rectangular portion of the stress block (of constant stress = 0.446 fck) in this case is greater than Df (Figs. 5.10.7 a, b and c). The section is split into two parts: (i) rectangular web of width bw and effective depth d, and (ii) flange of width (bf - bw) and depth Df (Figs. 5.10.7 d and e). Total compressive force = Compressive force of rectangular beam of width bw and depth d + Compressive force of rectangular flange of width (bf - bw) and depth Df . Thus, total compressive force


C = 0.36 fck bw xu, max + 0.45 fck (bf - bw) Df (5.5) (Assuming the constant stress of concrete in the flange as 0.45 fck in place of 0.446 fck ,as per G-2.2 of IS 456), and the tensile force T = 0.87 fy Ast (5.6) The lever arm of the rectangular beam (web part) is (d - 0.42 xu, max) and the same for the flanged part is (d - 0.5 Df ). So, the total moment = Moment due to rectangular web part + Moment due to rectangular flange part or Mu = 0.36 fck bw xu, max (d - 0.42 xu, max ) + 0.45 fck (bf - bw) Df (d - Df /2) or Mu = 0.36(xu, max /d){1 - 0.42( xu, max/d)} fck bw d2 + 0.45 fck(bf - bw) Df(d - Df /2) (5.7) Equation 5.7 is given in G-2.2 of IS 456. (b) When Df /d > 0.2, (Figs. 5.10.8 a to e) In this case, the depth of rectangular portion of stress block is within the flange (Figs. 5.10.8 a, b and c). It is assumed that this depth of constant stress (0.45 fck) is yf, where yf = 0.15 xu, max + 0.65 Df, but not greater than Df (5.8) The above expression of yf is derived in sec. 5.10.4.5. As in the previous case (ii a), when Df /d does not exceed 0.2, equations of C, T and Mu are obtained from Eqs. 5.5, 6 and 7 by changing Df to yf. Thus, we have (Figs. 5.10.8 d and e)

C = 0.36 fck bw xu, max + 0.45 fck (bf - bw) yf (5.9) T = 0.87 fy Ast (5.10) The lever arm of the rectangular beam (web part) is (d - 0.42 xu, max ) and the same for the flange part is (d - 0.5 yf ). Accordingly, the expression of Mu is as follows:


Mu = 0.36(xu, max /d){1 - 0.42( xu, max/d)} fck bw d2 + 0.45 fck(bf - bw) yf(d - yf /2) (5.11) 5.10.4.3 Case (iii): When the neutral axis is in the web and the section is under-reinforced (xu > Df ), (Figs. 5.10.9 and 10 a to e) (a) When Df / xu does not exceed 0.43, (Figs. 5.10.9 a to e) Since Df does not exceed 0.43 xu and h (depth of fibre where the strain is 0.002) is at a depth of 0.43 xu, the entire flange will be under a constant stress of 0.45 fck (Figs. 5.10.9 a, b and c). The equations of C, T and Mu can be written in the same manner as in sec. 5.10.4.2, case (ii a). The final forms of the equations are obtained from Eqs. 5.5, 6 and 7 by replacing xu, max by xu. Thus, we have (Figs. 5.10.9 d and e)

C = 0.36 fck bw xu + 0.45 fck (bf - bw) Df (5.12) T = 0.87 fy Ast (5.13) Mu = 0.36(xu /d){1 - 0.42( xu /d)} fck bw d2 + 0.45 fck(bf - bw) Df (d - Df /2) (5.14) (b) When Df / xu > 0.43, (Figs. 5.10.10 a to e) Since Df > 0.43 xu and h (depth of fibre where the strain is 0.002) is at a depth of 0.43 xu, the part of the flange having the constant stress of 0.45 fck is assumed as yf (Fig. 5.10.10 a, b and c). The expressions of yf , C, T and Mu can be written from Eqs. 5.8, 9, 10 and 11 of sec. 5.10.4.2, case (ii b), by replacing xu,max by xu. Thus, we have (Fig. 5.10.10 d and e) yf = 0.15 xu + 0.65 Df, but not greater than Df (5.15)

C = 0.36 fck bw xu + 0.45 fck (bf - bw) yf (5.16) T = 0.87 fy Ast (5.17) Mu = 0.36(xu /d){1 - 0.42( xu /d)} fck bw d2 + 0.45 fck(bf - bw) yf (d - yf /2) (5.18)


5.10.4.4 Case (iv): When the neutral axis is in the web and the section is over-reinforced (xu > Df ), (Figs. 5.10.7 and 8 a to e) For the over-reinforced beam, the depth of neutral axis xu is more than xu, max as in rectangular beams. However, xu is restricted up to xu,max. Therefore, the corresponding expressions of C, T and Mu for the two situations (a) when Df / d does not exceed 0.2 and (b) when Df / d > 0.2 are written from Eqs. 5.5 to 5.7 and 5.9 to 5.11, respectively of sec. 5.10.4.2 (Figs. 5.10.7 and 8). The expression of yf for (b) is the same as that of Eq. 5.8. (a) When Df /d does not exceed 0.2 (Figs. 5.10.7 a to e) The equations are: C = 0.36 fck bw xu, max + 0.45 fck (bf - bw) Df (5.5) T = 0.87 fy Ast (5.6) Mu = 0.36(xu, max /d){1 - 0.42( xu, max/d)} fck bw d2 + 0.45 fck(bf - bw) Df(d - Df /2) (5.7) (b) When Df /d > 0.2 (Figs. 5.10.8 a to e) yf = 0.15 xu, max + 0.65 Df, but not greater than Df (5.8)

C = 0.36 fck bw xu, max + 0.45 fck (bf - bw) yf (5.9) T = 0.87 fy Ast (5.10) Mu = 0.36(xu, max /d){1 - 0.42( xu, max/d)} fck bw d2 + 0.45 fck(bf - bw) yf(d - yf /2) (5.11) It is clear from the above that the over-reinforced beam will not have additional moment of resistance beyond that of the balanced one. Moreover, it will prevent steel failure. It is, therefore, recommended either to re-design or to go for doubly reinforced flanged beam than designing over-reinforced flanged beam.


5.10.4.5 Derivation of the equation to determine yf , Eq. 5.8, Fig. 5.10.11

Whitney's stress block has been considered to derive Eq. 5.8. Figure 5.10.11 shows the two stress blocks of IS code and of Whitney. yf = Depth of constant portion of the stress block when Df /d > 0.2. As yf is a function of xu and Df and let us assume yf = A xu + B Df (5.19) where A and B are to be determined from the following two conditions: (i) yf = 0.43 xu , when Df = 0.43 xu (5.20) (ii) yf = 0.8 xu , when Df = xu (5.21) Using the conditions of Eqs. 5.20 and 21 in Eq. 5.19, we get A = 0.15 and B = 0.65. Thus, we have yf = 0.15 xu + 0.65 Df (5.8) 5.10.5 Practice Questions and Problems with Answers Q.1: Why do we consider most of the beams as T or L-beams between the

supports and rectangular beams over the support of continuous span?


A.1: Sec. 5.10.1, first paragraph. Q.2: Draw cross-section of a beam with top slab and show the actual width and

effective width of the T-beam. A.2: Fig. 5.10.2 b. Q.3: State the requirements with figures as per IS 456 which ensure the

combined action of the part of the slab and the rib of flanged beams. A.3: Sec. 5.10.2.1(a) and (b), Figure 5.10.3 (a and b). Q.4: Define “effective width” of flanged beams. A.4: Effective width is an imaginary width of the flange over which the

compressive stress is assumed to be uniform to give the same compressive force as it would have been in case of the actual width with the true variation of compressive stress (Fig. 5.10.4 of text).

Q.5: Write the expressions of effective widths of T and L-beams and isolated beams. A.5: Sec. 5.10.2.2. Q.6: Name the four different cases of flanged beams. A.6: The four different cases are: (i) When the neutral axis is in the flange (xu < Df) (discussed in sec. 5.10.4.1).

(ii) When the neutral axis is in the web and the section is balanced (xu,max > Df). It has two situations: (a) when Df /d does not exceed 0.2 and (b) when Df /d > 0.2 (discussed in sec. 5.10.4.2).

(iii) When the neutral axis is in the web and the section is under-reinforced

(xu,max > xu > Df). It has two situations: (a) when Df /xu does not exceed 0.43 and (b) when Df /xu > 0.43 (discussed in sec. 5.10.4.3).

(iv) When the neutral axis is in the web and the section is over-reinforced

(xu > xu,max> Df). It has two situations: (a) when Df /d does not exceed 0.2 and (b) when Df /d > 0.2 (discussed in sec. 5.10.4.4).

Q.7: (a) Derive the following equation: yf = 0.15 xu,max + 0.65 Df


(b) State when this equation is to be used. (c) What is the limiting value of yf ? A.7: (a) For derivation of the equation, see sec. 5.10.4.5. (b) This equation gives the depth of flange over which the stress is

constant at 0.45 fck (i.e. strain is more than 0.002) when the neutral axis is in web. This occurs when Df /d > 0.2 for balanced beam and when Df /xu > 0.43 for under-reinforced beams.

(c) Limiting value of yf is Df. 5.10.6 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 5.10.7 Test 10 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Why do we consider most of the beams as T or L- beams between the

supports and rectangular beams over the support of continuous span? (5 marks)

A.TQ.1: Sec. 5.10.1, first paragraph.

TQ.2: Define “effective width” of flanged beams. (5 marks)

A.TQ.2: Effective width is a convenient hypothetical width of the flange over which the compressive stress is assumed to be uniform to give the same compressive force as it would have been in case of the actual width with the true variation of compressive stress (Fig. 5.10.4 of text).

TQ.3: State the requirements with figures as per IS 456 which ensure the

combined action of the part of the slab and the rib of flanged beams. (10 marks)

A.TQ.3: Sec. 5.10.2.1(a) and (b), Figure 5.10.3 (a and b). TQ.4: Write the expressions of effective widths of T and L-beams and isolated beams.

(10 marks) A.TQ.4: Sec. 5.10.2.2. TQ.5: Name the four different cases of flanged beams.

(10 marks) A.TQ.5: The four different cases are (i) When the neutral axis is in the flange (xu < Df) (discussed in sec. 5.10.4.1).

(ii) When the neutral axis is in the web and the section is balanced. It has two situations: (a) when Df /d does not exceed 0.2 and (b) when Df /d > 0.2 (discussed in sec. 5.10.4.2).


(iii) When the neutral axis is in the web and the section is under-reinforced. It has two situations: (a) when Df /xu does not exceed 0.43 and (b) when Df /xu > 0.43 (discussed in sec. 5.10.4.3).

(iv) When the neutral axis is in the web and the section is over-reinforced.

It has two situations: (a) when Df /d does not exceed 0.2 and (b) when Df /d > 0.2 (discussed in sec. 5.10.4.4).

TQ.6: (a) Derive the following equation: yf = 0.15 xu,max + 0.65 Df (b) State when this equation is to be used. (c) What is the limiting value of yf ?

(5 + 3 + 2 = 10 marks) A.TQ.6: (a) For derivation of the equation, see sec. 5.10.4.5. (b) This equation gives the depth of flange over which the stress is

constant at 0.45 fck (i.e. strain is more than 0.002) when the neutral axis is in web. This occurs when Df /d > 0.2 for balanced beam and when Df /xu > 0.43 for under-reinforced beams.

(c) Limiting value of yf is Df.

5.10.8 Summary of this Lesson This lesson illustrates the practical situations when slabs are cast integrally with the beams to form either T and L-beams or rectangular beams. The concept of effective width of the slab to form a part of the beam has been explained. The requirements as per IS 456 have been illustrated so that the considered part of the slab may become effective as a beam. Expressions of effective widths for different cases of T and L-beams are given. Four sets of governing equations for determining C, T and Mu are derived for four different cases. These equations form the basis of analysis and design of singly and doubly reinforced T and L- beams.


Module 5


Problems


Lesson 11

Flanged Beams – Numerical Problems


Instructional Objectives:

At the end of this lesson, the student should be able to:

• identify the two types of numerical problems – analysis and design types, • apply the formulations to analyse the capacity of a flanged beam, • determine the limiting moment of resistance quickly with the help of tables

of SP-16.

5.11.1 Introduction Lesson 10 illustrates the governing equations of flanged beams. It is now necessary to apply them for the solution of numerical problems. Two types of numerical problems are possible: (i) Analysis and (ii) Design types. This lesson explains the application of the theory of flanged beams for the analysis type of problems. Moreover, use of tables of SP-16 has been illustrated to determine the limiting moment of resistance of sections quickly for the three grades of steel. Besides mentioning the different steps of the solution, numerical examples are also taken up to explain their step-by-step solutions. 5.11.2 Analysis Type of Problems The dimensions of the beam bf, bw, Df, d, D, grades of concrete and steel and the amount of steel Ast are given. It is required to determine the moment of resistance of the beam. Step 1: To determine the depth of the neutral axis xu The depth of the neutral axis is determined from the equation of equilibrium C = T. However, the expression of C depends on the location of neutral axis, Df /d and Df /xu parameters. Therefore, it is required to assume first that the xu is in the flange. If this is not the case, the next step is to assume xu in the web and the computed value of xu will indicate if the beam is under-reinforced, balanced or over-reinforced.


Other steps:

After knowing if the section is under-reinforced, balanced or over-reinforced, the respective parameter Df/d or Df/xu is computed for the under-reinforced, balanced or over-reinforced beam. The respective expressions of C, T and Mu, as established in Lesson 10, are then employed to determine their values. Figure 5.11.1 illustrates the steps to be followed.


5.11.3 Numerical Problems (Analysis Type)

Ex.1: Determine the moment of resistance of the T-beam of Fig. 5.11.2. Given data: bf = 1000 mm, Df = 100 mm, bw = 300 mm, cover = 50 mm, d = 450 mm and Ast = 1963 mm2 (4- 25 T). Use M 20 and Fe 415. Step 1: To determine the depth of the neutral axis xu Assuming xu in the flange and equating total compressive and tensile forces from the expressions of C and T (Eq. 3.16 of Lesson 5) as the T-beam can be treated as rectangular beam of width bf and effective depth d, we get:

mm 100 mm 98.44 (20) (1000) 0.36

(1963) (415) 0.87 36.0 0.87

<===ckf

styu fb

Afx

So, the assumption of xu in the flange is correct. xu, max for the balanced rectangular beam = 0.48 d = 0.48 (450) = 216 mm. It is under-reinforced since xu < xu,max. Step 2: To determine C, T and Mu From Eqs. 3.9 (using b = bf) and 3.14 of Lesson 4 for C and T and Eq. 3.23 of Lesson 5 for Mu, we have:


C = 0.36 bf xu fck (3.9) = 0.36 (1000) (98.44) (20) = 708.77 kN T = 0.87 fy Ast (3.14) = 0.87 (415) (1963) = 708.74 kN

0.87 (1 - ) (3.23)

(1963) (415) 0.87 (415) (1963) (450) {1 - } = 290.06 kNm(20) (1000) (450)

st yu y st

ck f

A fM f A d

f b d=

=

This problem belongs to the case (i) and is explained in sec. 5.10.4.1 of Lesson 10.

Ex.2: Determine Ast,lim and Mu,lim of the flanged beam of Fig. 5.11.3. Given data are: bf = 1000 mm, Df = 100 mm, bw = 300 mm, cover = 50 mm and d = 450 mm. Use M 20 and Fe 415. Step 1: To determine Df/d ratio For the limiting case xu = xu,max = 0.48 (450) = 216 mm > Df. The ratio Df/d is computed.


Df/d = 100/450 = 0.222 > 0.2 Hence, it is a problem of case (ii b) and discussed in sec. 5.10.4.2 b of Lesson 10. Step 2: Computations of yf , C and T First, we have to compute yf from Eq.5.8 of Lesson 10 and then employ Eqs. 5.9, 10 and 11 of Lesson 10 to determine C, T and Mu, respectively. yf = 0.15 xu,max + 0.65 Df = 0.15 (216) + 0.65 (100) = 97.4 mm. (from Eq. 5.8) C = 0.36 fck bw xu,max + 0.45 fck (bf - bw) yf (5.9) = 0.36 (20) (300) (216) + 0.45 (20) (1000 - 300) (97.4) = 1,080.18 kN. T = 0.87 fy Ast = 0.87 (415) Ast (5.10) Equating C and T, we have

22 mm 2,991.77

N/mm (415) 0.87N (1000) (1080.18) ==stA

Provide 4-28 T (2463 mm2) + 3-16 T (603 mm2) = 3,066 mm2

Step 3: Computation of Mu

2 0.36( ) {1 - 0.42( )}

+ 0.45 ( ) ( /2) (5.11)

u, max u, maxu, lim ck w

ck f w f f

x xM f b d

d df b - b y d - y

=

2

0.36 (0.48) {1 - 0.42 (0.48)} (20) (300) (450) + 0.45 (20) (1000 - 300) (97.4) (450 - 97.4/2) = 413.87 kNm

=


Ex.3: Determine the moment of resistance of the beam of Fig. 5.11.4 when Ast = 2,591 mm2 (4- 25 T and 2- 20 T). Other parameters are the same as those of Ex.1: bf = 1,000 mm, Df = 100 mm, bw = 300 mm, cover = 50 mm and d = 450 mm. Use M 20 and Fe 415. Step 1: To determine xu Assuming xu to be in the flange and the beam is under-reinforced, we have from Eq. 3.16 of Lesson 5:

mm 100 mm 129.93 (20) (1000) 0.36

(2591) (415) 0.87 36.0 0.87

>===ckf

styu fb

Afx

Since xu > Df, the neutral axis is in web. Here, Df/d = 100/450 = 0.222 > 0.2. So, we have to substitute the term yf from Eq. 5.15 of Lesson 10, assuming Df / xu > 0.43 in the equation of C = T from Eqs. 5.16 and 17 of sec. 5.10.4.3 b of Lesson 10. Accordingly, we get:

0.36 fck bw xu + 0.45 fck (bf - bw) yf = 0.87 fy Ast or 0.36 (20) (300) (xu) + 0.45 (20) (1000 - 300) {0.15 xu + 0.65 (100)}

= 0.87 (415) (2591) or xu = 169.398 mm < 216 mm (xu,max = 0.48 xu = 216 mm)


So, the section is under-reinforced. Step 2: To determine Mu Df /xu = 100/169.398 = 0.590 > 0.43 This is the problem of case (iii b) of sec. 5.10.4.3 b. The corresponding equations are Eq. 5.15 of Lesson 10 for yf and Eqs. 5.16 to 18 of Lesson 10 for C, T and Mu, respectively. From Eq. 5.15 of Lesson 10, we have: yf = 0.15 xu + 0.65 Df = 0.15 (169.398) + 0.65 (100) = 90.409 mm From Eq. 5.18 of Lesson 10, we have Mu = 0.36(xu /d){1 - 0.42( xu /d)} fck bw d2 + 0.45 fck(bf - bw) yf (d - yf /2) or Mu = 0.36 (169.398/450) {1 - 0.42 (169.398/450)} (20) (300) (450) (450) + 0.45 (20) (1000 - 300) (90.409) (450 - 90.409/2) = 138.62 + 230.56 = 369.18 kNm.

Ex.4: Determine the moment of resistance of the flanged beam of Fig. 5.11.5 with Ast = 4,825 mm2 (6- 32 T). Other parameters and data are the same as those of Ex.1: bf = 1000 mm, Df = 100 mm, bw = 300 mm, cover = 50 mm and d = 450 mm. Use M 20 and Fe 415.


Step 1: To determine xu Assuming xu in the flange of under-reinforced rectangular beam we have from Eq. 3.16 of Lesson 5:

fckf

styu D

fb Af

x mm 241.95 (20) (1000) 0.36

(4825) (415) 0.87 36.0 0.87

>===

Here, Df/d = 100/450 = 0.222 > 0.2. So, we have to determine yf from Eq. 5.15 and equating C and T from Eqs. 5.16 and 17 of Lesson 10. yf = 0.15 xu + 0.65 Df (5.15)

0.36 fck bw xu + 0.45 fck (bf - bw) yf = 0.87 fy Ast (5.16 and 5.17)

or 0.36 (20) (300) (xu) + 0.45 (20) (1000 - 300) {0.15 xu + 0.65 (100)}

= 0.87 (415) (4825) or 2160 xu + 945 xu = - 409500 + 1742066 or xu = 1332566/3105 = 429.17 mm

xu,max = 0.48 (450) = 216 mm Since xu > xu,max, the beam is over-reinforced. Accordingly.

xu = xu, max = 216 mm. Step 2: To determine Mu This problem belongs to case (iv b), explained in sec.5.10.4.4 b of Lesson 10. So, we can determine Mu from Eq. 5.11 of Lesson 10. Mu = 0.36(xu, max /d){1 - 0.42(xu, max /d)} fck bw d2 + 0.45 fck(bf - bw) yf (d - yf /2) (5.11) where yf = 0.15 xu, max + 0.65 Df = 97.4 mm (5.8) From Eq. 5.11, employing the value of yf = 97.4 mm, we get: Mu = 0.36 (0.48) {1 - 0.42 (0.48)} (20) (300) (450) (450)


+ 0.45 (20) (1000 - 300) (97.4) (450 - 97.4/2) = 167.63 + 246.24 = 413.87 kNm It is seen that this over-reinforced beam has the same Mu as that of the balanced beam of Example 2. 5.11.4 Summary of Results of Examples 1-4

The results of four problems (Exs. 1-4) are given in Table 5.1 below. All

the examples are having the common data except Ast. Table 5.1 Results of Examples 1-4 (Figs. 5.11.2 – 5.11.5) Ex. No.

Ast(mm2)

Case Section No.

Mu(kNm)

Remarks

1 1,963 (i) 5.10.4.1 290.06 xu = 98.44 mm < xu, max (= 216 mm), xu < Df (= 100 mm), Under-reinforced, (NA in the flange).

2 3,066 (ii b) 5.10.4.2 (b)

413.87 xu = xu, max = 216 mm, Df /d = 0.222 > 0.2, Balanced, (NA in web).

3 2,591 (iii b) 5.10.4.3 (b)

369.18 xu = 169.398 mm < xu, max(= 216 mm), Df /xu = 0.59 > 0.43, Under-reinforced, (NA in the web).

4 4,825 (iv b) 5.10.4.4 (b)

413.87 xu = 241.95 mm > xu, max (= 216 mm), Df /d = 0.222 > 0.2, Over-reinforced, (NA in web).

It is clear from the above table (Table 5.1), that Ex.4 is an over-reinforced flanged beam. The moment of resistance of this beam is the same as that of balanced beam of Ex.2. Additional reinforcement of 1,759 mm2 (= 4,825 mm2 – 3,066 mm2) does not improve the Mu of the over-reinforced beam. It rather prevents the beam from tension failure. That is why over-reinforced beams are to be avoided. However, if the Mu has to be increased beyond 413.87 kNm, the flanged beam may be doubly reinforced. 5.11.5 Use of SP-16 for the Analysis Type of Problems Using the two governing parameters (bf /bw) and (Df /d), the Mu,lim of balanced flanged beams can be determined from Tables 57-59 of SP-16 for the


three grades of steel (250, 415 and 500). The value of the moment coefficient Mu,lim /bwd2fck of Ex.2, as obtained from SP-16, is presented in Table 5.2 making linear interpolation for both the parameters, wherever needed. Mu,lim is then calculated from the moment coefficient. Table 5.2 Mu,lim of Example 2 using Table 58 of SP-16 Parameters: (i) bf /bw = 1000/300 = 3.33 (ii) Df /d = 100/450 = 0.222

(Mu,lim /bw d2 fck) in N/mm2

bf /bwDf /d 3 4 3.33

0.22 0.309 0.395 0.23 0.314 0.402

0.222 0.31* 0.3964* 0.339* * by linear interpolation

So, from Table 5.2, 0.339 2lim , =

ckw

u

f dbM

Mu,lim = 0.339 bw d2 fck = 0.339 (300) (450) (450) (20) 10-6 = 411.88 kNm Mu,lim as obtained from SP-16 is close to the earlier computed value of Mu,lim = 413.87 kNm (see Table 5.1). 5.11.6 Practice Questions and Problems with Answers


Q.1: Determine the moment of resistance of the simply supported doubly reinforced flanged beam (isolated) of span 9 m as shown in Fig. 5.11.6. Assume M 30 concrete and Fe 500 steel.

A.1: Solution of Q.1:

Effective width bf = mm 1200 300 4 )(9000/1500

9000 4 )(

=++

=++ w

o

o b/bll

Step 1: To determine the depth of the neutral axis Assuming neutral axis to be in the flange and writing the equation C = T, we have: 0.87 fy Ast = 0.36 fck bf xu + (fsc Asc – fcc Asc) Here, = 65/600 = 0.108 = 0.1 (say). We, therefore, have fdd /'

sc = 353 N/mm2 . From the above equation, we have:

mm 120 mm 191.48 )1200( )30( 36.0

(1030)} (30) 0.446 - (1030) {(353) - (6509) (500) 0.87 >==ux

So, the neutral axis is in web. Df /d = 120/600 = 0.2 Assuming Df /xu < 0.43, and Equating C = T 0.87 fy Ast = 0.36 fck bw xu + 0.446 fck (bf – bw) Df + (fsc – fcc) Asc

0.87 (500) (6509) - 1030{353 - 0.446 (30)}- 0.446 (30) (1200 - 300) (120) 0 36 30 300

= 319.92 276 mm ( = 276 mm)

u

u ,max

x. ( ) ( )

x

=

>

So, xu = xu,max = 276 mm (over-reinforced beam). Df /xu = 120/276 = 0.4347 > 0.43 Let us assume Df /xu > 0.43. Now, equating C = T with yf as the depth of flange having constant stress of 0.446 fck. So, we have: yf = 0.15 xu + 0.65 Df = 0.15 xu + 78 0.36 fck bw xu + 0.446 fck (bf – bw) yf + Asc (fsc – fcc) = 0.87 fy Ast


0.36 (30) (300) xu + 0.446 (30) (900) (0.15 xu + 78) = 0.87 (500) (6509) – 1030 {353 – 0.446 (30)} or xu = 305.63 mm > xu,max. (xu,max = 276 mm) The beam is over-reinforced. Hence, xu = xu,max = 276 mm. This is a problem of case (iv), and we, therefore, consider the case (ii) to find out the moment of resistance in two parts: first for the balanced singly reinforced beam and then for the additional moment due to compression steel. Step 2: Determination of xu,lim for singly reinforced flanged beam Here, Df /d = 120/600 = 0.2, so yf is not needed. This is a problem of case (ii a) of sec. 5.10.4.2 of Lesson 10. Employing Eq. 5.7 of Lesson 10, we have: Mu,lim = 0.36 (xu,max /d) {1 – 0.42 (xu,max /d)} fck bw d2

+ 0.45 fck (bf – bw) Df (d – Df /2) = 0.36(0.46) {1 – 0.42(0.46)} (30) (300) (600) (600) + 0.45(30) (900) (120) (540) = 1,220.20 kNm

mm

62

0.87 {1 - 0.42 (

(1220.20) (10 5,794.6152 mm0 87 500 600 0 8068

u ,list ,li

y u,max

MA

f d x / d )}

)( . ) ( ) ( ) ( . )

=

= =

Step 3: Determination of Mu2 Total Ast = 6,509 mm2, Ast,lim = 5,794.62 mm2

Ast2 = 714.38 mm2 and Asc = 1,030 mm2

It is important to find out how much of the total Asc and Ast2 are required effectively. From the equilibrium of C and T forces due to additional steel (compressive and tensile), we have: (Ast2) (0.87) (fy) = (Asc) (fsc) If we assume Asc = 1,030 mm2


222 mm 714.38 mm 835.84

)500( 87.0(353) 1030 >==stA , (714.38 mm2 is the total

Ast2 provided). So, this is not possible. Now, using Ast2 = 714.38 mm2 , we get Asc from the above equation.

2(714.38) (0.87) (500) 880.326 1,030 mm353scA = = < , (1,030 mm2 is

the total Asc provided). kNm 167.807 60) - (600 (353) (880.326) )' - ( 2 === ddfAM scscu Total moment of resistance = Mu,lim + Mu2 = 1,220.20 + 167.81 = 1,388.01 kNm Total Ast required = Ast,lim + Ast2 = 5,794.62 + 714.38 = 6,509.00 mm2 , (provided Ast = 6,509 mm2) Asc required = 880.326 mm2 (provided 1,030 mm2). 5.11.7 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 5.11.8 Test 11 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Determine Mu,lim of the flanged beam of Ex. 2 (Fig. 5.11.3) with the help of SP-16 using (a) M 20 and Fe 250, (b) M 20 and Fe 500 and (c) compare the results with the Mu,lim of Ex. 2 from Table 5.2 when grades of concrete and steel are M 20 and Fe 415, respectively. Other data are: bf = 1000 mm, Df = 100 mm, bw = 300 mm, cover = 50 mm and d = 450 mm.

(10 X 3 = 30 marks) A.TQ.1: From the results of Ex. 2 of sec. 5.11.5 (Table 5.2), we have: Parameters: (i) bf /bw = 1000/300 = 3.33 (ii) Df /d = 100/450 = 0.222 For part (a): When Fe 250 is used, the corresponding table is Table 57 of SP-16. The computations are presented in Table 5.3 below: Table 5.3 (Mu,lim /bw d2 fck) in N/mm2 Of TQ.1 (PART a for M 20 and Fe 250)


bf /bwDf /d 3 4 3.33

0.22 0.324 0.411 0.23 0.330 0.421

0.222 0.3252* 0.413* 0.354174* • by linear interpolation Mu,lim /bw d2 fck = 0.354174 = 0.354 (say)


So, Mu,lim = (0.354) (300) (450) (450) (20) N mm = 430.11 kNm For part (b): When Fe 500 is used, the corresponding table is Table 59 of SP-16. The computations are presented in Table 5.4 below: Table 5.4 (Mu,lim /bw d2 fck) in N/mm2 Of TQ.1 (PART b for M 20 and Fe 500)


bf /bwDf /d 3 4 3.33

0.22 0.302 0.386 0.23 0.306 0.393

0.222 0.3028* 0.3874* 0.330718* * by linear interpolation Mu,lim /bw d2 fck = 0.330718 = 0.3307 (say) So, Mu,lim = (0.3307) (300) (450) (450) (20) mm = 401.8 kNm

For part (c): Comparison of results of this problem with that of Table 5.2 (M 20

and Fe 415) is given below in Table 5.5. Table 5.5 Comparison of results of Mu,lim

Sl. No.

Grade of Steel

Mu,lim (kNm)

1 Fe 250 430.11 2 Fe 415 411.88 3 Fe 500 401.80

It is seen that Mu,lim of the beam decreases with higher grade of steel for a particular grade of concrete. TQ.2: With the aid of SP-16, determine separately the limiting moments of

resistance and the limiting areas of steel of the simply supported isolated, singly reinforced and balanced flanged beam of Q.1 as shown in Fig. 5.11.6 if the span = 9 m. Use M 30 concrete and three grades of steel, Fe 250, Fe 415 and Fe 500, respectively. Compare the results obtained above with that of Q.1 of sec. 5.11.6, when balanced.

(15 + 5 = 20 marks) A.TQ.2: From the results of Q.1 sec. 5.11.6, we have: Parameters: (i) bf /bw = 1200/300 = 4.0 (ii) Df /d = 120/600 = 0.2


For Fe 250, Fe 415 and Fe 500, corresponding tables are Table 57, 58 and 59, respectively of SP-16. The computations are done accordingly. After computing the limiting moments of resistance, the limiting areas of steel are determined as explained below. Finally, the results are presented in Table 5.6 below:

mm

0.87 {1 - 0.42 (

u ,list ,li

y u,max

MA

f d x / d=

)}

Table 5.6 Values of Mu,lim in N/mm2 Of TQ.2 Grade of Fe / Q.1 of sec. 5.11.6

(Mu,lim/bw d2 fck) (N/mm2 )

Mu,lim (kNm) Ast,lim (mm2)

Fe 250 0.39 1, 263.60 12,455.32 Fe 415 0.379 1, 227.96 7,099.78 Fe 500 0.372 1, 205.28 5,723.76 Q.1 of sec. 5.11.6 (Fe 415)

1, 220.20 5,794.62

The maximum area of steel allowed is .04 b D = (.04) (300) (660) = 7,920 mm2 . Hence, Fe 250 is not possible in this case. 5.11.9 Summary of this Lesson This lesson mentions about the two types of numerical problems (i) analysis and (ii) design types. In addition to explaining the steps involved in solving the analysis type of numerical problems, several examples of analysis type of problems are illustrated explaining all steps of the solutions both by direct computation method and employing SP-16. Solutions of practice and test problems will give readers the confidence in applying the theory explained in Lesson 10 in solving the numerical problems.


Module 5


Problems


Lesson 12

Flanged Beams – Numerical Problems

(Continued)


Instructional Objectives: At the end of this lesson, the student should be able to: • identify the two types of problems – analysis and design types, • apply the formulations to design the flanged beams. 5.12.1 Introduction Lesson 10 illustrates the governing equations of flanged beams and Lesson 11 explains their applications for the solution of analysis type of numerical problems. It is now necessary to apply them for the solution of design type, the second type of the numerical problems. This lesson mentions the different steps of the solution and solves several numerical examples to explain their step-by-step solutions. 5.12.2 Design Type of Problems We need to assume some preliminary dimensions of width and depth of flanged beams, spacing of the beams and span for performing the structural analysis before the design. Thus, the assumed data known for the design are: Df, bw, D, effective span, effective depth, grades of concrete and steel and imposed loads. There are four equations: (i) expressions of compressive force C, (ii) expression of the tension force T, (iii) C = T and (iv) expression of Mu in terms of C or T and the lever arm {M = (C or T ) (lever arm)}. However, the relative dimensions of Df, D and xu and the amount of steel (under-reinforced, balanced or over-reinforced) influence the expressions. Accordingly, the respective equations are to be employed assuming a particular situation and, if necessary, they need to be changed if the assumed parameters are found to be not satisfactory. The steps of the design problems are as given below. Step 1: To determine the factored bending moment Mu Step 2: To determine the Mu,lim of the given or the assumed section The beam shall be designed as under-reinforced, balanced or doubly reinforced if the value of Mu is less than, equal to or more than Mu,lim. The design of over-reinforced beam is to be avoided as it does not increase the bending moment carrying capacity beyond Mu,lim either by increasing the depth or designing a doubly reinforced beam.


Step 3: To determine xu, the distance of the neutral axis, from the expression of Mu Here, it is necessary to assume first that xu is in the flange. Later on, it may be necessary to calculate xu if the value is found to be more than Df . This is to be done assuming first that Df /xu < 0.43 and then Df /xu > 0.43 separately. Step 4: To determine the area(s) of steel For doubly reinforced beams Ast = Ast,lim + Ast2 and Asc are to be obtained, while only Ast is required to be computed for under-reinforced and balanced beams. These are calculated employing C = T (for Ast and Ast, lim) and the expression of Mu2 to calculate Ast2 and Asc. Step 5: It may be necessary to check the xu and Ast once again after Step 4 It is difficult to prescribe all the relevant steps of design problems. Decisions are to be taken judiciously depending on the type of problem. For the design of a balanced beam, it is necessary to determine the effective depth in Step 3 employing the expression of bending moment Mu. For such beams and for under-reinforced beams, it may be necessary to estimate the Ast approximately immediately after Step 2. This value of Ast will facilitate to determine xu. 5.12.3 Numerical Problems Four numerical examples are solved below explaining the steps involved in the design problems.


Ex.5: Design the simply supported flanged beam of Fig. 5.12.1, given the following: Df = 100 mm, D = 750 mm, bw = 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m, cover = 90 mm, d = 660 mm and imposed loads = 5 kN/m2. Fe 415 and M 20 are used. Solution: Step 1: Computation of factored bending moment Weight of slab per m2 = (0.1) (1) (1) (25) = 2.5 kN/m2

So, Weight of slab per m = (4) (2.5) = 10.00 kN/m Dead loads of web part of the beam = (0.35) (0.65) (1) (25) = 5.6875 kN/m Imposed loads = (4) (5) = 20 kN/m Total loads = 30 + 5.6875 = 35.6875 kN/m

Factored Bending moment = (1.5) kNm 963.5625 8

(12) (12) (35.6875)=

Step 2: Computation of xu,lim Effective width of flange =(lo/6) + bw+ 6 Df = (12000/6) + 350 + 600 = 2,950 mm. xu,max = 0.48 d = 0.48 (660) = 316.80 mm. This shows that the neutral axis is in the web of this beam. Df /d = 100/660 = 0.1515 < 0.2, and Df /xu = 100/316.8 = 0.316 < 0.43 The expression of Mu,lim is obtained from Eq. 5.7 of Lesson 10 (case ii a of sec. 5.10.4.2) and is as follows: Mu,lim = 0.36(xu,max /d){1 - 0.42 (xu,max /d)} fck bw d2 + 0.45 fck (bf - bw) Df (d - Df /2) = 0.36(0.48) {1 - 0.42(0.48)} (20) (350) (650) (650) + 0.45 (20) (2950 - 350) (100) (660 - 50) = 1,835.43 kNm The design moment Mu = 963.5625 kNm is less than Mu,lim. Hence, one under-reinforced beam can be designed.


Step 3: Determination of xu Since the design moment Mu is almost 50% of Mu,lim, let us assume the neutral axis to be in the flange. The area of steel is to be calculated from the moment equation (Eq. 3.23 of Lesson 5), when steel is ensured to reach the design stress fd = 0.87 (415) = 361.05 N/mm2. It is worth mentioning that the term b of Eq. 3.23 of Lesson 5 is here bf as the T- beam is treated as a rectangular beam when the neutral axis is in the flange.

⎭⎬⎫

⎩⎨⎧

−=dbf

fAdAf.M

ck

yststyu 1870

(3.23) Here, all but Ast are known. However, this will give a quadratic equation of Ast and the lower one of the two values will be provided in the beam. The above equation gives:

2stA - 93831.3253 + 379416711.3 = 0 stA

which gives the lower value of Ast as: = 4,234.722097 mmstA 2. The reason of selecting the lower value of Ast is explained in sec 3.6.4.8 of Lesson 6 in the solution of Design Problem 3.1. Then, employing Eq. 3.16 of Lesson 5, we get

ck

styu fb.

Af.x

360870

=

(3.16) or xu = 71.98 mm. Again, employing Eq. 3.24 of Lesson 5, we can determine xu first and then Ast from Eq. 3.16 or 17 of Lesson 5, as explained in the next step. Eq. 3.24 of Lesson 5 gives: Mu = 0.36(xu /d) {1 - 0.42(xu /d)} fck bf d2 = 0.36 (xu) {1 - 0.42 (xu /d)} fck bf d 963.5625 (106) = 0.36 (xu) {1 - 0.42 (xu /660)} (20) (2950) (660) or xu = 72.03 mm.


The two values of xu are the same. It is thus seen that, the value of xu can be determined either first finding the value of Ast, from Eq. 3.23 of Lesson 5 or directly from Eq. 3.24 of Lesson 5 first and then the value of Ast can be determined. Step 4: Determination of Ast Equating C = T, we have from Eq. 3.17 of Lesson 5:

dbf

Afdx

fck

styu

36.0 0.87

=

20.36 0.36 (20) (2950) (72.03) 4,237.41 mm0 87 0.87 (415)

ck f ust

y

f b xA

. f= = =

Minimum Ast = (0.85/fy) bw d = (0.85/415) (350) (660) = 473.13 mm2

Maximum Ast = 0.04 bw D = (0.04) (350) (660) = 9,240 mm2

Hence, Ast = 4,237.41 mm2 is o.k. Provide 6 - 28 T (= 3694 mm2) + 2-20 T (= 628 mm2) to have total Ast = 4,322 mm2. Ex.6: Design a beam in place of the beam of Ex.5 (Fig. 5.12.1) if the imposed loads are increased to 12 kN/m2. Other data are: Df = 100 mm, bw = 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m simply supported and cover = 90 mm. Use Fe 415 and M 20. Solution: As in Ex.5, bf = 2,950 mm. Step 1: Computation of factored bending moment Weight of slab/m2 = 2.5 kN/m2 (as in Ex.1) Imposed loads = 12.0 kN/m2 (given) Total loads = 14.5 kN/m2

Total weight of slab and imposed loads = 14.5 (4) = 58.0 kN/m Dead loads of the beam = 0.65 (0.35) (25) = 5.6875 kN/m Total loads = 63.6875 kN/m


(Mu)factored = 1 5 (63.6875) (12) (12) 1,719.5625 kNm8

.=

Step 2: Determination of Mu,lim Mu,lim of the beam of Ex.5 = 1,835.43 kNm. The factored moment of this problem (1,719.5625 kNm) is close to the value of Mu,lim of the section. Step 3: Determination of d Assuming Df /d < 0.2, we have from Eq. 5.7 of Lesson 10, Mu = 0.36(xu,max /d){1 - 0.42 (xu,max /d)} fck bw d2 + 0.45 fck (bf - bw) Df (d - Df /2)

1719.5625 (106) = 0.36(0.48) {1 - 0.42(0.48)} (20) (350) d2

+ 0.45 (2600) (20) (100) (d - 50) Solving the above equation, we get d = 624.09 mm, giving total depth = 624.09 + 90 = 715 mm (say).

Since the dead load of the beam is reduced due to decreasing the depth of the beam, the revised loads are calculated below: Loads from the slab = 58.0 kN/m Dead loads (revised) = 0.615 (0.35) (25) = 5.38125 kN/m Total loads = 63.38125 kN/m

factored1.5 (63.38125) (12) (12) 1,711.29 kNm

8u( M ) = =

Approximate value of Ast:

6

21711.29 (10 8,243.06 mm0 87 (415) (625 - 50)

0 87 ( - 2

ust

fy

M )AD .

. f d )= = =


Step 4: Determination of Ast (Fig. 5.12.2) xu = xu,max = 0.48 (625) = 300 mm Equating T and C (Eq. 5.5 of Lesson 10), we have: 0.87 fy Ast = 0.36 xu,max bw fck + 0.45 fck (bf - bw) Df

or 20.36 (300) (350) (20) 0.45 (20) (2600) (100) 8,574.98 mm0.87 (415)stA +

= =

Maximum Ast = 0.04 b D = 0.04 (350) (715) = 10,010.00 mm2

Minimum Ast = (0.85/fy) bw d = (0.85/415) (350) (625) = 448.05 mm2

Hence, Ast = 8,574.98 mm2 is o.k. So, provide 8-36 T + 2-18 T = 8143 + 508 = 8,651 mm2

Step 5: Determination of xu Using Ast = 8,651 mm2 in the expression of T = C (Eq. 5.5 of Lesson 10), we have: 0.87 fy Ast = 0.36 xu bw fck + 0.45 fck (bf - bw) Df

or ckw

wfckstyu fb

- bbf A fx

36.0)( 0.45 - 0.87

=


mm) 300 ( 310.89 (20) (350) 0.36

(100) (2600) (20) 0.45 - (8651) (415) 0.87 max, =>== ux

So, Ast provided is reduced to 8-36 + 2-16 = 8143 + 402 = 8,545 mm2. Accordingly,

mm) 300 ( mm 295.703 (20) (350) 0.36

(100) (2600) (20) 0.45 - (8545) (415) 0.87 max, =<== uu xx

Step 6: Checking of Mu Df /d = 100/625 = 0.16 < 0.2 Df /xu = 100/215.7 = 0.33 < 0.43. Hence, it is a problem of case (iii a) and Mu can be obtained from Eq. 5.14 of Lesson 10. So, Mu = 0.36(xu /d) {1 - 0.42(xu /d)} fck bf d2 + 0.45 fck (bf - bw) (Df) (d - Df /2) = 0.36 (295.703/625) {1 - 0.42 (295.703/625)} (20) (350) (625) (625) + 0.45 (20) (2600) (100) (625 - 50) = 1,718.68 kNm > (Mu)design (= 1,711.29 kNm) Hence, the design is o.k. Ex.7: Determine the tensile reinforcement Ast of the flanged beam of Ex.5 (Fig. 5.12.1) when the imposed loads = 12 kN/m2. All other parameters are the same as those of Ex.5: Df = 100 mm, D = 750 mm, bw = 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m, simply supported, cover = 90 mm and d = 660 mm. Use Fe 415 and M 20. Solution: Step 1: Computation of factored bending moment Mu Dead loads of the slab (see Ex.5) = 2.5 kN/m2

Imposed loads = 12.0 kN/m2

Total loads = 14.5 kN/m2

Loads/m = 14.5 (4) = 58.0 kN/m Dead loads of beam = 0.65 (0.35) (25) = 5.6875 kN/m


Total loads = 63.6875 kN/m Factored Mu = (1.5) (63.6875) (12) (12)/8 = 1,719.5625 kNm. Step 2: Determination of Mu,lim From Ex.5, the Mu,lim of this beam = 1,835.43 kNm. Hence, this beam shall be designed as under-reinforced. Step 3: Determination of xu Assuming xu to be in the flange, we have from Eq. 3.24 of Lesson 5 and considering b = bf,

Mu = 0.36xu {1 - 0.42(xu /d)} fck bf d 1719.5625 (106) = 0.36 xu {1 - 0.42 (xu /660)} (20) (2950) (550) Solving, we get xu = 134.1 > 100 mm So, let us assume that the neutral axis is in the web and Df /xu < 0.43, from Eq. 5.14 of Lesson 10 (case iii a of sec. 5.10.4.3), we have: Mu = 0.36(xu /d) {1 - 0.42(xu /d)} fck bw d2 + 0.45 fck (bf - bw) (Df) (d - Df /2) = 0.36 xu {1 - 0.42 (xu /660)} (20) (350) (660) + 0.45 (20) (2600) (100) (660 - 50) Substituting the value of Mu = 1,719.5625 kNm in the above equation and simplifying, xu

2 - 1571.43 xu + 276042 = 0 Solving, we have xu = 201.5 mm Df /xu = 100/201.5 = 0.496 > 0.43. So, we have to use Eq. 5.15 and 5.18 of Lesson 10 for yf and Mu (case iii b of sec. 5.10.4.3). Thus, we have: Mu = 0.36 xu {1 - 0.42( xu /d)} fck bw d + 0.45 fck (bf - bw) yf (d - yf /2) where, yf = (0.15 xu + 0.65 Df) So, Mu = 0.36 xu {1 - 0.42 (xu/660)} (20) (350) (660)


+ 0.45 (20) (2600) (0.15 xu + 65) (660 - 0.075 xu - 32.5) or 1719.5625 (106) = 3.75165 (106) xu - 795.15 xu

2 + 954.4275 (106) Solving, we get xu = 213.63 mm. Df /xu = 100/213.63 = 0.468 > 0.43. Hence, o.k.

Step 4: Determination of Ast Equating C = T from Eqs. 5.16 and 5.17 of Lesson 10 (case iii b of sec. 5.10.4.3), we have: 0.87 fy Ast = 0.36 fck bw xu + 0.45 fck (bf – bw) yf where, yf = 0.15 xu + 0.65 Df Here, using xu = 213.63 mm, Df = 100 mm, we get

yf = 0.15 (213.63) + 0.65 (100) = 97.04 mm

So, 0.36 (20) (350) (213.63) 0.45 (20) (2600) (97.04) 0.87 (415)stA +

= = 7,780.32 mm2

Minimum Ast = (0.85/fy) (bw) (d) = 0.85 (350) (660)/(415) = 473.13 mm2

Maximum Ast = 0.04 bw D = 0.04 (350) (750) = 10,500 mm2

Hence, Ast = 7,780.32 mm2 is o.k.


Provide 6-36 T + 3-28 T (6107 + 1847 = 7,954 mm2). Please refer to Fig. 5.12.3. Step 5: Checking of xu and Mu using Ast = 7,954 mm2

From T = C (Eqs. 5.16 and 5.17 of Lesson 10), we have

0.87 fy Ast = 0.36 fck bw xu + 0.45 fck (bf – bw) yf where, yf = 0.15 xu + 0.65 Df or 0.87 (415) (7954) = 0.36 (20) (350) xu + 0.45 (20) (2600) (0.15 xu + 0.65 Df) or xu = 224.01 mm Df /xu = 100/224.01 = 0.446 > 0.43. Accordingly, employing Eq. 5.18 of Lesson 10 (case iii b of sec. 5.10.4.3), we have: So, Mu = 0.36 xu {1 - 0.42( xu /d)} fck bw d + 0.45 fck (bf - bw) yf (d - yf /2) = 0.36 (224.01){1 - 0.42 (224.01/660)} (20) (350) (660) + 0.45 (20) (2600) {(0.15) 224.01 + 65} {(660) - 0.15 (112) - 32.5} = 1,779.439 kNm > 1,719.5625 kNm Hence, o.k.


Ex.8: Design the flanged beam of Fig. 5.12.4, given in following: Df = 100 mm, D = 675 mm, bw = 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m simply supported, cover = 90 mm, d = 585 mm and imposed loads = 12 kN/m2. Use Fe 415 and M 20. Step 1: Computation of factored bending moment, Mu Weight of slab/m2 = (0.1) (25) = 2.5 kN/m2


Total loads = 14.5 kN/m2

Total weight of slab + imposed loads/m = 14.5 (4) = 58 kN/m Dead loads of beam = 0.575 (0.35) (25) = 5.032 kN/m Total loads = 63.032 kN/m Factored Mu = (1.5) (63.032) (12) (12)/8 = 1,701.87 kNm Step 2: Determination of Mu,lim Assuming the neutral axis to be in the web, Df /xu < 0.43 and Df /d = 100 / 585 = 0.17 < 0.2, we consider the case (ii a) of sec. 5.10.4.2 of Lesson 10 to get the following: Mu,lim = 0.36 (xu,max /d) {1 – 0.42 (xu,max /d)} fck bw d2

+ 0.45 fck (bf – bw) Df (d – Df /2) = 0.36(0.48) {1 – 0.42 (0.48)} (20) (350) (585) (585) + 0.45(20) (2600) (100) (585 – 50) = 1,582.4 kNm Since, factored Mu > Mu,lim, the beam is designed as doubly reinforced. Mu2 = Mu - Mu,lim = 1701.87 - 1582.4 = 119.47 kNm


Step 3: Determination of area of steel

Ast,lim is obtained equating T = C (Eqs. 5.5 and 6 of Lesson 10). 0.87 fy (Ast,lim) = 0.36 bw (xu,max /d) d fck + 0.45 fck (bf - bw) Df

or y

fwfckcku,wlist f

D - bbfd fdxbA

0.87 )( 0.45 )/( 0.36

maxm,

+=

= (415) 0.87

(100) (2600) (20) 0.45 (20) (585) (0.48) (350) 36.0 + = 8,440.98

mm2

)'( ) - (

2

d - dffM

Accsc

usc = (Eq. 4.4 of Lesson 8).

where fsc = 353 N/mm2 for d'/d = 0.1 fcc = 0.446 fck = 0.446 (20) = 8.92 N/mm2 Mu2 = 119.47 (106) Nmm d' = 58.5 mm d = 585 mm


Using the above values in the expression of Asc (Eq. 4.4 of Lesson 8), we get Asc = 659.63 mm2

y

ccscscst f

- ffAA

87.0)(

2 = (Eqs. 4.4 and 4.5 of Lesson 8).

Substituting the values of Asc, fsc, fcc and fy we get Ast2 = 628.48 mm2

Total Ast = Ast,lim + Ast2 = 8,440.98 + 628.48 = 9,069.46 mm2

Maximum Ast = 0.04 bw D = 0.04 (350) (675) = 9,450 mm2

and minimum Ast = (0.85/fy) bw d = (0.85/415) (350) (585) = 419.37 mm2

Hence, Ast = 9, 069.46 mm2 is o.k. Provide 8-36 T + 3-20 T = 8143 + 942 = 9,085 mm2 for Ast and 1-20 + 2-16 = 314 + 402 = 716 mm2 for Asc (Fig. 5.12.5). Step 4: To check for xu and Mu (Fig. 5.12.5) Assuming xu in the web and Df /xu < 0.43 and using T = C (case ii a of sec. 5.10.4.2 of Lesson 10 with additional compression force due to compression steel), we have:

0.87 fy Ast = 0.36 bw xu fck + 0.45 (bf - bw) fck Df + Asc (fsc - fcc)

or 0.87 (415) (9085) = 0.36 (350) xu (20) + 0.45 (2600) (20) (100) + 716 {353 - 0.45 (20)} This gives xu = 275.33 mm. xu,max = 0.48 (d) = 0.48 (585) = 280.8 mm. So, xu < xu,max, Df /xu = 100/275.33 = 0.363 < 0.43 and Df /d = 100/585 = 0.17 < 0.2. The assumptions, therefore, are correct. So, Mu can be obtained from Eq. 5.14 of sec. 5.10.4.3 of Lesson 10 with additional moment due to compression steel, as given below:


So, Mu = 0.36 bw xu fck (d - 0.42 xu) + 0.45 (bf - bw) fck Df (d - Df /2) + Asc (fsc - fcc) (d - d') = 0.36 (350) (275.33) (20) {585 - 0.42 (275.33)} + 0.45 (2600) (20) (100) (585 - 50) + 716 (344) (585 - 58.5) = 325.66 + 1251.9 + 129.67 = 1,707.23 kNm Factored moment = 1,701.87 kNm < 1,707.23 kNm. Hence, o.k. 5.12.4 Practice Questions and Problems with Answers

Q.1: Determine the steel reinforcement of a simply supported flanged beam

(Fig. 5.12.6) of Df = 100 mm, D = 700 mm, cover = 50 mm, d = 650 mm, bw = 300 mm, spacing of the beams = 4,000 mm c/c, effective span = 10 m and imposed loads = 10 kN/m2. Use M 20 and Fe 415.

A.1: Solution: Step 1: Computation of (Mu)factored Weight of slab = (0.1) (25) = 2.5 kN/m2


_________ 12.5 kN/m2


Total loads per m = (12.5) (4) = 50 kN/m Dead loads of beam = (0.3) (0.6) (25) = 4.50 kN/m Total loads = 54.50 kN/m Factored Mu = (1.5) (54.50) (10) (10)/8 = 1,021.87 kNm Step 2: Determination of Mu,lim Effective width of the flange bf = lo/6 + bw + 6 Df = (10,000/6) + 300 + 600 = 2,567 mm. xu,max = 0.48 d = 0.48 (650) = 312 mm Hence, the balanced neutral axis is in the web of the beam. Df /d = 100/650 = 0.154 < 0.2 Df /xu = 100/312 = 0.32 < 0.43 So, the full depth of flange is having a stress of 0.446 fck. From Eq. 5.7 of Lesson 10 (case ii a of sec. 5.10.4.2), we have, Mu,lim = 0.36 (xu,max /d) {1 – 0.42 (xu,max /d)} fck bw d2

+ 0.45 fck (bf – bw) Df (d – Df /2) = 0.36(0.48) {1 – 0.42(0.48)} (20) (300) (650) (650) + 0.45(20) (2267) (100) (650 – 50) = 1573.92 kNm > Mu (= 1021.87 kNm) So, the beam will be under-reinforced one. Step 3: Determination of xu Assuming xu is in the flange, we have from Eq. 3.24 of Lesson 5 (rectangular beam when b = bf ). Mu = 0.36 (xu /d) {1 – 0.42 (xu /d)} fck bf d2

= 0.36 xu {1 – 0.42 (xu /d)} fck bf d 1021.87 (106) = 0.36 xu {1 – 0.42(xu/650)} (20) (2567) (650)


xu = 89.55 mm2 < 100 mm (Hence, the neutral axis is in the flange.) Step 4: Determination of Ast Equating C = T, we have from Eq. 3.17 of Lesson 5:

db f.

A f.

dx

fck

styu

360870

=

or 20.36 0.36 (20) (2567) (89.55) 4,584.12 mm0.87 0.87 (415)

ck f ust

y

f b xA

f= = =

Minimum 2mm 399.39 415

(650) (300) 0.85 )( )(0.85/ === dbfA wyst

Maximum Ast = 0.04 bw D = (0.04) (300) (700) = 8,400 mm2

So, Ast = 4,584.12 mm2 is o.k. Provide 6-28 T + 2-25 T = 3694 + 981 = 4,675 mm2 (Fig. 5.12.6). 5.12.5 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 5.12.6 Test 12 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions.

TQ.1: Determine the steel reinforcement Ast of the simply supported flanged

beam of Q.1 (Fig. 5.12.6) having Df = 100 mm, D = 700 mm, cover = 50 mm, d = 650 mm, bw = 300 mm, spacing of the beams = 4,000 mm c/c, effective span = 12 m and imposed loads = 10 kN/m2. Use M 20 and Fe 415.


A.TQ.1: Solution: Step 1: Computation of (Mu)factored Total loads from Q.1 of sec. 5.12.4 = 54.50 kN/m Factored Mu = (1.5) (54.50) (12) (12)/8 = 1,471.5 kNm Step 2: Determination of Mu,lim

Effective width of flange = lo/6 + bw + 6 Df

= (12000/6) + 300 + 600 = 2,900 mm (Fig. 5.12.7) xu,max = 0.48 d = 0.48 (650) = 312 mm Hence, the balanced neutral axis is in the web. Df /d = 100/650 = 0.154 < 0.2 Df /xu = 100/312 = 0.32 < 0.43 So, the full depth of flange is having constant stress of 0.446 fck. From Eq. 5.7 of Lesson 10 (case ii a of sec. 5.10.4.2), we have Mu,lim = 0.36 (xu,max /d) {1 – 0.42 (xu,max /d)} fck bw d2

+ 0.45 fck (bf – bw) Df (d – Df /2) = 0.36(0.48) {1 – 0.42(0.48)} (20) (300) (650) (650) + 0.45(20) (2600) (100) (650 – 50) = 1,753.74 kNm > 1,471.5 kNm So, the beam will be under-reinforced. Step 3: Determination of xu Assuming xu to be in the flange, we have from Eq. 3.24 of Lesson 5 (singly reinforced rectangular beam when b = bf ): Mu = 0.36 xu {1 – 0.42 (xu /d)} fck bf d or 1471.5 (106) = 0.36 (xu) {1 – 0.42 (xu /650)} (20) (2900) (650) or xu

2 – 1547.49 xu + 167.81 (103) = 0


Solving, we have xu = 117.34 mm > 100 mm So, neutral axis is in the web. Assuming Df /xu < 0.43, we have from Eq. 5.14 of Lesson 10 (case iii a of sec. 5.10.4.3), Mu = 0.36 xu {1 – 0.42 (xu /d)} fck bw d + 0.45 fck (bf – bw) Df (d – Df /2) = 0.36 xu {1 – 0.42 (xu/650)} (20) (300) (650) + 0.45(20) (2600) (100) (650 – 50) or xu

2 – 1547.62 xu + 74404.7 = 0 Solving, we have xu = 49.67 < 100 mm However, in the above when it is assumed that the neutral axis is in the flange xu is found to be 117.34 mm and in the second trial when xu is assumed in the web xu is seen to be 49.67 mm. This indicates that the full depth of the flange will not have the strain of 0.002, neutral axis is in the web and Df /xu is more than 0.43. So, we have to use Eq. 5.18 of Lesson 10, with the introduction of yf from Eq. 5.15 of Lesson 10. Assuming Df /xu > 0.43, from Eqs. 5.15 and 5.18 of Lesson 10 (case iii b of sec. 5.10.4.3), we have: Mu = 0.36 xu {1 – 0.42 (xu /d)} fck bw d + 0.45 fck (bf – bw) yf (d – yf /2) where, yf = (0.15 xu + 0.65 Df) So, Mu = 0.36 xu {1 – 0.42 (xu/650)} (20) (300) (650) + 0.45(20) (2600) (0.15 xu + 0.65) (650 – 0.075 xu – 0.325 xu) or, 1471.5 (106) = - 1170.45 xu

2 + 3.45735 xu + 939.2175 (106) Solving, we get xu = 162.9454 mm. This shows that the assumption of Df /xu > 0.43 is correct as Df /xu = 100 / 162.9454 = 0.614. Step 4: Determination of Ast Equating C = T from Eqs. 5.16 and 5.17 of Lesson 10 (case iii b of sec. 5.10.4.3), we have


0.87 fy Ast = 0.36 fck bw xu + 0.45 fck (bf – bw) yf

or (415) 0.87

65} (162.9454) {0.15 (2600) (20) 0.45 (162.9454) (30) (20) 0.36 ++=stA

= 974.829 + 5,796.81 = 6,771.639 mm2

Minimum Ast = (0.85/fy) (bw) (d) = 0.85 (300) (650)/415 = 399.39 mm2

Maximum Ast = 0.04 (bw) (D) = 0.04 (300) (700) = 8,400 mm2

So, Ast = 6,771.639 is o.k. Provide 2-36 T + 6-32 T = 2035 + 4825 = 6,860 mm2 > 6,771.639 mm2 (Fig. 5.12.7).

5.12.7 Summary of this Lesson This lesson explains the steps involved in solving the design type of numerical problems. Further, several examples of design type of numerical problems are illustrated explaining the steps of their solutions. Solutions of practice problems and test problems will give the readers confidence in applying the theory explained in Lesson 10 in solving the numerical problems.


Module 6

Shear, Bond, Anchorage,

Development Length and Torsion


Lesson 13

Limit State of Collapse in Shear Version 2 CE IIT, Kharagpur


At the end of this lesson, the student should be able to: • name and explain the three different failure modes of reinforced concrete

beams under the combined effects of bending moment and shear force, • define nominal shear stress τv of rectangular and T-beams of uniform and

varying depths under the combined effects of bending moment and shear force,

• name the two parameters on which the design shear strength of concrete depends,

• find out the maximum shear stress of concrete beams τcmax with shear reinforcement,

• locate the critical sections for shear in beams, • explain when and why do we consider enhanced shear strength of concrete, • explain why the minimum shear reinforcement is provided in any beam, • determine the amount of minimum shear reinforcement to be provided in

any beam, • specify the three different ways of providing shear reinforcement in a beam, • design the shear reinforcement in a beam for each of the three methods

mentioned above, • design the shear reinforcement closed to the support of a beam, • specify the conditions to be satisfied for the curtailment of tension

reinforcement when designing shear reinforcement, • place the vertical stirrups in a beam. 6.13.1 Introduction This lesson explains the three failure modes due to shear force in beams and defines different shear stresses needed to design the beams for shear. The critical sections for shear and the minimum shear reinforcement to be provided in beams are mentioned as per IS 456. The design of shear reinforcement has been illustrated in Lesson 14 through several numerical problems including the curtailment of tension reinforcement in flexural members.


6.13.2 Failure Modes due to Shear

Bending in reinforced concrete beams is usually accompanied by shear, the exact analysis of which is very complex. However, experimental studies confirmed the following three different modes of failure due to possible combinations of shear force and bending moment at a given section (Figs. 6.13.1a to c): (i) Web shear (Fig. 6.13.1a) (ii) Flexural tension shear (Fig. 6.13.1b) (iii) Flexural compression shear (Fig. 6.13.1c) Web shear causes cracks which progress along the dotted line shown in Fig. 6.13.1a. Steel yields in flexural tension shear as shown in Fig. 6.13.1b, while concrete crushes in compression due to flexural compression shear as shown in Fig. 6.13.1c. An in-depth presentation of the three types of failure modes is beyond the scope here. Only the salient points needed for the routine design of beams in shear are presented here.


6.13.3 Shear Stress The distribution of shear stress in reinforced concrete rectangular, T and L-beams of uniform and varying depths depends on the distribution of the normal stress. However, for the sake of simplicity the nominal shear stress τv is considered which is calculated as follows (IS 456, cls. 40.1 and 40.1.1):


(i) In beams of uniform depth (Figs. 6.13.2a and b):

db

Vuv

=τ

(6.1) where Vu = shear force due to design loads,

b = breadth of rectangular beams and breadth of the web bw for flanged beams, and

d = effective depth. (ii) In beams of varying depth:

dbd

MV u

u

v

tan

βτ

±=

(6.2) where τv, Vu, b or bw and d are the same as in (i), Mu = bending moment at the section, and β = angle between the top and the bottom edges. The positive sign is applicable when the bending moment Mu decreases numerically in the same direction as the effective depth increases, and the negative sign is applicable when the bending moment Mu increases numerically in the same direction as the effective depth increases. 6.13.4 Design Shear Strength of Reinforced Concrete Recent laboratory experiments confirmed that reinforced concrete in beams has shear strength even without any shear reinforcement. This shear strength (τc) depends on the grade of concrete and the percentage of tension steel in beams. On the other hand, the shear strength of reinforced concrete with the reinforcement is restricted to some maximum value τcmax depending on the grade of concrete. These minimum and maximum shear strengths of reinforced concrete (IS 456, cls. 40.2.1 and 40.2.3, respectively) are given below: 6.13.4.1 Design shear strength without shear reinforcement (IS 456, cl. 40.2.1)


Table 19 of IS 456 stipulates the design shear strength of concrete τc for different grades of concrete with a wide range of percentages of positive tensile steel reinforcement. It is worth mentioning that the reinforced concrete beams must be provided with the minimum shear reinforcement as per cl. 40.3 even when τv is less than τc given in Table 6.1. Table 6.1 Design shear strength of concrete, τc in N/mm2

Grade of concrete

(100 As /b d)

M 20 M 25 M 30 M 35 M40 and above

≤ 0.15 0.25 0.50

0.28 0.36 0.48

0.29 0.36 0.49

0.29 0.37 0.50

0.29 0.37 0.50

0.30 0.38 0.51

0.75 1.00 1.25

0.56 0.62 0.67

0.57 0.64 0.70

0.59 0.66 0.71

0.59 0.67 0.73

0.60 0.68 0.74

1.50 1.75 2.00

0.72 0.75 0.79

0.74 0.78 0.82

0.76 0.80 0.84

0.78 0.82 0.86

0.79 0.84 0.88

2.25 2.50 2.75

≥ 3.00

0.81 0.82 0.82 0.82

0.85 0.88 0.90 0.92

0.88 0.91 0.94 0.96

0.90 0.93 0.96 0.99

0.92 0.95 0.98 1.01

In Table 6.1, As is the area of longitudinal tension reinforcement which continues at least one effective depth beyond the section considered except at support where the full area of tension reinforcement may be used provided the detailing is as per IS 456, cls. 26.2.2 and 26.2.3. 6.13.4.2 Maximum shear stress τcmax with shear reinforcement (cls. 40.2.3,

40.5.1 and 41.3.1) Table 20 of IS 456 stipulates the maximum shear stress of reinforced concrete in beams τcmax as given below in Table 6.2. Under no circumstances, the nominal shear stress in beams τv shall exceed τcmax given in Table 6.2 for different grades of concrete. Table 6.2 Maximum shear stress, τcmax in N/mm2

Grade of concrete

M 20 M 25 M 30 M 35 M 40 and above

Τcmax,

N/mm2

2.8

3.1

3.5

3.7

4.0


6.13.5 Critical Section for Shear

Clauses 22.6.2 and 22.6.2.1 stipulate the critical section for shear and are as follows: For beams generally subjected to uniformly distributed loads or where the principal load is located further than 2d from the face of the support, where d is the effective depth of the beam, the critical sections depend on the conditions of supports as shown in Figs. 6.13.3 a, b and c and are mentioned below. (i) When the reaction in the direction of the applied shear introduces tension (Fig. 6.13.3a) into the end region of the member, the shear force is to be computed at the face of the support of the member at that section. (ii) When the reaction in the direction of the applied shear introduces compression into the end region of the member (Figs. 6.13.3b and c), the shear force computed at a distance d from the face of the support is to be used for the design of sections located at a distance less than d from the face of the support. The enhanced shear strength of sections close to supports, however, may be considered as discussed in the following section.


6.13.6 Enhanced Shear Strength of Sections Close to Supports (cl. 40.5 of IS 456)

Figure 6.13.4 shows the shear failure of simply supported and cantilever beams without shear reinforcement. The failure plane is normally inclined at an angle of 30o to the horizontal. However, in some situations the angle of failure is more steep either due to the location of the failure section closed to a support or for some other reasons. Under these situations, the shear force required to produce failure is increased. Such enhancement of shear strength near a support is taken into account by increasing the design shear strength of concrete to (2dτc/av) provided that the design shear stress at the face of the support remains less than the value of τcmax given in Table 6.2 (Table 20 of IS 456). In the above expression of the enhanced shear strength d = effective depth of the beam,

τc = design shear strength of concrete before the enhancement as given in Table 6.1 (Table 19 of IS 456),

av = horizontal distance of the section from the face of the support (Fig. 6.13.4). Similar enhancement of shear strength is also to be considered for sections closed to point loads. It is evident from the expression (2dτc /av) that when av is equal to 2d, the enhanced shear strength does not come into picture. Further, to increase the effectivity, the tension reinforcement is recommended to be extended on each side of the point where it is intersected by a possible failure plane for a distance at least equal to the effective depth, or to be provided with an equivalent anchorage.


6.13.7 Minimum Shear Reinforcement (cls. 40.3, 26.5.1.5 and 26.5.1.6 of IS 456) Minimum shear reinforcement has to be provided even when τv is less than τc given in Table 6.1 as recommended in cl. 40.3 of IS 456. The amount of minimum shear reinforcement, as given in cl. 26.5.1.6, is given below. The minimum shear reinforcement in the form of stirrups shall be provided such that:

yv

sv

fsbA

0.870.4

≥

(6.3) where Asv = total cross-sectional area of stirrup legs effective in shear, sv = stirrup spacing along the length of the member,

b = breadth of the beam or breadth of the web of the web of flanged beam bw, and

fy = characteristic strength of the stirrup reinforcement in N/mm2 which

shall not be taken greater than 415 N/mm2. The above provision is not applicable for members of minor structural importance such as lintels where the maximum shear stress calculated is less than half the permissible value. The minimum shear reinforcement is provided for the following:

(i) Any sudden failure of beams is prevented if concrete cover bursts and the bond to the tension steel is lost.

(ii) Brittle shear failure is arrested which would have occurred without

shear reinforcement. (iii) Tension failure is prevented which would have occurred due to

shrinkage, thermal stresses and internal cracking in beams.

(iv) To hold the reinforcement in place when concrete is poured.

(v) Section becomes effective with the tie effect of the compression steel.


Further, cl. 26.5.1.5 of IS 456 stipulates that the maximum spacing of

shear reinforcement measured along the axis of the member shall not be more than 0.75 d for vertical stirrups and d for inclined stirrups at 45o, where d is the effective depth of the section. However, the spacing shall not exceed 300 mm in any case. 6.13.8 Design of Shear Reinforcement (cl. 40.4 of IS 456) When τv is more than τc given in Table 6.1, shear reinforcement shall be provided in any of the three following forms: (a) Vertical stirrups, (b) Bent-up bars along with stirrups, and (c) Inclined stirrups. In the case of bent-up bars, it is to be seen that the contribution towards shear resistance of bent-up bars should not be more than fifty per cent of that of the total shear reinforcement. The amount of shear reinforcement to be provided is determined to carry a shear force Vus equal to Vus = Vu – τc b d (6.4) where b is the breadth of rectangular beams or bw in the case of flanged beams. The strengths of shear reinforcement Vus for the three types of shear reinforcement are as follows: (a) Vertical stirrups:

v

svyus s

dAfV

0.87 =

(6.5) (b) For inclined stirrups or a series of bars bent-up at different cross-sections:

)cos (sin 0.87

αα +=v

svyus s

dAfV

(6.6)


(c) For single bar or single group of parallel bars, all bent-up at the same cross-section: αsin 0.87 vsvyus sAfV = (6.7) where Asv = total cross-sectional area of stirrup legs or bent-up bars within a distance sv, sv = spacing of stirrups or bent-up bars along the length of the member, τv = nominal shear stress, τc = design shear strength of concrete,

b = breadth of the member which for the flanged beams shall be taken as the breadth of the web bw,

fy = characteristic strength of the stirrup or bent-up reinforcement which

shall not be taken greater than 415 N/mm2,

α = angle between the inclined stirrup or bent-up bar and the axis of the member, not less than 45o, and

d = effective depth. The following two points are to be noted:

(i) The total shear resistance shall be computed as the sum of the resistance for the various types separately where more than one type of shear reinforcement is used.

(ii) The area of stirrups shall not be less than the minimum specified in

cl. 26.5.1.6. 6.13.9 Shear Reinforcement for Sections Close to Supports As stipulated in cl. 40.5.2 of IS 456, the total area of the required shear reinforcement As is obtained from: As = av b (τv – 2d τc /av)/0.87 fy and ≥ 0.4 av b/0.87 fy (6.8)


For flanged beams, b will be replaced by bw, the breadth of the web of flanged beams. This reinforcement should be provided within the middle three quarters of av, where av is less than d, horizontal shear reinforcement will be effective than vertical. Alternatively, one simplified method has been recommended in cl. 40.5.3 of IS 456 and the same is given below. The following method is for beams carrying generally uniform load or where the principal load is located further than 2d from the face of support. The shear stress is calculated at a section a distance d from the face of support. The value of τc is calculated in accordance with Table 6.1 and appropriate shear reinforcement is provided at sections closer to the support. No further check for shear at such sections is required. 6.13.10 Curtailment of Tension Reinforcement in

Flexural Members (cl. 26.2.3.2 of IS 456) Curtailment of tension reinforcement is done to provide the required reduced area of steel with the reduction of the bending moment. However, shear force increases with the reduction of bending moment. Therefore, it is necessary to satisfy any one of following three conditions while terminating the flexural reinforcement in tension zone: (i) The shear stress τv at the cut-off point should not exceed two-thirds of the permitted value which includes the shear strength of the web reinforcement. Accordingly, τv ≤ (2/3) (τc + Vus /b d) or Vus ≥ (1.5 τv - τc) b d (6.9) (ii) For each of the terminated bars, additional stirrup area should be provided over a distance of three-fourth of effective depth from the cut-off point. The additional stirrup area shall not be less than 0.4 b s/fy, where b is the breadth of rectangular beams and is replaced by bw, the breadth of the web for flanged beams, s = spacing of additional stirrups and fy is the characteristic strength of stirrup reinforcement in N/mm2. The value of s shall not exceed d/(8 βb), where βb is the ratio of area of bars cut-off to the total area of bars at that section, and d is the effective depth.


(iii) For bars of diameters 36 mm and smaller, the continuing bars provide double the area required for flexure at the cut-off point. The shear stress should not exceed three-fourths that permitted. Accordingly, τv ≤ (3/4) (τc + Vus /b d) or Vus ≥ (1.33 τv - τc) b d (6.10) In the above expression b is the breadth of the rectangular beams which will be bw in the case of flanged beams. 6.13.11 Placement of Stirrups

The stirrups in beams shall be taken around the outer-most tension and compression bars. In T and L-beams, the stirrups will pass around longitudinal bars located close to the outer face of the flange. In the rectangular beams, two holder bars of diameter 10 or 12 mm are provided if there is no particular need for compression reinforcement (Fig. 6.13.5). 6.13.12 Practice Questions and Problems with Answers Q.1: Name and explain the three different failure modes of reinforced concrete

beams under the combined effects of bending moment and shear force.


A.1: Sec. 6.13.2 Q.2: Define nominal shear stress τv of rectangular and T-beams of (i) uniform

depth and (ii) varying depth subjected to bending moment and shear force.

A.2: Sec. 6.13.3 Q.3: What is meant by “Design shear strength of concrete τc” ? A.3: Sec. 6.13.4 Q.4: On what parameters τc of beams without shear reinforcement depends ?

How do you get τc for different grades of concrete ? A.4: τc depends on (i) grade of concrete and (ii) percentage of tensile steel in

the beam. Table 19 of cl. 40.2.1 of IS 456 gives the values of τc and the same table is

presented in Table 6.1 of sec. 6.13.4.1 of this lesson. Q.5: How do you know the maximum shear stress of concrete beams τcmax

with shear reinforcement ? A.5: Sec. 6.13.4.2 Q.6: How do you determine the critical sections for shear in a beam ? A.6: Sec. 6.13.5 Q.7: When and why do we consider enhanced shear strength of concrete ? A.7: Sec. 6.13.6 Q.8: How do we determine the minimum shear reinforcement in rectangular

and T-beams ? Why do we provide the minimum shear reinforcement ? A.8: Sec. 6.13.7 Q.9: What are the three different ways to provide shear reinforcement ? Explain

the method of design of each of them. A.9: Sec. 6.13.8


Q.10: How do we design the shear reinforcement close to the support of a beam ? A.10: Sec. 6.13.9 Q.11: State the conditions to be satisfied for the curtailment of tension

reinforcement when designing the shear reinforcement. A.11: Sec. 6.13.10 Q.12: How do we place the vertical stirrups in a beam ? A.12: Sec. 6.13.11 6.13.13 References:

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 6.13.14 Test 13 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Define nominal shear stress τv of rectangular and T-beams of (i)

uniform depth and (ii) varying depth subjected to bending moment and shear force.

(5 marks)

A.TQ.1: Sec. 6.13.3 TQ.2: How do you determine the critical sections for shear in a beam ? (5 marks) A.TQ.2: Sec. 6.13.5 TQ.3: When and why do we consider enhanced shear strength of concrete? (5

marks) A.TQ.3: Sec. 6.13.6 TQ.4: How do we determine the minimum shear reinforcement in rectangular

and T-beams? Why do we provide the minimum shear reinforcement ? (5 marks)

A.TQ.4: Sec. 6.13.7 TQ.5: What are the three different ways to provide shear reinforcement ?

Explain the method of design of each of them. (5 marks) A.TQ.5: Sec. 6.13.8 TQ.6: How do we design the shear reinforcement close to the support of a beam?

(5 marks) A.TQ.6: Sec. 6.13.9 TQ.7: State the conditions to be satisfied for the curtailment of tension

reinforcement when designing the shear reinforcement. (5 marks)


A.TQ.7: Sec. 6.13.10 6.13.15 Summary of this Lesson Learning the different failure modes, the shear stresses and the design procedure of beams subjected to shear, this lesson explains the design procedure with special reference to curtailment of tension reinforcement in flexural members. Solutions of problems as illustrated in Lesson 14 and given in the practice problems and test, students will be thoroughly conversant with the design of rectangular and T-beams subjected to shear following the limit state of collapse as recommended by IS 456..


Module 6




Lesson 14

Limit State of Collapse in Shear – Numerical

Problems Version 2 CE IIT, Kharagpur

Instructional Objectives: At the end of this lesson, the student should be able to: • solve specific numerical problems of rectangular and T-beams for the

complete design of shear reinforcement as per the stipulations of IS 456. 6.14.1 Introduction Lesson 13 explains the three failure modes due to shear force in beams and defines different shear stresses needed to design the beams for shear. The critical sections for shear and the minimum shear reinforcement to be provided in beams are mentioned as per IS 456. In this lesson, the design of shear reinforcement has been illustrated through several numerical problems including the curtailment of tension reinforcement in flexural members. 6.14.2 Numerical Problems

Problem 1: Determine the shear reinforcement of the simply supported beam of effective span 8 m whose cross-section is shown in Fig. 6.14.1. Factored shear force is 250 kN. Use M 20 and Fe 415.


Solution 1:

Here, Ast = 2-25T + 2-20T gives the percentage of tensile reinforcement

1.43 (450) 250

(1609) 100 ==p

From Table 6.1 of Lesson 13, τc = 0.67 + 0.036 = 0.706 N/mm2 (by linear interpolation). Employing Eq. 6.1 of Lesson 13,

τv = 23

N/mm 2.22 (450) 250

)(10 250

==db

Vu and τcmax = 2.8 N/mm2 (from Table

6.2 of Lesson 13). Hence, τc < τv < τcmax. So, shear reinforcement is needed for the shear force (Eq. 6.4 of Lesson 13). Vus = Vu – τc b d = 250 – 0.706 (250) (450) (10-3) = 170.575 kN Providing 8 mm, 2 legged vertical stirrups, we have Asv = 2 (50) = 100 mm2

Hence, spacing of the stirrups as obtained from Eq. 6.5 of Lesson 13:

mm, 95.25 170575

(450) (100) (415) 0.87 0.87

===us

svyv V

dAfs say 95 mm.

For 10 mm, 2 legged vertical stirrups, (Asv = 157 mm2), spacing

mm 149.54 170575

(450) (157) (415) 0.87 ==vs

According to cl. 26.5.1.5 of IS 456, the maximum spacing of the stirrups = 0.75 d = 0.75 (450) = 337.5 mm = 300 mm (say). Minimum shear reinforcement (cl. 26.5.1.6 of IS 456) is obtained from (Eq.6.3 of sec. 6.13.7 of Lesson 13):


)( 0.87

0.4 )( yv

sv

fsbA

≥

From the above, Asv(minimum) = 2mm 40.16 (415) 0.87

(145) (250) 0.4 87.0 4.0

==y

v

fsb .

So, we select 10 mm, 2 legged stirrups @ 145 mm c/c. Problem 2:

Design the bending and shear reinforcement of the tapered cantilever beam of width b = 300 mm and as shown in Fig. 6.14.2 using M 20 and Fe 415 (i) without any curtailment of bending reinforcement and (ii) redesign the bending and shear reinforcement if some of the bars are curtailed at section 2-2 of the beam. Use SP-16 for the design of bending reinforcement. Solution 2:

kNm 459.375 2

(3.5) (3.5) 75 1-1section at ==uM

26

2 N/mm 6.125 )500( )500( 300

)(10 459.375

==db

M u


Solution 1:

Here, Ast = 2-25T + 2-20T gives the percentage of tensile reinforcement

1.43 (450) 250

(1609) 100 ==p

From Table 6.1 of Lesson 13, τc = 0.67 + 0.036 = 0.706 N/mm2 (by linear interpolation). Employing Eq. 6.1 of Lesson 13,

τv = 23

N/mm 2.22 (450) 250

)(10 250

==db

Vu and τcmax = 2.8 N/mm2 (from Table

6.2 of Lesson 13). Hence, τc < τv < τcmax. So, shear reinforcement is needed for the shear force (Eq. 6.4 of Lesson 13). Vus = Vu – τc b d = 250 – 0.706 (250) (450) (10-3) = 170.575 kN Providing 8 mm, 2 legged vertical stirrups, we have Asv = 2 (50) = 100 mm2

Hence, spacing of the stirrups as obtained from Eq. 6.5 of Lesson 13:

mm, 95.25 170575

(450) (100) (415) 0.87 0.87

===us

svyv V

dAfs say 95 mm.

For 10 mm, 2 legged vertical stirrups, (Asv = 157 mm2), spacing

mm 149.54 170575

(450) (157) (415) 0.87 ==vs

According to cl. 26.5.1.5 of IS 456, the maximum spacing of the stirrups = 0.75 d = 0.75 (450) = 337.5 mm = 300 mm (say). Minimum shear reinforcement (cl. 26.5.1.6 of IS 456) is obtained from (Eq.6.3 of sec. 6.13.7 of Lesson 13):


Reinforcing bars of 3-28T + 1-16T give 2048 mm2. Asc at section 2-2 = (0.686) (400) (300)/100 = 823.2 mm2.

Reinforcing bars of 2-20T + 2-12T give 854 mm2.

Though it is better to use 4-28T as Ast and 2-20T + 2-16 as Asc with proper curtailment from the practical aspects of construction, here the bars are selected to have areas close to the requirements for the academic interest only. Figure 6.14.4 shows the reinforcement at sec. 2-2.

Case (i): No curtailment (all bars of bending reinforcement are continued): Bending moment Mu at section 2-2 = 234.375 kNm Shear force Vu at section 2-2 = 187.5 kN Effective depth d at section 2-2 = 450 – 50 = 400 mm, and

Width b = 300 mm tanβ = (550 – 200)/3500 = 0.1 Clause 40.1.1 of IS 456 gives (Eq.6.2 of sec. 6.13.3 of Lesson 13):


23

N/mm 1.074 (400) (300)

(0.1)/0.4} 234.375 - {187.5 )(10

tan ==

−=

dbd

MV u

u

v

βτ

(Here, the negative sign is used as the bending moment increases numerically in the same direction as the effective depth increases.) Continuing 4-28T and 3-16T bars (= 3066 mm2), we get p = 3066 (100)/300 (400) = 2.555 % Table 6.1 of Lesson 13 gives τc = 0.82 N/mm2 < τv (= 1.074 N/mm2). Hence, shear reinforcement is needed for shear force obtained from Eq. 6.4 of Lesson 13: Vus = Vu - τv b d = 1.875 – 0.82 (300) (400) (10-3) kN = 89.1 kN From cl.40.4 of IS 456, we have (Eq.6.5 of sec. 6.13.8 of Lesson 13),

v

svyus s

dAfV

87.0 =

where Asv = 100 mm2 for 8 mm, 2 legged vertical stirrups. This gives sv = 162.087 mm (f y = 415 N/mm2). IS 456, cl. 26.5.1.6 gives the spacing considering minimum shear reinforcement (Eq.6.3 of sec. 6.13.7 of Lesson 13):

bAf

s svyv 4.0

87.0 ≤

or sv ≤ 300.875 mm Hence, provide 8 mm, 2 legged vertical stirrups @ 150 mm c/c, as shown in Fig. 6.14.3. Case (ii): With curtailment of bars: Clause 26.2.3.2 of IS 456 stipulates that any one of the three conditions is to be satisfied for the termination of flexural reinforcement in tension zone (see sec. 6.13.10 of Lesson 13). Here, two of the conditions are discussed. (a) Condition (i):

),

( 32

dbVus

cv +≤ ττ which gives Eq. 6.9 of Lesson 13 as


Vus ≥ (1.5 τv – τc) b d After the curtailment, at section 2-2 Ast = 2048 mm2 (3-28T + 1-16T bars), gives p = 2048 (100)/300 (400) ≅ 1.71 %. Table 6.1 of Lesson 13 gives τc = 0.7452 N/mm2 when p = 1.71% (making liner interpolation). Now from Eq. 6.2 of Lesson 13:

23

N/mm 1.074 (400) (300)

(0.1)/0.4} 234.375 - {187.5 )(10

tan ==

−=

dbd

MV u

u

v

βτ

So, Vus = {1.5 (1.074) – 0.7452} (300) (400) (10-3) kN = 103.896 kN which gives the spacing of stirrups (Eq. 6.5):

us

svyv V

dAfs

87.0 ≤

Using 8 mm, 2 legged vertical stirrups (Asv = 100 mm2), we have: sv = 0.87 (415) (100) (400)/(103.896) (103) = 139.005 mm Hence, provide 8 mm, 2 legged vertical stirrups @ 130 mm c/c, as shown in Fig. 6.14.4.


(b) Condition (ii):

Additional stirrup area for a distance of 0.75 d {= 0.75 (400) = 300 mm} = 0.4 b s/fy, where spacing s is not greater than (d/8βb), where βb = cut off bar area/total bar area = 2048/3066 = 0.67. Since, additional stirrups are of lower diameter, mild steel bars are preferred with fy = 250 N/mm2. Maximum spacing s = d/8βb = 400/8 (0.67) = 75 mm. Excess area = 0.4 b s/fy = 0.4 (300) (75)/250 = 36 mm2.

Provide 6 mm, 2 legged mild steel vertical stirrups (56 mm2) @ 75 mm c/c for a distance of 300 mm, i.e., five numbers of stirrups (additional), as shown in Fig. 6.14.5. Problem 3:

Design the flexural and shear reinforcement of the simply supported T-beam (Fig. 6.14.6) of effective span 8 m placed @ 4.2 m c/c and subjected to a total factored load of 150 kN/m. Use M 30, Fe 415 and SP-16 tables for the design of flexural reinforcement. Solution 3: Design of flexural reinforcement with SP-16: Effective width of flange bf = lo/6 + bw + 6 Df = 8000/6 + 300 + 6(120) = 2353 mm > 2100 mm (breadth of the web plus half the sum of the clear distance to the adjacent beams on either side). So, bf = 2100 mm. (Mu)factored = 150 (8) (8)/(8) = 1200 kNm, at the mid-span.


(Vu)factored = 150 (8)/(2) = 600 kN, at the support.

The bending moment and shear force diagrams are shown in Fig. 6.14.7. At the mid-span

6

2

1200 (10 0.37 300 (600) (600) (30)

u

w ck

M )b d f

= =

Df /d = 120/600 = 0.2 and bf /bw = 2100/300 = 7 Table 58 of SP-16 (fy = 415 N/mm2) gives:


0.62 2lim, =

ck

u

fdbM

(when bf /bw = 7.0 and Df /d = 0.2) > 0.37 in

this case. Hence, o.k.

26

mm 6155 (540) (415) 87.0

)(10 1200 /2) - ( 87.0

===fy

ust ddf

MA

Provide 7-32T + 1-28T (= 6245 mm2) bars at mid-span and up to section 5-5 (Fig. 6.14.8, sec. 5-5). The flexural reinforcement is cranked up and the reinforcement diagrams are shown at five sections in Fig. 6.14.8.


Design of shear reinforcement: The details of calculations are shown below for the section 1-1 in six steps. Results of all four sections are presented in Table 6.3. Step 1: Ast at section 1-1 is determined (= 3217 mm2 = 4-32T) from Fig. 6.14.7 to calculate p = Ast (100)/bw d = 3217 (100)/300 (600) = 1.79%. From Table 6.1 of Lesson 13, τc is determined for p = 1.79% as 0.81 N/mm2. Table 6.2 of Lesson 13 gives τc,max for M 30 = 3.5 N/mm2. Step 2: Vu is 600 kN at section 1-1 from Fig.6.14.7 to get τv from Eq.6.1 of Lesson 13 as:

23

N/mm 3.33 (600) (300)

)(10 (600)

===db

V

w

uvτ

Here, τv is less than τcmax. Step 3: The magnitude of shear force for which shear reinforcement is needed (say, Vreinf.) is determined from Fig. 6.14.7. For section 1-1, from Eq. 6.4 of Lesson 13, Vreinf = Vu – τc b d = 600 – 0.81 (300) (600) = 454.2 kN. The magnitude of shear force taken by bent up bar(s) is obtained from Eq. 6.7 of Lesson 13, Vbent = 0.87 fy Asv sinα = 0.87 (415) (804) (1/√2) (10-3) = 206.5 kN. This force should not be greater than 0.5 (Vreinf), which, at this section, is 227.1 kN (vide sec. 6.13.8 of Lesson 13). The magnitude of the shear force for the design of vertical stirrup = Vus = Vreinf – Vbent = 454.2 – 206.5 = 247.7 kN. Step 4: Assume the diameter of vertical stirrup bars as 10 mm, 2 legged (Asv = 157 mm2). The spacing of vertical stirrups sv = 0.87 fy Asv d/Vus = 0.87 (415) (157) (600)/247.7 (103) = 137.3 mm c/c. Please refer to Eq.6.5 of Lesson 13.


Step 5: Check sv considering minimum shear reinforcement from cl. 26.5.1.6 of IS 456 as (see Eq.6.3 of Lesson 13):

sv ≤ 0.87 fy Asv /0.4 bw ≤ 0.87 (415) (157) /0.4 (300) ≤ 472 mm Further, cl. 26.5.1.5 stipulates the maximum spacing = 0.75 d on 300 mm. Here, the maximum spacing = 300 mm. Step 6: So, provide 10 mm, 2 legged stirrups @ 135 mm c/c. The results of all the sections are given below: Table 6.3 Design of stirrups using 10 mm 2 legged vertical stirrups, (τcmax = 3.5

N/mm2)

Step Values of Sec. 1-1 Sec. 2-2 Sec. 3-3 Sec. 4-4 Ast (mm2) 3,217 4,021 4,825 5,629

p (%) 1.79 2.23 2.68 3.13

1 τc (N/mm2) 0.81 0.877 0.93 0.96

Vu (kN) 600 525 450 375 2

τv (N/mm2) 3.33 2.92 2.50 2.08

Vreinf (kN) 454.2 367.14 282.6 202.2 Vbent (kN) 206.5*

< 0.5 Vreinf

206.5 < 183.57

206.5 < 141.3

157.24** < 101.1

3

Vus (kN) 247.7 183.57 141.3 101.1 4 sv (mm) 137.3 185.27 240.7 336.4 5 Min sv (mm) ≤ 300 ≤ 300 ≤ 300 ≤ 300 6 Provide sv (mm) 135 180 240 300

* diameter of bent up bar = 32 mm ** diameter of bent up bar = 28 mm To avoid several spacings for the practical consideration, provide stirrups @ 130 mm c/c for first 1 m, @ 230 mm for next 1 m and then @ 300 mm up to the mid-span in a symmetric manner (Fig. 6.14.8, secs. 1-1 to 5-5).


6.14.3 Practice Questions and Problems with Answers Q.1: Check if shear reinforcement is needed for the beam shown in Fig. 6.14.9.

If so, design the shear reinforcement using M 20 and mild steel (Fe 250). The tensile reinforcement is of Fe 415. Factored shear force = 300 kN.

A.1:

Ast = 4-25T = 1963 mm2

p = 1963/300(500) = 1.31% τc from Table 6.1 of sec. 6.13.4.1 of Lesson 13 = 0.68 N/mm2

From Table 6.2 of sec. 6.13.4.2 of Lesson 13 τcmax = 2.8 N/mm2

τv = Vu /b d (Eq.6.1 of Lesson 13) = 300000/300(500) = 2.0 N/mm2 Since τc < τv < τcmax, shear reinforcement is needed. From Eq.6.4 of Lesson 13: Vus = Vu – τc b d = 300 – 0.68(300)(500)(10-3) = 198 kN Alternative 1: Providing 10 mm, 2 legged stirrups (Asv = 157 mm2), the spacing sv = 0.87(250)(157)(500)/198000 = 86.23 mm. Provide 10 mm, 2 legged stirrups @ 85 mm c/c. Please refer to Eq.6.5 of Lesson 13. The required area of stirrups spaced @ 85 mm c/c to satisfy the minimum shear reinforcement (cl.


26.5.1.6 of IS 456) is obtained from Eq. 6.3 of Lesson 13 as: Asv = 0.4(300)(85)/0.87(250) = 46.89 mm2 < 157 mm2 . Alternative 2: Providing 12 mm, 2 legged stirrups (Asv = 226 mm2), the spacing sv = 0.87(250)(226)(500)/198000 = 124.13 mm. Provide 12 mm, 2 legged stirrups @ 120 mm c/c. The required area of stirrups spaced @ 120 mm c/c to satisfy the minimum shear reinforcement (cl. 26.5.1.6 of IS 456) is obtained from Eq. 6.3 of Lesson 13 as: Asv = 0.4(300)(120)/0.87(250) = 66.21 mm2 < 226 mm2 . Further, the maximum spacing (cl. 26.5.1.5 of IS 456 and sec. 6.13.7 of Lesson 13) = 0.75 d = 0.75(500) = 375 mm. Hence, both are possible, though 12 mm @ 120 mm c/c is desirable since the other spacing of 85 mm c/c is very close (Fig. 6.14.9). 6.14.4 References:

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 6.14.5 Test 14 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ. 1: The T-beam of Fig. 6.14.10 has a factored shear force of 400 kN.

Determine the diameter and spacing of vertical stirrups at a section where two 25 mm diameter bent up bears are also available for the shear resistance. Use M 20 and Fe 415.

(50 marks)

A.TQ. 1:

Ast = 4-25φ = 1963 mm2


p = 1963(100)/300(550) = 1.19% τc = 0.658 N/mm2 (Table 6.1 of Lesson 13) Vu = 400 kN τv (Eq.6.1 of Lesson 13) = 400/165 = 2.43 N/mm2 < τcmax of 2.8 N/mm2.

Hence, o.k. Vreinf. = Vu – τc b d = 400 – 0.658(300)(0.55) = 291.43 kN (please

refer to Eq.6.4 of Lesson 13) Vbent (2-25 mm diameter) = 0.87 fy Asv sinα = 0.87(415)(981)(1/√2) = 250.48 kN,

subjected to the maximum value of 0.5(Vreinf.) = 145.71 kN (see sec. 6.13.8 of Lesson 13)

Vus = 145.72 kN

Using 10 mm, 2 legged vertical stirrups (Asv = 157 mm2), the spacing, obtained from Eq.6.5 of Lesson 13, sv = 0.87 fy Asv d/Vus = 0.87 (415) (157) (550)/145720 = 213.95 mm c/c.

For the minimum shear of reinforcement as per cl. 26.5.1.6 of IS 456, using 10 mm, 2 legged vertical stirrups, the spacing as obtained from Eq.6.3 of Lesson 13:

sv ≤ 0.87 fy Asv /0.4 bw ≤ 0.87 (415) (157) /0.4 (300) ≤ 472 mm

Again, cl. 26.5.1.5 of IS 456 stipulates the maximum spacing = 0.75 d = 0.75(550) = 412.5 mm (see sec. 6.13.7 of Lesson 13). Hence, provide 10 mm, 2 legged vertical stirrups @ 210 mm c/c (Fig.

6.14.10). 6.14.6 Summary of this Lesson Learning the different failure modes, the shear stresses and the design procedure of beams subjected to shear in Lesson 13, this lesson explains the design through several numerical problems with special reference to curtailment of tension reinforcement in flexural members. Solution of problems given in the


practice problems and test, students will be thoroughly conversant with the design of rectangular and T-beams subjected to shear following the limit state of collapse.


Module 6




Lesson 15

Bond, Anchorage, Development Length

and Splicing


Instruction Objectives: At the end of this lesson, the student should be able to:

• understand the importance of bond and why is it essential to provide between steel and concrete,

• explain the development length,

• understand the need for anchoring the tensile bars,

• justify the superiority of deformed bars to smooth mild steel bars,

• define the design bond stress,

• understand the reason for different values of design bond stresses of plain

bars and deformed bars in tension and compression,

• determine the development length of a single bar and bars bundled in contact,

• derive the expression to check the development length of bars in tension,

• specify the salient points of anchoring bars in tension, compression and

shear,

• determine the bearing stress and check the same,

• take adequate precautions when the direction of reinforcement changes,

• specify the salient points of splicing and welding of reinforcement to make it longer,

• apply the theory for designing beams in different situations as may arise.

6.15.1 Introduction The bond between steel and concrete is very important and essential so that they can act together without any slip in a loaded structure. With the perfect bond between them, the plane section of a beam remains plane even after bending. The length of a member required to develop the full bond is called the anchorage length. The bond is measured by bond stress. The local bond stress varies along a member with the variation of bending moment. The average value


throughout its anchorage length is designated as the average bond stress. In our calculation, the average bond stress will be used. Thus, a tensile member has to be anchored properly by providing additional length on either side of the point of maximum tension, which is known as ‘Development length in tension’. Similarly, for compression members also, we have ‘Development length Ld in compression’. It is worth mentioning that the deformed bars are known to be superior to the smooth mild steel bars due to the presence of ribs. In such a case, it is needed to check for the sufficient development length Ld only rather than checking both for the local bond stress and development length as required for the smooth mild steel bars. Accordingly, IS 456, cl. 26.2 stipulates the requirements of proper anchorage of reinforcement in terms of development length Ld only employing design bond stress τbd. 6.15.2 Design Bond Stress τbd (a) Definition The design bond stress τbd is defined as the shear force per unit nominal surface area of reinforcing bar. The stress is acting on the interface between bars and surrounding concrete and along the direction parallel to the bars. This concept of design bond stress finally results in additional length of a bar of specified diameter to be provided beyond a given critical section. Though, the overall bond failure may be avoided by this provision of additional development length Ld, slippage of a bar may not always result in overall failure of a beam. It is, thus, desirable to provide end anchorages also to maintain the integrity of the structure and thereby, to enable it carrying the loads. Clause 26.2 of IS 456 stipulates, “The calculated tension or compression in any bar at any section shall be developed on each side of the section by an appropriate development length or end anchorage or by a combination thereof.” (b) Design bond stress – values Section 6.15.1 mentions that the local bond stress varies along the length of the reinforcement while the average bond stress gives the average value throughout its development length. This average bond stress is still used in the working stress method and IS 456 has mentioned about it in cl. B-2.1.2. However, in the limit state method of design, the average bond stress has been designated as design bond stress τbd and the values are given in cl. 26.2.1.1. The same is given below as a ready reference.


Table 6.4 τbd for plain bars in tension

Grade of concrete

M 20 M 25 M 30 M 35 M 40 and above

Design Bond

Stress τbd in N/mm2

1.2

1.4

1.5

1.7

1.9

For deformed bars conforming to IS 1786, these values shall be increased by 60 per cent. For bars in compression, the values of bond stress in tension shall be increased by 25 per cent. 6.15.3 Development Length

(a) A single bar Figure 6.15.1(a) shows a simply supported beam subjected to uniformly distributed load. Because of the maximum moment, the Ast required is the maximum at x = L/2. For any section 1-1 at a distance x < L/2, some of the


tensile bars can be curtailed. Let us then assume that section 1-1 is the theoretical cut-off point of one bar. However, it is necessary to extend the bar for a length Ld as explained earlier. Let us derive the expression to determine Ld of this bar. Figure 6.15.1(b) shows the free body diagram of the segment AB of the bar. At B, the tensile force T trying to pull out the bar is of the value T = (π φ 2 σs /4), where φ is the nominal diameter of the bar and σs is the tensile stress in bar at the section considered at design loads. It is necessary to have the resistance force to be developed by τbd for the length Ld to overcome the tensile force. The resistance force = π φ (Ld) (τbd). Equating the two, we get π φ (Ld) (τbd) = (π φ 2 σs /4) (6.11) Equation 6.11, thus gives

bd

dLτσφ

4

s=

(6.12) The above equation is given in cl. 26.2.1 of IS 456 to determine the development length of bars. The example taken above considers round bar in tension. Similarly, other sections of the bar should have the required Ld as determined for such sections. For bars in compression, the development length is reduced by 25 per cent as the design bond stress in compression τbd is 25 per cent more than that in tension (see the last lines below Table 6.4). Following the same logic, the development length of deformed bars is reduced by 60 per cent of that needed for the plain round bars. Tables 64 to 66 of SP-16 present the development lengths of fully stressed plain and deformed bars (when σs = 0.87 fy) both under tension and compression. It is to be noted that the consequence of stress concentration at the lugs of deformed bars has not been taken into consideration. (b) Bars bundled in contact The respective development lengths of each of the bars for two, three or four bars in contact are determined following the same principle. However, cl. 26.2.1.2 of IS 456 stipulates a simpler approach to determine the development length directly under such cases and the same is given below:


“The development length of each bar of bundled bars shall be that for the individual bar, increased by 10 per cent for two bars in contact, 20 per cent for three bars in contact and 33 per cent for four bars in contact.” However, while using bundled bars the provision of cl. 26.1.1 of IS 456 must be satisfied. According to this clause:

• In addition to single bar, bars may be arranged in pairs in contact or in

groups of three or four bars bundled in contact. • Bundled bars shall be enclosed within stirrups or ties to ensure the bars

remaining together.

• Bars larger than 32 mm diameter shall not be bundled, except in columns.

Curtailment of bundled bars should be done by terminating at different points spaced apart by not less than 40 times the bar diameter except for bundles stopping at support (cl. 26.2.3.5 of IS 456). 6.15.4 Checking of Development Lengths of Bars in Tension The following are the stipulation of cl. 26.2.3.3 of IS 456. (i) At least one-third of the positive moment reinforcement in simple members and one-fourth of the positive moment reinforcement in continuous members shall be extended along the same face of the member into the support, to a length equal to Ld/3. (ii) Such reinforcements of (i) above shall also be anchored to develop its design stress in tension at the face of the support, when such member is part of the primary lateral load resisting system. (iii) The diameter of the positive moment reinforcement shall be limited to a diameter such that the Ld computed for σs = fd in Eq. 6.12 does not exceed the following:

ofd LVML

d )( 1

when s+≤=σ

(6.13) where M1 = moment of resistance of the section assuming all reinforcement at

the section to be stressed to fd, fd = 0.87 fy, V = shear force at the section due to design loads,


Lo = sum of the anchorage beyond the centre of the support and the

equivalent anchorage value of any hook or mechanical anchorage at simple support. At a point of inflection, Lo is limited to the effective depth of the member or 12φ , whichever is greater, and

φ = diameter of bar. It has been further stipulated that M1/V in the above expression may be increased by 30 per cent when the ends of the reinforcement are confined by a compressive reaction. 6.15.5 Derivation of the Limiting Ld (Eq. 6.13)

At the face of simple support and at the points of inflection in continuous beams, the tensile capacity to be developed is normally small although the rate of change of tensile stress in the bars is high. These bond stresses are designated as flexural bond stress. Figure 6.15.2(a) shows the tensile bar AB near the support of a beam and Fig. 6.15.2(b) explains the stresses and forces in the free body diagram of the


segment CD of the bar. The tensile force at C (TC) is greater than that at D (TD). Considering z as the lever arm (distance between the centres of gravity of tensile and compressive force), we have MC = TC (z) (6.14) MD = TD (z) Again TC - TD = bdx τπφ )d( (6.15) From Eqs.6.14 and 6.15,

C D - (dbdM M x )

zπφτ=

which gives

bdzxM τπφ d

d=

or z

Vbd

φπ

τ =

(6.16) Equation 6.16 gives the flexural bond stress in the tension reinforcement at any section. If there are N bars of equal size, we have

∑

=)0(

z

Vbdτ

(6.17) where ∑ = φπ 0 N = total perimeter of all bars in tension at the section. Again, for N bars of equal diameter, we get from Eq. 6.11: yststbdd fAALN 0.87 s == στφπ (when ys f 0.87 =σ ), which gives:


∑

=)0(

87.0

d

stybd L

Afτ

(6.18) Equations 6.17 and 6.18 give:

∑

=∑ )0(

)0(

87.0z

VL

Af

d

sty

or VMLd

1 =

(6.19) where 1 0.87 y stM f A z= is the moment of resistance at the section and V = shear force at the section From Eq.6.19, we find that the ratio of M1/V at the section must be equal to or greater than Ld if the design bond stress τbd is to be restricted within limit. The stipulation of additional Lo in the expression of Eq.6.13 is for additional safety. The meaning of Lo has been mentioned in sec. 6.15.4 (iii). 6.15.6 Anchoring Reinforcing Bars Section 6.15.2(a) mentions that the bars may be anchored in combination of providing development length to maintain the integrity of the structure. Such anchoring is discussed below under three sub-sections for bars in tension, compression and shear respectively, as stipulated in cl. 26.2.2 of IS 456.


(a) Bars in tension (cl. 26.2.2.1 of IS 456)

The salient points are:

• Derformed bars may not need end anchorages if the development length requirement is satisfied.

• Hooks should normally be provided for plain bars in tension.

• Standard hooks and bends should be as per IS 2502 or as given in Table

67 of SP-16, which are shown in Figs.6.15.3 a and b.

• The anchorage value of standard bend shall be considered as 4 times the diameter of the bar for each 45o bend subject to a maximum value of 16 times the diameter of the bar.

• The anchorage value of standard U-type hook shall be 16 times the

diameter of the bar. (b) Bars in compression (cl. 26.2.2.2 of IS 456) Here, the salient points are:


• The anchorage length of straight compression bars shall be equal to its development length as mentioned in sec. 6.15.3.

• The development length shall include the projected length of hooks, bends

and straight lengths beyond bends, if provided. (c) Bars in shear (cl. 26.2.2.4 of IS 456)

The salient points are:

• Inclined bars in tension zone will have the development length equal to that of bars in tension and this length shall be measured from the end of sloping or inclined portion of the bar.


• Inclined bars in compression zone will have the development length equal

to that of bars in tension and this length shall be measured from the mid-depth of the beam.

• For stirrups, transverse ties and other secondary reinforcement, complete

development length and anchorage are considered to be satisfied if prepared as shown in Figs.6.15.4.

6.15.7 Bearing Stresses at Bends (cl. 26.2.2.5 of IS 456) The bearing stress inside a bend is to be calculated from the expression:

Bearing stress = φr

Fbt

(6.20) where Fbt = tensile force due to design loads in a bar or group of bars, r = internal radius of the bend, and φ = size of the bar or bar of equivalent area in bundled bars. The calculated bearing stress of Eq.6.20 shall not exceed the following:

Calculated bearing stress a

fck

/ 2 1 1.5

φ+

>/

(6.21) where fck = characteristic cube strength of concrete

a = center to center distance between bars or groups of bars perpendicular to the plane of the bend. For bars adjacent to the face of the member, a shall be taken as cover plus size of the bar φ .

6.15.8 Change in Direction of Reinforcement (cl. 26.2.2.6 of IS 456) In some situations, the change in direction of tension or compression reinforcement induces a resultant force. This force may have a tendency to split the concrete and, therefore, should be taken up by additional links or stirrups. Normally, this aspect is taken care while detailing of bars is carried out.


6.15.9 Reinforcement Splicing (cl. 26.2.5 of IS 456) Reinforcement is needed to be joined to make it longer by overlapping sufficient length or by welding to develop its full design bond stress. They should be away from the sections of maximum stress and be staggered. IS 456 (cl. 26.2.5) recommends that splices in flexural members should not be at sections where the bending moment is more than 50 per cent of the moment of resistance and not more than half the bars shall be spliced at a section. (a) Lap Splices (cl. 26.2.5.1 of IS 456) The following are the salient points:

• They should be used for bar diameters up to 36 mm.

• They should be considered as staggered if the centre to centre distance of the splices is at least 1.3 times the lap length calculated as mentioned below.

• The lap length including anchorage value of hooks for bars in flexural

tension shall be Ld or 30φ , whichever is greater. The same for direct tension shall be 2Ld or 30φ , whichever is greater.

• The lap length in compression shall be equal to Ld in compression but

not less than 24φ .

• The lap length shall be calculated on the basis of diameter of the smaller bar when bars of two different diameters are to be spliced.

• Lap splices of bundled bars shall be made by splicing one bar at a time

and all such individual splices within a bundle shall be staggered. (b) Strength of Welds (cl. 26.2.5.2 of IS 456) The strength of welded splices and mechanical connections shall be taken as 100 per cent of the design strength of joined bars for compression splices. For tension splices, such strength of welded bars shall be taken as 80 per cent of the design strength of welded bars. However, it can go even up to 100 per cent if welding is strictly supervised and if at any cross-section of the member not more than 20 per cent of the tensile reinforcement is welded. For mechanical connection of tension splice, 100 per cent of design strength of mechanical connection shall be taken.


6.15.10 Numerical Problems

Problem 1: Determine the anchorage length of 4-20T reinforcing bars going into the support of the simply supported beam shown in Fig. 6.15.5. The factored shear force Vu = 280 kN, width of the column support = 300 mm. Use M 20 concrete and Fe 415 steel. Solution 1: Table 6.4 gives

τbd for M 20 and Fe 415 (with 60% increased) = 1.6(1.2) = 1.92 N/mm2

Eq.6.12 gives

(1) ...... 47.01 ) 0.87 (when 4(1.92)

0.87(415) 4

ss φσφ

τσφ

==== ybd

d fL

Eq.6.13 gives

ofd LVML

d )( 1

when s+≤=σ

Here, to find M1, we need xu


mm 209.94 (300) 0.36(20)(1256) 0.87(415)

0.36 0.87

===bf

Afx

ck

styu

xu,max = 0.48(500) = 240 mm Since xu < xu,max ; M1 = 0.87 fy Ast (d – 0.42 xu) or M1 = 0.87(415) (1256) {500 – 0.42(209.94)} = 187.754 kNm and V = 280 kN We have from Eq.6.13 above, with the stipulation of 30 per cent increase assuming that the reinforcing bars are confined by a compressive reaction:

(2) ...... )( 1.3 1od L

VML +≤

From Eqs.(1) and (2), we have

)( 1.3 01.47 1oL

VM

+≤φ

or zero. as assumed is if };)10(280

)10(754.187{ 1.3 01.47 3

6

oL≤φ

or mm 18.54 ≤φ Therefore, 20 mm diameter bar does not allow Lo = 0. Determination of Lo:

01.47 )( 1.3 1 φ≥+ oLVM

Minimum mm 68.485 )280

1877541.3( - 47.01(20) )( 1.3 - 01.47 1 ===VMLo φ

So, the bars are extended by 100 mm to satisfy the requirement as shown in Fig.6.15.6.


6.15.11 Practice Questions and Problems with Answers Q.1: Explain the importance of the bond and why is it essential to provide

between steel and concrete in beams? A.1: Sec. 6.15.1 para 1 Q.2: Define the design bond stress bdτ . A.2: Sec. 6.15.2(a) Q.3: State the percentage increase/decrease of design bond stress of deformed

bars in tension and compression with reference to the respective values of plain bars.

A.3: Three lines below Table 6.4 of sec. 6.15.2(b) Q.4: Derive the expression of determining the development length of a single

bar in tension. State the changes, if any, for the compression bars. A.4: Sec. 6.15.3(a). Q.5: How would you determine the development lengths of bars when two,

three or four bars are bundled in contact? State the salient points of the stipulations of IS 456 in this respect.

A.5: Sec. 6.15.3(b)


Q.6: Derive the limiting value of the development length for bars in tension having both bending moment and shear force. Explain the role of additional length Lo.

A.6: Sec. 6.15.5 and the final form is given in Eq.6.13. Q.7: State the salient points of the stipulations of IS 456 regarding anchoring

reinforcing bars in tension, compression and shear, respectively. A.7: Sec. 6.15.6(a), (b) and (c) are the respective answers. Q.8: Write down the expressions of calculated bearing stress at bends and its

limiting value. A.8: Eqs.6.20 and 6.21 are the answers. Q.9: State the additional measure to be taken when the reinforcing bars change

the direction. A.9: Sec. 6.15.8. Q.10: State the salient points of splicing and welding of reinforcing bars. A.10: Sec. 6.15.9(a) and (b), respectively, are the answers. Q.11: Check the bond requirement of the continuous beam of Fig.6.15.7 if the

factored shear force is 200 kN at the point of inflection. Assume M 20 and Fe 415.


A.11: Table 6.4 gives τbd for M 20 and Fe 415 as 1.92 N/mm2 (see Solution 1 of

Problem 1 of sec. 6.15.10). Eq.6.12 gives

(3) ...... 47.01 4(1.92)

0.87(415) 4

s φφτσφ

===bd

dL

Eq.6.13 gives

ofd LVML

d )( 1

when s+≤=σ

To find M1, we need xu

mm 157.46 (300) 0.36(20)(942) 0.87(415)

0.36 0.87

===bf

Afx

ck

styu

xu,max = 0.48(400) = 192 mm Since xu < xu,max ; M1 = 0.87 fy Ast (d – 0.42 xu)

= 0.87(415) (942) {400 – 0.42(157.46)} = 113.55 kNm ….. (4) From Eqs. (3) and (4), we have

(5) ...... )( 01.47 1oL

VM

+≤φ


At the point of inflection Lo is the maximum of d (= 400 mm) or mm). 240 ( 12 =φ With Lo = 400 mm, we can write from Eq.(5)

400 })10(200

)10(55.113{ 01.47 3

6

+≤φ

which gives ≤ φ 20.58 mm. Thus, use of mm 20 =φ satisfies the bond requirement. 6.15.12 References:

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 6.15.13 Test 15 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Derive the expression of determining the development length of a single

bar in tension. State the changes, if any, for the compression bars. (10 marks)

A.TQ.1: Sec. 6.15.3(a). TQ.2: Derive the limiting value of the development length for bars in tension

having both bending moment and shear force. Explain the role of additional length Lo.

(10 marks) A.TQ.2: Sec. 6.15.5 and the final form is given in Eq.6.13. TQ.3: State the salient points of the stipulations of IS 456 regarding anchoring

reinforcing bars in tension, compression and shear, respectively. (10 marks)

A.TQ.3: Sec. 6.15.6(a), (b) and (c) are the respective answers. TQ.4: Design the main tensile reinforcement of a singly reinforced rectangular

beam of width 300 mm and effective span of 8 m carrying a total factored load of 40 kN/m using M 20 and Fe 415 with the help of SP-16. Check the development length at the support if 50 per cent of the reinforcing bars are continued to the support. Assume width of the support as 300 mm. (20 marks)

A.TQ.4: Design of beam using SP-16: Factored mid-span bending moment = 40(8)(8)/8 = 320 kNm Factored shear at the support = 40(8)/2 = 160 kN Required moment of resistance per metre width = 320/0.3 = 1067 kNm/m


Chart 15 of SP-16 gives d = 63.25 cm and percentage of reinforcement = 0.9. Using the total depth D = 700 mm and effective depth as 650 mm, we get

the area of main tensile reinforcement at mid-span = 0.9(bd)/100 = 0.9(300)(650)/100 = 1755 mm2.

Let us provide 6-20T of area = 1885 mm2. Three 20T bars (50 per cent) are continued at the support. The value of

bdτ for M 20 and Fe 415 is obtained from Table 6.4 with 60 per cent increased as 1.92 N/mm2. Eq.6.12 gives

mm 940.23 92)(415)/4(1.(20)(0.87) )/4( s === bddL τσφ

(6) Eq.6.13 gives ofd LVML

ds )/( )( 1 when +≤=σ . To find M1, we need xu as

given below: xu = (0.87 fy Ast)/(0.36 fck b) = 0.87(415)(942.5)/(0.36)(20)(300) = 157.54 mm < xu,max {= 0.48(650) = 312 mm}

So, M1 = 0.87 fy Ast (d – 0.42 xu) = 0.87(415)(942.5){(650) – 0.42(157.54)} = 198.67 kNm We have from Eq.6.13 with the stipulation of increase of 30 per cent

assuming that the reinforcing bars are confined by a compressive reaction:

Ld ≤ 1.3(M1/V) + Lo (7) From Eqs.6 and 7 above, we have 940.23 ≤ 1.3(M1/V) + Lo (8)


Here, 1.3(M1/V) = 1.3(198.67)/160 = 1.614 m Hence, Lo is not required as Ld is satisfied when Lo = 0 (sec. Eq.8). 6.15.14 Summary of this Lesson This lesson explains the importance of bond between steel and concrete which is essential for the two materials to act together. To ensure the bond, additional length of the reinforcing bars is to be provided, which is known as the development length and calculated on the basis of design bond stresses. The design bond stress has been defined in this lesson and the values of the design bond stresses for different grades of concrete are given for mild steel plain and ribbed steel bars. The expression of determination of development length has been derived. The stipulations of IS 456 regarding anchoring bars in tension, compression and shear are discussed. Numerical problems solved in the lesson and other examples of practice problem and test question explain the calculations of satisfying the requirements of reinforcing bars under different situations with respect to bond.


Module 6




Lesson 16

Torsion in Beams - Limit State of Collapse


Instruction Objectives: At the end of this lesson, the student should be able to: • identify the beams and frames subjected to torsion, • name and explain the two types of torsion, • state the basis of approach of design for combined bending, shear and

torsion as per IS 456, • select the critical section for the design, • determine the equivalent shear and moment from the given factored bending

moment, shear and torsional moment, • define equivalent nominal shear stress, • state when do we provide minimum shear reinforcement in beams subjected

to combined bending moment, shear and torsional moment, • state when do we provide both longitudinal and transverse reinforcement in

beams subjected to combined bending moment, shear and torsional moment, • state when do we provide tensile, compressive and side face reinforcement,

respectively, in beams subjected to combined bending shear and torsional moment,

• design the beams subjected to combined bending, shear and torsional moment as per IS 456.

6.16.1 Introduction This lesson explains the presence of torsional moment along with bending moment and shear in reinforced concrete members with specific examples. The approach of design of such beams has been explained mentioning the critical section to be designed. Expressing the equivalent shear and bending moment, this lesson illustrates the step by step design procedure of beam under combined bending, shear and torsion. The requirements of IS 456 regarding the design are also explained. Numerical problems have been solved to explain the design of beams under combined bending, shear and torsion.


6.16.2 Torsion in Reinforced Concrete Members

On several situations beams and slabs are subjected to torsion in addition to bending moment and shear force. Loads acting normal to the plane of bending will cause bending moment and shear force. However, loads away from the plane of bending will induce torsional moment along with bending moment and shear. Space frames (Fig.6.16.1a), inverted L-beams as in supporting sunshades and canopies (Fig.6.16.1b), beams curved in plan (Fig.6.16.1c), edge beams of slabs (Fig.6.16.1d) are some of the examples where torsional moments are also present. Skew bending theory, space-truss analogy are some of the theories developed to understand the behaviour of reinforced concrete under torsion combined with bending moment and shear. These torsional moments are of two types:

(i) Primary or equilibrium torsion, and (ii) Secondary or compatibility torsion.

The primary torsion is required for the basic static equilibrium of most of

the statically determinate structures. Accordingly, this torsional moment must be considered in the design as it is a major component. The secondary torsion is required to satisfy the compatibility condition between members. However, statically indeterminate structures may have any of


the two types of torsions. Minor torsional effects may be ignored in statically indeterminate structures due to the advantage of having more than one load path for the distribution of loads to maintain the equilibrium. This may produce minor cracks without causing failure. However, torsional moments should be taken into account in the statically indeterminate structures if they are of equilibrium type and where the torsional stiffness of the members has been considered in the structural analysis. It is worth mentioning that torsion must be considered in structures subjected to unsymmetrical loadings about axes. Clause 41 of IS 456 stipulates the above stating that, "In structures, where torsion is required to maintain equilibrium, members shall be designed for torsion in accordance with 41.2, 41.3 and 41.4. However, for such indeterminate structures where torsion can be eliminated by releasing redundant restraints, no specific design for torsion is necessary, provided torsional stiffness is neglected in the calculation of internal forces. Adequate control of any torsional cracking is provided by the shear reinforcement as per cl. 40". 6.16.3 Analysis for Torsional Moment in a Member The behaviour of members under the effects of combined bending, shear and torsion is still a subject of extensive research. We know that the bending moments are distributed among the sharing members with the corresponding distribution factors proportional to their bending stiffness EI/L where E is the elastic constant, I is the moment of inertia and L is the effective span of the respective members. In a similar manner, the torsional moments are also distributed among the sharing members with the corresponding distribution factors proportional to their torsional stiffness GJ/L, where G is the elastic shear modulus, J is polar moment of inertia and L is the effective span (or length) of the respective members. The exact analysis of reinforced concrete members subjected to torsional moments combined with bending moments and shear forces is beyond the scope here. However, the codal provisions of designing such members are discussed below. 6.16.4 Approach of Design for Combined Bending, Shear and Torsion as per IS 456 As per the stipulations of IS 456, the longitudinal and transverse reinforcements are determined taking into account the combined effects of bending moment, shear force and torsional moment. Two impirical relations of equivalent shear and equivalent bending moment are given. These fictitious shear force and bending moment, designated as equivalent shear and equivalent bending moment, are separate functions of actual shear and torsion, and actual


bending moment and torsion, respectively. The total vertical reinforcement is designed to resist the equivalent shear Ve and the longitudinal reinforcement is designed to resist the equivalent bending moment Me1 and Me2, as explained in secs. 6.16.6 and 6.16.7, respectively. These design rules are applicable to beams of solid rectangular cross-section. However, they may be applied to flanged beams by substituting bw for b. IS 456 further suggests to refer to specialist literature for the flanged beams as the design adopting the code procedure is generally conservative. 6.16.5 Critical Section (cl. 41.2 of IS 456) As per cl. 41.2 of IS 456, sections located less than a distance d from the face of the support is to be designed for the same torsion as computed at a distance d, where d is the effective depth of the beam. 6.16.6 Shear and Torsion (a) The equivalent shear, a function of the actual shear and torsional moment is determined from the following impirical relation: Ve = Vu + 1.6(Tu/b) (6.22) where Ve = equivalent shear, Vu = actual shear, Tu = actual torsional moment, b = breadth of beam.

(b) The equivalent nominal shear stress veτ is determined from:

)/ ( bdVeve =τ (6.23)

However, veτ shall not exceed maxcτ given in Table 20 of IS 456 and Table 6.2 of Lesson 13. (c) Minimum shear reinforcement is to be provided as per cl. 26.5.1.6 of IS 456, if the equivalent nominal shear stress veτ obtained from Eq.6.23 does not exceed cτ given in Table 19 of IS 456 and Table 6.1 of Lesson 13. .


(d) Both longitudinal and transverse reinforcement shall be provided as per cl. 41.4 and explained below in sec. 6.16.7, if veτ exceeds cτ given in Table 19 of IS 456 and Table 6.1 of Lesson 13 and is less than maxcτ , as mentioned in (b) above. 6.16.7 Reinforcement in Members subjected to Torsion (a) Reinforcement for torsion shall consist of longitudinal and transverse reinforcement as mentioned in sec. 6.16.6(d). (b) The longitudinal flexural tension reinforcement shall be determined to resist an equivalent bending moment Me1 as given below: Me1 = Mu + Mt (6.24) where Mu = bending moment at the cross-section, and Mt = (Tu/1.7) {1 + (D/b)} (6.25) where Tu = torsional moment, D = overall depth of the beam, and b = breadth of the beam. (c) The longitudinal flexural compression reinforcement shall be provided if the numerical value of Mt as defined above in Eq.6.25 exceeds the numerical value of Mu. Such compression reinforcement should be able to resist an equivalent bending moment Me2 as given below: Me2 = Mt - Mu (6.26) The Me2 will be considered as acting in the opposite sense to the moment Mu.


(d) The transverse reinforcement consisting of two legged closed loops (Fig.6.16.2) enclosing the corner longitudinal bars shall be provided having an area of cross-section Asv given below:

1 1 1

0 87 2.5 0 87

u v u vsv

y y

T s V sAb d ( . f ) d ( . f )

= +

(6.27) However, the total transverse reinforcement shall not be less than the following: ) /(0.87 ) - ( cve yvsv fsbA ττ≥ (6.28) where Tu = torsional moment, Vu = shear force, sv = spacing of the stirrup reinforcement, b1 = centre to centre distance between corner bars in the direction of the width, d1 = centre to centre distance between corner bars,


b = breadth of the member, fy = characteristic strength of the stirrup reinforcement, veτ = equivalent shear stress as specified in Eqs.6.22 and 6.23, and

cτ = shear strength of concrete as per Table 19 of IS 456 and Table 6.1 of Lesson 13.

6.16.8 Requirements of Reinforcement Beams subjected to bending moment, shear and torsional moment should satisfy the following requirements: (a) Tension reinforcement (cl. 26.5.1.1 of IS 456) The minimum area of tension reinforcement should be governed by As /(bd) = 0.85/fy (6.29) where As = minimum area of tension reinforcement, b = breadth of rectangular beam or breadth of web of T-beam, d = effective depth of beam, fy = characteristic strength of reinforcement in N/mm2. The maximum area of tension reinforcement shall not exceed 0.04 bD, where D is the overall depth of the beam. (b) Compression reinforcement (cl. 26.5.1.2 of IS 456) The maximum area of compression reinforcement shall not exceed 0.04 bD. They shall be enclosed by stirrups for effective lateral restraint. (c) Side face reinforcement (cls. 26.5.1.3 and 26.5.1.7b) Beams exceeding the depth of 750 mm and subjected to bending moment and shear shall have side face reinforcement. However, if the beams are having torsional moment also, the side face reinforcement shall be provided for the overall depth exceeding 450 mm. The total area of side face reinforcement shall be at least 0.1 per cent of the web area and shall be distributed equally on two faces at a spacing not exceeding 300 mm or web thickness, whichever is less.


(d) Transverse reinforcement (cl. 26.5.1.4 of IS 456) The transverse reinforcement shall be placed around the outer-most tension and compression bars. They should pass around longitudinal bars located close to the outer face of the flange in T- and I-beams. (e) Maximum spacing of shear reinforcement (cl. 26.5.1.5 of IS 456)

The centre to centre spacing of shear reinforcement shall not be more than 0.75 d for vertical stirrups and d for inclined stirrups at 45o, but not exceeding 300 mm, where d is the effective depth of the section. (f) Minimum shear reinforcement (cl. 26.5.1.6 of IS 456) This has been discussed in sec. 6.13.7 of Lesson 13 and the governing equation is Eq.6.3 of Lesson 13. (g) Distribution of torsion reinforcement (cl. 26.5.1.7 of IS 456) The transverse reinforcement shall consist of rectangular close stirrups placed perpendicular to the axis of the member. The spacing of stirrups shall not be more than the least of x1, (x1 + y1)/4 and 300 mm, where x1 and y1 are the short and long dimensions of the stirrups (Fig.6.16.2). Longitudinal reinforcements should be placed as close as possible to the corners of the cross-section. (h) Reinforcement in flanges of T- and L-beams (cl. 26.5.1.8 of IS 456) For flanges in tension, a part of the main tensile reinforcement shall be distributed over the effective flange width or a width equal to one-tenth of the span, whichever is smaller. For effective flange width greater than one-tenth of the span, nominal longitudinal reinforcement shall be provided to the outer portion of the flange.


6.16.9 Numerical Problems Problem 1

Determine the reinforcement required of a ring beam (Fig.6.16.3) of b = 400 mm, d = 650 mm, D = 700 mm and subjected to factored Mu = 200 kNm, factored Tu = 50 kNm and factored Vu = 100 kN. Use M 20 and Fe 415 for the design. Solution 1 The solution of the problem is illustrated in seven steps below. Step 1: Check for the depth of the beam From Eq.6.22, we have the equivalent shear Ve = Vu + 1.6(Tu/b) = 100 + 1.6(50/0.4) = 300 kN From Eq.6.2.3, the equivalent shear stress )/ ( bdVeve =τ = 300/(0.4)(0.65) = 1.154 N/mm2

From Table 6.2 of Lesson 13 (Table 20 of IS 456), maxcτ = 2.8 N/mm2. Hence, the section does not need any revision. Step 2: Check if shear reinforcement shall be required. Assuming percentage of tensile steel as 0.5, Table 6.1 of Lesson 13 (Table 19 of IS 456) gives cτ = 0.48 N/mm2 < veτ < maxcτ . So, both longitudinal and transverse reinforcement shall be required.


Step 3: Longitudinal tension reinforcement

From Eqs.6.24 and 6.25, we have, Me1 = Mu + Mt = Mu + (Tu/1.7) {1 + (D/b)} = 200 + (50/1.7) {1 + (700/400)} = 200 + 80.88 = 280.88 kNm Me1/bd2 = (280.88)(106)/(400)(650)(650) = 1.66 N/mm2

From Table 2 of SP-16, corresponding to Mu/bd2 = 1.66 N/mm2, we have by linear interpolation pt = 0.5156. So, Ast = 0.5156(400)(650)/100 = 1340.56 mm2. Provide 2-25T and 2-16T = 981 + 402 = 1383 mm2. This gives percentage of tensile reinforcement = 0.532, for which cτ from Table 6.1 of Lesson 13 is 0.488 N/mm2. From Eq.6.29, minimum percentage of tension reinforcement = (0.85/fy)(100) = 0.205 and from sec.6.16.8, the maximum percentage of tension reinforcement is 4.0. So, 2-25T and 2-16T bars satisfy the requirements (Fig. 6.16.4).


Step 4: Longitudinal compression reinforcement Here, in this problem, the numerical value of Mt (= 80.88 kNm) is less than that of Mu (200 kNm). So, as per sec. 6.16.7c, longitudinal compression reinforcement shall not be required. Step 5: Longitudinal side face reinforcement Side face reinforcement shall be provided as the depth of the beam exceeds 450 mm. Providing 2-10 mm diameter bars (area = 157 mm2) at the mid-depth of the beam and one on each face (Fig.6.16.4), the total area required as per sec.6.16.8c, 0.1(400)(300)/100 = 120 mm2 < 157 mm2. Hence o.k. Step 6: Transverse reinforcement

Providing two legged, 10 mm diameter stirrups (area = 157 mm2), we have (Fig.6.16.5) d1 = 700 - 50 - 50 = 600 mm b1 = 400 - 2(25 + 10 + 12.5) = 305 mm From Eq.6.27, we have 1 1 10 87 / ( / ( / 2.5 y sv v u u. f A s T b d ) V d= + )


Using the numerical values of Tu, b1, d1 and Vu, we have …. (1) 0 87 / 339.89 N/mmy sv v. f A s = Again from Eq.6.28, we have …. (2) 0 87 / (1.154 - 0.48) 400 269.6 N/mmy sv v. f A s ≥ ≥ So, Eq.(1) is governing and we get for 2 legged 10 mm stirrups (Asv = 157 mm2), sv = 0.87(415)(157)/339.89 = 166.77 mm Step 7: Check for sv Figure 6.16.4 shows the two legged 10 mm diameter stirrups for which x1 = 340 mm and y1 = 628.5 mm. The maximum spacing sv should be the least of x1, (x1 + y1)/4 and 300 mm (Figs. 6.16.4 and 5). Here, x1 = 340 mm, (x1 + y1)/4 = 242.12 mm. So, provide 2 legged 10 mm T stirrups @ 160 mm c/c. 6.16.10 Practice Questions and Problems with Answers Q.1: Explain the situations when torsional moments remain present in beams

and frames. A.1: The first paragraph of sec. 6.16.2. Q.2: Explain and differentiate between primary and secondary types of torsion. A.2: Third, fourth and fifth paragraphs of sec. 6.16.2. Q.3: Write expressions of equivalent shear and equivalent bending moment. A.3: Eqs.6.22, 24 and 25. Q.4: When do you provide minimum shear reinforcement in beam subjected to

bending moment, shear and torsional moment? A.4: Sec. 6.16.6(c)


Q.5: Explain the situations when do you provide longitudinal tension, compression and side face reinforcement in beam subjected to bending moment, shear and torsional moment.

A.5: Sec. 6.16.7a, b and c parts. Q.6: Illustrate the steps of designing transverse reinforcement in beams

subjected to bending moment, shear and torsional moment. A.6: Sec. 6.16.7d.

Q.7: A reinforced concrete rectangular beam (Fig.6.16.6) of b = 300 mm, d =

600 mm and D = 650 mm is subjected to factored shear force Vu = 70 kN in one section. Assuming the percentage of tensile reinforcement as 0.5 in that section, determine the factored torsional moment that the section can resist if (a) no additional reinforcement for torsion is provided, (b) maximum steel for torsion is provided in that section, and (c) determine the reinforcement needed for the case (b). Assume M 30 concrete, Fe 500 for longitudinal and Fe 415 for transverse reinforcing steel bars.

A.7: (a) When no additional reinforcement for torsion is provided in that section. For M 30 concrete with 0.5 per cent tensile reinforcement, Table 6.1 of Lesson 13 (Table 19 of IS 456) gives cτ = 0.5 N/mm2 and Table 6.2 of Lesson 13 (Table 20 of IS 456 gives maxcτ = 3.5 N/mm2.


Since no additional reinforcement for torsion will be provided, we will take c ττ =ve = 0.5 N/mm2. From Eq.6.23, we have

Ve = )( bdveτ = 0.5(300)(600) N = 90 kN From Eq.6.22, we have Tu = (Ve - Vu) (b/1.6) = (90 - 70) (0.3/1.6) = 3.75 kNm So, that section of the beam can resist factored torsional moment of 3.75 kNm if no additional reinforcement is provided. (b) When maximum steel is provided for torsion in that section. In this case, cmax ττ =ve = 3.5 N/mm2. Using this value in Eq.6.23, we get Ve = )( bdveτ = (3.5)(300)(600) N = 630 kN From Eq.6.22, we get Tu = (Ve - Vu) (b/1.6) = (630 - 70) (0.3/1.6) = 105 kNm This section, therefore, can resist a factored torsional moment of 105 kNm when maximum torsional reinforcement is provided. (c) Determination of maximum torsional reinforcement for case (b) Step 1: Determination of the Mu when 0.5 per cent tensile reinforcement is assumed. From Table 4 of SP-16, for Fe 500 and M 30 with pt = 0.5 per cent, we have Mu = (1.993)(300)(600)2 Nmm = 215.244 kNm (using linear interpolation).


Step 2: Tension and compression reinforcement

From Eqs.6.24 and 6.25, we get Me1 = Mu + Mt = Mu + (Tu/1.7) {1 + (D/b)} = 215.244 + (105/1.7){1 + (650/300)}= 215.244 + 195.588 = 410.832 kNm Here, the numerical value of Mt (= 195.588 kNm) is less than that of Mu (= 215.244 kNm). So, no compression reinforcement is needed. Table 4 of SP-16 is used to determine tension reinforcement with Mu/bd2 = 410.832/(0.3)(0.6)(0.6) = 3.804 N/mm. From Table 4, we get pt = 1.064 by linear interpolation, which gives Ast = 1.064(300)(600)/100 = 1915.2 mm2. Provide 4-25T (1963 mm2) as shown in Fig.6.16.7. Now, pt = 1963(100)/(300)(600) = 1.09, for which cτ = 0.678 N/mm2 (Table 6.1 of Lesson 13). Step 3: Side face reinforcement Since the depth of the beam exceeds 450 mm, we provide side face reinforcement with two 10 mm bars (area = 157 mm2) near the mid-depth of the beam, one on each side to get the spacing of the bar 280 mm (Fig.6.16.7). Area required to satisfy = 0.1(300)(280)/(100) = 84 mm2 < 157 mm2, hence, o.k.


Step 4: Transverse reinforcement

Assuming 2 legged 12 mm dia stirrup (area = 226 mm2) of Fe 415, we have from Fig.6.16.8, d1 = 557 mm and b1 = 201 mm. From Eq.6.27,

111 5.2

87.0

dV

dbT

sAf uu

v

svy +=

= {105(106)/(201)(557)} + {70(103)/(2.5)(557)} = 988.13 N/mm From Eq.6.28 the least value of the above is - ve c( ) bτ τ = (3.5 - 0.678)(300) = 846.6 Nm2. So, from 0.87 fy Asv/sv = 988.13 N/mm, we get sv = 0.87(415)(226)/988.13 = 82.57 mm Step 5: Check for sv Figure 6.16.8 shows the stirrups of 12 mm diameter two legged for which x1 = 238 mm and y1 = 587.5 mm. The maximum spacing should be the least of x1, (x1 + y1)/4 and 300 mm. Here, x1 = 238 mm, (x1 + y1)/4 = 206.375 mm and 300 mm. So, the spacing of 80 mm c/c is o.k. Provide 12 mm, 2 legged stirrups @ 80 mm c/c, as shown in Fig.6.16.8.


6.16.11 References References:
















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi.

6.16.12 Test 16 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. Each question carries five marks.


TQ.1: Explain and differentiate between primary and secondary types of

torsion. (10 marks) A.TQ.1: Third, fourth and fifth paragraphs of sec. 6.16.2. TQ.2: Explain the situations when do you provide longitudinal tension,

compression and side face reinforcement in beam subjected to bending moment, shear and torsional moment. (10 marks)

A.TQ.2: Sec. 6.16.7a, b and c parts. TQ.3: Illustrate the steps of designing transverse reinforcement in beams

subjected to bending moment, shear and torsional moment. (10 marks)

A.TQ.3: Sec. 6.16.7d

TQ.4: The beam of Fig.6.16.9 has factored bending moment Mu = 70 kNm,

factored shear force Vu = 100 kN and factored torsional moment Tu = 60 kNm at one section. Design the reinforcement of that section assuming that the torsion is fully taken by the web. Assume M 30 concrete, Fe 500 for longitudinal and Fe 415 for transverse reinforcing steel bars.

(20 marks) A.TQ.4:


Step 1: Checking of depth of the beam From Eq.6.22, the equivalent shear

Ve = Vu + 1.6(Tu/b) = 100 + 1.6(60/0.3) = 420 kN From Eq.6.23, we get )/ ( bdVeve =τ = 420/(0.3)(0.5) kN/m2 = 2.8 N/mm2

Table 6.2 of Lesson 13 gives maxcτ = 3.5 N/mm2 > veτ , So the depth is satisfying. Step 2: Check if shear reinforcement is required Assuming percentage of tensile steel as 0.5, Table 6.1 of Lesson 13 gives

cτ = 0.5 N/mm2. Hence, both longitudinal and transverse reinforcements shall be provided. Step 3: Longitudinal tension reinforcement

From Eqs.6.24 and 6.25, we have,

Me1 = Mu + Mt = Mu + (Tu/1.7) {1 + (D/b)}


= 70 + (60/1.7) {1 + (550/300)} = 70 + 100 = 170 kNm. Me1/bd2 = 170(106)/(300)(500)(500) = 2.267 N/mm2

Table 4 of SP-16 or M 30, Fe 500 and Mu/bd2 = 2.267 N/mm2, we have by linear interpolation pt = 0.577. Ast = 0.577(300)(500)/100 = 865.5 mm2. Minimum Ast (Eq.6.29) = 0.85 bd/fy = 0.85(300)(500)/(500) = 255 mm2. Maximum Ast = 0.04 bD = 0.04(300)(550) = 6600 mm2. So, Ast = 865.5 mm2 is acceptable. Provide 3-20T (area = 942 mm2) giving pt = (942)(100)/(300)(500) = 0.628% (Fig. 6.16.10). Table 6.1 of Lesson 13 gives cτ = 0.546 N/mm2 (for pt = 0.628%). Step 4: Longitudinal compression reinforcement Here, the numerical value of Mt (= 100 kNm) is greater than that of Mu (= 70 kNm), as computed in Step 3. So, compression reinforcement shall be provided for Me2 = Mt - Mu = 100 - 70 = 30 kNm. Me2/bd2 = 30(106)/(300)(500)(500) = 0.4 N/mm. Table 4 of SP-16 gives pc = 0.093 to have compression steel reinforcement Asc = 0.093(300)(500)/100 = 139.5 mm2. Maximum area compression steel = 0.04 bD = 0.04(300)(550) = 6600 mm2. Hence, Asc = 139.5 mm2 is satisfying. Provide 2-12T (area = 226 mm2) as compression steel (Fig. 6.16.10). Step 5: Side face reinforcement Since the depth of the beam exceeds 450 mm, provide 2-10 mm T (area = 157 mm2) near the mid-depth, one on each side of the beam with maximum spacing = 230 mm (Fig.6.16.9). Area required = 0.1(300)(230)/100 = 69 mm2 < 157 mm2. Hence, two 10 mmT bars as shown in Fig.6.16.10 is o.k. Step 6: Transverse reinforcement Using 10 mm T, 2 legged stirrups, we have d1 = 459 mm and b1 = 210 mm (Fig.6.16.10). From Eq.6.27, we have,

111 5.2

87.0

dV

dbT

sAf uu

v

svy +=


= 60(106)/(210)(459)} + 100(103)/(2.5)(459)} = 709.61 N/mm2

However, Eq.6.28 gives the least value of 0.87 fy As/sv = - ve c( ) bτ τ = (2.8 - 0.546)(300) = 676.2 N/mm. So, 0.87 fy Asv/sv = 709.61 gives sv = 0.87 fy Asv/709.61 = 0.87(415)(157)/709.61 = 79.88 mm Step 7: Checking of sv

Figure 6.16.11 shows x1 = 300 - 2(25 + 5) = 240 mm and y1 = 550 - 30 - 35 = 485 mm. Accordingly, the maximum spacing is the least of x1, (x1 + y1)/4 or 300 mm. Here, the least value is 181.25 mm. Hence sv = 76 mm is acceptable (Fig.6.16.10). 6.16.13 Summary of this Lesson This lesson explains the different situations when beams and frames are subjected to combined bending, shear and torsional moment. Briefly discussing about the analysis of such beams, design of beams as per IS 456 has been illustrated with numerical examples. The requirements of IS 456 have been specified. Solutions of practice and test problems will help the students in understanding the theory and designing the beams subjected to combined bending moment, shear and torsional moment.


Module 7

Limit State of Serviceability


Lesson 17

Limit State of Serviceability


Instruction Objectives:


• explain the need to check for the limit state of serviceability after designing the structures by limit state of collapse,

• differentiate between short- and long-term deflections,

• state the influencing factors to both short- and long-term deflections,

• select the preliminary dimensions of structures to satisfy the requirements

as per IS 456,

• calculate the short- and long-term deflections of designed beams. 7.17.1 Introduction Structures designed by limit state of collapse are of comparatively smaller sections than those designed employing working stress method. They, therefore, must be checked for deflection and width of cracks. Excessive deflection of a structure or part thereof adversely affects the appearance and efficiency of the structure, finishes or partitions. Excessive cracking of concrete also seriously affects the appearance and durability of the structure. Accordingly, cl. 35.1.1 of IS 456 stipulates that the designer should consider all relevant limit states to ensure an adequate degree of safety and serviceability. Clause 35.3 of IS 456 refers to the limit state of serviceability comprising deflection in cl. 35.3.1 and cracking in cl. 35.3.2. Concrete is said to be durable when it performs satisfactorily in the working environment during its anticipated exposure conditions during service. Clause 8 of IS 456 refers to the durability aspects of concrete. Stability of the structure against overturning and sliding (cl. 20 of IS 456), and fire resistance (cl. 21 of IS 456) are some of the other importance issues to be kept in mind while designing reinforced concrete structures. This lesson discusses about the different aspects of deflection of beams and the requirements as per IS 456. In addition, lateral stability of beams is also taken up while selecting the preliminary dimensions of beams. Other requirements, however, are beyond the scope of this lesson. 7.17.2 Short- and Long-term Deflections As evident from the names, short-term deflection refers to the immediate deflection after casting and application of partial or full service loads, while the long-term deflection occurs over a long period of time largely due to shrinkage


and creep of the materials. The following factors influence the short-term deflection of structures: (a) magnitude and distribution of live loads, (b) span and type of end supports, (c) cross-sectional area of the members, (d) amount of steel reinforcement and the stress developed in the reinforcement, (e) characteristic strengths of concrete and steel, and (f) amount and extent of cracking.

The long-term deflection is almost two to three times of the short-term deflection. The following are the major factors influencing the long-term deflection of the structures. (a) humidity and temperature ranges during curing, (b) age of concrete at the time of loading, and

(c) type and size of aggregates, water-cement ratio, amount of compression reinforcement, size of members etc., which influence the creep and shrinkage of concrete.

7.17.3 Control of Deflection Clause 23.2 of IS 456 stipulates the limiting deflections under two heads as given below: (a) The maximum final deflection should not normally exceed span/250 due to all loads including the effects of temperatures, creep and shrinkage and measured from the as-cast level of the supports of floors, roof and all other horizontal members. (b) The maximum deflection should not normally exceed the lesser of span/350 or 20 mm including the effects of temperature, creep and shrinkage occurring after erection of partitions and the application of finishes. It is essential that both the requirements are to be fulfilled for every structure. 7.17.4 Selection of Preliminary Dimensions The two requirements of the deflection are checked after designing the members. However, the structural design has to be revised if it fails to satisfy any one of the two or both the requirements. In order to avoid this, IS 456 recommends the guidelines to assume the initial dimensions of the members which will generally satisfy the deflection limits. Clause 23.2.1 stipulates different span to effective depth ratios and cl. 23.3 recommends limiting slenderness of


beams, a relation of b and d of the members, to ensure lateral stability. They are given below: (A) For the deflection requirements Different basic values of span to effective depth ratios for three different support conditions are prescribed for spans up to 10 m, which should be modified under any or all of the four different situations: (i) for spans above 10 m, (ii) depending on the amount and the stress of tension steel reinforcement, (iii) depending on the amount of compression reinforcement, and (iv) for flanged beams. These are furnished in Table 7.1. (B) For lateral stability The lateral stability of beams depends upon the slenderness ratio and the support conditions. Accordingly cl. 23.3 of IS code stipulates the following: (i) For simply supported and continuous beams, the clear distance between the lateral restraints shall not exceed the lesser of 60b or 250b2/d, where d is the effective depth and b is the breadth of the compression face midway between the lateral restraints. (ii) For cantilever beams, the clear distance from the free end of the cantilever to the lateral restraint shall not exceed the lesser of 25b or 100b2/d. Table 7.1 Span/depth ratios and modification factors Sl. No.

Items Cantilever Simply supported

Continuous

1 Basic values of span to effective depth ratio for spans up to 10 m

7 20 26

2 Modification factors for spans > 10 m

Not applicable as deflection calculations are to be done.

Multiply values of row 1 by 10/span in metres.

3 Modification factors depending on area and stress of steel

Multiply values of row 1 or 2 with the modification factor from Fig.4 of IS 456.

4 Modification factors depending as area of compression steel

Further multiply the earlier respective value with that obtained from Fig.5 of IS 456.

5 Modification factors for flanged beams

(i) Modify values of row 1 or 2 as per Fig.6 of IS 456. (ii) Further modify as per row 3 and/or 4 where


reinforcement percentage to be used on area of section equal to bf d.

7.17.5 Calculation of Short-Term Deflection Clause C-2 of Annex C of IS 456 prescribes the steps of calculating the short-term deflection. The code recommends the usual methods for elastic deflections using the short-term modulus of elasticity of concrete Ec and effective moment of inertia Ieff given by the following equation:

; but 1 2 - ( 1

reff r eff

r w

IgrI I I

. M / M )( z / d )( x / d )( b / b )= ≤

−I≤

(7.1) where Ir = moment of inertia of the cracked section,

Mr = cracking moment equal to (fcr Igr)/yt , where fcr is the modulus of rupture of concrete, Igr is the moment of inertia of the gross section about the centroidal axis neglecting the reinforcement, and yt is the distance from centroidal axis of gross section, neglecting the reinforcement, to extreme fibre in tension,

M = maximum moment under service loads, z = lever arm, x = depth of neutral axis, d = effective depth, bw = breadth of web, and b = breadth of compression face. For continuous beams, however, the values of Ir, Igr and Mr are to be modified by the following equation:

o121

1 ) - (1 2 XkXXkX e +⎥⎦

⎤⎢⎣⎡ +

=

(7.2) where Xe = modified value of X, X1, X2 = values of X at the supports,


Xo = value of X at mid span, k1 = coefficient given in Table 25 of IS 456 and in Table 7.2 here, and X = value of Ir, Igr or Mr as appropriate. Table 7.2 Values of coefficient k1

k1 0.5 or

less

0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4

k2

0 0.03 0.08 0.16 0.30 0.50 0.73 0.91 0.97 1.0

Note: k2 is given by (M1 + M2)/(MF1 + MF2), where M1 and M2 = support moments, and MF1 and MF2 = fixed end moments. 7.17.6 Deflection due to Shrinkage Clause C-3 of Annex C of IS 456 prescribes the method of calculating the deflection due to shrinkage csα from the following equation: (7.3)

2cs3 lkcs ψα =

where k3 is a constant which is 0.5 for cantilevers, 0.125 for simply supported members, 0.086 for members continuous at one end, and 0.063 for fully continuous members; csψ is shrinkage curvature equal to k4 csε /D where csε is the ultimate shrinkage strain of concrete. For csε , cl. 6.2.4.1 of IS 456 recommends an approximate value of 0.0003 in the absence of test data. 1.0 - 0.25for 1.0, /) - 0.72( 4 <≤≤= cttct pppppk

1.0 - for 1.0, /) - 0.65( ≥≤= cttct ppppp (7.4) where pt = 100Ast/bd and pc = 100Asc/bd, D is the total depth of the section, and l is the length of span.


7.17.7 Deflection Due to Creep Clause C-4 of Annex C of IS 456 stipulates the following method of calculating deflection due to creep. The creep deflection due to permanent loads

)( permccα is obtained from the following equation: 1(perm))(1)( - ααα permccpermcc = (7.5) where )(1 permccα = initial plus creep deflection due to permanent loads obtained

using an elastic analysis with an effective modulus of elasticity,

), 1/( θθ+= cce EE being the creep coefficient, and )(1 permα = short-term deflection due to permanent loads using Ec. 7.17.8 Numerical Problems

Problem 1: Figures 7.17.1 and 2 present the cross-section and the tensile steel of a simply supported T-beam of 8 m span using M 20 and Fe 415 subjected to dead


load of 9.3 kN/m and imposed loads of 10.7 kN/m at service. Calculate the short- and long-term deflections and check the requirements of IS 456. Solution 1: Step 1: Properties of plain concrete section Taking moment of the area about the bottom of the beam

mm 429.48 300)(100) - (2234 (300)(600)

550)300)(100)( - (2234 (300)(300)(600) =++

=ty

49333

mm (10) (11.384) 3

)1934(70.52 - 3

2)2234(170.5 3

)300(429.48 =+=grI

This can also be computed from SP-16 as explained below: Here, bf /bw = 7.45, Df /D = 0.17. Using these values in chart 88 of SP-16, we get k1 = 2.10. Igr = k1bw D3/12 = (2.10)(300)(600)3/12 = (11.384)(10)9 mm4

Step 2: Properties of the cracked section (Fig.7.17.2)

ckcr ff 0.7 = (cl. 6.2.2 of IS 456) = 3.13 N/mm2


Mr = fcr Igr /yt = 3.13(11.384)(10)9/429.48 = 82.96 kNm Es = 200000 N/mm2

Ec = 5000 ckf (cl. 6.2.3.1 of IS 456) = 22360.68 N/mm2

m = Es /Ec = 8.94 Taking moment of the compressive concrete and tensile steel about the neutral axis, we have (Fig.7.17.2) bf x2/2 = m Ast (d – x) gives (2234)(x2/2) = (8.94)(1383)(550 – x) or x2 + 11.07 x – 6087.92 = 0 . Solving the equation, we get x = 72.68 mm. z = lever arm = d – x/3 = 525.77 mm

4923

mm 3.106(10) 72.68) - (5508.94(1383) 3

)2234(72.68 =+=rI

M = wl2/8 = (9.3 + 10.7)(8)(8)/8 = 160 kNm

)( ) - (1 - 2.1

bb

dx

dz

MM

IIwr

reff = …. (Eq. 7.1)

= rr II 0.875

)2234300( )

55072.68 - (1 )

550525.77( )

16082.96( - 1.2

= . But greffr III ≤≤

So, Ieff = Ir = 3.106(10)9 mm4. Step 3: Short-term deflection (sec. 7.17.5) ckc fE 5000 = (cl. 6.2.3.1 of IS 456) = 22360.68 N/mm2

Short-term deflection = (5/384) wl4/EcIeff = (5)(20)(8)4(1012)/(384)(22360.68)(3.106)(109) = 15.358 mm (1)


Step 4: Deflection due to shrinkage (sec. 7.17.6) 0.6599 0.84 0.72(0.84) /) - 0.72( 4 === tct pppk -7

cs4 3.2995(10) .0003)/600(0.6599)(0 / === Dkcs εψ k3 = 0.125 (from sec. 7.17.6) (Eq. 7.3) = (0.125)(3.2995)(10)2

cs3 lkcs ψα = -7(64)(106) = 2.64 mm (2) Step 5: Deflection due to creep (sec. 7.17.7) Equation 7.5 reveals that the deflection due to creep )( permccα can be obtained after calculating )(1 permccα and )(1 permα . We calculate )(1 permccα in the next step. Step 5a: Calculation of )(1 permccα Assuming the age of concrete at loading as 28 days, cl. 6.2.5.1 of IS 456 gives θ = 1.6. So, Ecc = Ec /(1 + θ ) = 22360.68/(1 + 1.6) = 8600.2615 N/mm2 and m = Es /Ecc = 200000/8600.2615 = 23.255 Step 5b: Properties of cracked section


Taking moment of compressive concrete and tensile steel about the neutral axis (assuming at a distance of x from the bottom of the flange as shown in Fig.7.17.3): 2234(100)(50 + x ) = (23.255)(1383)(450 - x ) or x = 12.92 mm which gives x = 112.92 mm. Accordingly, z = lever arm = d – x/3 = 512.36 mm. Ir = 2234(100)3/12 + 2234(100)(62.92)2 + 23.255(1383)(550 – 112.92)2

+ 300(12.92)3/3 = 7.214(109) mm4

Mr = 82.96 kNm (see Step 2) M = wperm l2/8 = 9.3(8)(8)/8 = 74.4 kNm.

rr

eff III 0.918 )

2234300( )

550112.92 - (1 )

550512.36( )

74.482.96( - )2.1(

==

However, to satisfy Ir ≤ Ieff ≤ Igr, Ieff should be equal to Igr. So, Ieff = Igr = 11.384(109). For the value of Igr please see Step 1. Step 5c: Calculation of )(1 permccα )(1 permccα = 5wl4/384(Ecc)(Ieff) = 5(9.3)(8)4(10)12/384(8600.2615)(11.384)(109) = 5.066 mm (3) Step 5d: Calculation of )(1 permα )(1 permα = 5wl4/384(Ec)(Ieff) = 5(9.3)(8)4(10)12/384(22360.68)(11.384)(109) = 1.948 mm (4) Step 5e: Calculation of deflection due to creep )( permccα = )(1 permccα - )(1 permα


= 5.066 – 1.948 = 3.118 mm (5) It is important to note that the deflection due to creep )( permccα can be obtained even without computing )(1 permccα . The relationship of )( permccα and

)(1 permα is given below. )( permccα = )(1 permccα - )(1 permα = {5wl4/384(Ec)(Ieff)} {(Ec /Ecc) – 1} = )(1 permα (θ ) Hence, the deflection due to creep, for this problem is: )( permccα = )(1 permα (θ ) = 1.948(1.6) = 3.116 mm Step 6: Checking of the requirements of IS 456 The two requirements regarding the control of deflection are given in sec. 7.17.3. They are checked in the following: Step 6a: Checking of the first requirement The maximum allowable deflection = 8000/250 = 32 mm

The actual final deflection due to all loads

= 15.358 (see Eq.1 of Step 3) + 2.64 (see Eq.2 of Step 4) + 3.118 (see Eq.5 of Step 5e) = 21.116 mm < 32 mm. Hence,

o.k. Step 6b: Checking of the second requirement The maximum allowable deflection is the lesser of span/350 or 20 mm. Here, span/350 = 22.86 mm. So, the maximum allowable deflection = 20 mm. The actual final deflection = 1.948 (see Eq.4 of Step 5d) + 2.64 (see Eq.2 of Step 4) + 3.118 (see Eq.5 of step 5e) = 7.706 mm < 20 mm. Hence, o.k. Thus, both the requirements of cl.23.2 of IS 456 and as given in sec. 7.17.3 are satisfied.


7.17.9 Practice Questions and Problems with Answers Q.1: Why is it essential to check the structures, designed by the limit state of

collapse, by the limit state of serviceability? A.1: See sec. 7.17.1. Q.2: Explain short- and long-term deflections and the respective influencing

factors of them. A.2: See sec. 7.17.2. Q.3: State the stipulations of IS 456 regarding the control of deflection. A.3: See sec. 7.17.3. Q.4: How would you select the preliminary dimensions of structures to satisfy (i)

the deflection requirements, and (ii) the lateral stability ? A.4: See secs. 7.17.4 A for (i) and B for (ii). Q.5: Check the preliminary cross-sectional dimensions of Problem 1 of sec.

7.17.8 (Fig.7.17.1) if they satisfy the requirements of control of deflection. The spacing of the beam is 3.5 m c/c. Other data are the same as those of Problem 1 of sec. 7.17.8.

A.5: Step 1: Check for the effective width bf = lo /6 + bw + 6Df or spacing of the beam, whichever is less. Here, bf = (8000/6) + 300 + 6(100) = 2234 < 3500. Hence, bf = 2234 mm is o.k. Step 2: Check for span to effective depth ratio (i) As per row 1 of Table 7.1, the basic value of span to effective depth ratio is 20. (ii) As per row 2 of Table 7.1, the modification factor is 1 since the span 8 m < 10 m. (iii) As per row 5 of Table 7.1, the modification factor for the flanged beam is to be obtained from Fig. 6 of IS 456 for which the ratio of web width to flange width


= 300/2234 = 0.134. Figure 6 of IS 456 gives the modification factor as 0.8. So, the revised span to effective depth ratio = 20(0.8) = 16. (iv) Row 3 of Table 7.1 deals with the area and stress of tensile steel. At the preliminary stage these values are to be assumed. However, for this problem the area of steel is given as 1383 mm2 (2-25T + 2-16T), for which pt = Ast(100)/bf d = 1383(100)/(2234)(550) = 0.112. fs = 0.58 fy (area of cross-section of steel required)/(area of cross-section of steel provided) = 0.58(415)(1) = 240.7 (assuming that the provided steel is the same as required, which is a rare case). Figure 4 of IS 456 gives the modification factor as 1.8. So, the revised span to effective depth ratio = 16(1.8) = 28.8. (v) Row 4 is concerning the amount of compression steel. Here, compression steel is not there. So, the modification factor = 1. Therefore, the final span to effective depth ratio = 28.8.

Accordingly, effective depth of the beam = 8000/28.8 = 277.8 mm < 550 mm. Hence, the dimensions of the cross-section are satisfying the requirements. 7.17.9 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 7.17.11 Test 17 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Explain short- and long-term deflections and the respective influencing

factors of them. (10 marks)

A.TQ.1: See sec. 7.17.2.


TQ.2: Check the preliminary dimensions of a singly reinforced rectangular cantilever beam of span 4 m (Fig.7.17.4) using M 20 and Fe 415. (15 marks)

A.TQ.2: (i) From row 1 of Table 7.1, the basic value of span to effective depth ratio is 7. (ii) Modification factor for row 2 is 1 as this is a singly reinforced beam. (iii) Assuming pt as 0.6 and area of steel to be provided is the same as area of steel required, fs = 0.58(415(1) = 240.7 N/mm2. From Fig. 4 of IS 456, the modification factor = 1.18. Hence, the revised span to effective depth ratio is 7(1.18) = 8.26. (iv) Modification factors for rows 4 and 5 are 1 as there is no compression steel and this being a rectangular beam. Hence, the preliminary effective depth needed = 4000/8.26 = 484.26 mm < 550 mm. Hence, o.k. TQ.3: Determine the tensile steel of the cantilever beam of TQ 2 (Fig. 7.17.4)

subjected to service imposed load of 11.5 kN/m using M 20 and Fe 415. Use Sp-16 for the design. Calculate short- and long-term deflections and check the requirements of IS 456 regarding the deflection. (25 marks)

A.TQ.3: Determination of tensile steel of the beam using SP-16: Dead load of the beam = 0.3(0.6)(25) kN/m = 4.5 kN/m Service imposed loads = 11.5 kN/m Total service load = 16.0 kN/m Factored load = 16(1.5) = 24 kN/m Mu = 24(4)(4)/2 = 192 kNm For this beam of total depth 600 mm, let us assume d = 550 mm. Mu /bd2 = 192/(0.3)(0.55)(0.55) = 2115.70 kN/m2

Table 2 of SP-16 gives the corresponding pt = 0.678 + 0.007(0.015)/0.02 = 0.683 Again, for Mu per metre run as 192/0.3 = 640 kNm/m, chart 15 of SP-16 gives pt = 0.68 when d = 550 mm.


With pt = 0.683, Ast = 0.683(300)(500)/100 = 1126.95 mm2. Provide 4-20T to have 1256 mm2. This gives provided pt = 0.761%. Calculation of deflection Step 1: Properties of concrete section yt = D/2 = 300 mm, Igr = bD3/12 = 300(600)3/12 = 5.4(109) mm4

Step 2: Properties of cracked section

fcr = 0.7 20 (cl. 6.2.2 of IS 456) = 3.13 N/mm2

yt = 300 mm Mr = fcr Igr /yt = 3.13(5.4)(109)/300 = 5.634(107) Nmm Es = 200000 N/mm2

Ec = 5000 ckf (cl. 6.2.3.1 of IS 456) = 22360.68 N/mm2

m = Es /Ec = 8.94 Taking moment of the compressive concrete and tensile steel about the neutral axis (Fig.7.17.5): 300 x2/2 = (8.94)(1256)(550 – x) or x2 + 74.86 x – 41171.68 = 0


This gives x = 168.88 mm and z = d – x/3 = 550 – 168.88/3 = 493.71 mm. Ir = 300(168.88)3/3 + 8.94(1256)(550 – 168.88)2 = 2.1126(109) mm4

M = wl2/2 = 20(4)(4)/2 = 160 kNm

49 mm )(10 2.1548 1.02 )(1 )

550168.88 - (1 )

550493.71( )

165.634( - )2.1(

=== rr

eff III

This satisfies Ir ≤ Ieff ≤ Igr. So, Ieff = 2.1548(109) mm4. Step 3: Short-term deflection (sec. 7.17.5) Ec = 22360.68 N/mm2 (cl. 6.2.3.1 of IS 456) Short-term deflection = wl4/8EcIeff

= 20(44)(1012)/8(22360.68)(2.1548)(109) = 13.283 mm So, short-term deflection = 13.283 mm (1) Step 4: Deflection due to shrinkage (sec. 7.17.6) 0.664 761.0/)761.00.72( 4 ==k -7

cs4 3.32(10) 0003)/600(0.664)(0. / === Dkcs εψ k3 = 0.5 (from sec. 7.17.6) = (0.5)(3.32)(10)2

cs3 lkcs ψα = -7(16)(106) = 2.656 mm (2) Step 5: Deflection due to creep (sec. 7.17.7) Step 5a: Calculation of )(1 permccα Assuming the age of concrete at loading as 28 days, cl. 6.2.5.1 of IS 456 gives


θ = 1.6

So, Ecc = Ec /(1 + θ ) = 8600.2615 N/mm2

m = Es /Ecc = 200000/8600.2615 = 23.255 Step 5b: Properties of cracked section

From Fig.7.17.6, taking moment of compressive concrete and tensile steel about the neutral axis, we have: 300 x2/2 = (23.255)(1256)(550 - x) or x2 + 194.72 x – 107097.03 = 0 solving we get x = 244.072 mm z = d – x/3 = 468.643 mm Ir = 300(244.072)3/3 + (23.255)(1256)(550 – 468.643)2

= 1.6473(10)9 mm4


Mr = 5.634( 107) Nmm (see Step 2) M = wperm l2/2 = 4.5(42)/2 = 36 kNm

49 mm )3.5888(10 2.1786 )(1 )

550244.072 - (1 )

550468.643( )

3.65.634( - 2.1

=== rr

eff III

Since this satisfies Ir ≤ Ieff ≤ Igr, we have, Ieff = 3.5888(109) mm4. For the value of Igr please see Step 1. Step 5c: Calculation of )(1 permccα )(1 permccα = (wperm)( l4)/(8Ecc Ieff) = 4.5(4)4(10)12/8(8600.2615)(3.5888)(109) = 4.665 mm (3) Step 5d: Calculation of )(1 permα )(1 permα = (wperm)( l4)/(8Ec Ieff) = 4.5(4)4(10)12/8(22360.68)(3.5888)(109) = 1.794 mm (4) Step 5e: Calculation of deflection due to creep )( permccα = )(1 permccα - )(1 permα = 4.665 – 1.794 = 2.871 mm (5) Moreover: )( permccα = )(1 permccα (θ ) gives )( permccα = 1.794(1.6) = 2.874 mm. Step 6: Checking of the two requirements of IS 456 Step 6a: First requirement Maximum allowable deflection = 4000/250 = 16 mm

The actual deflection = 13.283 (Eq.1 of Step 3) + 2.656 (Eq.2 of Step 4)


+ 2.871 (Eq.5 of Step 5e) = 18.81 > Allowable 16 mm.

Step 6b: Second requirement The allowable deflection is lesser of span/350 or 20 mm. Here, span/350 = 11.428 mm is the allowable deflection. The actual deflection = 1.794 (Eq.4 of Step 5d) + 2.656 (Eq.2 of Step 4) + 2.871 (Eq.5 of step 5e) = 7.321 mm < 11.428 mm. Remarks: Though the second requirement is satisfying, the first requirement is not satisfying. However the extra deflection is only 2.81 mm, which can be made up by giving camber instead of revising the section. 7.17.12 Summary of this Lesson This lesson illustrates the importance of checking the structures for the limit state of serviceability after designing by the limit state of collapse. The short- and long-term deflections along with their respective influencing factors are explained. The code requirements for the control of deflection and the necessary guidelines for the selection of dimensions of cross-section are stated. Numerical examples, solved as illustrative example and given in the practice problem and test will help the students in understanding the calculations clearly for their application in the design problems.


Module 8

Reinforced Concrete Slabs


Lesson 18

One-way Slabs



• state the names of different types of slabs used in construction, • identify one-way and two-way slabs stating the limits of ly /lx ratios for one

and two-way slabs,

• explain the share of loads by the supporting beams of one- and two-way slabs when subjected to uniformly distributed vertical loads,

• explain the roles of the total depth in resisting the bending moments,

shear force and in controlling the deflection,

• state the variation of design shear strength of concrete in slabs of different depths with identical percentage of steel reinforcement,

• assume the depth of slab required for the control of deflection for different

support conditions,

• determine the positive and negative bending moments and shear force,

• determine the amount of reinforcing bars along the longer span,

• state the maximum diameter of a bar that can be used in a particular slab of given depth,

• decide the maximum spacing of reinforcing bars along two directions of

one-way slab,

• design one-way slab applying the design principles and following the stipulated guidelines of IS 456,

• draw the detailing of reinforcing bars of one-way slabs after the design.


8.18.1 Introduction


Slabs, used in floors and roofs of buildings mostly integrated with the supporting beams, carry the distributed loads primarily by bending. It has been mentioned in sec. 5.10.1 of Lesson 10 that a part of the integrated slab is considered as flange of T- or L-beams because of monolithic construction. However, the remaining part of the slab needs design considerations. These slabs are either single span or continuous having different support conditions like fixed, hinged or free along the edges (Figs.8.18.1a,b and c). Though normally these slabs are horizontal, inclined slabs are also used in ramps, stair cases and inclined roofs (Figs.8.18.2 and 3). While square or rectangular plan forms are normally used, triangular, circular and other plan forms are also needed for different functional requirements. This lesson takes up horizontal and rectangular /square slabs of buildings supported by beams in one or both directions and subjected to uniformly distributed vertical loadings. The other types of slabs, not taken up in this module, are given below. All these slabs have additional requirements depending on the nature and magnitude of loadings in respective cases.

(a) horizontal or inclined bridge and fly over deck slabs carrying heavy concentrated loads,

(b) horizontal slabs of different plan forms like triangular, polygonal or

circular,

(c) flat slabs having no beams and supported by columns only,

(d) inverted slabs in footings with or without beams,

(e) slabs with large voids or openings,

(f) grid floor and ribbed slabs.


8.18.2 One-way and Two-way Slabs

Figures 8.18.4a and b explain the share of loads on beams supporting solid slabs along four edges when vertical loads are uniformly distributed. It is evident from the figures that the share of loads on beams in two perpendicular directions depends upon the aspect ratio ly /lx of the slab, lx being the shorter span. For large values of ly, the triangular area is much less than the trapezoidal area (Fig.8.18.4a). Hence, the share of loads on beams along shorter span will gradually reduce with increasing ratio of ly /lx. In such cases, it may be said that the loads are primarily taken by beams along longer span. The deflection profiles of the slab along both directions are also shown in the figure. The deflection profile is found to be constant along the longer span except near the edges for the slab panel of Fig.8.18.4a. These slabs are designated as one-way slabs as


they span in one direction (shorter one) only for a large part of the slab when ly /lx > 2. On the other hand, for square slabs of ly /lx = 1 and rectangular slabs of ly /lx up to 2, the deflection profiles in the two directions are parabolic (Fig.8.18.4b). Thus, they are spanning in two directions and these slabs with ly /lx up to 2 are designated as two-way slabs, when supported on all edges. It would be noted that an entirely one-way slab would need lack of support on short edges. Also, even for ly /lx < 2, absence of supports in two parallel edges will render the slab one-way. In Fig. 8.18.4b, the separating line at 45 degree is tentative serving purpose of design. Actually, this angle is a function of ly /lx . This lesson discusses the analysis and design aspects of one-way slabs. The two-way slabs are taken up in the next lesson. 8.18.3 Design Shear Strength of Concrete in Slabs Experimental tests confirmed that the shear strength of solid slabs up to a depth of 300 mm is comparatively more than those of depth greater than 300 mm. Accordingly, cl.40.2.1.1 of IS 456 stipulates the values of a factor k to be multiplied with cτ given in Table 19 of IS 456 for different overall depths of slab. Table 8.1 presents the values of k as a ready reference below: Table 8.1 Values of the multiplying factor k Overall depth of slab (mm)

300 or more

275 250 225 200 175 150 or less

k 1.00 1.05 1.10 1.15 1.20 1.25 1.30

Thin slabs, therefore, have more shear strength than that of thicker slabs. It is the normal practice to choose the depth of the slabs so that the concrete can resist the shear without any stirrups for slab subjected to uniformly distributed loads. However, for deck slabs, culverts, bridges and fly over, shear reinforcement should be provided as the loads are heavily concentrated in those slabs. Though, the selection of depth should be made for normal floor and roof slabs to avoid stirrups, it is essential that the depth is checked for the shear for these slabs taking due consideration of enhanced shear strength as discussed above depending on the overall depth of the slabs.


8.18.4 Structural Analysis As explained in sec. 8.18.2, one-way slabs subjected to mostly uniformly distributed vertical loads carry them primarily by bending in the shorter direction. Therefore, for the design, it is important to analyse the slab to find out the bending moment (both positive and negative) depending upon the supports. Moreover, the shear forces are also to be computed for such slabs. These internal bending moments and shear forces can be determined using elastic method of analysis considering the slab as beam of unit width i.e. one metre (Fig.8.18.1a). However, these values may also be determined with the help of the coefficients given in Tables 12 and 13 of IS 456 in cl.22.5.1. It is worth mentioning that these coefficients are applicable if the slab is of uniform cross-section and subjected to substantially uniformly distributed loads over three or more spans and the spans do not differ by more than fifteen per cent of the longer span. It is also important to note that the average of the two values of the negative moment at the support should be considered for unequal spans or if the spans are not equally loaded. Further, the redistribution of moments shall not be permitted to the values of moments obtained by employing the coefficients of bending moments as given in IS 456. For slabs built into a masonry wall developing only partial restraint, the negative moment at the face of the support should be taken as Wl/24, where W is the total design loads on unit width and l is the effective span. The shear coefficients, given in Table 13 of IS 456, in such a situation, may be increased by 0.05 at the end support as per cl.22.5.2 of IS 456. 8.18.5 Design Considerations The primary design considerations of both one and two-way slabs are strength and deflection. The depth of the slab and areas of steel reinforcement are to be determined from these two aspects. The detailed procedure of design of one-way slab is taken up in the next section. However, the following aspects are to be decided first. (a) Effective span (cl.22.2 of IS 456) The effective span of a slab depends on the boundary condition. Table 8.2 gives the guidelines stipulated in cl.22.2 of IS 456 to determine the effective span of a slab. Table 8.2 Effective span of slab (cl.22.2 of IS 456) Sl.No. Support condition Effective span

1 Simply supported not built integrally with its supports

Lesser of (i) clear span + effective depth of slab, and (ii) centre to centre of supports


2 Continuous when the width of the support is < 1/12th of clear span

Do

3 Continuous when the width of the support is > lesser of 1/12th of clear span or 600 mm (i) for end span with one end fixed and the other end continuous or for intermediate spans, (ii) for end span with one end free and the other end continuous, (iii) spans with roller or rocker bearings.

(i) Clear span between the supports (ii) Lesser of (a) clear span + half the effective depth of slab, and (b) clear span + half the width of the discontinuous support (iii) The distance between the centres of bearings

4 Cantilever slab at the end of a continuous slab

Length up to the centre of support

5 Cantilever span Length up to the face of the support + half the effective depth

6 Frames Centre to centre distance (b) Effective span to effective depth ratio (cls.23.2.1a-e of IS 456) The deflection of the slab can be kept under control if the ratios of effective span to effective depth of one-way slabs are taken up from the provisions in cl.23.2.1a-e of IS 456. These stipulations are for the beams and are also applicable for one-way slabs as they are designed considering them as beam of unit width. These provisions are explained in sec.3.6.2.2 of Lesson 6. (c) Nominal cover (cl.26.4 of IS 456) The nominal cover to be provided depends upon durability and fire resistance requirements. Table 16 and 16A of IS 456 provide the respective values. Appropriate value of the nominal cover is to be provided from these tables for the particular requirement of the structure. (d) Minimum reinforcement (cl.26.5.2.1 of IS 456) Both for one and two-way slabs, the amount of minimum reinforcement in either direction shall not be less than 0.15 and 0.12 per cents of the total cross-sectional area for mild steel (Fe 250) and high strength deformed bars (Fe 415 and Fe 500)/welded wire fabric, respectively.


(e) Maximum diameter of reinforcing bars (cl.26.5.2.2) The maximum diameter of reinforcing bars of one and two-way slabs shall not exceed one-eighth of the total depth of the slab. (f) Maximum distance between bars (cl.26.3.3 of IS 456) The maximum horizontal distance between parallel main reinforcing bars shall be the lesser of (i) three times the effective depth, or (ii) 300 mm. However, the same for secondary/distribution bars for temperature, shrinkage etc. shall be the lesser of (i) five times the effective depth, or (ii) 450 mm. 8.18.6 Design of One-way Slabs The procedure of the design of one-way slab is the same as that of beams. However, the amounts of reinforcing bars are for one metre width of the slab as to be determined from either the governing design moments (positive or negative) or from the requirement of minimum reinforcement. The different steps of the design are explained below. Step 1: Selection of preliminary depth of slab The depth of the slab shall be assumed from the span to effective depth ratios as given in section 3.6.2.2 of Lesson 6 and mentioned here in sec.8.18.5b. Step 2: Design loads, bending moments and shear forces The total factored (design) loads are to be determined adding the estimated dead load of the slab, load of the floor finish, given or assumed live loads etc. after multiplying each of them with the respective partial safety factors. Thereafter, the design positive and negative bending moments and shear forces are to be determined using the respective coefficients given in Tables 12 and 13 of IS 456 and explained in sec.8.18.4 earlier. Step 3: Determination/checking of the effective and total depths of slabs The effective depth of the slab shall be determined employing Eq.3.25 of sec.3.5.6 of Lesson 5 and is given below as a ready reference here, Mu,lim = R,lim bd2 …. (3.25) where the values of R,lim for three different grades of concrete and three different grades of steel are given in Table 3.3 of Lesson 5 (sec.3.5.6). The value of b shall be taken as one metre.


The total depth of the slab shall then be determined adding appropriate nominal cover (Table 16 and 16A of cl.26.4 of IS 456) and half of the diameter of the larger bar if the bars are of different sizes. Normally, the computed depth of the slab comes out to be much less than the assumed depth in Step 1. However, final selection of the depth shall be done after checking the depth for shear force. Step 4: Depth of the slab for shear force Theoretically, the depth of the slab can be checked for shear force if the design shear strength of concrete is known. Since this depends upon the percentage of tensile reinforcement, the design shear strength shall be assumed considering the lowest percentage of steel. The value of cτ shall be modified after knowing the multiplying factor k from the depth tentatively selected for the slab in Step 3. If necessary, the depth of the slab shall be modified. Step 5: Determination of areas of steel Area of steel reinforcement along the direction of one-way slab should be determined employing Eq.3.23 of sec.3.5.5 of Lesson 5 and given below as a ready reference. Mu = 0.87 fy Ast d {1 – (Ast)(fy)/(fck)(bd)} …. (3.23) The above equation is applicable as the slab in most of the cases is under-reinforced due to the selection of depth larger than the computed value in Step 3. The area of steel so determined should be checked whether it is at least the minimum area of steel as mentioned in cl.26.5.2.1 of IS 456 and explained in sec.8.18.5d. Alternatively, tables and charts of SP-16 may be used to determine the depth of the slab and the corresponding area of steel. Tables 5 to 44 of SP-16 covering a wide range of grades of concrete and Chart 90 shall be used for determining the depth and reinforcement of slabs. Tables of SP-16 take into consideration of maximum diameter of bars not exceeding one-eighth the depth of the slab. Zeros at the top right hand corner of these tables indicate the region where the percentage of reinforcement would exceed pt,lim. Similarly, zeros at the lower left and corner indicate the region where the reinforcement is less than the minimum stipulated in the code. Therefore, no separate checking is needed for the allowable maximum diameter of the bars or the computed area of steel exceeding the minimum area of steel while using tables and charts of SP-16. The amount of steel reinforcement along the large span shall be the minimum amount of steel as per cl.26.5.2.1 of IS 456 and mentioned in sec.8.18.5d earlier.


Step 6: Selection of diameters and spacings of reinforcing bars (cls.26.5.2.2 and 26.3.3 of IS 456)

The diameter and spacing of bars are to be determined as per cls.26.5.2.2 and 26.3.3 of IS 456. As mentioned in Step 5, this step may be avoided when using the tables and charts of SP-16. 8.18.7 Detailing of Reinforcement


Figures 8.18.5a and b present the plan and section 1-1 of one-way continuous slab showing the different reinforcing bars in the discontinuous and continuous ends (DEP and CEP, respectively) of end panel and continuous end of adjacent panel (CAP). The end panel has three bottom bars B1, B2 and B3 and four top bars T1, T2, T3 and T4. Only three bottom bars B4, B5 and B6 are shown in the adjacent panel. Table 8.3 presents these bars mentioning the respective zone of their placement (DEP/CEP/CAP), direction of the bars (along x or y) and the resisting moment for which they shall be designed or if to be provided on the basis of minimum reinforcement. These bars are explained below for the three types of ends of the two panels. Table 8.3 Steel bars of one-way slab (Figs.8.18.5a and b) Sl.No. Bars Panel Along Resisting moment

1

2

B1, B2

B3

DEP

DEP

x y

+ 0.5 Mx for each, Minimum steel

3

4

B4, B5

B6

CAP

CAP

x y

+ 0.5 Mx for each, Minimum steel

5

6

T1, T2

T3

CEP

DEP

x x

- 0.5 Mx for each, + 0.5 Mx

7 T4

DEP y Minimum steel

Notes: (i) DEP = Discontinuous End Panel (ii) CEP = Continuous End Panel (iii) CAP = Continuous Adjacent Panel (i) Discontinuous End Panel (DEP)

• Bottom steel bars B1 and B2 are alternately placed such that B1 bars are curtailed at a distance of 0.25 lx1 from the adjacent support and B2 bars are started from a distance of 0.15lx1 from the end support. Thus, both B1 and B2 bars are present in the middle zone covering 0.6lx1, each of which is designed to resist positive moment 0.5Mx. These bars are along the direction of x and are present from one end to the other end of ly.

• Bottom steel bars B3 are along the direction of y and cover the entire

span lx1 having the minimum area of steel. The first bar shall be placed at a distance not exceeding s/2 from the left discontinuous support, where s is the spacing of these bars in y direction.

• Top bars T3 are along the direction of x for resisting the negative

moment which is numerically equal to fifty per cent of positive Mx. These


bars are continuous up to a distance of 0.1lx1 from the centre of support at the discontinuous end.

• Top bars T4 are along the direction of y and provided up to a distance of

0.1lx1 from the centre of support at discontinuous end. These are to satisfy the requirement of minimum steel.

(ii) Continuous End Panel (CEP)

• Top bars T1 and T2 are along the direction of x and cover the entire ly. They are designed for the maximum negative moment Mx and each has a capacity of -0.5Mx. Top bars T1 are continued up to a distance of 0.3lx1, while T2 bars are only up to a distance of 0.15lx1.

• Top bars T4 are along y and provided up to a distance of 0.3lx1 from the

support. They are on the basis of minimum steel requirement. (iii) Continuous Adjacent Panel (CAP)

• Bottom bars B4 and B5 are similar to B1 and B2 bars of (i) above. • Bottom bars B6 are similar to B3 bars of (i) above.

Detailing is an art and hence structural requirement can be satisfied by more than one mode of detailing each valid and acceptable. 8.18.8 Numerical Problems (a) Problem 8.1 Design the one-way continuous slab of Fig.8.18.6 subjected to uniformly distributed imposed loads of 5 kN/m2 using M 20 and Fe 415. The load of floor finish is 1 kN/m2. The span dimensions shown in the figure are effective spans. The width of beams at the support = 300 mm.


Solution of Problem 8.1 Step 1: Selection of preliminary depth of slab The basic value of span to effective depth ratio for the slab having simple support at the end and continuous at the intermediate is (20+26)/2 = 23 (cl.23.2.1 of IS 456). Modification factor with assumed p = 0.5 and fs = 240 N/mm2 is obtained as 1.18 from Fig.4 of IS 456. Therefore, the minimum effective depth = 3000/23(1.18) = 110.54 mm. Let us take the effective depth d = 115 mm and with 25 mm cover, the total depth D = 140 mm. Step 2: Design loads, bending moment and shear force Dead loads of slab of 1 m width = 0.14(25) = 3.5 kN/m Dead load of floor finish =1.0 kN/m Factored dead load = 1.5(4.5) = 6.75 kN/m Factored live load = 1.5(5.0) = 7.50 kN/m Total factored load = 14.25 kN/m Maximum moments and shear are determined from the coefficients given in Tables 12 and 13 of IS 456. Maximum positive moment = 14.25(3)(3)/12 = 10.6875 kNm/m Maximum negative moment = 14.25(3)(3)/10 = 12.825 kNm/m Maximum shear Vu = 14.25(3)(0.4) = 17.1 kN Step 3: Determination of effective and total depths of slab From Eq.3.25 of sec. 3.5.6 of Lesson 5, we have Mu,lim = R,lim bd2 where R,lim is 2.76 N/mm2 from Table 3.3 of sec. 3.5.6 of Lesson 5. So, d = {12.825(106)/(2.76)(1000)}0.5 = 68.17 mm


Since, the computed depth is much less than that determined in Step 1, let us keep D = 140 mm and d = 115 mm. Step 4: Depth of slab for shear force Table 19 of IS 456 gives cτ = 0.28 N/mm2 for the lowest percentage of steel in the slab. Further for the total depth of 140 mm, let us use the coefficient k of cl. 40.2.1.1 of IS 456 as 1.3 to get c ττ kc = = 1.3(0.28) = 0.364 N/mm2. Table 20 of IS 456 gives maxcτ = 2.8 N/mm2. For this problem bdVuv / =τ = 17.1/115 = 0.148 N/mm2. Since, maxc cv τττ << , the effective depth d = 115 mm is acceptable. Step 5: Determination of areas of steel From Eq.3.23 of sec. 3.5.5 of Lesson 5, we have Mu = 0.87 fy Ast d {1 – (Ast)(fy)/(fck)(bd)} (i) For the maximum negative bending moment 12825000 = 0.87(415)(Ast)(115){1 – (Ast)(415)/(1000)(115)(20)} or - 5542.16 A2

stA st + 1711871.646 = 0 Solving the quadratic equation, we have the negative Ast = 328.34 mm2

(ii) For the maximum positive bending moment 10687500 = 0.87(415) Ast(115) {1 – (Ast)(415)/(1000)(115)(20)} or - 5542.16 A2

stA st + 1426559.705 = 0 Solving the quadratic equation, we have the positive Ast = 270.615 mm2

Alternative approach: Use of Table 2 of SP-16 (i) For negative bending moment

Mu/bd2 = 0.9697


Table 2 of SP-16 gives: ps = 0.2859 (by linear interpolation). So, the area of negative steel = 0.2859(1000)(115)/100 = 328.785 mm2. (ii) For positive bending moment

Mu/bd2 = 0.8081 Table 2 of SP-16 gives: ps = 0.23543 (by linear interpolation). So, the area of positive steel = 0.23543(1000)(115)/100 = 270.7445 mm2. These areas of steel are comparable with those obtained by direct computation using Eq.3.23. Distribution steel bars along longer span ly Distribution steel area = Minimum steel area = 0.12(1000)(140)/100 = 168 mm2. Since, both positive and negative areas of steel are higher than the minimum area, we provide:

(a) For negative steel: 10 mm diameter bars @ 230 mm c/c for which Ast = 341 mm2 giving ps = 0.2965.

(b) For positive steel: 8 mm diameter bars @ 180 mm c/c for which Ast = 279 mm2 giving ps = 0.2426

(c) For distribution steel: Provide 8 mm diameter bars @ 250 mm c/c for

which Ast (minimum) = 201 mm2. Step 6: Selection of diameter and spacing of reinforcing bars The diameter and spacing already selected in step 5 for main and distribution bars are checked below: For main bars (cl. 26.3.3.b.1 of IS 456), the maximum spacing is the lesser of 3d and 300 mm i.e., 300 mm. For distribution bars (cl. 26.3.3.b.2 of IS 456), the maximum spacing is the lesser of 5d or 450 mm i.e., 450 mm. Provided spacings, therefore, satisfy the requirements. Maximum diameter of the bars (cl. 26.5.2.2 of IS 456) shall not exceed 140/8 = 17 mm is also satisfied with the bar diameters selected here.


Figure 8.18.7 presents the detailing of the reinforcement bars. The abbreviation B1 to B3 and T1 to T4 are the bottom and top bars, respectively which are shown in Fig.8.18.5 for a typical one-way slab. The above design and detailing assume absence of support along short edges. When supports along short edges exist and there is eventual clamping top reinforcement would be necessary at shorter supports also. 8.18.9 Practice Questions and Problems with Answers Q.1: State the names of different types of slabs used in construction. A.1: See sec. 8.18.1. Q.2: (a) State the limit of the aspect ratio of ly/lx of one- and two-way slabs. (b) Explain the share of loads by the supporting beams in one- and two-

way slabs. A.2: (a) The aspect ratio ly/lx (lx is the shorter one) is from 1 to 2 for two-way

slabs and beyond 2 for one-way slabs. (b) See sec. 8.18.2. Q.3: How to determine the design shear strength of concrete in slabs of

different depths having the same percentage of reinforcement? A.3: See sec. 8.18.3.


Q.4: State span to depth ratios of one-way slabs for different support conditions to be considered for the control of deflection.

A.4: See sec. 8.18.5b. Q.5: State the minimum amounts of reinforcing bars to be provided in slabs? A.5: See sec. 8.18.5d. Q.6: State the maximum diameter of a bar to be used in slabs. A.6: See sec. 8.18.5e. Q.7: How do we determine the effective depth of a slab for a given factored

moment? A.7: See sec. 8.18.6, Step 3, Eq.3.25. Q.8: How do we determine the area of steel to be provided for a given

factored moment? A.8: See sec. 8.18.6, Step 5, Eq.3.23. Q.9: How do we determine the amount of steel in the longer span direction? A.9: Minimum amount of steel shall be provided for temperature, shrinkage etc.

as per cl. 26.5.2.1 of IS 456. These are known as distribution bars. Q.10: Design the cantilever panel of the one-way slab shown in Fig.8.18.8

subjected to uniformly distributed imposed loads 5 kN/m2 using M 20 and Fe 415. The load of floor finish is 0.75 kN/m2. The span dimensions shown in the figure are effective spans. The width of the support is 300 mm.

A.10:


Step 1: Selection of preliminary depth of slab Basic value of span to depth ratio (cl. 23.2.1 of IS 456) = 7 Modification factor = 1.18 (see Problem 8.1) Minimum effective depth = 1850/7(1.18) = 223.97 mm Assume d = 225 mm and D = 250 mm. Step 2: Design loads, bending moment and shear force Factored dead loads = (1.5)(0.25)(25) = 9.375 kN/m Factored load of floor finish = (1.5)(0.75) = 1.125 kN/m Factored live loads = (1.5)(5) = 7.5 kN/m Total factored loads = 18.0 kN/m Maximum negative moment = 18.0(1.85)(1.85)(0.5) = 30.8025 kNm/m Maximum shear force = 18.0(1.85) = 33.3 kN/m Step 3: Determination of effective and total depths of slab From Eq.3.25, Step 3 of sec. 8.18.6, we have d = {30.8025(106)/2.76(103)}0.5 = 105.64 mm, considering the value of R = 2.76 N/mm2 from Table 3.3 of sec. 3.5.5 of Lesson 5. This depth is less than assumed depth of slab in Step 1. Hence, assume d = 225 mm and D = 250 mm. Step 4: Depth of slab for shear force Using the value of k = 1.1 (cl. 40.2.1.1 of IS 456) for the slab of 250 mm depth, we have cτ (from Table 19 of IS 456) = 1.1(0.28) = 0.308 N/mm2. Table 20 of IS 456 gives maxcτ = 2.8 N/mm2. Here, The depth of the slab is safe in shear as

.N/mm 0.148 33.3/225 / 2=== bdVuvτ. maxccv τττ <<

Step 5: Determination of areas of steel (using table of SP-16) Table 44 gives 10 mm diameter bars @ 200 mm c/c can resist 31.43 kNm/m > 30.8025 kNm/m. Fifty per cent of the bars should be curtailed at a


distance of larger of Ld or 0.5lx. Table 65 of SP-16 gives Ld of 10 mm bars = 470 mm and 0.5lx = 0.5(1850) = 925 mm from the face of the column. The curtailment distance from the centre line of beam = 925 + 150 = 1075, say 1100 mm. The above, however, is not admissible as the spacing of bars after the curtailment exceeds 300 mm. So, we provide 10 mm @ 300 c/c and 8 mm @ 300 c/c. The moment of resistance of this set is 34.3 kNm/m > 30.8025 kNm/m (see Table 44 of SP-16).

Figure 8.18.9 presents the detailing of reinforcing bars of this problem. 8.18.10 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 8.18.11 Test 18 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: (a) State the limit of the aspect ratio of ly/lx of one- and two-way slabs. (b) Explain the share of loads by the supporting beams in one- and two-

way slabs. (10 marks) A.TQ.1: (a) The aspect ratio ly/lx (lx is the shorter one) is from 1 to 2 for two-way

slabs and beyond 2 for one-way slabs. (b) See sec. 8.18.2. TQ.2: How to determine the design shear strength of concrete in slabs of

different depths having the same percentage of reinforcement? (10 marks)

A.TQ.2: See sec. 8.18.3. TQ3: Determine the areas of steel, bar diameters and spacings in the two

directions of a simply supported slab of effective spans 3.5 m x 8 m (Figs.8.18.10a and b) subjected to live loads of 4 kN/m2 and the load of


floor finish is 1 kN/m2. Use M 20 and Fe 415. Draw the diagram showing the detailing of reinforcement. (30 marks)

A.TQ.3:


This is one-way slab as ly/lx = 8/3.5 = 2.285 > 2. The calculations are shown in different steps below: Step 1: Selection of preliminary depth of slab Clause 23.2.1 stipulates the basic value of span to effective depth ratio of 20. Using the modification factor of 1.18 from Fig.4 of IS 456, with p = 0.5 per cent and fs = 240 N/mm2, we have the span to effective depth ratio = 20(1.18) = 23.6. So, the minimum effective depth of slab = 3500/23.6 = 148.305 mm. Let us take d = 150 mm and D = 175 mm. Step 2: Design loads, bending moment and shear force Factored dead loads of slab = (1.5)(0.175)(25) = 6.5625 kN/m Factored load of floor finish = (1.5)(1) = 1.5 kN/m Factored live load = (1.5)(4) = 6.0 kN/m Total factored load = 14.0625 kN/m Maximum positive bending moment = 14.0625(3.5)(3.5)/8 = 21.533 kNm/m Maximum shear force = 14.0625(3.5)(0.5) = 24.61 kN/m Step 3: Determination/checking of the effective and total depths of slab Using Eq.3.25 as explained in Step 3 of sec. 8.18.6, we have d = {21.533(106)/(2.76)(103)}0.5 = 88.33 mm < 150 mm, as assumed in Step 1. So, let us keep d = 150 mm and D = 175 mm. Step 4: Depth of the slab for shear force With the multiplying factor k = 1.25 for the depth as 175 mm (vide Table 8.1 of this lesson) and cτ = 0.28 N/mm2 from Table 19 of IS 456, we have cτ = 1.25(0.28) = 0.35 N/mm2.

Table 20 of IS 456 gives maxcτ = 2.8 N/mm2. For this problem: .N/mm 0.1641 50))/(1000)(124.61(1000 2==vτ

Thus, the effective depth of slab as 150 mm is safe as . maxccv τττ <<


Step 5: Determination of areas of steel Table 41 of SP-16 gives 8 mm diameter bars @ 120 mm c/c have 22.26 kNm/m > 21.533 kNm/m. Hence, provide 8 mm T @ 120 mm c/c as main positive steel bars along the short span of 3.5 mm. The minimum amount of reinforcement (cl. 26.5.2.1 of IS 456) = 0.12(175)(1000)/100 = 210 mm2. Provide 6 mm diameter bars @ 120 mm c/c (236 mm2) along the large span of 8m. Figure 8.18.10b shows the detailing of reinforcing bars. 8.18.12 Summary of this Lesson This lesson mentions the different types of slabs used in construction and explains the differences between one and two-way slabs. Illustrating the principles of design as strength and deflection, the methods of determining the bending moments and shear forces are explained. The stipulated guidelines of assuming the preliminary depth, maximum diameter of reinforcing bars, nominal covers, spacing of reinforcements, minimum amount of reinforcing bars etc. are illustrated. The steps of the design of one-way slabs are explained. Design problems are solved to illustrate the application of design guidelines. Further, the detailing of reinforcement bars are explained for typical one-way slab and for the numerical problems solved in this lesson.


Module 8

Reinforced Concrete Slabs


Lesson 19

Two-way Slabs




• determine the shear force of two-way slabs subjected to uniformly distributed loads,

• state the two types of two-way slabs mentioning the differences between

them,

• determine the preliminary depth of two-way slabs of different support conditions from the span to effective depth ratios as given in IS 456,

• explain the provisions of torsion reinforcing bars at two types of corners of

a restrained two-way slab,

• design the two types of two-way slabs applying the different methods explained in this lesson and draw the detailing of reinforcing bars.

8.19.1 Introduction Lesson 18 explains the various types of slabs with different support conditions, plan forms, horizontal/inclined etc. Moreover, sec. 8.18.2 of Lesson 18 illustrates the sharing of uniformly distributed loads to the supporting beams of both one and two-way slabs including the profiles of deflection (Figs.8.18.4a and b). It is, thus, understood that two-way slabs span in both directions having the aspect ratio of ly/lx up to 2, considering lx as the shorter span. This lesson presents the different aspects of analysis and design of two-way slabs. Many of the stipulations of IS 456 are the same as those of one-way slabs. While mentioning the common stipulations with their respective section in Lesson 18, this lesson presents other relevant requirements regarding the analysis, design and detailing of two-way slabs. Numerical problems are also solved to illustrate the applications of the theory in the design of two-way slabs. 8.19.2 Two-way Slabs Two-way slabs subjected mostly to uniformly distributed loads resist them primarily by bending about both the axis. However, as in the one-way slab, the depth of the two-way slabs should also be checked for the shear stresses to avoid any reinforcement for shear. Moreover, these slabs should have sufficient depth for the control deflection. Thus, strength and deflection are the requirements of design of two-way slabs.


8.19.3 Design Shear Strength of Concrete Design shear strength of concrete in two-way slabs is to be determined incorporating the multiplying factor k from Table 8.1 of Lesson 18 in the same manner as discussed in sec. 8.18.3 of Lesson 18. 8.19.4 Structural Analysis

8.19.4.1 Computation of shear force

Shear forces are computed following the procedure stated below with reference to Fig.8.19.1. The two-way slab of Fig. 8.19.1 is divided into two trapezoidal and two triangular zones by drawing lines from each corner at an angle of 45o. The loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be beam 2-3. The shear forces per unit width of the strips aa and bb are highest at the ends of strips. Moreover, the length of half the strip bb is equal to the length of the strip aa. Thus, the shear forces in both strips are equal and we can write, Vu = W (lx/2) (8.1) where W = intensity of the uniformly distributed loads. The nominal shear stress acting on the slab is then determined from bdVuv / =τ (8.2)


8.19.4.2 Computation of bending moments Two-way slabs spanning in two directions at right angles and carrying uniformly distributed loads may be analysed using any acceptable theory. Pigeoud’s or Wester-guard’s theories are the suggested elastic methods and Johansen’s yield line theory is the most commonly used in the limit state of collapse method and suggested by IS 456 in the note of cl. 24.4. Alternatively, Annex D of IS 456 can be employed to determine the bending moments in the two directions for two types of slabs: (i) restrained slabs, and (ii) simply supported slabs. The two methods are explained below: (i) Restrained slabs Restrained slabs are those whose corners are prevented from lifting due to effects of torsional moments. These torsional moments, however, are not computed as the amounts of reinforcement are determined from the computed areas of steel due to positive bending moments depending upon the intensity of torsional moments of different corners. This aspect has been explained in Step 7 of sec. 8.19.6. Thus, it is essential to determine the positive and negative bending moments in the two directions of restrained slabs depending on the various types of panels and the aspect ratio ly/lx.

Restrained slabs are considered as divided into two types of strips in each direction: (i) one middle strip of width equal to three-quarters of the respective length of span in either directions, and (ii) two edge strips, each of width equal to one-eighth of the respective length of span in either directions. Figures 8.19.2a and b present the two types of strips for spans lx and ly separately.


The maximum positive and negative moments per unit width in a slab are determined from (8.3)

2 xxx lwM α=

(8.4)

2 yxy lwM α=

where yx αα and are coefficients given in Table 26 of IS 456, Annex D, cl. D-1.1. Total design load per unit area is w and lengths of shorter and longer spans are represented by lx and ly, respectively. The values of yx αα and , given in Table 26 of IS 456, are for nine types of panels having eight aspect ratios of ly/lx from one to two at an interval of 0.1. The above maximum bending moments are applicable only to the middle strips and no redistribution shall be made. Tension reinforcing bars for the positive and negative maximum moments are to be provided in the respective middle strips in each direction. Figure 8.19.2 shows the positive and negative coefficients yx αα and . The edge strips will have reinforcing bars parallel to that edge following the minimum amount as stipulated in IS 456. The detailing of all the reinforcing bars for the respective moments and for the minimum amounts as well as torsional requirements are discussed in sec. 8.19.7(i). (ii) Simply supported slabs The maximum moments per unit width of simply supported slabs, not having adequate provision to resist torsion at corners and to prevent the corners from lifting, are determined from Eqs.8.3 and 8.4, where yx αα and are the respective coefficients of moments as given in Table 27 of IS 456, cl. D-2. The notations Mx, My, w, lx and ly are the same as mentioned below Eqs.8.3 and 8.4 in (i) above. The detailing of reinforcing bars for the respective moments is explained in sec. 8.19.7(ii).


The coefficients yx αα and of simply supported two-way slabs are derived from the Grashoff-Rankine formula which is based on the consideration of the same deflection at any point P (Fig.8.19.3) of two perpendicular interconnected strips containing the common point P of the two-way slab subjected to uniformly distributed loads. 8.19.5 Design Considerations The design considerations mentioned in sec. 8.18.5 of Lesson 18 in (a), (c), (d), (e) and (f) are applicable for the two-way slabs also. However, the effective span to effective depth ratio is different from those of one-way slabs. Accordingly, this item for the two-way slabs is explained below. Effective span to effective depth ratio (cl. 24.1 of IS 456) The following are the relevant provisions given in Notes 1 and 2 of cl. 24.1.

• The shorter of the two spans should be used to determine the span to effective depth ratio.

• For spans up to 3.5 m and with mild steel reinforcement, the span to

overall depth ratios satisfying the limits of vertical deflection for loads up to 3 kN/m2 are as follows:

Simply supported slabs 35

Continuous slabs 40


• The same ratios should be multiplied by 0.8 when high strength deformed bars (Fe 415) are used in the slabs.

8.19.6 Design of Two-way Slabs The procedure of the design of two-way slabs will have all the six steps mentioned in sec. 8.18.6 for the design of one-way slabs except that the bending moments and shear forces are determined by different methods for the two types of slab. While the bending moments and shear forces are computed from the coefficients given in Tables 12 and 13 (cl. 22.5) of IS 456 for the one-way slabs, the same are obtained from Tables 26 or 27 for the bending moment in the two types of two-way slabs and the shear forces are computed from Eq.8.1 for the two-way slabs. Further, the restrained two-way slabs need adequate torsional reinforcing bars at the corners to prevent them from lifting. There are three types of corners having three different requirements. Accordingly, the determination of torsional reinforcement is discussed in Step 7, as all the other six steps are common for the one and two-way slabs. Step 7: Determination of torsional reinforcement

Three types of corners, C1, C2 and C3, shown in Fig.8.19.4, have three different requirements of torsion steel as mentioned below. (a) At corner C1 where the slab is discontinuous on both sides, the torsion reinforcement shall consist of top and bottom bars each with layers of bar placed parallel to the sides of the slab and extending a minimum distance of one-fifth of the shorter span from the edges. The amount of reinforcement in each of the four layers shall be 75 per cent of the area required for the maximum mid-span moment in the slab. This provision is given in cl. D-1.8 of IS 456.


(b) At corner C2 contained by edges over one of which is continuous, the torsional reinforcement shall be half of the amount of (a) above. This provision is given in cl. D-1.9 of IS 456. (c) At corner C3 contained by edges over both of which the slab is continuous, torsional reinforcing bars need not be provided, as stipulated in cl. D-1.10 of IS 456. 8.19.7 Detailing of Reinforcement As mentioned in sec. 8.19.6, Step 5 of sec. 8.18.6 explains the two methods of determining the required areas of steel required for the maximum positive and negative moments. The two methods are (i) employing Eq.3.23 as given in Step 5 of sec. 8.18.6 or (ii) using tables and charts of SP-16. Thereafter, Step 7 of sec. 8.19.6 explains the method of determining the areas steel for corners of restrained slab depending on the type of corner. The detailing of torsional reinforcing bars is explained in Step 7 of sec. 8.19.6. In the following, the detailings of reinforcing bars for (i) restrained slabs and (ii) simply supported slabs are discussed separately for the bars either for the maximum positive or negative bending moments or to satisfy the requirement of minimum amount of steel. (i) Restrained slabs The maximum positive and negative moments per unit width of the slab calculated by employing Eqs.8.3 and 8.4 as explained in sec. 8.19.4.2(i) are applicable only to the respective middle strips (Fig.8.19.2). There shall be no redistribution of these moments. The reinforcing bars so calculated from the maximum moments are to be placed satisfying the following stipulations of IS 456.


• Bottom tension reinforcement bars of mid-span in the middle strip shall

extent in the lower part of the slab to within 0.25l of a continuous edge, or 0.15l of a discontinuous edge (cl. D-1.4 of IS 456). Bars marked as B1, B2, B5 and B6 in Figs.8.19.5 a and b are these bars.

• Top tension reinforcement bars over the continuous edges of middle strip

shall extend in the upper part of the slab for a distance of 0.15l from the support, and at least fifty per cent of these bars shall extend a distance of 0.3l (cl. D-1.5 of IS 456). Bars marked as T2, T3, T5 and T6 in Figs.8.19.5 a and b are these bars.

• To resist the negative moment at a discontinuous edge depending on the

degree of fixity at the edge of the slab, top tension reinforcement bars equal to fifty per cent of that provided at mid-span shall extend 0.1l into the span (cl. D-1.6 of IS 456). Bars marked as T1 and T4 in Figs.8.19.5 a and b are these bars.


• The edge strip of each panel shall have reinforcing bars parallel to that edge satisfying the requirement of minimum amount as specified in sec. 8.18.15d of Lesson 18 (cl. 26.5.2.1 of IS 456) and the requirements for torsion, explained in Step 7 of sec. 8.19.6 (cls. D-1.7 to D-1.10 of IS 456). The bottom and top bars of the edge strips are explained below.

• Bottom bars B3 and B4 (Fig.8.19.5 a) are parallel to the edge along lx for

the edge strip for span ly, satisfying the requirement of minimum amount of steel (cl. D-1.7 of IS 456).

• Bottom bars B7 and B8 (Fig.8.19.5 b) are parallel to the edge along ly for

the edge strip for span lx, satisfying the requirement of minimum amount of steel (cl. D-1.7 of IS 456).

• Top bars T7 and T8 (Fig.8.19.5 a) are parallel to the edge along lx for the

edge strip for span ly, satisfying the requirement of minimum amount of steel (cl. D-1.7 of IS 456).

• Top bars T9 and T10 (Fig.8.19.5 b) are parallel to the edge along ly for

the edge strip for span lx, satisfying the requirement of minimum amount of steel (cl. D-1.7 of IS 456).

The detailing of torsion bars at corners C1 and C2 is explained in Fig.8.19.7 of Problem 8.2 in sec. 8.19.8. The above explanation reveals that there are eighteen bars altogether comprising eight bottom bars (B1 to B8) and ten top bars (T1 to T10). Tables 8.4 and 8.5 present them separately for the bottom and top bars, respectively, mentioning the respective zone of their placement (MS/LDES/ACES/BDES to designate Middle Strip/Left Discontinuous Edge Strip/Adjacent Continuous Edge Strip/Bottom Discontinuous Edge Strip), direction of the bars (along x or y), the resisting moment for which they shall be determined or if to be provided on the basis of minimum reinforcement clause number of IS 456 and Fig. No. For easy understanding, plan views in (a) and (b) of Fig.8.19.5 show all the bars separately along x and y directions, respectively. Two sections (1-1 and 2-2), however, present the bars shown in the two plans. Torsional reinforcements are not included in Tables 8.4 and 8.5 and Figs.8.19.5 a and b. Table 8.4 Details of eight bottom bars

Sl.No. Bars Into Along Resisting Moment

Cl.No. of IS 456

Fig.No.

1 B1, B2 MS x Max. + Mx

D-1.3,1.4 8.19.5a, c, d

2 B3 LDES x Min. Steel

D-1.7 8.19.5a, c


3 B4 ACES x Min. Steel

D-1.7 8.19.5a, c

4 B5, B6 MS y Max. + My

D-1.3,1.4 8.19.5b, c, d

5 B7 BDES y Min. Steel

D-1.7 8.19.5b, d

6 B8 ACES y Min. Steel

D-1.7 8.19.5b, d

Notes: (i) MS = Middle Strip (ii) LDES = Left Discontinuous Edge Strip (iii) ACES = Adjacent Continuous Edge Strip (iv) BDES = Bottom Discontinuous Edge Strip Table 8.5 Details of eight top bars

Sl.No. Bars Into Along Resisting Moment

Cl.No. of IS 456

Fig.No.

1 T1 BDES x + 0.5 Mx D-1.6 8.19.5a, d

2 T2, T3 ACES x - 0.5 Mx for each

D-1.5 8.19.5a, d

3 T4 LDES y + 0.5 My D-1.6 8.19.5b, c

4 T5, T6 ACES y -0.5 My for each

D-1.5 8.19.5b, c

5 T7 LDES x Min. Steel

D-1.7 8.19.5a, c

6 T8 ACES x Min. Steel

D-1.7 8.19.5a, c

7 T9 LDES y Min. Steel

D-1.7 8.19.5b, d

8 T10 ACES y Min. Steel

D-1.7 8.19.5b, d

Notes: (i) MS = Middle Strip (ii) LDES = Left Discontinuous Edge Strip (iii) ACES = Adjacent Continuous Edge Strip (iv) BDES = Bottom Discontinuous Edge Strip


(ii) Simply supported slabs

Figures 8.19.6 a, b and c present the detailing of reinforcing bars of simply supported slabs not having adequate provision to resist torsion at corners and to prevent corners from lifting. Clause D-2.1 stipulates that fifty per cent of the tension reinforcement provided at mid-span should extend to the supports. The remaining fifty per cent should extend to within 0.1lx or 0.1ly of the support, as appropriate.


8.19.8 Numerical Problems Problem 8.2

Design the slab panel 1 of Fig.8.19.7 subjected to factored live load of 8 kN/m2 in addition to its dead load using M 20 and Fe 415. The load of floor finish is 1 kN/m2. The spans shown in figure are effective spans. The corners of the slab are prevented from lifting. Solution of Problem 8.2 Step 1: Selection of preliminary depth of slab The span to depth ratio with Fe 415 is taken from cl. 24.1, Note 2 of IS 456 as 0.8 (35 + 40) / 2 = 30. This gives the minimum effective depth d = 4000/30 = 133.33 mm, say 135 mm. The total depth D is thus 160 mm. Step 2: Design loads, bending moments and shear forces Dead load of slab (1 m width) = 0.16(25) = 4.0 kN/m2


Dead load of floor finish (given) = 1.0 kN/m2

Factored dead load = 1.5(5) = 7.5 kN/m2

Factored live load (given) = 8.0 kN/m2

Total factored load = 15.5 kN/m2

The coefficients of bending moments and the bending moments Mx and My per unit width (positive and negative) are determined as per cl. D-1.1 and Table 26 of IS 456 for the case 4, “Two adjacent edges discontinuous” and presented in Table 8.6. The ly / lx for this problem is 6/4 = 1.5. Table 8.6 Maximum bending moments of Problem 8.2

Short span Long span For xα Mx (kNm/m) yα My (kNm/m)

Negative moment at continuous edge

0.075 18.6 0.047 11.66

Positive moment at mid-span

0.056 13.89 0.035 8.68

Maximum shear force in either direction is determined from Eq.8.1 (Fig.8.19.1) as Vu = w(lx/2) = 15.5 (4/2) = 31 kN/m Step 3: Determination/checking of the effective depth and total depth of slab Using the higher value of the maximum bending moments in x and y directions from Table 8.6, we get from Eq.3.25 of Lesson 5 (sec. 3.5.5): Mu,lim = R,lim bd2

or d = [(18.6)(106)/{2.76(103)}] 1/2 = 82.09 mm, where 2.76 N/mm2 is the value of R,lim taken from Table 3.3 of Lesson 5 (sec. 3.5.5). Since, this effective depth is less than 135 mm assumed in Step 1, we retain d = 135 mm and D = 160 mm.


Step 4: Depth of slab for shear force Table 19 of IS 456 gives the value of cτ = 0.28 N/mm2 when the lowest percentage of steel is provided in the slab. However, this value needs to be modified by multiplying with k of cl. 40.2.1.1 of IS 456. The value of k for the total depth of slab as 160 mm is 1.28. So, the value of cτ is 1.28(0.28) = 0.3584 N/mm2. Table 20 of IS 456 gives maxcτ = 2.8 N/mm2. The computed shear stress

vτ = Vu/bd = 31/135 = 0.229 N/mm2. Since, vτ < cτ < maxcτ , the effective depth of the slab as 135 mm and the total depth as 160 mm are safe. Step 5: Determination of areas of steel The respective areas of steel in middle and edge strips are to be determined employing Eq.3.23 of Step 5 of sec. 8.18.6 of Lesson 18. However, in Problem 8.1 of Lesson 18, it has been shown that the areas of steel computed from Eq.3.23 and those obtained from the tables of SP-16 are in good agreement. Accordingly, the areas of steel for this problem are computed from the respective Tables 40 and 41 of SP-16 and presented in Table 8.7. Table 40 of SP-16 is for the effective depth of 150 mm, while Table 41 of SP-16 is for the effective depth of 175 mm. The following results are, therefore, interpolated values obtained from the two tables of SP-16. Table 8.7 Reinforcing bars of Problem 8.2

Short span lx Long span lyParticulars Table No.

Mx (kNm/m)

Dia. & spacing

Table No.

My (kNm/m)

Dia. & spacing

Top steel for negative moment

40,41 18.68 > 18.6

10 mm @ 200 mm c/c

40,41 12.314 > 11.66

8 mm @ 200 mm

c/c Bottom steel for positive moment

40,41 14.388 > 13.89

8 mm @ 170 mm c/c

40,41 9.20 > 8.68

8 mm @ 250 mm

c/c The minimum steel is determined from the stipulation of cl. 26.5.2.1 of IS 456 and is As = (0.12/100)(1000)(160) = 192 mm2


and 8 mm bars @ 250 mm c/c (= 201 mm2) is acceptable. It is worth mentioning that the areas of steel as shown in Table 8.7 are more than the minimum amount of steel. Step 6: Selection of diameters and spacings of reinforcing bars The advantages of using the tables of SP-16 are that the obtained values satisfy the requirements of diameters of bars and spacings. However, they are checked as ready reference here. Needless to mention that this step may be omitted in such a situation. Maximum diameter allowed, as given in cl. 26.5.2.2 of IS 456, is 160/8 = 20 mm, which is more that the diameters used here. The maximum spacing of main bars, as given in cl. 26.3.3(1) of IS 456, is the lesser of 3(135) and 300 mm. This is also satisfied for all the bars. The maximum spacing of minimum steel (distribution bars) is the lesser of 5(135) and 450 mm. This is also satisfied.


Figures 8.19.8 and 9 present the detailing of reinforcing bars. Step 7: Determination of torsional reinforcement


Torsional reinforcing bars are determined for the three different types of corners as explained in sec. 8.19.6 (Fig.8.19.4). The length of torsional strip is 4000/5 = 800 mm and the bars are to be provided in four layers. Each layer will have 0.75 times the steel used for the maximum positive moment. The C1 type of corners will have the full amount of torsional steel while C2 type of corners will have half of the amount provided in C1 type. The C3 type of corners do not need any torsional steel. The results are presented in Table 8.8 and Figs.8.19.10 a, b and c. Table 8.8 Torsional reinforcement bars of Problem 8.2

Dimensions along No. of bars along Type x (mm) y (mm)

Bar diameter & spacing x y

Cl. no. of IS 456

C1 800 800 8 mm @ 200 mm c/c

5 5 D-1.8

C2 800 1600 8 mm @ 250 mm c/c

5 8 D-1.9

C2 1600 800 8 mm @ 250 mm c/c

8 5 D-1.9

8.19.9 Practice Questions and Problems with Answers Q.1: How do you determine the shear force of a two-way slab subjected to

uniformly distributed loads?


A.1: See sec. 8.19.4.1. Q.2: Name the two types of two-way slabs. A.2: The two types of two-way slabs are: (i) restrained slabs and (ii) simply

supported slabs. Q.3: What is the difference in the design of the two types of slabs of Q.2? A.3: The restrained slabs are those whose corners are prevented from lifting

and accordingly, there are torsional reinforcing bars in the two types of corners. The simply supported slabs do not have adequate provision to resist torsion at corners and to prevent the corners from lifting. So, torsional reinforcing bars are not provided in these slabs.

Q.4: State span to depth ratios of two-way slabs for different support

conditions to be considered for the control of deflection. A.4: See sec. 8.19.5. Q.5: Explain the provisions of torsional reinforcing bars in restrained type of two-

way slabs. A.5: Step 7 of sec. 8.19.6. Q.6:


Design a two-way simply supported slab of Fig.8.19.11, not having

adequate provision to resist torsion at corners and to prevent the corners from lifting. The factored live load is 6 kN/m2 and the load of the floor finish in 1 kN/m2. The spans shown in the figure are effective spans. Use M 20 and Fe 415. The width of the support is 300 mm.

A.6: Step 1: Selection of preliminary depth of slab As per cl.24.1, Note 2, the span to effective depth ratio = 0.8(35) = 28. The minimum effective depth = d = 4200/28 = 150 mm and, therefore, D = 175 mm. Step 2: Design loads, bending moments and shear forces Factored dead load of the slab = 1.5(0.175)(25) = 6.5625 kN/m2

Factored load of floor finish = 1.5(1) = 1.5 kN/m2

Factored live loads = 6.0 kN/m2

Total factored loads = 14.0625 kN/m2


For this slab ly/lx = 5880/4200 = 1.4, Table 27 of IS 456 gives xα = 0.099 and yα = 0.051. + Mx = xα w lx2 = (0.099)(14.0625)(4.2)(4.2) = 24.558 kNm/m + My = yα w lx2 = (0.051)(14.0625)(4.2)(4.2) = 12.651 kNm/m Vu = 0.5 w lx = 0.5(14.0625)(4.2) = 29.531 kN/m Step 3: Determination/checking of the effective depth and total depth of slab d = {24.558(103)/2.76}0.5 = 94.328 mm < 150 mm So, we keep d = 150 mm and D = 175 mm. Step 4: Depth of slab for shear forces Table 19 of IS 456 gives cτ = 0.28 N/mm2. Clause 40.2.1.1 of IS 456 gives k = 1.25 for D = 175 mm. So, cτ = (1.25)(0.28) = 0.35 N/mm2. Table 20 of IS 456 gives maxcτ = 2.8 N/mm2. For this problem vτ = Vu/bd = 29.531/150 = 0.1968 N/mm2. Since vτ < cτ < maxcτ , the depth is safe. Step 5: Determination of areas of steel The positive steel in the two directions and the minimum steel are furnished below in Table 8.9. These are the results obtained from the use of Table 41 of SP-16. Table 8.9 Reinforcing bars of Problem Q.6

Short span lx Long span lyParticulars SP

Table No.

Mx (kNm/m)

Dia. & spacing

SP Table No.

My (kNm/m)

Dia. & spacing

Positive steel 41 26.40 > 24.558

8 mm @ 100 mm c/c (503

mm2)

41 12.93 > 12.651

6 mm @ 120 mm c/c (236

mm2) Minimum steel = 0.12(1750)

96 Min. steel

6 mm @ 120 mm c/c (236

96 Min. steel 6 mm @ 120 mm c/c (236


= 210 mm2 mm2) > 210 mm2

mm2) > 210 mm2

Figures 8.19.12a and b show the detailing of reinforcing bars. 8.19.10 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 8.19.11 Test 19 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Explain the provisions of torsional reinforcing bars in restrained type of

two-way slabs. (20 marks)

A.TQ.1: Step 7 of sec. 8.19.6. TQ.2: Design the interior panel (Panel 2) of Problem 8.2 (Fig.8.19.7). Other data

are the same as those of Problem 8.2. (30 marks)

A.TQ.2: Let us keep the effective and total depths of the slab as 135 mm and

160 mm, respectively (see Problem 8.2). The total factored load = 15.5 kN/m2 (see Problem 8.2). The coefficients of bending moments and the bending moments Mx and My (positive and negative) per unit width are determined as per cl. D-1.1 and Table 26 of IS 456 for the case 1 (interior panel) and presented in Table 8.10. The ly/lx for this problem is 1.5.


Table 8.10 Bending moments of Problem TQ.2

Short span Long span For xα Mx (kNm/m) yα My (kNm/m)

Negative moment at continuous edge

0.053 13.144 0.032 7.936

Positive moment at mid-span

0.041 10.168 0.024 5.952

The maximum shear force in either direction is the same as that of Problem 8.2 = 31 kN/m. Since the bending moments are much less than those of Problem 8.2, the effective depth of 135 m and total depth of 160 mm are safe. In Step 4 of solution of Problem 8.2, this depth has been found to be safe in shear. So, the depths 135 mm and 160 mm are safe. Step 5: Determination of areas of steel The areas of steel using the table of SP-16 are presented in Table 8.11. The maximum diameter and spacing of bars are not needed to check separately as the results obtained from tables of SP-16 already take into consideration these aspects. Table 8.11 Reinforcing bars of Problem TQ.2

Short span lx Long span lyParticulars Table No.

Mx (kNm/m)

Dia. & spacing

Table No.

My (kNm/m)

Dia. & spacing

Top steel for negative moment

40,41 13.618 > 13.144

8 T @ 180 mm c/c

40 9.2 > 7.936

8 T @ * 250 mm

c/c Bottom steel for positive moment

40,41 10.758 > 10.168

8 T @ 230 mm c/c

40 9.2 > 5.952

8 T @ * 250 mm

c/c

* Note: The areas of bars are selected to satisfy the minimum amount of steel. The minimum steel will be the same as that of Problem 8.2 i.e., 8 mm diameter @ 250 mm c/c. Since this is an internal panel, torsional reinforcing bars are not needed at any of the four corners.


The detailing of the bars can be drawn following the same of Problem 8.2 shown in Figs.8.19.8 and 9. 8.19.12 Summary of this Lesson This lesson explains the differences of the methods of computing shear force and bending moments of the restrained and simply supported two-way slabs. The design methods of the two types of slabs are explained avoiding the common steps of the design of one-way slabs as explained in Lesson 18. Separate diagrams for the detailing of reinforcing bars are presented to illustrate them. Torsional bars at the two types of corners of restrained slabs are illustrated. Numerical examples are solved to design and detail the reinforcements of both types of two-way slabs. Solutions of practice problems and test problems will help in understanding the complete design and detailing of the two types of two-way slabs. It would be seen that detailing would need great care and thoroughness.


Module 9

Staircases Version 2 CE IIT, Kharagpur

Lesson 20

Types and Design of Staircases




• classify the different types of staircases based on geometrical configurations,

• name and identify the different elements of a typical flight,

• state the general guidelines while planning a staircase,

• determine the dimensions of trade, riser, depth of slab etc. of a staircase,

• classify the different staircases based on structural systems,

• explain the distribution of loadings and determination of effective spans of

stairs,

• analyse different types of staircases including the free-standing staircases in a simplified manner,

• design the different types of staircases as per the stipulations of IS 456.

9.20.1 Introduction Staircase is an important component of a building providing access to different floors and roof of the building. It consists of a flight of steps (stairs) and one or more intermediate landing slabs between the floor levels. Different types of staircases can be made by arranging stairs and landing slabs. Staircase, thus, is a structure enclosing a stair. The design of the main components of a staircase-stair, landing slabs and supporting beams or wall – are already covered in earlier lessons. The design of staircase, therefore, is the application of the designs of the different elements of the staircase.


9.20.2 Types of Staircases

Figures 9.20.1a to e present some of the common types of staircases based on geometrical configurations:

(a) Single flight staircase (Fig. 9.20.1a)


(b) Two flight staircase (Fig. 9.20.1b) (c) Open-well staircase (Fig. 9.20.1c) (d) Spiral staircase (Fig. 9.20.1d) (e) Helicoidal staircase (Fig. 9.20.1e)

Architectural considerations involving aesthetics, structural feasibility and

functional requirements are the major aspects to select a particular type of the staircase. Other influencing parameters of the selection are lighting, ventilation, comfort, accessibility, space etc.

9.20.3 A Typical Flight

Figures 9.20.2a to d present plans and sections of a typical flight of different possibilities. The different terminologies used in the staircase are given below:


(a) Tread: The horizontal top portion of a step where foot rests (Fig.9.20.2b) is known as tread. The dimension ranges from 270 mm for residential buildings and factories to 300 mm for public buildings where large number of persons use the staircase. (b) Nosing: In some cases the tread is projected outward to increase the space. This projection is designated as nosing (Fig.9.20.2b). (c) Riser: The vertical distance between two successive steps is termed as riser (Fig.9.20.2b). The dimension of the riser ranges from 150 mm for public buildings to 190 mm for residential buildings and factories. (d) Waist: The thickness of the waist-slab on which steps are made is known as waist (Fig.9.20.2b). The depth (thickness) of the waist is the minimum thickness perpendicular to the soffit of the staircase (cl. 33.3 of IS 456). The steps of the staircase resting on waist-slab can be made of bricks or concrete. (e) Going: Going is the horizontal projection between the first and the last riser of an inclined flight (Fig.9.20.2a). The flight shown in Fig.9.20.2a has two landings and one going. Figures 9.2b to d present the three ways of arranging the flight as mentioned below:

(i) waist-slab type (Fig.9.20.2b), (ii) tread-riser type (Fig.9.20.2c), or free-standing staircase, and (iii) isolated tread type (Fig.9.20.2d).

9.20.4 General Guidelines The following are some of the general guidelines to be considered while planning a staircase:

• The respective dimensions of tread and riser for all the parallel steps should be the same in consecutive floor of a building.

• The minimum vertical headroom above any step should be 2 m.

• Generally, the number of risers in a flight should be restricted to twelve.

• The minimum width of stair (Fig.9.20.2a) should be 850 mm, though it is

desirable to have the width between 1.1 to 1.6 m. In public building, cinema halls etc., large widths of the stair should be provided.


9.20.5 Structural Systems

Different structural systems are possible for the staircase, shown in Fig. 9.20.3a, depending on the spanning direction. The slab component of the stair spans either in the direction of going i.e., longitudinally or in the direction of the steps, i.e., transversely. The systems are discussed below: (A) Stair slab spanning longitudinally Here, one or more supports are provided parallel to the riser for the slab bending longitudinally. Figures 9.20.3b to f show different support arrangements of a two flight stair of Fig.9.20.3a:


(i) Supported on edges AE and DH (Fig.9.20.3b) (ii) Clamped along edges AE and DH (Fig.9.20.3c) (iii) Supported on edges BF and CG (Fig.9.20.3d) (iv) Supported on edges AE, CG (or BF) and DH (Fig.9.20.3e) (v) Supported on edges AE, BF, CG and DH (Fig.9.20.3f)

Cantilevered landing and intermediate supports (Figs.9.20.3d, e and f) are

helpful to induce negative moments near the supports which reduce the positive moment and thereby the depth of slab becomes economic.

In the case of two flight stair, sometimes the flight is supported between the landings which span transversely (Figs.9.20.4a and b). It is worth mentioning that some of the above mentioned structural systems are statically determinate while others are statically indeterminate where deformation conditions have to taken into account for the analysis.


Longitudinal spanning of stair slab is also possible with other configurations including single flight, open-well helicoidal and free-standing staircases. (B) Stair slab spanning transversely

Here, either the waist slabs or the slab components of isolated tread-slab and trade-riser units are supported on their sides or are cantilevers along the width direction from a central beam. The slabs thus bend in a transverse vertical plane. The following are the different arrangements:

(i) Slab supported between two stringer beams or walls (Fig.9.20.5a) (ii) Cantilever slabs from a spandreal beam or wall (Fig.9.20.5b) (iii) Doubly cantilever slabs from a central beam (Fig.9.20.5c)


9.20.6 Effective Span of Stairs The stipulations of clause 33 of IS 456 are given below as a ready reference regarding the determination of effective span of stair. Three different cases are given to determine the effective span of stairs without stringer beams. (i) The horizontal centre-to-centre distance of beams should be considered as the effective span when the slab is supported at top and bottom risers by beams spanning parallel with the risers. (ii) The horizontal distance equal to the going of the stairs plus at each end either half the width of the landing or one meter, whichever is smaller when the stair slab is spanning on to the edge of a landing slab which spans parallel with the risers. See Table 9.1 for the effective span for this type of staircases shown in Fig.9.20.3a. Table 9.1 Effective span of stairs shown in Fig.9.20.3a

Sl. No. x y Effective span in metres1 < 1 m < 1 m G + x + y 2 < 1 m ≥ 1 m G + x + 1 3 ≥ 1 m < 1 m G + y + 1 4 ≥ 1 m ≥ 1 m G + 1 + 1

Note: G = Going, as shown in Fig. 9.20.3a 9.20.7 Distribution of Loadings on Stairs


Figure 9.20.6 shows one open-well stair where spans partly cross at right angle. The load in such stairs on areas common to any two such spans should be taken as fifty per cent in each direction as shown in Fig.9.20.7. Moreover, one 150 mm strip may be deducted from the loaded area and the effective breadth of the section is increased by 75 mm for the design where flights or landings are embedded into walls for a length of at least 110 mm and are designed to span in the direction of the flight (Fig.9.20.7). 9.20.8 Structural Analysis Most of the structural systems of stair spanning longitudinally or transversely are standard problems of structural analysis, either statically determinate or indeterminate. Accordingly, they can be analysed by methods of analysis suitable for a particular system. However, the rigorous analysis is difficult and involved for a trade-riser type or free standing staircase where the slab is repeatedly folded. This type of staircase has drawn special attraction due to its aesthetic appeal and, therefore, simplified analysis for this type of staircase spanning longitudinally is explained below. It is worth mentioning that certain idealizations are made in the actual structures for the applicability of the simplified analysis. The designs based on the simplified analysis have been found to satisfy the practical needs.


Figure 9.20.8a shows the simply supported trade-riser staircase. The uniformly distributed loads are assumed to act at the riser levels (Fig.9.20.8b). The bending moment and shear force diagrams along the treads and the bending moment diagram along the risers are shown in Figs.9.20.8c, d and e, respectively. The free body diagrams of CD, DE and EF are shown in Figs.9.20.8f, g and h, respectively. It is seen that the trade slabs are subjected to varying bending moments and constant shear force (Fig.9.20.8f). On the other hand the riser slabs are subjected to a constant bending moment and axial force


(either compressive or tensile). The assumption is that the riser and trade slabs are rigidly connected. It has been observed that both trade and riser slabs may be designed for bending moment alone as the shear stresses in trade slabs and axial forces in riser slabs are comparatively low. The slab thickness of the trade and risers should be kept the same and equal to span/25 for simply supported and span/30 for continuous stairs.

Figure 9.20.9a shows an indeterminate trade-riser staircase. Here, the analysis can be done by adding the effect of the support moment MA (Fig.9.20.9b) with the results of earlier simply supported case. However, the value of MA can be determined using the moment-area method. The free body diagrams of two vertical risers BC and DE are show in Figs.9.20.9c and d, respectively. 9.20.9 Illustrative Examples Two typical examples of waist-slab and trade-riser types spanning longitudinally are taken up here to illustrate the design.


Example 9.1: Design the waist-slab type of the staircase of Fig.9.20.10. Landing slab A is supported on beams along JK and PQ, while the waist-slab and landing slab B are spanning longitudinally as shown in Fig.9.20.10. The finish loads and live loads are 1 kN/m2 and 5 kN/m2, respectively. Use riser R = 160 mm, trade T = 270 mm, concrete grade = M 20 and steel grade = Fe 415. Solution: With R = 160 mm and T = 270 mm, the inclined length of each step = {(160)2 + (270)2}½ = 313.85 mm. (A) Design of going and landing slab B Step 1: Effective span and depth of slab


The effective span (cls. 33.1b and c) = 750 + 2700 + 1500 + 150 = 5100 mm. The depth of waist slab = 5100/20 = 255 mm. Let us assume total depth of 250 mm and effective depth = 250 – 20 – 6 = 224 mm (assuming cover = 20 mm and diameter of main reinforcing bar = 12 mm). The depth of landing slab is assumed as 200 mm and effective depth = 200 – 20 – 6 = 174 mm. Step 2: Calculation of loads (Fig.9.20.11, sec. 1-1) (i) Loads on going (on projected plan area) (a) Self-weight of waist-slab = 25(0.25)(313.85)/270 = 7.265 kN/m2

(b) Self-weight of steps = 25(0.5)(0.16) = 2.0 kN/m2

(c) Finishes (given) = 1.0 kN/m2 (d) Live loads (given) = 5.0 kN/m2

Total = 15.265 kN/m2

Total factored loads = 1.5(15.265) = 22.9 kN/m2

(ii) Loads on landing slab A (50% of estimated loads) (a) Self-weight of landing slab = 25(0.2) = 5 kN/m2

(b) Finishes (given) = 1 kN/m2

(c) Live loads (given) = 5 kN/m2

Total = 11 kN/m2

Factored loads on landing slab A = 0.5(1.5)(11) = 8.25 kN/m2

(iii) Factored loads on landing slab B = (1.5)(11) = 16.5 kN/m2

The loads are drawn in Fig.9.20.11. Step 3: Bending moment and shear force (Fig. 9.20.11) Total loads for 1.5 m width of flight = 1.5{8.25(0.75) + 22.9(2.7) + 16.5(1.65)} = 142.86 kN


VC = 1.5{8.25(0.75)(5.1 – 0.375) + 22.9(2.7)(5.1 – 0.75 – 1.35) + 16.5(1.65)(1.65)(0.5)}/5.1 = 69.76 kN VD = 142.86 – 69.76 = 73.1 kN The distance x from the left where shear force is zero is obtained from: x = {69.76 – 1.5(8.25)(0.75) + 1.5(22.9)(0.75)}/(1.5)(22.9) = 2.51 m The maximum bending moment at x = 2.51 m is

= 69.76(2.51) – (1.5)(8.25)(0.75)(2.51 – 0.375)

- (1.5)(22.9)(2.51 – 0.75)(2.51 – 0.75)(0.5) = 102.08 kNm. For the landing slab B, the bending moment at a distance of 1.65 m from D = 73.1(1.65) – 1.5(16.5)(1.65)(1.65)(0.5) = 86.92 kNm Step 4: Checking of depth of slab From the maximum moment, we get d = {102080/2(2.76)}½ = 135.98 mm < 224 mm for waist-slab and < 174 mm for landing slabs. Hence, both the depths of 250 mm and 200 mm for waist-slab and landing slab are more than adequate for bending. For the waist-slab, vτ = 73100/1500(224) = 0.217 N/mm2. For the waist-slab of depth 250 mm, k = 1.1 (cl. 40.2.1.1 of IS 456) and from Table 19 of IS 456, cτ = 1.1(0.28) = 0.308 N/mm2. Table 20 of IS 456, maxcτ = 2.8 N/mm2. Since vτ < cτ < maxcτ , the depth of waist-slab as 250 mm is safe for shear. For the landing slab, vτ = 73100/1500(174) = 0.28 N/mm2. For the landing slab of depth 200 mm, k = 1.2 (cl. 40.2.1.1 of IS 456) and from Table 19 of IS 456, cτ = 1.2(0.28) = 0.336 N/mm2 and from Table 20 of IS 456, maxcτ = 2.8 N/mm2. Here also vτ < cτ < maxcτ , so the depth of landing slab as 200 mm is safe for shear. Step 5: Determination of areas of steel reinforcement


(i) Waist-slab: Mu/bd2 = 102080/(1.5)224(224) = 1.356 N/mm2. Table 2 of SP-16 gives p = 0.411. The area of steel = 0.411(1000)(224)/(100) = 920.64 mm2. Provide 12 mm diameter @ 120 mm c/c (= 942 mm2/m). (ii) Landing slab B: Mu/bd2 at a distance of 1.65 m from VD (Fig. 9.20.11) = 86920/(1.5)(174)(174) = 1.91 N/mm2. Table 2 of SP-16 gives: p = 0.606. The area of steel = 0.606(1000) (174)/100 = 1054 mm2/m. Provide 16 mm diameter @ 240 mm c/c and 12 mm dia. @ 240 mm c/c (1309 mm2) at the bottom of landing slab B of which 16 mm bars will be terminated at a distance of 500 mm from the end and will continue up to a distance of 1000 mm at the bottom of waist slab (Fig. 9.20.12). Distribution steel: The same distribution steel is provided for both the slabs as calculated for the waist-slab. The amount is = 0.12(250) (1000)/100 = 300 mm2/m. Provide 8 mm diameter @ 160 mm c/c (= 314 mm2/m). Step 6: Checking of development length and diameter of main bars Development length of 12 mm diameter bars = 47(12) = 564 mm, say 600 mm and the same of 16 mm dia. Bars = 47(16) = 752 mm, say 800 mm. (i) For waist-slab


M1 for 12 mm diameter @ 120 mm c/c (= 942 mm2) = 942(102.08)/920.64 = 104.44 kNm. With V (shear force) = 73.1 kN, the diameter of main bars ≤ {1.3(104440)/73.1}/47 39.5 mm. Hence, 12 mm diameter is o.k. ≤ (ii) For landing-slab B M1 for 16 mm diameter @ 120 mm c/c (= 1675 mm2) = 1675(102.08)/1650.88 = 103.57 kNm. With V (shear force) = 73.1 kN, the diameter of main bars {1.3(103570)/73.1}/47 = 39.18 mm. Hence, 16 mm diameter is o.k.

≤

The reinforcing bars are shown in Fig.9.20.12 (sec. 1-1). (B) Design of landing slab A Step 1: Effective span and depth of slab The effective span is lesser of (i) (1500 + 1500 + 150 + 174), and (ii) (1500 + 1500 + 150 + 300) = 3324 mm. The depth of landing slab = 3324/20 = 166 mm, < 200 mm already assumed. So, the depth is 200 mm.

Step 2: Calculation of loads (Fig.9.20.13) The following are the loads: (i) Factored load on landing slab A(see Step 2 of A @ 50%) = 8.25 kN/m2

(ii) Factored reaction VC (see Step 3 of A) = 69.76 kN as the total load of

one flight (iii) Factored reaction VC from the other flight = 69.76 kN Thus, the total load on landing slab A

= (8.25)(1.5)(3.324) + 69.76 + 69.76 = 180.65 kN


Due to symmetry of loadings, VE = VF = 90.33 kN. The bending moment is maximum at the centre line of EF. Step 3: Bending moment and shear force (width = 1500 mm) Maximum bending moment = (180.65)(3.324)/8 = 75.06 kNm Maximum shear force = 0.5(180.65) = 90.33 kN Step 4: Checking of depth of slab In Step 3 of A, it has been observed that 135.98 mm is the required depth for bending moment = 102.08 kNm. So, the depth of 200 mm is safe for this bending moment of 75.06 kNm. However, a check is needed for shear force. vτ = 90330/1500(174) = 0.347 N/mm2 > 0.336 N/mm2

The above value of cτ = 0.336 N/mm2 for landing slab of depth 200 mm has been obtained in Step 4 of A. However, here cτ is for the minimum tensile steel in the slab. The checking of depth for shear shall be done after determining the area of tensile steel as the value of vτ is marginally higher. Step 5: Determination of areas of steel reinforcement

For Mu/bd2 = 75060/(1.5)(174)(174) = 1.65 N/mm2, Table 2 of SP-16 gives p = 0.512. The area of steel = (0.512)(1000)(174)/100 = 890.88 mm2/m. Provide 12 mm diameter @ 120 mm c/c (= 942 mm2/m). With this area of steel p = 942(100)/1000(174) = 0.541.


Distribution steel = The same as in Step 5 of A i.e., 8 mm diameter @ 160 mm c/c. Step 6: Checking of depth for shear Table 19 and cl. 40.2.1.1 gives: cτ = (1.2)(0.493) = 0.5916 N/mm2. vτ = 0.347 N/mm2 (see Step 3 of B) is now less than cτ (= 0.5916 N/mm2). Since, vτ < cτ < maxcτ , the depth of 200 mm is safe for shear. The reinforcing bars are shown in Fig. 9.20.14. Example 9.2:

Design a trade-riser staircase shown in Fig.9.20.15 spanning longitudinally. Landing slabs are supported on beams spanning transversely. The dimensions of riser and trade are 160 mm and 270 mm, respectively. The finish loads and live loads are 1 kN/m2 and 5 kN/m2, respectively. Use M 20 and Fe 415. Solution: The distribution of loads on landings common to two spans perpendicular to each other shall be done as per cl. 33.2 of IS 456 (50% in each direction), since the going is supported on landing slabs which span transversely. The effective span in the longitudinal direction shall be taken as the distance between two centre lines of landings. (A) Design of going


Step 1: Effective span and depth of slab

Figure 9.20.16 shows the arrangement of the landings and going. The effective span is 4200 mm. Assume the thickness of trade-riser slab = 4200/25 = 168 mm, say 200 mm. The thickness of landing slab is also assumed as 200 mm. Step 2: Calculation of loads (Fig. 9.20.17)


The total loads including self-weight, finish and live loads on projected area of going (1500 mm x 2465 mm) is first determined to estimate the total factored loads per metre run. (i) Self-weight of going (a) Nine units of (0.2)(0.36)(1.5) @ 25(9) = 24.3 kN (b) One unit of (0.27)(0.36)(1.5) @ 25(1) = 3.645 kN (c) Nine units of (0.07)(0.2)(1.5) @ 25(9) = 4.725 kN (ii) Finish loads @ 1 kN/m2 = (1.5)(2.465)(1) = 3.6975 kN (iii) Live loads @ 5 kN/m2 = (1.5)(2.465)(5) = 18.4875 kN Total = 54.855 kN

Factored loads per metre run = 1.5(54.855)/2.465 = 33.38 kN/m (iv) Self-weight of landing slabs per metre run = 1.5(0.2)(25) = 7.5 kN/m (v) Live loads on landings = (1.5)(5) = 7.5 kN/m (vi) Finish loads on landings = (1.5)(1) = 1.5 kN/m Total = 16.5 kN/m Factored loads = 1.5(16.5) = 24.75 kN/m Due to common area of landings only 50 per cent of this load should be considered. So, the loads = 12.375 kN/m. The loads are shown in Fig.9.20.17. Step 3: Bending moment and shear force Total factored loads = 33.38(2.465) + 12.375(0.85 + 0.885) = 103.75 kN VC = {12.375(0.85)(4.2 – 0.425) + 33.38(2.465)(0.885 + 1.2325) + 12.375(0.885)(0.885)(0.5)}/4.2 = 52.09 kN VD = 103.75 – 52.09 = 51.66 kN The distance x from the left support where shear force is zero is now determined:


52.09 – 12.375(0.85) – 32.38(x – 0.85) = 0 or x = {52.09 – 12.375(0.85) + 33.38(0.85)}/33.38 = 2.095 m Maximum factored bending moment at x = 2.095 m is 52.09(2.095) – 12.375(0.85)(2.095 – 0.425) – 33.38(2.095 – 0.85)(2.095 – 0.85)(0.5) = 65.69 kNm Step 4: Checking of depth of slab From the maximum bending moment, we have

d = {(65690)/(1.5)(2.76)}½ = 125.97 mm < 174 mm From the shear force Vu = VA, we get vτ = 52090/(1500)(174) = 0.199 N/mm2. From cl. 40.2.1.1 and Table 19 of IS 456, we have cτ = (1.2)(0.28) = 0.336 N/mm2. Table 20 of IS 456 gives maxcτ = 2.8 N/mm2. Since, vτ < cτ < maxcτ , the depth of 200 mm is accepted. Step 5: Determination of areas of steel reinforcement

Mu/bd2 = 65.69(106)/(1500)(174)(174) = 1.446 N/mm2


Table 2 of SP-16 gives, p = 0.4416, to have Ast = 0.4416(1000)(174)/100 = 768.384 mm2/m. Provide 12 mm diameter bars @ 140 mm c/c (= 808 mm2) in form of closed ties (Fig.9.20.18). Distribution bars: Area of distribution bars = 0.12(1000)(200)/100 = 240 mm2/m. Provide 8 mm diameter bars @ 200 mm c/c. The reinforcing bars are shown in Fig. 9.20.18. (B) Design of landing slab

Step 1: Effective span and depth of slab With total depth D = 200 mm and effective depth d = 174 mm, the effective span (cl. 22.2a) = lesser of (1500 + 150 + 1500 + 174) and (1500 + 150 + 1500 + 300) = 3324 mm. Step 2: Calculation of loads (Fig.9.20.19) (i) Factored load of landing slab A = 50% of Step 2 (iv to vi) @ 12.375 kN/m = 12.375(3.324 = 41.1345 kN (ii) Factored reaction VC from one flight (see Step 3) = 52.09 kN (iii) Factored reaction VC from other flight = 52.09 kN Total factored load = 145.32 kN. Due to symmetry of loads, VG = VH = 72.66 kN. The bending moment is maximum at the centre line of GH. Step 3: Bending moment and shear force (width b = 1500 mm) Maximum bending moment = 145.32(3.324)/8 = 60.38 kNm


Maximum shear force VG = VH = 145.32/2 = 72.66 kN Step 4: Checking of depth of slab From bending moment: d = {60380/(1.5)(2.76)}½ = 120.77 mm < 174 mm. Hence o.k. From shear force: vτ = 72660/(1500)(174) = 0.278 N/mm2

From Step 4 of A: cτ = 0.336 N/mm2, maxcτ = 2.8 N/mm2. Hence, the depth is o.k. for shear also. Step 5: Determination of areas of steel reinforcement

Mu/bd2 = 60380/(1.5)(174)(174) = 1.33 N/mm2

Table 2 of SP-16 gives, p = 0.4022. So, Ast = 0.4022(1000)(174)/100 = 699.828 mm2. Provide 12 mm diameter bars @ 160 mm c/c (= 707 mm2). Distribution steel area = (0.12/100)(1000)(200) = 240 mm2

Provide 8 mm diameter @ 200 mm c/c (= 250 mm2). Step 6: Checking of development length

The moment M1 for 12 mm @ 160 mm c/c (707 mm2) = (707/699.828)60.38 = 60.998 kNm.


The shear force V = 72.66 kN. The diameter of the bar should be less/equal to {(1.3)(60.998)(106)/72.66(103)}/47 = 23.2 mm. Hence 12 mm diameter bars are o.k. Use Ld = 47(12) = 564 mm, = 600 mm (say). The reinforcing bars are shown in Fig. 9.20.20. 9.20.10 Practice Questions and Problems with Answers Q.1: Name five types of staircases based on geometrical configurations. A.1: See sec. 9.20.2. Q.2: Draw a typical flight and show: (a) trade, (b) nosing, (c) riser, (d) waist and

(e) going. A.2: Figures 9.20.2a, b and c (sec. 9.20.3) Q.3: Mention four general considerations for the design of a staircase. A.3: See sec. 9.20.4. Q.4: Draw schematic diagrams of different types of staircases based on different

structural systems. A.4: Figures 9.20.3a, b, c, d, e; Figs.9.20.4a and b; Figs.9.20.5a, b and c (sec.9.20.5). Q.5: Explain the method of determining the effective spans of stairs. A.5: See sec. 9.20.6 (Table 9.1 also). Q.6: Explain the distribution of loadings of open-well stairs and for those where

the landings are embedded in walls. A.6: See sec. 9.20.7. Q.7: Illustrate the simplified analysis of longitudinally spanning free-standing staircases. A.7: See sec. 9.20.8.


Q.8: Design the open-well staircase of Fig.9.20.21. The dimensions of risers and

trades are 160 mm and 270 mm, respectively. The finish loads and live loads are 1 kN/m2 and 5 kN/m2, respectively. Landing A has a beam at the edge while other landings (B and C) have brick walls. Use concrete of grade M 20 and steel of grade Fe 415.

A.8: Solution: In this case landing slab A is spanning longitudinally along sec. 11 of Fig.9.20.21. Landing slab B is common to spans of sec. 11 and sec. 22, crossing at right angles. Distribution of loads on landing slab B shall be made 50 per cent in each direction (cl. 33.2 of IS 456). The effective span for sec. 11 shall be from the centre line of edge beam to centre line of brick wall, while the effective span for sec. 22 shall be from the centre line of landing slab B to centre line of landing slab C (cl. 33.1b of IS 456). (A) Design of landing slab A and going (sec. 11 of Fig.9.20.21) Step 1: Effective span and depth of slab


The effective span = 150 + 2000 + 1960 + 1000 = 5110 mm. The depth of waist slab is assumed as 5110/20 = 255.5 mm, say 250 mm. The effective depth = 250 – 20 – 6 = 224 mm. The landing slab is also assumed to have a total depth of 250 mm and effective depth of 224 mm.

Step 2: Calculation of loads (Fig.9.20.22) (i) Loads on going (on projected plan area) (a) Self weight of waist slab = 25(0.25)(313.85/270) = 7.265 kN/m2

(b) Self weight of steps = 25(0.5)(0.16) = 2.0 kN/m2

(c) Finish loads (given) = 1.0 kN/m2

(d) Live loads (given) = 5.0 kN/m2

Total = 15.265 kN/m2

So, the factored loads = 1.5(15.265) = 22.9 kN/m2

(ii) Landing slab A (a) Self weight of slab = 25(0.25) = 6.25 kN/m2

(b) Finish loads = 1.00 kN/m2

(c) Live loads = 5.00 kN/m2

Total = 12.25 kN/m2

Factored loads = 1.5(12.25) = 18.375 kN/m2

(iii) Landing slab B = 50 per cent of loads of landing slab A = 9.187 kN/m2


The total loads of (i), (ii) and (iii) are shown in Fig.9.22. Total loads (i) going = 22.9(1.96)(2) = 89.768 kN Total loads (ii) landing slab A = 18.375(2.15)(2) = 79.013 kN Total loads (iii) landing slab B = 9.187(1.0)(2) = 18.374 kN Total loads = 187.155 kN The loads are shown in Fig. 9.20.22. Step 3: Bending moment and shear force (width = 2.0 m, Fig. 9.20.22) VP = {79.013(5.11 – 1.075) + 89.768(5.11 – 3.13) + 18.374(0.5)}/5.11 = 98.97 kN VJ = 187.155 – 98.97 = 88.185 kN The distance x where the shear force is zero is obtained from: 98.97 – 79.013 – 22.9(2)(x – 2.15) = 0 or x = 2.15 + (98.97 – 79.013)/22.9(2) = 2.586 m Maximum bending moment at x = 2.586 m (width = 2 m) = 98.97(2.586) – 79.013 – (22.9)(2)(0.436)(0.436)(0.5) = 161.013 kNm Maximum shear force = 98.97 kN Step 4: Checking of depth From the maximum moment d = {161.013(103)/2(2.76)}½ = 170.8 mm < 224 mm. Hence o.k. From the maximum shear force, vτ = 98970/2000(224) = 0.221 N/mm2. For the depth of slab as 250 mm, k = 1.1(cl. 40.2.1.1 of IS 456) and cτ = 1.1(0.28) = 0.308 N/mm2 (Table 19 of IS 456). maxcτ = 2.8 N/mm2 (Table 20 of IS 456). Since, vτ < cτ < maxcτ , the depth of slab as 250 mm is safe. Step 5: Determination of areas of steel reinforcement


Mu/bd2 = 161.013(103)/2(224)(224) = 1.60 N/mm2. Table 2 of SP-16 gives p = 0.494, to have Ast = 0.494(1000)(224)/100 = 1106.56 mm2/m. Provide 12 mm diameter bars @ 100 mm c/c (= 1131 mm2/m) both for landings and waist slab. Distribution reinforcement = 0.12(1000)(250)/100 = 300 mm2/m. Provide 8 mm diameter @ 160 mm c/c (= 314 mm2). Step 6: Checking of development length

Development length of 12 mm diameter bars 7(12) = 564 mm. Provide Ld = 600 mm. For the slabs M1 for 12 mm diameter @ 100 mm c/c = (1131)(161.013)/1106.56 = 164.57 kNm. Shear force = 98.97 kN. Hence, 47 φ ≤ 1.3(164.57)/98.97 ≤ 2161.67 mm or the diameter of main bar φ 45.99 mm. Hence, 12 mm diameter is o.k. The reinforcing bars are shown in Fig.9.20.23.

≤

(B) Design of landing slabs B and C and going (sec. 22 of Fig.9.20.21) Step 1: Effective span and depth of slab The effective span from the centre line of landing slab B to the centre line of landing slab C = 1000 + 1960 + 1000 = 3960 mm. The depths of waist slab and landing slabs are maintained as 250 mm like those of sec. 11.


Step 2: Calculation of loads (Fig.9.20.24)

(i) Loads on going (Step 2(i) of A) = 22.9 kN/m2

(ii) Loads on landing slab B (Step 2(iii)) = 9.187 kN/m2

(iii) Loads on landing slab C (Step 2(iii)) = 9.187 kN/m2

Total factored loads are: (i) Going = 22.9(1.96)(2) = 89.768 kN (ii) Landing slab A = 9.187(1.0)(2) = 18.374 kN (iii) Landing slab B = 9.187(1.0)(2) = 18.374 kN Total = 126.506 kN The loads are shown in Fig.9.20.24. Step 3: Bending moment and shear force (width = 2.0 m, Fig.9.20.24) The total load is 126.506 kN and symmetrically placed to give VG = VH = 63.253 kN. The maximum bending moment at x = 1.98 m (centre line of the span 3.96 m = 63.253(1.98) – 18.374(1.98 – 0.5) – 22.9(2)(0.98)(0.98)(0.5) = 76.05 kNm. Maximum shear force = 63.253 kN. Since the maximum bending moment and shear force are less than those of the other section (maximum moment = 161.013 kNm and maximum shear force = 98.97 kN), the depth of 250 mm here is o.k. Accordingly, the amount of reinforcing bars are determined. Step 4: Determination of areas of steel reinforcement


Mu/bd2 = 76.05(103)/2(224)(224) = 0.76 N/mm2. Table 2 of SP-16 gives

p = 0.221. The area of steel = (0.221)(1000)(224)/100 = 495.04 mm2. Providing 12 mm diameter @ 220 mm c/c gives 514 mm2, however let us provide 12 mm diameter @ 200 mm c/c (565 mm2) as it is easy to detail with 12 mm diameter @ 100 mm c/c for the other section. Distribution bars are same as for sec. 11 i.e., 8 mm diameter @ 160 mm c/c. Step 5: Checking of development length For the slab reinforcement 12 mm dia. @ 200 mm c/c, M1 = (565)(76.05)/495.04 = 86.80 kNm, V = 63.25 kN. So, the diameter of main bar φ

{(1.3)(86.80)(10≤ 3)/(63.25)}/47, i.e., ≤ 37.96 mm. Hence, 12 mm diameter bars are o.k. Distribution steel shall remain the same as in sec. 11, i.e., 8 mm diameter @ 160 mm c/c. The reinforcing bars are shown in Fig.9.20.25. Figures 9.20.23 and 9.20.25 show the reinforcing bars considered separately. However, it is worth mentioning that the common areas (landing B and C) will have the bars of larger areas of either section eliminating the lower bars of other section. 9.20.11 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 9.20.12 Test 20 with Solutions Maximum Marks = 50, Maximum Time = 1 hour Answer all questions. TQ.1: Draw a typical flight and show: (a) trade, (b) nosing, (c) riser, (d) waist and

(e) going. (5 marks)

A.TQ.1: Figures 9.20.2a, b and c (sec. 9.20.3)


TQ.2: Draw schematic diagrams of different types of staircases based on different structural systems. (10 marks)

A.TQ.2: Figures 9.20.3a, b, c, d, e; Figs.9.20.4a and b; Figs.9.20.5a, b and c (sec.9.20.5). TQ.3: Illustrate the simplified analysis of longitudinally spanning free-standing

staircases. (10 marks)

A.TQ.3: See sec. 9.20.8. TQ.4: Design the staircase of illustrative example 9.1 of Fig.9.20.10 if supported

on beams along KQ and LR only making both the landings A and B as cantilevers. Use the finish loads = 1 kN/m2, live loads = 5 kn/m2, riser R = 160 mm, trade T = 270 mm, grade of concrete = M 20 and grade of steel = Fe 415.

(25 marks) A.TQ.4:


Solution: The general arrangement is shown in Fig.9.20.26. With R = 160 mm and T = 270 mm, the inclined length of each step = {(160)2 + (270)2}½ = 313.85 mm. The structural arrangement is that the going is supported from beams along KQ and LR and the landings A and B are cantilevers. Step 1: Effective span and depth of slab As per cl. 33.1a of IS 456, the effective span of going = 3000 mm and as per cl. 22.2c of IS 456, the effective length of cantilever landing slabs = 1350 mm. The depth of waist slab and landing is kept at 200 mm (greater of 3000/20 and 1350/7). The effective depth = 200 – 20 – 6 = 174 mm. Step 2: Calculation of loads

(i) Loads on going (on projected plan area) (a) Self weight of waist-slab = 25(0.20)(313.85)/270 = 6.812 kN/m2

(b) Self weight of steps = 25(0.5)(0.16) = 2.0 kN/m2

(c) Finishes (given) = 1.0 kN/m2

(d) Live loads (given) = 5.0 kN/m2

Total: 14.812 kN/m2

Total factored loads = 1.5(14.812) = 22.218 kN/m2

(ii) Loads on landing slabs A and B (a) Self weight of landing slabs = 25(0.2) = 5 kN/m2

(b) Finishes (given) = 1 kN/m2

(c) Live loads (given) = 5 kN/m2

Total: 11 kN/m2

Total factored loads = 1.5(11) = 16.5 kN/m2. The total loads are shown in Fig.9.20.27.


Step 3: Bending moments and shear forces Here, there are two types of loads: (i) permanent loads consisting of self-weights of slabs and finishes for landings and self-weights of slab, finishes and steps for going, and (ii) live loads. While the permanent loads will be acting everywhere all the time, the live loads can have several cases. Accordingly, five different cases are listed below. The design moments and shear forces will be considered taking into account of the values in each of the cases,. The different cases are (Fig.9.20.28): (i) Permanent loads on going and landing slabs (ii) Live loads on going and landing slabs (iii) Live loads on landing slab A only (iv) Live loads on going only (v) Live loads on landing slabs A and B only The results of VQ, VR, negative bending moment at Q and positive bending moment at T (Fig.9.20.27) are summarized in Table 9.2. It is seen from Table 9.2 that the design moments and shear forces are as follows: (a) Positive bending moment = 25.195 kNm at T for load cases (i) and (iv). (b) Negative bending moment = -22.553 kNm at Q for load cases (i) and (ii). (c) Maximum shear force = 83.4 kN at Q and R for load cases (i) and (ii). Table 9.2 Values of reaction forces and bending moments for different cases of loadings (Example: TQ.4, Figs. 9.20.26 to 9.20.28)

Case VQ (kN) VR (kN) Negative moment at Q

(kNm)

Positive moment at T

(kNm) Permanent loads on going and landings

+51.34

+51.34

-12.302

+12.535

Live loads on going and landings

+32.06

+32.06

-10.251

+2.404

Live loads on landing A only

+18.605

-3.417 -10.251 -5.13

Live loads on +16.875 +16.875 0 +2.66


going only Live loads on landings A and B only

+15.1875 +15.1875 -10.251 -10.251

Critical positive M (i) + (iv)

+68.215 +68.215 -12.302 +25.195

Critical negative M (i) + (ii)

+83.40 +83.40 -22.553 +14.939

Step 4: Checking of depth of slab The depth is checked for the positive moment of 25.195 kNm as the two depths are the same. The effective depth of slab d = {25.195(106)/1500(2.76)}½ = 78 mm < 174 mm. Hence o.k. The nominal shear stress vτ = 83400/1500(174) = 0.3195 N/mm2. Using the value of k = 1.2 (cl. 40.2.11 of IS 456) and from Table 19 of IS 456, we get

cτ = 1.2(0.28) = 0.336 N/mm2 and maxcτ = 2.8 N/mm2 from Table 20 of IS 456. Since, vτ < cτ < maxcτ , the depth of 200 m is safe against shear. Step 5: Determination of areas of steel reinforcement


(i) Waist slab: Mu/bd2 = 25195/1.5(174)(174) = 0.555 N/mm2. Table 2 of SP-16 gives: p = 0.165. Accordingly, Ast = 0.165(1000)(174)/100 = 287.1 mm2/m. Provide 8 mm dia. bars @ 150 mm c/c (= 335 mm2 ). (ii) Landing slab: Since the difference of positive and negative bending moments is not much, same reinforcement bars i.e., 8 mm diameter @ 150 mm c/c is used as positive and negative steel bars of waist and landing slabs. Distribution bars: 0.12(200)(1000)/100 = 240 mm2/m. The same bar i.e., 8 mm diameter @ 150 mm c/c can be used as distribution bar. The extra amount is useful to take care of change of ending moments due to different cases of loadings. The reinforcement bars are shown in Fig.9.20.29. 9.20.13 Summary of this Lesson This lesson explains the different types of staircases based on geometrical considerations. The different terminologies commonly used in a typical flight are mentioned. Important guidelines to be considered at the planning stage of the staircase are discussed. The classification of staircases based on structural system is explained. The distribution of loadings, determination of effective spans and selection of preliminary discussions of trade, riser and depth of slabs are illustrated. Simplified analysis procedures of staircases including free-standing staircase are explained. Several numerical problems are solved in the illustrative examples, practice problem and test, which will help to understand the applications of all the guidelines and the design of different types of staircases.


Module 10

Compression Members Version 2 CE IIT, Kharagpur

Lesson 21

Definitions, Classifications, Guidelines and

Assumptions




• define effective length, pedestal, column and wall, • classify the columns based on types of reinforcement, loadings and

slenderness ratios,

• identify and explain the functions of bracing in a braced column,

• determine the minimum and maximum percentage of longitudinal reinforcement,

• determine the minimum numbers and diameter of bars in rectangular and

circular columns,

• determine the longitudinal reinforcement in a pedestal,

• determine the type, pitch and diameter of lateral ties of columns after determining the longitudinal steel,

• state the assumptions in the design of compression member by limit state

of collapse,

• determine the strain distribution lines of a compression member subjected to axial load with or without the moments about one or both the axes,

• explain the need of the minimum eccentricity to be considered in the

design of compression members. 10.21.1 Introduction

Compression members are structural elements primarily subjected to axial compressive forces and hence, their design is guided by considerations of strength and buckling. Figures 10.21.1a to c show their examples: pedestal, column, wall and strut. While pedestal, column and wall carry the loads along its length l in vertical direction, the strut in truss carries loads in any direction. The letters l, b and D represent the unsupported vertical length, horizontal lest lateral dimension, width and the horizontal longer lateral dimension, depth. These compression members may be made of bricks or reinforced concrete. Herein, reinforced concrete compression members are only discussed.


This module is intended to explain the definition of some common terminologies and to illustrate the design of compression members and other related issues. This lesson, however, explain the definitions and classifications of columns depending on different aspects. Further, the recommendations of IS 456 to be followed in the design are discussed regarding the longitudinal and lateral reinforcing bars. The assumptions made in the design of compression member by limit sate of collapse are illustrated. 10.21.2 Definitions (a) Effective length: The vertical distance between the points of inflection of the compression member in the buckled configuration in a plane is termed as effective length le of that compression member in that plane. The effective length is different from the unsupported length l of the member, though it depends on the unsupported length and the type of end restraints. The relation between the effective and unsupported lengths of any compression member is le = k l (10.1) where k is the ratio of effective to the unsupported lengths. Clause 25.2 of IS 456 stipulates the effective lengths of compression members (vide Annex E of IS 456). This parameter is needed in classifying and designing the compression members. (b) Pedestal: Pedestal is a vertical compression member whose effective length le does not exceed three times of its least horizontal dimension b (cl. 26.5.3.1h, Note). The other horizontal dimension D shall not exceed four times of b (Fig.10.21.1a). (c) Column: Column is a vertical compression member whose unsupported length l shall not exceed sixty times of b (least lateral dimension), if restrained at the two ends. Further, its unsupported length of a cantilever column shall not exceed 100b2/D, where D is the larger lateral dimension which is also restricted up to four times of b (vide cl. 25.3 of IS 456 and Fig.10.21.1b).

(d) Wall: Wall is a vertical compression member whose effective height Hwe to thickness t (least lateral dimension) shall not exceed 30 (cl. 32.2.3 of IS 456). The larger horizontal dimension i.e., the length of the wall L is more than 4t (Fig.10.21.1c).


10.21.3 Classification of Columns Based on Types of Reinforcement


Based on the types of reinforcement, the reinforced concrete columns are classified into three groups: (i) Tied columns: The main longitudinal reinforcement bars are enclosed within closely spaced lateral ties (Fig.10.21.2a). (ii) Columns with helical reinforcement: The main longitudinal reinforcement bars are enclosed within closely spaced and continuously wound spiral reinforcement. Circular and octagonal columns are mostly of this type (Fig.10.21.2b). (iii) Composite columns: The main longitudinal reinforcement of the composite columns consists of structural steel sections or pipes with or without longitudinal bars (Fig.20.21.2c and d). Out of the three types of columns, the tied columns are mostly common with different shapes of the cross-sections viz. square, rectangular, T-, L-, cross etc. Helically bound columns are also used for circular or octagonal shapes of cross-sections. Architects prefer circular columns in some specific situations for the functional requirement. This module, accordingly takes up these two types (tied and helically bound) of reinforced concrete columns.


10.21.4 Classification of Columns Based on Loadings


Columns are classified into the three following types based on the loadings:


(i) Columns subjected to axial loads only (concentric), as shown in Fig.20.21.3a.

(ii) Columns subjected to combined axial load and uniaxial bending, as shown in Fig.10.21.3b.

(iii) Columns subjected to combined axial load and bi-axial bending, as

shown in Fig.10.21.3c.

Figure 10.21.4 shows the plan view of a reinforced concrete rigid frame having columns and inter-connecting beams in longitudinal and transverse directions. From the knowledge of structural analysis it is well known that the bending moments on the left and right of columns for every longitudinal beam will be comparable as the beam is continuous. Similarly, the bending moments at the two sides of columns for every continuous transverse beam are also comparable (neglecting small amounts due to differences of l1, l2, l3 and b1, b2, b3, b4). Therefore, all internal columns (C1a to C1f) will be designed for axial force only. The side columns (C2a to C2j) will have axial forces with uniaxial bending moment, while the four corner columns (C3a to C3d) shall have axial forces with bi-axial bending moments. Thus, all internal columns (C1a to C1f), side columns


(C2a to C2j) and corner columns (C3a to C3d) are the columns of type (i), (ii) and (iii), respectively. It is worth mentioning that pure axial forces in the inside columns is a rare case. Due to rigid frame action, lateral loadings and practical aspects of construction, there will be bending moments and horizontal shear in all the inside columns also. Similarly, side columns and corner columns will have the column shear along with the axial force and bending moments in one or both directions, respectively. The effects of shear are usually neglected as the magnitude is very small. Moreover, the presence of longitudinal and transverse reinforcement is sufficient to resist the effect of column shear of comparatively low magnitude. The effect of some minimum bending moment, however, should be taken into account in the design even if the column is axially loaded. Accordingly, cls. 39.2 and 25.4 of IS 456 prescribes the minimum eccentricity for the design of all columns. In case the actual eccentricity is more than the minimum, that should be considered in the design. 10.21.5 Classification of Columns Based on Slenderness Ratios Columns are classified into the following two types based on the slenderness ratios: (i) Short columns (ii) Slender or long columns


Figure 10.21.5 presents the three modes of failure of columns with different slenderness ratios when loaded axially. In the mode 1, column does not undergo any lateral deformation and collapses due to material failure. This is known as compression failure. Due to the combined effects of axial load and moment a short column may have material failure of mode 2. On the other hand, a slender column subjected to axial load only undergoes deflection due to beam-column effect and may have material failure under the combined action of direct load and bending moment. Such failure is called combined compression and bending failure of mode 2. Mode 3 failure is by elastic instability of very long column even under small load much before the material reaches the yield stresses. This type of failure is known as elastic buckling. The slenderness ratio of steel column is the ratio of its effective length le to its least radius of gyration r. In case of reinforced concrete column, however, IS 456 stipulates the slenderness ratio as the ratio of its effective length le to its least lateral dimension. As mentioned earlier in sec. 10.21.2(a), the effective length le is different from the unsupported length, the rectangular reinforced concrete column of cross-sectional dimensions b and D shall have two effective lengths in the two directions of b and D. Accordingly, the column may have the possibility of buckling depending on the two values of slenderness ratios as given below: Slenderness ratio about the major axis = lex/D Slenderness ratio about the minor axis = ley/b Based on the discussion above, cl. 25.1.2 of IS 456 stipulates the following: A compression member may be considered as short when both the slenderness ratios lex/D and ley/b are less than 12 where lex = effective length in respect of the major axis, D = depth in respect of the major axis, ley = effective length in respect of the minor axis, and b = width of the member. It shall otherwise be considered as a slender compression member. Further, it is essential to avoid the mode 3 type of failure of columns so that all columns should have material failure (modes 1 and 2) only. Accordingly, cl. 25.3.1 of IS 456 stipulates the maximum unsupported length between two restraints of a column to sixty times its least lateral dimension. For cantilever columns, when one end of the column is unrestrained, the unsupported length is restricted to 100b2/D where b and D are as defined earlier.


10.21.6 Braced and unbraced columns

It is desirable that the columns do not have to resist any horizontal loads due to wind or earthquake. This can be achieved by bracing the columns as in the case of columns of a water tank or tall buildings (Figs.10.21.6a and b). Lateral tie members for the columns of water tank or shear walls for the columns of tall buildings resist the horizontal forces and these columns are called braced columns. Unbraced columns are supposed to resist the horizontal loads also. The bracings can be in one or more directions depending on the directions of the lateral loads. It is worth mentioning that the effect of bracing has been taken into account by the IS code in determining the effective lengths of columns (vide Annex E of IS 456). 10.21.7 Longitudinal Reinforcement The longitudinal reinforcing bars carry the compressive loads along with the concrete. Clause 26.5.3.1 stipulates the guidelines regarding the minimum and maximum amount, number of bars, minimum diameter of bars, spacing of bars etc. The following are the salient points: (a) The minimum amount of steel should be at least 0.8 per cent of the gross cross-sectional area of the column required if for any reason the provided area is more than the required area.


(b) The maximum amount of steel should be 4 per cent of the gross cross-sectional area of the column so that it does not exceed 6 per cent when bars from column below have to be lapped with those in the column under consideration. (c) Four and six are the minimum number of longitudinal bars in rectangular and circular columns, respectively. (d) The diameter of the longitudinal bars should be at least 12 mm. (e) Columns having helical reinforcement shall have at least six longitudinal bars within and in contact with the helical reinforcement. The bars shall be placed equidistant around its inner circumference. (f) The bars shall be spaced not exceeding 300 mm along the periphery of the column. (g) The amount of reinforcement for pedestal shall be at least 0.15 per cent of the cross-sectional area provided. 10.21.8 Transverse Reinforcement Transverse reinforcing bars are provided in forms of circular rings, polygonal links (lateral ties) with internal angles not exceeding 135o or helical reinforcement. The transverse reinforcing bars are provided to ensure that every longitudinal bar nearest to the compression face has effective lateral support against buckling. Clause 26.5.3.2 stipulates the guidelines of the arrangement of transverse reinforcement. The salient points are:

(a) Transverse reinforcement shall only go round corner and alternate bars if the longitudinal bars are not spaced more than 75 mm on either side (Fig.10.21.7).


(b) Longitudinal bars spaced at a maximum distance of 48 times the diameter of the tie shall be tied by single tie and additional open ties for in between longitudinal bars (Fig.10.21.8).

(c) For longitudinal bars placed in more than one row (Fig.10.21.9): (i) transverse reinforcement is provided for the outer-most row in accordance with (a) above, and (ii) no bar of the inner row is closer to the nearest compression face than three times the diameter of the largest bar in the inner row.

(d) For longitudinal bars arranged in a group such that they are not in contact and each group is adequately tied as per (a), (b) or (c) above, as


appropriate, the transverse reinforcement for the compression member as a whole may be provided assuming that each group is a single longitudinal bar for determining the pitch and diameter of the transverse reinforcement as given in sec.10.21.9. The diameter of such transverse reinforcement should not, however, exceed 20 mm (Fig.10.21.10). 10.21.9 Pitch and Diameter of Lateral Ties (a) Pitch: The maximum pitch of transverse reinforcement shall be the least of the following:

(i) the least lateral dimension of the compression members; (ii) sixteen times the smallest diameter of the longitudinal

reinforcement bar to be tied; and

(iii) 300 mm.

(b) Diameter: The diameter of the polygonal links or lateral ties shall be not less than one-fourth of the diameter of the largest longitudinal bar, and in no case less than 6 mm. 10.21.10 Helical Reinforcement (a) Pitch: Helical reinforcement shall be of regular formation with the turns of the helix spaced evenly and its ends shall be anchored properly by providing one and a half extra turns of the spiral bar. The pitch of helical reinforcement shall be determined as given in sec.10.21.9 for all cases except where an increased load on the column is allowed for on the strength of the helical reinforcement. In such cases only, the maximum pitch shall be the lesser of 75 mm and one-sixth of the core diameter of the column, and the minimum pitch shall be the lesser of 25 mm and three times the diameter of the steel bar forming the helix. (b) Diameter: The diameter of the helical reinforcement shall be as mentioned in sec.10.21.9b. 10.21.11 Assumptions in the Design of Compression

Members by Limit State of Collapse It is thus seen that reinforced concrete columns have different classifications depending on the types of reinforcement, loadings and slenderness ratios. Detailed designs of all the different classes are beyond the scope here. Tied and helically reinforced short and slender columns subjected to axial loadings with or without the combined effects of uniaxial or biaxial bending


will be taken up. However, the basic assumptions of the design of any of the columns under different classifications are the same. The assumptions (i) to (v) given in sec.3.4.2 of Lesson 4 for the design of flexural members are also applicable here. Furthermore, the following are the additional assumptions for the design of compression members (cl. 39.1 of IS 456).

(i) The maximum compressive strain in concrete in axial compression is taken as 0.002.

(ii) The maximum compressive strain at the highly compressed

extreme fibre in concrete subjected to axial compression and bending and when there is no tension on the section shall be 0.0035 minus 0.75 times the strain at the least compressed extreme fibre.


The assumptions (i) to (v) of section 3.4.2 of Lesson 4 and (i) and (ii) mentioned above are discussed below with reference to Fig.10.21.11a to c presenting the cross-section and strain diagrams for different location of the neutral axis. The discussion made in sec. 3.4.2 of Lesson 4 regarding the assumptions (i), (iii), (iv) and (v) are applicable here also. Assumption (ii) of sec.3.4.2 is also applicable here when kD, the depth of neutral axis from the highly compressed right edge is within the section i.e., k < 1. The corresponding strain profile IN in Fig.10.21.11b is for particular value of P and M such that the maximum compressive strain is 0.0035 at the highly compressed right edge and tensile strain develops at the opposite edge. This strain profile is very much similar to that of a beam in flexure of Lesson 4. The additional assumption (i) of this section refers to column subjected axial load P only resulting compressive strain of maximum (constant) value of 0.002 and for which the strain profile is EF in Fig.10.21.11b. The neutral axis is at infinity (outside the section). Extending the assumption of the strain profile IN (Fig.10.21.11b), we can draw another strain profile IH (Fig.10.21.11c) having maximum compressive strain of 0.0035 at the right edge and zero strain at the left edge. This strain profile 1H along with EF are drawn in Fig.10.21.11c to intersect at V. From the two similar triangles EVI and GHI, we have EV/GH = 0.0015/0.0035 = 3/7, which gives EV = 3D/7 (10.2) The point V, where the two profiles intersect is assumed to act as a fulcrum for the strain profiles when the neutral axis lies outside the section. Another strain profile JK drawn on this figure passing through the fulcrum V and whose neutral axis is outside the section. The maximum compressive strain GJ of this profile is related to the minimum compressive strain HK as explained below. GJ = GI – IJ = GI – 0.75 HK, as we can write IJ in term of HK from two similar triangles JVI and HVK: IJ/HK = VE/VF = 0.75. The value of the maximum compressive strain GJ for the profile JK is, therefore, 0.0035 minus 0.75 times the strain HK on the least compressed edge. This is the assumption (ii) of this section (cl. 39.1b of IS 456).


10.21.12 Minimum Eccentricity Section 10.21.4 illustrates that in practical construction, columns are rarely truly concentric. Even a theoretical column loaded axially will have accidental eccentricity due to inaccuracy in construction or variation of materials etc. Accordingly, all axially loaded columns should be designed considering the minimum eccentricity as stipulated in cl. 25.4 of IS 456 and given below (Fig.10.21.3c) ex min ≥ greater of )l/500 + D/30) or 20 mm (10.3) ey min ≥ greater of )l/500 + b/30) or 20 mm where l, D and b are the unsupported length, larger lateral dimension and least lateral dimension, respectively. 10.21.13 Practice Questions and Problems with Answers Q.1: Define effective length, pedestal, column and wall. A.1: See sec. 10.21.2. Q.2: Classify the columns based on types of reinforcement. A.2: See sec. 10.21.3 Q.3: Classify the columns based on loadings. A.3: See sec. 10.21.4. Q.4: Classify the columns based on slenderness ratios. A.4: See Sec. 10.21.5 Q.5: Explain braced and unbraced columns. A.5: See sec. 10.21.6. Q.6: Answer the following: (a) What are the minimum and maximum amounts of longitudinal

reinforcement in a column?


(b) What are the minimum numbers of longitudinal bars in rectangular and circular columns?

(c) What is the amount of longitudinal reinforcement in a pedestal? (d) What is the maximum pitch of transverse reinforcement in a column? (e) What is the diameter of lateral ties in a column? A.6: (a) 0.8% and 4% (b) 4 and 6 (c) 0.15% of cross-sectional area of the pedestal (d) See sec. 10.21.9(a) (e) See sec. 10.21.9(b). Q.7: Explain the assumptions of determining the strain distribution lines in a

column subjected to axial force and biaxial bending. A.7: See sec. 10.21.11(i) and (ii). Q.8: State the minimum eccentricity of a rectangular column for designing. A.8: See sec. 10.21.12. 10.21.14 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 10.21.15 Test 21 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions carrying equal marks. TQ.1: Define effective length, pedestal, column and wall. A.TQ.1: See sec. 10.21.2. TQ.2: Classify the columns separately based on loadings and slenderness ratios. A.TQ.2: See secs. 10.21.4 and 5. TQ.3: Explain braced and unbraced columns. A.TQ.3: See sec. 10.21.6. TQ.4: Answer the following: (a) What are the minimum and maximum amounts of longitudinal

reinforcement in a column? (b) What are the minimum numbers of longitudinal bars in rectangular

and circular columns?


(c) What is the amount of longitudinal reinforcement in a pedestal? (d) What is the maximum pitch of transverse reinforcement in a column? (e) What is the diameter of lateral ties in a column? A.TQ.4: (a) 0.8% and 4% (b) 4 and 6 (c) 0.15% of cross-sectional area of the pedestal (d) See sec. 10.21.9(a) (e) See sec. 10.21.9(b). TQ.5: Explain the assumptions of determining the strain distribution lines in a

column subjected to axial force and biaxial bending. A.TQ.5: See sec. 10.21.11(i) and (ii). 10.21.16 Summary of this Lesson This lesson defines the effective length, pedestal, column and wall. Three different classifications of columns based on types of reinforcement, loadings slenderness ratio are explained. The need and functions of bracings are illustrated. The guidelines of IS 456 are discussed regarding the types, arrangement, minimum numbers and diameter of bars, pitch and other aspects of longitudinal and transverse reinforcement of columns. The assumptions needed for the design of compression members are illustrated. The determination of strain distribution lines are explained depending on the location of the neutral axis. The need for considering the minimum eccentricity and its amount are explained.


Module 10


Lesson 22

Short Axially Loaded Compression Members




• state additional assumptions regarding the strengths of concrete and steel for the design of short axially loaded columns,

• specify the values of design strengths of concrete and steel,

• derive the governing equation for the design of short and axially loaded

tied columns,

• derive the governing equation for the design of short and axially loaded spiral columns,

• derive the equation to determine the pitch of helix in spiral columns,

• apply the respective equations to design the two types of columns by

direct computation,

• use the charts of SP-16 to design these two types of columns subjected to axial loads as per IS code.

10.22.1 Introduction Tied and helically bound are the two types of columns mentioned in sec.10.21.3 of Lesson 21. These two types of columns are taken up in this lesson when they are short and subjected to axially loads. Out of several types of plan forms, only rectangular and square cross-sections are covered in this lesson for the tied columns and circular cross-section for the helically bound columns. Axially loaded columns also need to be designed keeping the provision of resisting some moments which normally is the situation in most of the practical columns. This is ensured by checking the minimum eccentricity of loads applied on these columns as stipulated in IS 456. Moreover, the design strengths of concrete and steel are further reduced in the design of such columns. The governing equations of the two types of columns and the equation for determining the pitch of the helix in continuously tied column are derived and explained. The design can be done by employing the derived equation i.e., by direct computation or by using the charts of SP-16. Several numerical examples are solved to explain the design of the two types of columns by direct computation and using the charts of SP-16.


10.22.2 Further Assumptions Regarding the Strengths of Concrete and Steel All the assumptions required for the derivation of the governing equations are given in sec.10.21.11 of Lesson 21. The stress-strain diagrams of mild steel (Fe 250) and cold worked deformed bars (Fe 415 and Fe 500) are given in Figs.1.2.3 and 4, respectively of Lesson 2. The stress block of compressive part of concrete is given in Fig.3.4.1.9 of Lesson 4, which is used in the design of beam by limit state of collapse. The maximum design strength of concrete is shown as constant at 0.446 fck when the strain ranges from 0.002 to 0.0035. The maximum design stress of steel is 0.87 fy. Sections 10.21.4 and 12 of Lesson 21 explain that all columns including the short axially loaded columns shall be designed with a minimum eccentricity (cls. 25.4 and 39.2 of IS 456). Moreover, the design strengths of concrete and steel are further reduced to 0.4 fck and 0.67 fy, respectively, to take care of the minimum eccentricity of 0.05 times the lateral dimension, as stipulated in cl.39.3 of IS 456. It is noticed that there is not attempt at strain compatibility. Also the phenomenon of creep has not been directly considered. ex min greater of (l/500 + D/30) or 20 mm ≥ (10.3) ey min greater of (l/500 + b/30) or 20 mm ≥ The maximum values of lex/D and ley/b should not exceed 12 in a short column as per cl.25.1.2 of IS 456. For a short column, when the unsupported length l = lex (for the purpose of illustration), we can assume l = 12 D (or 12b when b is considered). Thus, we can write the minimum eccentricity = 12D/500 + D/30 = 0.057D, which has been taken as 0.05D or 0.05b as the maximum amount of eccentricity of a short column. It is, therefore, necessary to keep provision so that the short columns can resist the accidental moments due to the allowable minimum eccentricity by lowering the design strength of concrete by ten per cent from the value of 0.446fck, used for the design of flexural members. Thus, we have the design strength of concrete in the design of short column as (0.9)(0.446fck) = 0.4014fck, say 0.40 fck. The reduction of the design strength of steel is explained below. For mild steel (Fe 250), the design strength at which the strain is 0.002 is fy/1.15 = 0.87fy. However, the design strengths of cold worked deformed bars (Fe 415 and Fe 500) are obtained from Fig.1.2.4 of Lesson 2 or Fig.23A of IS 456. Table A of SP-16 presents the stresses and corresponding strains of Fe 415 and Fe 500. Use of Table A of SP-16 is desirable as it avoids error while reading from figures (Fig.1.2.4 or Fig.23A, as mentioned above). From Table A of SP-16, the


corresponding design strengths are obtained by making linear interpolation. These values of design strengths for which the strain is 0.002 are as follows: (i) Fe 415: {0.9fyd + 0.05fyd(0.002 – 0.00192)/(0.00241 – 0.00192)} = 0.908fyd = 0.789fy (ii) Fe 500: {0.85fyd + 0.05fyd(0.002 – 0.00195)/(0.00226 – 0.00195)} = 0.859fyd = 0.746fy A further reduction in each of three values is made to take care of the minimum eccentricity as explained for the design strength of concrete. Thus, the acceptable design strength of steel for the three grades after reducing 10 per cent from the above mentioned values are 0.783fy, 0.710fy and 0.671fy for Fe 250, Fe 415 and Fe 500, respectively. Accordingly, cl. 39.3 of IS 456 stipulates 0.67fy as the design strength for all grades of steel while designing the short columns. Therefore, the assumed design strengths of concrete and steel are 0.4fck and 0.67fy, respectively, for the design of short axially loaded columns. 10.22.3 Governing Equation for Short Axially Loaded Tied Columns Factored concentric load applied on short tied columns is resisted by concrete of area Ac and longitudinal steel of areas Asc effectively held by lateral ties at intervals (Fig.10.21.2a of Lesson 21). Assuming the design strengths of concrete and steel are 0.4fck and 0.67fy, respectively, as explained in sec. 10.22.2, we can write Pu = 0.4fck Ac + 0.67fy Asc (10.4) where Pu = factored axial load on the member, fck = characteristic compressive strength of the concrete, Ac = area of concrete, fy = characteristic strength of the compression reinforcement, and Asc = area of longitudinal reinforcement for columns. The above equation, given in cl. 39.3 of IS 456, has two unknowns Ac and Asc to be determined from one equation. The equation is recast in terms of Ag, the gross area of concrete and p, the percentage of compression reinforcement employing


Asc = pAg/100 (10.5) Ac = Ag(1 – p/100) (10.6) Accordingly, we can write Pu/Ag = 0.4fck + (p/100) (0.67fy – 0.4fck) (10.7) Equation 10.7 can be used for direct computation of Ag when Pu, fck and fy are known by assuming p ranging from 0.8 to 4 as the minimum and maximum percentages of longitudinal reinforcement. Equation 10.4 also can be employed to determine Ag and p in a similar manner by assuming p. This method has been illustrated with numerical examples and is designated as Direct Computation Method. On the other hand, SP-16 presents design charts based on Eq.10.7. Each chart of charts 24 to 26 of SP-16 has lower and upper sections. In the lower section, Pu/Ag is plotted against the reinforcement percentage p(= 100As/Ag) for different grades of concrete and for a particular grade of steel. Thus, charts 24 to 26 cover the three grades of steel with a wide range of grades of concrete. When the areas of cross-section of the columns are known from the computed value of Pu/Ag, the percentage of reinforcement can be obtained directly from the lower section of the chart. The upper section of the chart is a plot of Pu/Ag versus Pu for different values of Ag. For a known value of Pu, a horizontal line can be drawn in the upper section to have several possible Ag values and the corresponding Pu/Ag values. Proceeding vertically down for any of the selected Pu/Ag value, the corresponding percentage of reinforcement can be obtained. Thus, the combined use of upper and lower sections of the chart would give several possible sizes of the member and the corresponding Asc without performing any calculation. It is worth mentioning that there may be some parallax error while using the charts. However, use of chart is very helpful while deciding the sizes of columns at the preliminary design stage with several possible alternatives. Another advantage of the chart is that, the amount of compression reinforcement obtained from the chart are always within the minimum and maximum percentages i.e., from 0.8 to 4 per cent. Hence, it is not needed to examine if the computed area of steel reinforcement is within the allowable range as is needed while using Direct Computation Method. This method is termed as SP-16 method while illustrating numerical examples.


10.22.4 Governing Equation of Short Axially Loaded Columns with Helical Ties Columns with helical reinforcement take more load than that of tied columns due to additional strength of spirals in contributing to the strength of columns. Accordingly, cl. 39.4 recommends a multiplying factor of 1.05 regarding the strength of such columns. The code further recommends that the ratio of volume of helical reinforcement to the volume of core shall not be less than 0.36 (Ag/Ac – 1) (fck/fy), in order to apply the additional strength factor of 1.05 (cl. 39.4.1). Accordingly, the governing equation of the spiral columns may be written as Pu = 1.05(0.4 fck Ac + 0.67 fy Asc) (10.8) All the terms have been explained in sec.10.22.3. Earlier observations of several investigators reveal that the effect of containing holds good in the elastic stage only and it gets lost when spirals reach the yield point. Again, spirals become fully effective after spalling off the concrete cover over the spirals due to excessive deformation. Accordingly, the above two points should be considered in the design of such columns. The first point is regarding the enhanced load carrying capacity taken into account by the multiplying factor of 1.05. The second point is maintaining specified ratio of volume of helical reinforcement to the volume of core, as specified in cl.39.4.1 and mentioned earlier. The second point, in fact, determines the pitch p of the helical reinforcement, as explained below with reference to Fig.10.21.2b of Lesson 21. Volume of helical reinforcement in one loop = spspc aD ) - ( φπ (10.9) Volume of core = (10.10)

pDc )4/( 2π

where Dc = diameter of the core (Fig.10.21.2b) spφ = diameter of the spiral reinforcement (Fig.10.21.2b) asp = area of cross-section of spiral reinforcement p = pitch of spiral reinforcement (Fig.10.21.2b) To satisfy the condition of cl.39.4.1 of IS 456, we have


)/( 1) - /0.36( } )4//{(} ) - ({ 2

yckcgcspspc ffAApDaD ≥πφπ which finally gives (10.11)

ckcyspspc fDDfaDp ) - /( ) - 11.1( 22φ≤

Thus, Eqs.10.8 and 11 are the governing equations to determine the diameter of column, pitch of spiral and area of longitudinal reinforcement. It is worth mentioning that the pitch p of the spiral reinforcement, if determined from Eq.10.11, automatically satisfies the stipulation of cl.39.4.1 of IS 456. However, the pitch and diameter of the spiral reinforcement should also satisfy cl. 26.5.3.2 of IS 456:2000. 10.22.5 Illustrative Examples Problem 1: Design the reinforcement in a column of size 400 mm x 600 mm subjected to an axial load of 2000 kN under service dead load and live load. The column has an unsupported length of 4.0 m and effectively held in position and restrained against rotation in both ends. Use M 25 concrete and Fe 415 steel. Solution 1: Step 1: To check if the column is short or slender Given l = 4000 mm, b = 400 mm and D = 600 mm. Table 28 of IS 456 = lex = ley = 0.65(l) = 2600 mm. So, we have lex/D = 2600/600 = 4.33 < 12 ley/b = 2600/400 = 6.5 < 12 Hence, it is a short column. Step 2: Minimum eccentricity ex min = Greater of (lex/500 + D/30) and 20 mm = 25.2 mm ey min = Greater of (ley/500 + b/30) and 20 mm = 20 mm 0.05 D = 0.05(600) = 30 mm > 25.2 mm (= ex min)


0.05 b = 0.05(400) = 20 mm = 20 mm (= ey min) Hence, the equation given in cl.39.3 of IS 456 (Eq.10.4) is applicable for the design here. Step 3: Area of steel Fro Eq.10.4, we have Pu = 0.4 fck Ac + 0.67 fy Asc …. (10.4) 3000(103) = 0.4(25){(400)(600) – Asc} + 0.67(415) Asc which gives, Asc = 2238.39 mm2

Provide 6-20 mm diameter and 2-16 mm diameter rods giving 2287 mm2 (> 2238.39 mm2) and p = 0.953 per cent, which is more than minimum percentage of 0.8 and less than maximum percentage of 4.0. Hence, o.k. Step 4: Lateral ties The diameter of transverse reinforcement (lateral ties) is determined from cl.26.5.3.2 C-2 of IS 456 as not less than (i) φ /4 and (ii) 6 mm. Here, φ = largest bar diameter used as longitudinal reinforcement = 20 mm. So, the diameter of bars used as lateral ties = 6 mm. The pitch of lateral ties, as per cl.26.5.3.2 C-1 of IS 456, should be not more than the least of

(i) the least lateral dimension of the column = 400 mm

(ii) sixteen times the smallest diameter of longitudinal reinforcement bar to be tied = 16(16) = 256 mm

(iii) 300 mm


Let us use p = pitch of lateral ties = 250 mm. The arrangement of longitudinal and transverse reinforcement of the column is shown in Fig. 10.22.1. Problem 2:

Design the column of Problem 1 employing the chart of SP-16. Solution 2: Steps 1 and 2 are the same as those of Problem 1. Step 3: Area of steel Pu/Ag = 3000(103)/(600)(400) = 12.5 N/mm2

From the lower section of Chart 25 of SP-16, we get p = 0.95% when Pu/Ag = 12.5 N/mm2 and concrete grade is M 25. This gives Asc = 0.95(400)(600)/100 = 2288 mm2. The results of both the problems are in good agreement. Marginally higher value of Asc while using the chart is due to parallax error while reading the value from the chart. Here also, 6-20 mm diameter bars + 2-16 mm diameter bars (Asc provided = 2287 mm2) is o.k., though it is 1 mm2 less. Step 4 is the same as that of Problem 1. Figure 10.22.1, thus, is also the figure showing the reinforcing bars (longitudinal and transverse reinforcement) of this problem (same column as that of Problem 1). Problem 3: Design a circular column of 400 mm diameter with helical reinforcement subjected to an axial load of 1500 kN under service load and live load. The column has an unsupported length of 3 m effectively held in position at both ends but not restrained against rotation. Use M 25 concrete and Fe 415 steel.


Solution 3: Step 1: To check the slenderness ratio Given data are: unsupported length l = 3000 mm, D = 400 mm. Table 28 of Annex E of IS 456 gives effective length le = l = 3000 mm. Therefore, le/D = 7.5 < 12 confirms that it is a short column. Step 2: Minimum eccentricity emin = Greater of (l/500 + D/30) or 20 mm = 20 mm

0.05 D = 0.05(400) = 20 mm As per cl.39.3 of IS 456, emin should not exceed 0.05D to employ the equation given in that clause for the design. Here, both the eccentricities are the same. So, we can use the equation given in that clause of IS 456 i.e., Eq.10.8 for the design. Step 3: Area of steel From Eq.10.8, we have Pu = 1.05(0.4 fck Ac + 0.67 fy Asc) … (10.8) Ac = Ag – Asc = 125714.29 - Asc Substituting the values of Pu, fck, Ag and fy in Eq.10.8, 1.5(1500)(103) = 1.05{0.4(25)(125714.29 – Asc) + 0.67(415) Asc} we get the value of Asc = 3304.29 mm2. Provide 11 nos. of 20 mm diameter bars (= 3455 mm2) as longitudinal reinforcement giving p = 2.75%. This p is between 0.8 (minimum) and 4 (maximum) per cents. Hence o.k. Step 4: Lateral ties It has been mentioned in sec.10.22.4 that the pitch p of the helix determined from Eq.10.11 automatically takes care of the cl.39.4.1 of IS 456. Therefore, the pitch is calculated from Eq.10.11 selecting the diameter of helical reinforcement from cl.26.5.3.2 d-2 of IS 456. However, automatic satisfaction of cl.39.4.1 of IS 456 is also checked here for confirmation. Diameter of helical reinforcement (cl.26.5.3.2 d-2) shall be not less than greater of (i) one-fourth of the diameter of largest longitudinal bar, and (ii) 6 mm.


Therefore, with 20 mm diameter bars as longitudinal reinforcement, the diameter of helical reinforcement = 6 mm. From Eq.10.11, we have Pitch of helix p ≤ 11.1(Dc - spφ ) asp fy/(D2 - … (10.11) ckc fD )2

where Dc = 400 – 40 – 40 = 320 mm, spφ = 6 mm, asp = 28 mm2, fy = 415 N/mm2, D = 400 mm and fck = 25 N/mm2. So, p ≤ 11.1(320 – 6) (28) (415)/(4002 – 3202) (25) ≤ 28.125 mm As per cl.26.5.3.2 d-1, the maximum pitch is the lesser of 75 mm and 320/6 = 53.34 mm and the minimum pitch is lesser of 25 mm and 3(6) = 18 mm. We adopt pitch = 25 mm which is within the range of 18 mm and 53.34 mm. So, provide 6 mm bars @ 25 mm pitch forming the helix. Checking of cl. 39.4.1 of IS 456 The values of helical reinforcement and core in one loop are obtained from Eqs.10.8 and 9, respectively. Substituting the values of Dc, spφ , asp and pitch p in the above two equations, we have Volume of helical reinforcement in one loop = 27632 mm3 and Volume of core in one loop = 2011428.571 mm3. Their ratio = 27632/2011428.571 = 0.0137375 0.36(Ag/Ac – 1) (fck/fy) = 0.012198795

It is, thus, seen that the above ratio (0.0137375) is not less than 0.36(Ag/Ac – 1) (fck/fy).


Hence, the circular column of diameter 400 mm has eleven longitudinal bars of 20 mm diameter and 6 mm diameter helix with pitch p = 25 mm. The reinforcing bars are shown in Fig.10.22.2. 10.22.6 Practice Questions and Problems with Answers Q.1: State and explain the values of design strengths of concrete and steel to be

considered in the design of axially loaded short columns. A.1: See sec. 10.22.2. Q.2: Derive the governing equation for determining the dimensions of the

column and areas of longitudinal bars of an axially loaded short tied column.

A.2: See sec. 10.22.2. Q.3: Derive the governing equation for determining the diameter and areas of

longitudinal bars of an axially loaded circular spiral short column. A.3: First and second paragraph of sec. 10.22.4. Q.4: Derive the expression of determining the pitch of helix in a short axially

loaded spiral column which satisfies the requirement of IS 456. A.4: See third paragraph onwards up to the end of sec. 10.22.4. Q.5: Design a short rectangular tied column of b = 300 mm having the maximum

amount of longitudinal reinforcement employing the equation given in


cl.39.3 of IS 456, to carry an axial load of 1200 kN under service dead load and live load using M 25 and Fe 415. The column is effectively held in position at both ends and restrained against rotation at one end. Determine the unsupported length of the column.

A.5: Step 1: Dimension D and area of steel Asc

Substituting the values of Pu = 1.5(1200) = 1800 kN and Asc = 0.04(300)D in Eq.10.4, we have 1800(103) = 0.4(25)(300D)(1 – 0.04) + 0.67(415)(0.04)(300D) we get D = 496.60 mm. Use 300 mm x 500 mm column. Asc = 0.04(300)(500) = 6000 mm2, provide 4-36 mm diameter + 4-25 mm diameter bars to give 4071 + 1963 = 6034 mm2 > 6000 mm2. Step 2: Lateral ties

Diameter of lateral ties shall not be less than the larger of (i) 36/4 = 9 mm and (ii) 6 mm. Use 10 mm diameter bars as lateral ties. Pitch of the lateral ties p shall not be more than the least of (i) 300 mm, (ii) 16(25) = 400 mm and (iii) 300 mm. So, provide 10 mm diameter bars @ 300 mm c/c. The reinforcement bars are shown in Fig.10.22.3.


The centre to centre distance between two corner longitudinal bas along 500 mm direction is 500 – 2(4) + 10 + 18) = 364 mm which is less than 48 (diameter of lateral tie). Hence, the arrangement is satisfying Fig.9 of cl. 26.5.3.2 b-2 of IS 456. Step 3: Unsupported length As per the stipulation in cl. 25.1.2 of IS 456, the column shall be considered as short if lex = 12(D) = 6000 mm and ley = 12(300) = 3600 mm. For the type of support conditions given in the problem, Table 28 of IS 456 gives unsupported length is the least of (i) l = lex/0.8 = 6000/0.8 = 7500 mm and (ii) ley/0.8 = 3600/0.8 = 4500 mm. Hence, the unsupported length of the column is 4.5 m if the minimum eccentricity clause (cl. 39.3) is satisfied, which is checked in the next step. Step 4: Check for minimum eccentricity According to cl. 39.4 of IS 456, the minimum eccentricity of 0.05b or 0.05D shall not exceed as given in cl. 25.4 of IS 456. Thus, we have (i) 0.05(500) = l/500 + 500/30 giving l = 4165 mm (ii) 0.05(300) = l/500 + 300/10 giving l = 2500 mm Therefore, the column shall have the unsupported length of 2.5 m. Q.6: (a) Suggest five alternative dimensions of square short column with the

minimum longitudinal reinforcement to carry a total factored axial load of 3000 kN using concrete of grades 20, 25, 30, 35 and 40 and Fe 415. Determine the respective maximum unsupported length of the column if it is effectively held in position at both ends but not restrained against rotation. Compare the given factored load of the column with that obtained by direct computation for all five alternative columns.

(b) For each of the five alternative sets of dimensions obtained in (a),

determine the maximum factored axial load if the column is having maximum longitudinal reinforcement (i) employing SP-16 and (ii) by direct computation.


A.6:

Solution of Part (a): Step 1: Determination of Ag and column dimensions b (= D) Chart 25 of SP-16 gives all the dimensions of five cases. The two input data are Pu = 3000 kN and 100 As/Ag = 0.8. In the lower section of Chart 25, one horizontal line AB is drawn starting from A where p = 0.8 (Fig.10.22.4) to meet the lines for M 20, 25, 30, 35 and 40 respectively. In Fig.10.22.4, B is the meeting point for M 20 concrete. Separate vertical lines are drawn from these points of intersection to meet another horizontal line CD from the point C where Pu = 3000 kN in the upper section of the figure. The point D is the intersecting point. D happens to be on line when Ag = 3000 cm2. Otherwise, it may be in between two liens with different values of Ag. For M 20, Ag = 3000 cm2. However, in case the point is in between two lines with different values of Ag, the particular Ag has to be computed by linear interpolation. Thus, all five values of Ag are obtained. The dimension b = D = 300000 = 550 mm. Other four values are obtained similarly. Table 10.1 presents the values of Ag and D along with other results as explained below.


Step 2: Unsupported length of each column The unsupported length l is determined from two considerations: (i) Clause 25.1.2 of IS 456 mentions that the maximum effective length lex is 12 times b or D (as b = D here for a square column). The unsupported length is related to the effective length depending on the type of support. In this problem Table 28 of IS 456 stipulates l = lex. Therefore, maximum value of l = 12 D. (ii) The minimum eccentricity of cl. 39.3 should be more than the same as given in cl. 25.4. Assuming them to be equal, we get l/500 + D/30 = D/20, which gives l = 8.33D. For the column using M 20 and Fe 415, the unsupported length = 8.33(550) = 4581 mm. All unsupported lengths are presented in Table 10.1 using the equation l = 8.33 D (1) Step 3: Area of longitudinal steel Step 1 shows that the area provided for the first case is 550 mm x 550 mm = 302500 mm2, slightly higher than the required area of 300000 mm2 for the practical aspects of construction. However, the minimum percentage of the longitudinal steel is to the calculated as 0.8 per cent of area required and not area provided (vide cl. 26.5.3.1 b of IS 456). Hence, for this case Asc = 0.8(300000)/100 = 2400 mm2. Provide 4-25 mm diameter + 4-12 mm diameter bars (area = 1963 + 452 = 2415 mm2). Table 10.1 presents this and other areas of longitudinal steel obtained in a similar manner. Step 4: Factored load by direct computation Equation 10.4 is employed to calculate the factored load by determining Ac from Ag and Ast. With a view to comparing the factored loads, we will use the values of Ag as obtained from the chart and not the Ag actually provided. From Eq.10.4, we have Pu from direct computation = 0.4(fck)(0.992 Ag) + 0.67(fy)(0.008)Ag or Pu = Ag(0.3968 fck + 0.00536 fy) (2) For the first case when Ag = 300000 mm2, fck = 20 N/mm2, and fy = 415 N/mm2, Eq.(2) gives Pu = 3048.12 kN. This value and other values of factored loads obtained from the direct computation are presented in Table 10.1.


Table 10.1 Results of Q.6a (Minimum Longitudinal Steel), given factored Pu = 3000 kN

Gross area of concrete (Ag)

Area of steel (Asc) Concrete grade Require

d (cm2) Provided (cm2)

b = D (cm) Require

d (cm2) Provided (cm2)

Bars

Pu by direct computation

l

(m)

M 20 3000 3025 55 24 24.15 4-25 + 4-12

3048.12 4.581

M 25 2500 2500 50 20 20.60 4-20 + 4-16

3036.10 4.165

M 30 2200 2209 47 17.60 17.85 2.25 + 4-16

3108.25 3.915

M 35 1800 1806 42.5 14.40 14.57 2-28 + 2-12

2900.23 3.540

M 40 1600 1600 40 12.80 13.06 2-20 + 6-12

2895.42 3.332

Solution of Part (b): Step 1: Determination of Pu Due to the known dimensions of the column section, the Ag is now known. With known Ag and reinforcement percentage 100As/Ag as 4 per cent, the factored Pu shall be determined. For the first case, when b = D = 550 mm, Ag = 302500 mm2. In Chart 25, we draw a horizontal line EF from E, where 100As/Ag = 4 in the lower section of the chart (see Fig.10.22.4) to meet the M 20 line at F. Proceeding vertically upward, the line FG intersects the line Ag = 302500 at G. A horizontal line towards left from G, say GH, meets the load axis at H where Pu = 5600 kN. Similarly, Pu for other sets are determined and these are presented in Table 10.2, except for the last case when M 40 is used, as this chart has ended at p = 3.8 per cent. Step 2: Area of longitudinal steel The maximum area of steel, 4 per cent of gross area of column = 0.04(550)(550) = 12100 mm2. Provide 12-36 mm diameter bars to have the actual area of steel = 12214 mm2 > 12100 mm2, as presented in Table 10.2. Step 3: Factored Pu from direct computation From Eq,10.4, as in Step 4 of the solution of Part (a) of this question, we have Pu = 0.4 fck (Ag – Asc) + 0.67 fy Asc (3)


Substituting the values of Ag and Asc actually provided, we get the maximum Pu of the same column when the longitudinal steel is the maximum. For the first case when Ag = 302500 mm2, Asc = 12214 mm2, fck = 20 N/mm2 and fy = 415 N/mm2, we get Pu = 5718.4 kN. This value along with other four values are presented in Table 10.2. Remarks: Tables 10.1 and 10.2 reveal that two sets of results obtained from charts of SP-16 and by direct computation methods are in good agreement. However, values obtained from the chart are marginally different from those obtained by direct computation both on the higher and lower sides. These differences are mainly due to personal error (parallax error) while reading the values with eye estimation from the chart. Table 10.2 Results of Q.6(b) (Maximum Longitudinal Steel) given the respective Ag

Area of steel (Asc) Pu = Factored load Concrete grade

b = D

(cm)

Gross concrete area (As) (cm2)

Required (cm2)

Provided (cm2)

Bars (No.

)

SP-chart (kN)

Direct Computation (kN)

M 20 55 3025 121 122.14 12-36

5600 5718.4

M 25 50 2500 100 101.06 8-36 + 4-25

5200 5208.9

M 30 47 2209 88.36 88.97 8-32 + 4-28

5000 5017.8

M 35 42.5 1806.25 72.25 73.69 12-28

4500 4474.5

M 40 40 1600 64 64.46 8-28 + 4-32

Not available

4249.2

Q.7: Design a short, helically reinforced circular column with minimum amount of

longitudinal steel to carry a total factored axial load of 3000 kN with the same support condition as that of Q.6, using M 25 and Fe 415. Determine its unsupported length. Compare the results of the dimension and area of longitudinal steel with those of Q.6(a) when M 25 and Fe 415 are used.

A.7: Step 1: Diameter of helically reinforced circular column


As per cl. 39.4 of IS 456, applicable for short spiral column, we get from Eq.10.8 Pu = 1.05(0.4 fck Ac + 0.67 fy Asc) …. (10.8) Given data are: Pu = 3000 kN, Ac = π /4 (D2)(0.992), Asc = 0.008(π /4) D2, fck = 25 N/mm2 and fy = 415 N/mm2. So, we have 3000(103) = 1.05(12.1444)(π /4)D2

giving D = 547.2 mm and Ag = 235264.2252 mm2. Provide diameter of 550 mm. Step 2: Area of longitudinal steel Providing 550 mm diameter, the required Ag has been exceeded. Clause 26.5.3.1b stipulates that the minimum amount of longitudinal bar shall be determined on the basis of area required and not area provided for any column. Accordingly, the area of longitudinal steel = 0.008 Ag = 0.008(235264.2252) = 1882.12 mm2. Provide 6-20 mm diameter bars (area = 1885 mm2) as longitudinal steel, satisfying the minimum number of six bar (cl. 26.5.3.1c of IS 456). Step 3: Helical reinforcement

Minimum diameter of helical reinforcement is greater of (i) 20/4 or (ii) 6 mm. So, provide 6 mm diameter bars for the helical reinforcement (cl. 26.5.3.2d-2 of IS 456). The pitch of the helix p is determined from Eq.10.11 as follows: p ≤ 11.1(Dc - spφ ) asp fy/(D2 - ) f2

cD ck …. (10.11)


Using Dc = 550 – 40 – 40 = 470 mm, spφ = 6 mm, asp = 28 mm2, D = 550 mm, fck = 25 N/mm2 and fy = 415 N/mm2, we get p ≤ 11.1(470 – 6)(28)(415)/(5502 – 4702)(25) ≤ 29.34 mm Provide 6 mm diameter bar @ 25 mm c/c as helix. The reinforcement bars are shown in Fig. 10.22.5. Though use of Eq.10.11 automatically checks the stipulation of cl. 39.4.1 of IS 456, the same is checked as a ready reference in Step 4 below. Step 4: Checking of cl. 39.4.1 of IS 456 Volume of helix in one loop = π (Dc - spφ ) asp …. (10.9) Volume of core = (π /4) (p) …. (10.10) 2

cD The ratio of Eq.10.9 and Eq.10.10 = 4(Dc - spφ ) asp/ p 2

cD = 4(470 – 6)(28)/(470)(470)(25) = 0.009410230874 This ratio should not be less than 0.36(Ag/Ac – 1)(fck/fy) = 0.36{(D2/ ) – 1)} (f2

cD ck/fy) = 0.008011039177 Hence, the stipulation of cl. 39.4.1 is satisfied. Step 5: Unsupported length The unsupported length shall be the minimum of the two obtained from (i) short column requirement given in cl. 25.1.2 of IS 456 and (ii) minimum eccentricity requirement given in cls. 25.4 and 39.3 of IS 456. The two values are calculated separately: (i) l = le = 12D = 12(550) = 6600 mm (ii) l/500 + D/30 = 0.05 D gives l = 4583.3 mm So, the unsupported length of this column = 4.58 m. Step 6: Comparison of results Table 10.3 presents the results of required and actual gross areas of concrete and area of steel bars, dimensions of column and number and diameter of longitudinal reinforcement of the helically reinforced circular and the square


columns of Q.6(a) when M 20 and Fe 415 are used for the purpose of comparison. Table 10.3 Comparison of results of circular and square columns with minimum

longitudinal steel (Pu = 3000 kN, M 25, Fe 415)

Gross concrete area Area of steel Column shape and Q.No.

Required (cm2)

Provided (cm2)

Dimension D (cm)

Required (cm2)

Provided (cm2)

Bar dia. and No.

(mm,No.)Circular

(Q.7) 2352.64 2376.78 55 18.82 18.85 6-20

Square (Q.6(a)

2500 2500 50 20 20.6 4-20 + 4-16

10.22.7 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 10.22.8 Test 22 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions carrying equal marks. TQ.1: Derive the expression of determining the pitch of helix in a short axially

loaded spiral column which satisfies the requirement of IS 456. (20 marks)

A.TQ.1: See third paragraph onwards up to the end of sec. 10.22.4. TQ.2: Design a square, short tied column of b = D = 500 mm to carry a total

factored load of 4000 kN using M 20 and Fe 415. Draw the reinforcement diagram.

(30 marks) A.TQ.2: Step 1: Area of longitudinal steel Assuming p as the percentage of longitudinal steel, we have Ac = (500)(500)(1 – 0.01p), Asc = (500)(500)(0.01p), fck = 20 N/mm2 and fy = 415 N/mm2. Using these values in Eq.10.4 Pu = 0.4 fck Ac + 0.67 fy Ast …. (10.4) or 4000000 = 0.4(20)(250000)(1 – 0.01p) + 0.67(415)(250000)(0.01p) we get p = 2.9624, which gives Asc = (2.9624)(500)(500)/100 = 7406 mm2. Use 4-36 + 4-25 + 4-22 mm diameter bars (4071 + 1963 + 1520) = 7554 mm2 > 7406 mm2 as longitudinal reinforcement.


Step 2: Lateral ties Diameter of tie is the greater of (i) 36/4 and (ii) 6 mm. Provide 10 mm diameter lateral ties. The pitch of the lateral ties is the least of (i) 500 mm, (ii) 16(22) = 352 mm, and (iii) 300 mm. Provide 10 mm diameter @ 300 mm c/c. The reinforcement bars are shown in Fig.10.22.6. It is evident that the centre to centre distance between two corner bars = 364 mm is less than 48 times the diameter of lateral ties = 480 mm (Fig.9 of cl. 26.5.3.2b-2 of IS 456). 10.22.9 Summary of this Lesson This lesson illustrates the additional assumptions made regarding the strengths of concrete and steel for the design of short tied and helically reinforced columns subjected to axial loads as per IS 456. The governing equations for determining the areas of cross sections of concrete and longitudinal steel are derived and explained. The equation for determining the pitch of the helix for circular columns is derived. Several numerical examples are solved to illustrate the applications of the derived equations and use of charts of SP-16 for the design of both tied and helically reinforced columns. The results of the same problem by direct computation are compared with those obtained by employing the charts of SP-16 are compared. The differences of results, if any, are discussed. Understanding the illustrative examples and solving the practice and test problems will explain the applications of the equation and use of charts of SP-16 for designing these two types of columns.


Module 10


Lesson 23

Short Compression Members under Axial Load with

Uniaxial Bending




• draw the strain profiles for different locations of the depth of the neutral axis,

• explain the behaviour of such columns for any one of the strain profiles,

• name and identify the three modes of failure of such columns,

• explain the interaction diagram and divide into the three regions indicating

three modes of failure,

• identify the three modes of failure from the depth of neutral axis,

• identify the three modes of failure from the eccentricity of the axial load,

• determine the area of compressive stress block, distance of the centroid of the area of the compressive stress block from the highly compressed edge when the neutral axis is within and outside the cross-section of the column,

• determine the compressive stress of concrete and tensile/compressive

stress of longitudinal steel for any location of the neutral axis within or outside the cross-section of the column,

• write the two equations of equilibrium,

• explain the need to recast the equations in non-dimensional form for their

use in the design of such columns. 10.23.1 Introduction

Short reinforced concrete columns under axial load with uniaxial bending behave in a different manner than when it is subjected to axial load, though columns subjected to axial load can also carry some moment that may appear during construction or otherwise. The behaviour of such columns and the three modes of failure are illustrated in this lesson. It is explained that the moment M, equivalent to the load P with eccentricity e (= M/P), will act in an interactive manner. A particular column with specific amount of longitudinal steel, therefore, can carry either a purely axial load Po (when M = 0), a purely moment Mo (when P = 0) or several pairs of P and M in an interactive manner. Hence, the needed interaction diagram of columns, which is a plot of P versus M, is explained


discussing different positions of neutral axis, either outside or within the cross-section of the column.

Depending on the position of the neutral axis, the column may or may not

have tensile stress to be taken by longitudinal steel. In the compression region however, longitudinal steel will carry the compression load along with the concrete as in the case of axially loaded column. 10.23.2 Behaviour of Short Columns under Axial Load and Uniaxial Moment Normally, the side columns of a grid of beams and columns are subjected to axial load P and uniaxial moment Mx causing bending about the major axis xx, hereafter will be written as M. The moment M can be made equivalent to the axial load P acting at an eccentricity of e (= M/P). Let us consider a symmetrically reinforced short rectangular column subjected to axial load Pu at an eccentricity of e to have Mu causing failure of the column. Figure 10.21.11b of Lesson 21 presents two strain profiles IN and EF. For the strain profile IN, the depth of the neutral axis kD is less than D, i.e., neutral axis is within the section resulting the maximum compressive strain of 0.0035 on the right edge and tensile strains on the left of the neutral axis forming cracks. This column is in a state of collapse for the axial force Pu and moment Mu for which IN is the strain profile. Reducing the eccentricity of the load Pu to zero, we get the other strain profile EF resulting in the constant compressive strain of 0.002, which also is another collapse load. This axial load Pu is different from the other one, i.e., a pair of Pu and Mu, for which IN is the profile. For the strain profile EF, the neutral axis is at infinity (k = α ). Figure 10.21.11c of Lesson 21 presents the strain profile EF with two more strain profiles IH and JK intersecting at the fulcrum point V. The strain profile IH has the neutral axis depth kD = D, while other strain profile JK has kD > D. The load and its eccentricity for the strain profile IH are such that the maximum compressive strain reaches 0.0035 at the right edge causing collapse of the column, though the strains throughout the depth is compressive and zero at the left edge. The strain profile JK has the maximum compressive strain at the right edge between 0.002 and 0.0035 and the minimum compressive strain at the left edge. This strain profile JK also causes collapse of the column since the maximum compressive strain at the right edge is a limiting strain satisfying assumption (ii) of sec. 10.21.10 of Lesson 21. The four strain profiles, IN, EF, JK and IH of Figs.10.21.11b and c, separately cause collapse of the same column when subjected to four different pairs of Pu and Mu. This shows that the column may collapse either due to a uniform constant strain throughout (= 0.002 by EF) or due to the maximum compressive strain at the right edge satisfying assumption (ii) of sec.10.21.10 of


Lesson 21 irrespective of the strain at the left edge (zero for IH and tensile for IN). The positions of the neutral axis and the eccentricities of the load are widely varying as follows: (i) For the strain profile EF, kD is infinity and the eccentricity of the load is zero.

(ii) For the strain profile JK, kD is outside the section (D < kD < α ), with appropriate eccentricity having compressive strain in the section.

(iii) For the strain profile IH, kD is just at the left edge of the section (kD =

D), with appropriate eccentricity having zero and 0.0035 compressive strains at the left and right edges, respectively.

(iv) For the strain profile IN, kD is within the section (kD < D), with

appropriate eccentricity having tensile strains on the left of the neutral axis and 0.0035 compressive strain at the right edge.


It is evident that gradual increase of the eccentricity of the load Pu from zero is changing the strain profiles from EF to JK, IH and then to IN. Therefore, we can accept that if we increase the eccentricity of the load to infinity, there will be only Mu acting on the column. Designating by Po as the load that causes collapse of the column when acting alone and Mo as the moment that also causes collapse when acting alone, we mark them in Fig.10.23.1 in the vertical and horizontal axes. These two points are the extreme points on the plot of Pu versus Mu, any point on which is a pair of Pu and Mu (of different magnitudes) that will cause collapse of the same column having the neutral axis either outside or within the column. The plot of Pu versus Mu of Fig.10.23.1 is designated as interaction diagram since any point on the diagram gives a pair of values of Pu and Mu causing collapse of the same column in an interactive manner. Following the same logic, several alternative column sections with appropriate longitudinal steel bars are also possible for a particular pair of Pu and Mu. Accordingly for the purpose of designing the column, it is essential to understand the different modes of failure of columns, as given in the next section. 10.23.3 Modes of Failure of Columns The two distinct categories of the location of neutral axis, mentioned in the last section, clearly indicate the two types of failure modes: (i) compression failure, when the neutral axis is outside the section, causing compression throughout the section, and (ii) tension failure, when the neutral axis is within the section developing tensile strain on the left of the neutral axis. Before taking up these two failure modes, let us discuss about the third mode of failure i.e., the balanced failure. (A) Balanced failure Under this mode of failure, yielding of outer most row of longitudinal steel near the left edge occurs simultaneously with the attainment of maximum compressive strain of 0.0035 in concrete at the right edge of the column. As a result, yielding of longitudinal steel at the outermost row near the left edge and crushing of concrete at the right edge occur simultaneously. The different yielding strains of steel are determined from the following: (i) For mild steel (Fe 250): yε = 0.87fy/Es (10.12) (ii) For cold worked deformed bars: yε = 0.87fy/Es + 0.002 (10.13)


The corresponding numerical values are 0.00109, 0.0038 and 0.00417 for Fe 250, Fe 415 and Fe 500, respectively. Such a strain profile is known a balanced strain profile which is shown by the strain profile IQ in Fig.10.23.2b with a number 5. This number is shown in Fig.10.23.1 lying on the interaction diagram causing


collapse of the column. The depth of the neutral axis is designated as kbD and shown in Fig.10.23.2b. The balanced strain profile IQ in Fig.10.23.2b also shows the strain yε whose numerical value would change depending on the grade of steel as mentioned earlier. It is also important to observe that this balanced profile IQ does not pass trough the fulcrum point V in Fig.10.23.2b, while other profiles 1, 2 and 3 i.e., EF, LM and IN pass through the fulcrum point V as none of them produce tensile strain any where in the section of the column. The neutral axis depth for the balanced strain profile IQ is less than D, while the same for the other three are either equal to or more than D. To have the balanced strain profile IQ causing balanced failure of the column, the required load and moment are designated as Pb and Mb , respectively and shown in Fig.10.23.1 as the coordinates of point 5. The corresponding eccentricity of the load Pb is defined by the notation eb (= Mb/Pb). The four parameters of the balanced failure are, therefore, Pb, Mb, eb and kb (the coefficient of the neutral axis depth kbD). (B) Compression failure Compression failure of the column occurs when the eccentricity of the load Pu is less than that of balanced eccentricity (e < eb) and the depth of the neutral axis is more than that of balanced failure. It is evident from Fig.10.23.2b that these strain profiles may develop tensile strain on the left of the neutral axis till kD = D. All these strain profiles having 1 > k > kb will not pass through the fulcrum point V. Neither the tensile strain of the outermost row of steel on the left of the neutral axis reaches yε . On the other hand, all strain profiles having kD greater than D pass through the fulcrum point V and cause compression failure (Fig.10.23.2b). The loads causing compression failure are higher than the balanced load Pb having the respective eccentricities less than that of the load of balanced failure. The extreme strain profile is EF marked by 1 in Fig.10.23.2b. Some of these points causing compression failure are shown in Fig.10.23.1 as 1, 2, 3 and 4 having k > kb, either within or outside the section. Three such strain profiles are of interest and need further elaboration. One of them is the strain profile IH (Fig.10.23.2b) marked by point 3 (Fig.10.23.1) for which kD = D. This strain profile develops compressive strain in the section with zero strain at the left edge and 0.0035 in the right edge as explained in sec. 10.23.2. Denoting the depth of the neutral axis by D and eccentricity of the load for this profile by eD, we observe that the other strain profiles LM and EF (Fig.10.23.2b), marked by 2 and 1 in Fig.10.23.1, have the respective kD > D and e < eD.


The second strain profile is EF (Fig.10.23.2b) marked by point 1 in Fig.10.23.1 is for the maximum capacity of the column to carry the axial load Po when eccentricity is zero and for which moment is zero and the neutral axis is at infinity. This strain profile has also been discussed earlier in sec.10.23.2. The third important strain profile LM, shown in Fig.10.23.2b and by point 2 in Figs.10.23.1 and 2, is also due to another pair of collapse Pm and Mm, having the capacity to accommodate the minimum eccentricity of the load, which hardly can be avoided in practical construction or for other reasons. The load Pm , as seen from Fig.10.23.1, is less than Po and the column can carry Pm and Mm in an interactive mode to cause collapse. Hence, a column having the capacity to carry the truly concentric load Po (when M = 0) shall not be allowed in the design. Instead, its maximum load shall be restricted up to Pm (< Po) along with Mm (due to minimum eccentricity). Accordingly, the actual interaction diagram to be used for the purpose of the design shall terminate with a horizontal line 22’ at point 2 of Fig.10.23.1. Point 2 on the interaction diagram has the capacity of Pm with Mm having eccentricity of em (= Mm/Pm) and the depth of the neutral axis is >> D (Fig.10.23.2b). It is thus seen that from points 1 to 5 (i.e., from compression failure to balanced failure) of the interaction diagram of Fig.10.23.1, the loads are gradually decreasing and the moments are correspondingly increasing. The eccentricities of the successive loads are also increasing and the depths of neutral axis are decreasing from infinity to finite but outside and then within the section up to kbD at balanced failure (point 5). Moreover, this region of compression failure can be subdivided into two zones: (i) zone from point 1 to point 2, where the eccentricity of the load is less than the minimum eccentricity that should be considered in the actual design as specified in IS 456, and (ii) zone from point 2 to point 5, where the eccentricity of the load is equal to or more than the minimum that is specified in IS 456. It has been mentioned also that the first zone from point 1 to point 2 should be avoided in the design of column. (C) Tension failure Tension failure occurs when the eccentricity of the load is greater than the balanced eccentricity eb. The depth of the neutral axis is less than that of the balanced failure. The longitudinal steel in the outermost row on the left of the neutral axis yields first. Gradually, with the increase of tensile strain, longitudinal steel of inner rows, if provided, starts yielding till the compressive strain reaches 0.0035 at the right edge. The line IR of Fig.10.23.2b represents such a profile for which some of the inner rows of steel bars have yielded and compressive strain has reached 0.0035 at the right edge. The depth of the neutral axis is designated by (kminD).


It is interesting to note that in this region of the interaction diagram (from 5 to 6 in Fig.10.23.1), both the load and the moment are found to decrease till point 6 when the column fails due to Mo acting alone. This important behaviour is explained below starting from the failure of the column due to Mo alone at point 6 of Fig.10.23.1. At point 6, let us consider that the column is loaded in simple bending to the point (when M = Mo) at which yielding of the tension steel begins. Addition of some axial compressive load P at this stage will reduce the previous tensile stress of steel to a value less than its yield strength. As a result, it can carry additional moment. This increase of moment carrying capacity with the increase of load shall continue till the combined stress in steel due to additional axial load and increased moment reaches the yield strength. 10.23.4 Interaction Diagram It is now understood that a reinforced concrete column with specified amount of longitudinal steel has different carrying capacities of a pair of Pu and Mu before its collapse depending on the eccentricity of the load. Figure 10.23.1 represents one such interaction diagram giving the carrying capacities ranging from Po with zero eccentricity on the vertical axis to Mo (pure bending) on the horizontal axis. The vertical axis corresponds to load with zero eccentricity while the horizontal axis represents infinite value of eccentricity. A radial line joining the origin O of Fig.10.23.1 to point 2 represents the load having the minimum eccentricity. In fact, any radial line represents a particular eccentricity of the load. Any point on the interaction diagram gives a unique pair of Pu and Mu that causes the state of incipient failure. The interaction diagram has three distinct zones of failure: (i) from point 1 to just before point 5 is the zone of compression failure, (ii) point 5 is the balanced failure and (iii) from point 5 to point 6 is the zone of tension failure. In the compression failure zone, small eccentricities produce failure of concrete in compression, while large eccentricities cause failure triggered by yielding of tension steel. In between, point 5 is the critical point at which both the failures of concrete in compression and steel in yielding occur simultaneously. The interaction diagram further reveals that as the axial force Pu becomes larger the section can carry smaller Mu before failing in the compression zone. The reverse is the case in the tension zone, where the moment carrying capacity Mu increases with the increase of axial load Pu. In the compression failure zone, the failure occurs due to over straining of concrete. The large axial force produces high compressive strain of concrete keeping smaller margin available for additional compressive strain line to bending. On the other hand, in the tension failure zone, yielding of steel initiates failure. This tensile yield stress reduces with the additional compressive stress due to additional axial load. As a


result, further moment can be applied till the combined stress of steel due to axial force and increased moment reaches the yield strength. Therefore, the design of a column with given Pu and Mu should be done following the three steps, as given below: (i) Selection of a trial section with assumed longitudinal steel,

(ii) Construction of the interaction diagram of the selected trial column section by successive choices of the neutral axis depth from infinity (pure axial load) to a very small value (to be found by trial to get P = 0 for pure bending),

(iii) Checking of the given Pu and Mu, if they are within the diagram. We will discuss later whether the above procedure should be followed or not. Let us first understand the corresponding compressive stress blocks of concrete for the two distinct cases of the depth of the neutral axis: (i) outside the cross-section and (ii) within the cross-section in the following sections. 10.23.5 Compressive Stress Block of Concrete when the

Neutral Axis Lies Outside the Section


Figure 10.23.3c presents the stress block for a typical strain profile JK having neutral axis depth kD outside the section (k > 1). The strain profile JK in Fig.10.23.3b shows that up to a distance of 3D/7 from the right edge (point AO), the compressive strain is 0.002 and, therefore, the compressive stress shall remain constant at 0.446f

≥ck. The remaining part of the column section of length

4D/7, i.e., up to the left edge, has reducing compressive strains (but not zero). The stress block is, therefore, parabolic from AO to H which becomes zero at U (outside the section). The area of the compressive stress block shall be obtained subtracting the parabolic area between AO to H from the rectangular area between G and H. To establish the expression of this area, it is essential to know the equation of the parabola between AO and U, whose origin is at AO. The positive coordinates of X and Y are measured from the point AO upwards and to the left, respectively. Let us assume that the general equation of the parabola as


X = aY2 + bY + c (10.14) The values of a, b and c are obtained as follows: (i) At Y = 0, X = 0, at the origin: gives c = 0 (ii) At Y = 0, dX/dY = 0, at the origin: gives b = 0 (iii) At Y = (kD – 3D/7), i.e., at point U, X = 0.446fck: gives a = 0.446fck/D2(k-3/7)2. Therefore, the equation of the parabola is: X = {0.446fck/D2(k – 3/7)2}Y2 (10.15) The value of X at the point H (left edge of the column), g is now determined from Eq.10.15 when Y = 4D/7, which gives g = 0.446 fck {4/(7k – 3)}2 (10.16) Hence, the area of the compressive stress block = 0.446 fck D [1 – (4/21){4/(7k – 3)}2] = C1 fck D (10.17) where C1 = 0.446[1 – (4/21){4/(7k – 3)}2] (10.18) Equation 10.17 is useful to determine the area of the stress block for any value of k > 1 (neutral axis outside the section) by substituting the value of C1 from Eq.10.18. The symbol C1 is designated as the coefficient for the area of the stress block. The position of the centroid of the compressive stress block is obtained by dividing the moment of the stress block about the right edge by the area of the stress block. The moment of the stress block is obtained by subtracting the moment of the parabolic part between AO and H about the right edge from the moment of the rectangular stress block of full depth D about the right edge. The expression of the moment of the stress block about the right edge is: 0.446 fck D(D/2) – (1/3)(4D/7) 0.446 fck {4/(7k – 3)}2 {3D/7 + (3/4)(4D/7)}


= 0.446 fck D2 [(1/2) – (8/49){4/(7k – 3)}2] (10.19) Dividing Eq.10.19 by Eq.10.17, we get the distance of the centroid from the right edge is: D[(1/2) – (8/49){4/(7k – 3)}2]/[1 – (4/21){4/(7k – 3)}2] (10.20) = C2 D (10.21) where C2 is the coefficient for the distance of the centroid of the compressive stress block of concrete measured from the right edge and is: C2 = [(1/2) – (8/49){4/(7k – 3)}2]/[1 – (4/21){4/(7k – 3)}2] (10.22) Table 10.4 presents the values of C1 and C2 for different values of k greater than 1, as given in Table H of SP-16. For a specific depth of the neutral axis, k is known. Using the corresponding values of C1 and C2 from Table 10.4, area of the stress block of concrete and the distance of centroid from the right edge are determined from Eqs.10.17 and 10.21, respectively. Table 10.4 Stress block parameters C1 and C2 when the neutral axis is outside the section

K C1 C21.00 1.05

0.361 0.374

0.416 0.432

1.10 1.20

0.384 0.399

0.443 0.458

1.30 1.40

0.409 0.417

0.468 0.475

1.50 2.00

0.422 0.435

0.480 0.491

2.50 3.00

0.440 0.442

0.495 0.497

4.00 0.444 0.499 It is worth mentioning that the area of the stress block is 0.446fckD and the distance of the centroid from the right edge is 0.5D, when k is infinite. Values of C1 and C2 at k = 4 are very close to those when k = ∞ . In fact, for the practical


interaction diagrams, it is generally adequate to consider values of k up to about 1.2. 10.23.6 Determination of Compressive Stress Anywhere in

the Section when the Neutral Axis Lies outside the Section

The compressive stress of concrete at any point between G and AO of Fig.10.23.3c is constant at 0.446fck as the strain in this zone is equal to or greater than 0.002. So, we can write fc = 0.446fck if 0.002 ≤≤ cε 0.0035 (10.23) However, compressive stress of concrete between AO and H is to be determined using the equation of parabola. Let us determine the concrete stress fc at a distance of Y from the origin AO. From Fig.10.23.3c, we have fc = 0.446 fck - gc (10.24) where gc is as shown in Fig.10.23.3c and obtained from Eq.10.15. Thus, we get fc = 0.446 fck – {0.446 fck/D2(k – 3/7)2}Y2

or fc = 0.446 fck {1 – Y2/(kD – 3D/7)2} (10.25) Designating the strain of concrete at this point by cε (Fig.10.23.3b), we have from similar triangles cε /0.002 = 1 – Y/(kD – 3D/7) which gives Y = {1 – (ε /0.002)}(kD – 3D/7) (10.26) Substituting the value of Y from Eq.10.26 in Eq.10.25, we have fc = 0.446 fck [2( cε /0.002) - ( cε /0.002)2], if 0 c ε≤ < 0.002 (10.27)


10.23.7 Compressive Stress Block of Concrete when the Neutral Axis is within the Section

Figure 10.23.4c presents the stress block for a typical strain profile IN having neutral axis depth = kD within the section (k < 1). The strain profile IN in Fig.10.23.4b shows that from a to AO, i.e., up to a distance of 3kD/7 from the right edge, the compressive strain is 0.002 and, therefore, the compressive ≥


stress shall remain constant at 0.446fck. From AO to U, i.e., for a distance of 4kD/7, the strain is reducing from 0.002 to zero and the stress in this zone is parabolic as shown in Fig.10.23.4c. The area of the stress block shall be obtained subtracting the parabolic area between AO and U from the total rectangular area between G and U. As in the case when the neutral axis is outside the section (sec.10.23.5), we have to establish the equation of the parabola with AO as the origin and the positive coordinates X and Y are measured from the point AO upwards for X and from the point AO to the left for Y, as shown in Fig.10.23.4c. Proceeding in the same manner as in sec.10.23.5 and assuming the same equation of the parabola as in Eq.10.14, the values of a, b and c are obtained as: (i) At Y = 0, X = 0, at the origin: gives c = 0 (ii) At Y = 0, dX/dY = 0, at the origin: gives b = 0 (iii) At U, (i.e., at Y = 4kD/7), X = 0.446 fck: gives a = 0.446 fck/(4kD/7)2. Therefore, the equation of the parabola OR is: X = {0.446 fck/(4kD/7)2}Y2 (10.28) The area of the stress block = 0.446 fck kD – (1/3) 0.446 fck (4kD/7) = 0.36 fck kD, the same as obtained earlier in Eq.3.9 of Lesson 4 for flexural members. Similarly, the distance of the centroid can be obtained by dividing the moment of area of stress block about the right edge by the area of the stress block. The result is the same as in Eq.3.12 for the flexural members. Therefore, we have Area of the stress block = 0.36 fck kD (10.29) The distance of the centroid of the stress block from the right edge = 0.42kD (10.30) Thus, the values of C1 and C2 of Eqs.10.17 and 10.21, respectively, are 0.36 and 0.42 when the neutral axis is within the section. It is to be noted that the coefficients C1 and C2 are multiplied by Dfck and D, respectively when the neutral axis is outside the section. However, they are to be multiplied here, when the neutral axis is within the section, by kDfck and kD, respectively. It is further to note that though the expressions of the area of stress block and the distance of the centroid of the stress block from the right edge are the same as those for the flexural members, the important restriction of the maximum depth of the neutral axis xumax in the flexural members is not applicable in case of column. By this restriction, the compression failure of the flexural members is


avoided. In case of columns, compression failure is one of the three modes of failure. 10.23.8 Determination of Compressive Stress Anywhere in

the Compressive Zone when the Neutral Axis is within the Section

The compressive stress at any point between G and AO of Fig.10.23.4c is constant at 0.446fck as the strain in this zone is equal to or greater than 0.002. So, we can write

fc = 0.446 fck if 0.002 ≤≤ cε 0.0035 …. (10.23) However, the compressive stress between AO and U is to be determined from the equation of the parabola. Let us determine the compressive stress fci at a distance of Y from the origin AO. From Fig.10.23.4c, we have fc = 0.446 fck - gc (10.31) where gc as shown in Fig.10.23.4c, is obtained from Eq.10.28. Thus, we get, fc = {0.446 fck – 0.446 fck (4kD/7)2}Y2 (10.32) Designating the strain of concrete at this point by cε (Fig.10.23.4b), we have from similar triangles cε /0.002 = 1 – Y/(4kD/7), which gives Y = {1 - cε /0.002}(4kD/7) (10.33) Substituting the value of Y from Eq.10.33 in Eq.10.32, we get the same equation, Eq.10.27 of sec.10.23.6, when the neutral axis is outside the section. Therefore, fc = 0.446 fck [2( cε /0.002) – ( cε /0.002)2] …. (10.26) From the point U to the left edge H of the cross-section of the column, the compressive stress is zero. Thus, we have fc = 0 if cε 0 ≤


fc = 0.446 fck if cε 0.002 ≥ fc = 0.446 fck{2( cε /0.002) – ( cε /0.002)2}, if 0 ≤ cε < 0.002 (10.34) 10.23.9 Tensile and Compressive Stresses of Longitudinal Steel Stresses are compressive in all the six rows (A1 to A6 of Figs.10.23.3a and c) of longitudinal steel provided in the column when the neutral axis depth kD D. However, they are tensile on the left side of the neutral axis and compressive on the right side of the neutral axis (Figs.10.23.4a and c) when kD < D. These compressive or tensile stresses of longitudinal steel shall be calculated from the strain

≥

siε at that position of the steel which is obtained from the strain profile considered for the purpose. It should be remembered that the linear strain profiles are based on the assumption that plane sections remain plane. Moreover, at the location of steel in a particular row, the strain of steel siε shall be the same as that in the adjacent concrete ciε . Thus, the strain of longitudinal steel can be calculated from the particular strain profile if the neutral axis is within or outside the cross-section of the column. The corresponding stresses of longitudinal steel are determined from the strain

sif

siε (which is the same as that of ciε in the adjacent concrete) from the respective stress-strain diagrams of mild steel (Fig.1.2.3 of Lesson 2) and High Yield Strength Deformed bars (Fig.1.2.4 of Lesson2). The values are summarized in Table 10.5 below as presented in Table A of SP-16. Table 10.5 Values of compressive or tensile from known values of sif siε of

longitudinal steel (Fe 250, Fe 415 and Fe 500)

Fe 250 Fe 415 Fe 500 Strain

siε Stress

(N/mm2) sif

Strain siε

Stress (N/mm2)

sif

Strain siε

Stress (N/mm2)

sif < 0.00109 siε (Es) < 0.00144 siε (Es) < 0.00174 siε (Es) ≥ 0.00109 217.5

(= 0.87 fy) 0.00144 288.7 0.00174 347.8

0.00163 306.7 0.00195 369.6 0.00192 324.8 0.00226 391.3

0.00241 342.8 0.00277 413.0


0.00276 351.8 0.00312 423.9 0.00380 360.9 0.00417 434.8

Notes: 1. Linear interpolation shall be done for intermediate values. 2. Strain at initial yield = fy/Es 3. Strain at final yield = fy/Es + 0.002 10.23.10 Governing Equations A column subjected to Pu and Mu (= Pu e) shall satisfy the two equations of equilibrium, viz., ∑V = 0 and ∑M = 0, taking moment of vertical forces about the centroidal axis of the column. The two governing equation are, therefore, Pu = Cc + Cs (10.35) Mu = Cc (appropriate lever arm) + Cs (appropriate lever arm) (10.36) where Cc = Force due to concrete in compression

Cs = Force due to steel either in compression when kD D or in tension and compression when kD < D

≥

However, two points are to be remembered while expanding the equation ∑V = 0. The first is that while computing the force of steel in compression, the force of concrete that is not available at the location of longitudinal steel has to be subtracted. The second point is that the total force of steel shall consist of the summation of forces in every row of steel having different stresses depending on the respective distances from the centroidal axis. These two points are also to be considered while expanding the other equation ∑M = 0. Moreover, negative sign should be used for the tensile force of steel on the left of the neutral axis when kD < D. It is now possible to draw the interaction diagram of a trial section for the given values of Pu and Mu following the three steps mentioned in sec.10.23.4. However, such an attempt should be avoided for the reason explained below. It has been mentioned in sec.10.23.2 that any point on the interaction diagram gives a pair of values of Pu and Mu causing collapse. On the other hand, it is also true that for the given Pu and Mu, several sections are possible. Drawing of interaction diagrams for all the trial sections is time consuming. Therefore, it is necessary to recast the interaction diagram selecting appropriate non-dimensional parameters instead of Pu versus Mu as has been explained in this lesson. Non-dimensional interaction diagram has the advantage of selecting alternative sections quickly for a given pair of Pu and Mu. It is worth mentioning


that all the aspects of the behaviour of column and the modes of failure shall remain valid in constructing the more versatile non-dimensional interaction diagram, which is taken up in Lesson 24. 10.23.11 Practice Questions and Problems with Answers Q.1: Draw four typical strain profiles of a short, rectangular and symmetrically

reinforced concrete column causing collapse subjected to different pairs of Pu and Mu when the depths of the neutral axis are (i) less than the depth of column D, (ii) equal to the depth of column D, (iii) D < kD < ∞ and (iv) kD = ∞ . Explain the behaviour of column for each of the four strain profiles.

A.1: See sec. 10.23.2. Q.2: Name and explain the three modes of failures of short, rectangular and

symmetrically reinforced concrete columns subjected to axial load Pu uniaxial moment Mu.

A.2: See sec.10.23.3. Q.3: Draw a typical interaction diagram, and explain the three zones

representing three modes of failure of a short, rectangular and symmetrically reinforced concrete column subjected to axial load Pu and uniaxial moment Mu.

A.3: See sec.10.23.4. Q.4: (a) Draw the compressive stress block of concrete of a short, rectangular

and symmetrically reinforced concrete column subjected to axial load Pu and uniaxial moment Mu, when the neutral axis lies outside the section.

(b) Derive expressions of determining the area of the compressive stress

block of concrete and distance of the centroid of the compressive stress block from the highly compressed right edge for a column of Q4(a).

A.4: See sec.10.23.5. Q.5: Derive expression of determining the stresses anywhere within the section

of a column of Q4. A.5: See sec.10.23.6. Q.6: (a) Draw the compressive stress block of concrete of a short, rectangular

and symmetrically reinforced concrete column subjected to axial load Pu and uniaxial moment Mu, when the neutral axis is within the section.


(b) Derive expressions of determining the area of the compressive stress block of concrete and distance of the centroid of the compressive stress block from the compressed right edge for a column of Q6(a).

A.6: See sec.10.23.7. Q.7: Derive expression of determining the compressive stress in the

compression zone of a column of Q6. A.7: See sec.10.23.8. Q.8: Explain the principle of determining the stresses (both tensile and

compressive) of longitudinal steel of a short, rectangular and symmetrically reinforced concrete column subjected to axial load Pu and uniaxial moment Mu.

A.8: See sec.10.23.9. Q.9: (a) Write the governing equations of equilibrium of a short, rectangular and

symmetrically reinforced concrete column subjected to axial load Pu and uniaxial moment Mu.

(b) Would you use the equations of equilibrium for the design of a short,

rectangular and symmetrically reinforced concrete column for a given pair of Pu and Mu? Justify your answer.

A.9: See sec.10.23.10. 10.23.12 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 10.23.13 Test 23 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Each of the following statements has four possible answers. Choose the

correct answer (2 x 5 = 20 marks)

(a) The designed axial load of a short column has the theoretical carrying

capacity before it collapses (i) P = Po only as obtained from the interaction diagram on the vertical axis. (ii) P = Designed axial load with the code stipulated minimum eccentricity only. (iii) A pair of Pb and Mb only. (iv) All of the above. A.TQ.1a: (iv)

(b) A short column in compression failure due to an axial load Pu and uniaxial moment Mu may have


(i) kD = 0 and e = 0 (ii) kD = ∞ and e = 0 (iii) kD = 0 and e = ∞ (iv) kD = ∞ and e = ∞ A.TQ.1b: (ii) (c) The fulcrum of the strain profile of a short column is a point through which (i) The strain profiles causing compression failure will pass. (ii) The strain profile causing balanced failure will pass.

(iii) The strain profiles having no tension and causing compression failure will pass.

(iv) The strain profiles causing tension failure will pass. A.TQ.1c: (iii)

(d) The maximum compressive strain of concrete in balanced failure of a short column subjected to Pb and Mb is

(i) 0.0035 (ii) 0.0035 minus 0.75 times the tensile strain of steel (iii) 0.002 (iv) None of the above A.TQ.1d: (i) TQ.2: (a) Draw the compressive stress block of concrete of a short, rectangular

and symmetrically reinforced concrete column subjected to axial load Pu and uniaxial moment Mu, when the neutral axis is within the section.

(b) Derive expressions of determining the area of the compressive stress

block of concrete and distance of the centroid of the compressive stress block from the compressed right edge for a column of TQ.2 (a). (10 + 20 = 30)


A.TQ.2: See sec.10.23.7. 10.23.14 Summary of this Lesson Illustrating the behaviour of short, rectangular and symmetrically reinforced rectangular columns under axial load Pu and uniaxial bending Mu, this lesson explains the three modes of failure and the interaction diagram of such columns. The different possible strain profiles, and the compressive stress blocks are drawn and explained when the neutral axis is within and outside the cross-section of the column. Determination of compressive stresses of concrete and tensile/compressive stresses of longitudinal steel are explained. The governing equations of equilibrium are introduced to illustrate the need for recasting them in non-dimensional form for the purpose of design of such columns.


Module 10


Lesson 24

Preparation of Design Charts




• identify a design chart and understand the differences between a design chart and interaction diagram of P and M,

• name the major design parameters of short columns subjected to axial

loads and uniaxial bending,

• state the design parameters assumed before the design,

• state the design parameter actually designed for the column,

• explain the roles of each of the design parameters in increasing the strength capacities of column,

• name the two non-dimensional design parameters to prepare the design

charts,

• derive the governing equations in four separate cases while preparing the design charts,

• mention the various points at which the values of the two non-dimensional

parameters are determined to prepare the design charts,

• prepare the design chart of any short and rectangular column subjected to axial loads and uniaxial moment.

10.24.1 Introduction

Lesson 23 illustrates the different steps of determining the capacities of a short, rectangular, reinforced with steel bars, concrete column. Several pairs of collapse strengths Pu and Mu are to be determined for a column with specific percentage of longitudinal steel bars assuming different positions of the neutral axis. A designer has to satisfy that each of the several pairs of Pu and Mu, obtained from the structural analysis, is less than or equal to the respective strengths in form of pairs of Pu and Mu obtained from determining the capacities for several locations of the neutral axis. Thus, the design shall involve several trials of a particular cross-section of a column for its selection. On the other hand, it is also possible to prepare non-dimensional interaction diagram selecting appropriate non-dimensional parameters. This would help to get several possible cross-sections with the respective longitudinal


steel bars. This lesson explains the preparation of such non-dimensional interaction diagrams which are also known as design charts. Similar design charts of circular and other types of cross-sections can be prepared following the same procedure as that of rectangular cross-section. However, the stress block parameters, explained in Lesson 23, are to be established separately by summing up the forces and moment of several strips by dividing the cross-section of columns into the strips. This lesson is restricted to columns of rectangular cross-section which are symmetrically reinforced. 10.24.2 Design Parameters The following are the four major design parameters to be determined for any column so that it has sufficient pairs of strengths (Pu and Mu) to resist all critical pairs obtained from the analysis: (i) dimensions b and D of the rectangular cross-section,

(ii) longitudinal steel reinforcing bars - percentage p, nature of distribution

(equally on two or four sides) and d'/D, (iii) grades of concrete and steel, and (iv) transverse reinforcement.

The roles and importance of each of the above four parameters are elaborated below: (i) Dimensions b and D of the rectangular cross-section The strength of column depends on the two dimensions b and D. However, preliminary dimensions of b and D are already assumed for the analysis of structure, which are usually indeterminate statically. In the subsequent redesign, these dimensions may be revised, if needed, inviting re-analysis with the revised dimensions. (ii) Longitudinal steel reinforcing bars It is a very important consideration to utilise the total area of steel bars effectively. The total area of steel, expressed in percentage p ranges from the minimum 0.8 to the maximum 4 per cent of the gross area of the cross-section. The bars may be distributed either equally on two sides or on all four sides judiciously having two or multiple rows of steel bars. The strain profiles of Fig.10.23.2 reveals that the rows of bars may be all in compression or both compression and tension depending on the location of the neutral axis. Accordingly, the total strength of the longitudinal bars is determined by adding all


the individual strengths of bars of different rows. The effective cover d', though depends on the nominal cover, has to be determined from practical considerations of housing all the steel bars. (iii) Grades of concrete and steel The dimensions b and D of the cross-section and the amount of longitudinal steel bars depend on the grades of concrete and steel. (iv) Transverse reinforcement The transverse reinforcement, provided in form of lateral ties or spirals, are important for the following advantages in (a) preventing premature / local buckling of the longitudinal bars,

(b) improving ductility and strength by the effect of confinement of the core

concrete, (c) holding the longitudinal bars in position during construction, and (d) providing resistance against shear and torsion, if present.

However, the transverse reinforcement does not have a major contribution in influencing the capacities of the column. Moreover, the design of transverse reinforcement involves selection of bar diameter and spacing following the stipulations in the design code. The bar diameter of the transverse reinforcement also depends on the bar diameter of longitudinal steel. Accordingly, the transverse reinforcement is designed after finalizing other parameters mentioned above. It is, therefore, clear that the design of columns mainly involves the determination of percentage of longitudinal reinforcement p, either assuming or knowing the dimensions b and D, grades of concrete and steel, distribution of longitudinal bars in two or multiple rows and d'/D ratio from the analysis or elsewhere. Needless to mention that any designed column should be able to resist several critical pairs of Pu and Mu obtained from the analysis of the structure. It is also a fact that several trials may be needed to arrive at the final selection revising any or all the assumed parameters. Accordingly, the design charts are prepared to give the results for the unknown parameter quickly avoiding lengthy calculations after selecting appropriate non-dimensional parameters. Based on the above considerations and making the design simple, quick and fairly accurate, the following are the two non-dimensional parameters:


For axial load: Pu/fckbD For moment: Mu/fckbD2

The characteristic strength of concrete fck has been associated with the non-dimensional parameters as the grade of concrete does not improve the strength of the column significantly. The design charts prepared by SP-16 are assuming the constant value of fck for M 20 to avoid different sets of design charts for different grades of concrete. However, separate design charts are presented in SP-16 for three grades of steel (Fe 250, Fe 415 and Fe 500), four values of d'/D (0.05, 0.1, 0.15 and 0.2) and two types of distribution of longitudinal steel (distributed equally on two and four sides). Accordingly there are twenty-four design charts for the design of rectangular columns. Twelve separate design charts are also presented in SP-16 for circular sections covering the above mentioned three grades of steel and for values of d'/D ratio. However, the unknown parameter p, the percentage of longitudinal reinforcement has been modified to p/fck in all the design charts of SP-16, so that for grades other than M 20, the more accurate value of p can be obtained by multiplying the p/fck with the actual grade of concrete used in the design of that column. However, this lesson explains that it is also possible to prepare design chart taking into consideration the actual grade of concrete. As mentioned earlier, the design charts are prepared getting the pairs of values of Pu and Mu in non-dimensional form from the equations of equilibrium for different locations of the neutral axis. We now take up the respective non-dimensional equations for four different cases as follows:

(a) When the neutral axis is at infinity, i.e., kD = ∞ , pure axial load is applied on the column.

(b) When the neutral axis is outside the cross-section of the column, i.e.,

> kD D. ∞ ≥ (c) When the neutral axis is within the cross-section of the column, i.e.,

kD < D. (d) When the column behaves like a steel beam.

10.24.3 Non-dimensional Equation of Equilibrium when k =

, (Pure Axial Load) ∞

Figures 10.23.2b and c of Lesson 23 present the strain profile EF and the corresponding stress block for this case. As the load is purely axial, we need to


express the terms Cc and Cs of Eq.10.35 of sec.10.23.10 of Lesson 23. The total compressive force due to concrete of constant stress of 0.446 fck is: Cc = 0.446 fck b D (10.37) However, proper deduction shall be made for the compressive force of concrete not available due to the replacement by steel bars while computing Cs.

The force of longitudinal steel bars in compression is now calculated. The

steel bars of area pbD/100 are subjected to the constant stress of fsc when the strain is 0.002. Subtracting the compressive force of concrete of the same area pbD/100, we have,

Cs = (pbD/100) (fsc - 0.446 fck)

(10.38)

Thus, we have from Eq.10.35 of sec.10.23.10 of Lesson 23 after substituting the expressions of Cc and Cs from Eqs.10.37 and 10.38, Pu = 0.446 fck b D + (pbD/100) (fsc - 0.446 fck) (10.39)

Dividing both sides of Eq.10.39 by fck bD, we have (Pu/fck bD) = 0.446 + (p/100 fck) (fsc - 0.446 fck) (10.40) Thus, Eq.10.40 is the only governing equation for this case to be considered. 10.24.4 Non-dimensional Equations of Equilibrium when Neutral Axis is Outside the Section (∞ > kD ≥ D) Figures 10.23.3b and c of Lesson 23 present the strain profile JK and the corresponding stress block for this case. The expressions of Cc, Cs and appropriate lever arms are determined to write the two equations of equilibrium (Eqs.10.35 and 36) of Lesson 23. While computing Cc, the area of parabolic stress block is determined employing the coefficient C1 from Table 10.4 of Lesson 23. Similarly, the coefficient C2, needed to write the moment equation, is obtained from Table 10.4 of Lesson 23. The forces and the corresponding lever arms of longitudinal steel bars are to be considered separately and added for each of the n rows of the longitudinal bars. Thus, we have the first equation as,


Pu = C1 fck bD +

(10.41)

) - ( )100/ (1

cisii

n

iffbDp∑

=

where C1 = coefficient for the area of stress block to be taken from Table 10.4

of Lesson 23, pi = Asi/bD where Asi is the area of reinforcement in the ith row, fsi = stress in the ith row of reinforcement, taken positive for compression

and negative for tension, fci = stress in concrete at the level of the ith row of reinforcement, and n = number of rows of reinforcement.

Here also, the deduction of the compressive force of concrete has been made for the concrete replaced by the longitudinal steel bars. Dividing both sides of Eq.10.41 by fckbD, we have

(Pu/fckbD) = C1 +

(10.42)

) - ( ) 100/ (1

cisicki

n

ifffp∑

=

Similarly, the moment equation (Eq.10.36) becomes,

Mu = C1 fckbD (D/2 - C2D) +

(10.43)

icisii

n

iyffbDp ) - ( )100/ (

1∑=

where C2 = coefficient for the distance of the centroid of the compressive stress

block of concrete measured from the highly compressed right edge and is taken from Table 10.4 of Lesson 23, and

yi = the distance from the centroid of the section to the ith row of

reinforcement, positive towards the highly compressed right edge and negative towards the least compressed left edge.

Dividing both sides of Eq.10.43 by fckbD2, we have

(Mu/fckbD2) = C1(0.5 - C2) + (y) - ( ) 100/ (1

cisicki

n

ifffp∑

=i/D)

(10.44)


Equations 10.42 and 10.44 are the two non-dimensional equations of equilibrium in this case when . DkD ≤<∞

10.24.5 Non-dimensional Equations of Equilibrium when the Neutral Axis is within the Section (kD < D)

The strain profile IN and the corresponding stress block of concrete are presented in Figs.10.23.4b and c for this case. Following the same procedure of computing Cc, Cs and the respective lever arms, we have the first equation as

Pu = 0.36 fck kbD +

(10.45)

) - ( )100/ (1

cisii

n

iffbDp∑

=

Dividing both sides of Eq.10.45 by fckbD, we have

Pu/fckbD = 0.36 k +

(10.46)

) - ( ) 100/ (1

cisicki

n

ifffp∑

=

and the moment equation (Eq.10.36) as

Mu = 0.36 fck kbD(0.5 - 0.42 k) D + (y) - ( )100/ (1

cisii

n

i

ffbDp∑=

i/D)

(10.47) Dividing both sides of Eq.10.47 by fckbD2, we have

(Mu/fckbD2) = 0.36 k(0.5 - 0.42 k) + (y) - ( ) 100/ (1

cisicki

n

ifffp∑

=i/D)

(10.48)

where k = Depth of the neutral axis/Depth of column, mentioned earlier in sec.10.21.10 and Fig.10.21.11 of Lesson 21.

Equations 10.46 and 10.48 are the two non-dimensional equations of equilibrium in this case. 10.24.6 Non-dimensional Equation of Equilibrium when the Column Behaves as a Steel Beam This is a specific situation when the column is subjected to pure moment Mu = Mo only (Point 6 of the interaction diagram in Fig.10.23.1 of Lesson 23).


Since the column has symmetrical longitudinal steel on both sides of the centroidal axis of the column, the column will resist the pure moment by yielding of both tensile and compressive steel bars (i.e., fsi = 0.87 fy = fyd). Thus, we have only one equation (Eq.10.36 of Lesson 23), which becomes

Mu = (y)(0.87 )100/ (1

yi

n

i

fbDp∑=

i/D)

(10.49) Dividing both sides of Eq.10.49 by fck bD2, we have

(Mu/fckbD2) = (y)(0.87 ) 100/(1

ycki

n

i

ffp∑=

i/D)

(10.50) Equation 10.50 is the equation of equilibrium in this case.

10.24.7 Preparation of Design Charts

Design charts are prepared employing the equations of four different

cases as given in secs.10.24.3 to 6. The advantage of employing the equations is that the actual grade of concrete can be taken into account, though it may not be worthwhile to follow this accurately. However, preparation of interaction diagram will help in understanding the behaviour of column with the change of neutral axis depth for the four cases mentioned in sec.10.24.2. The step by step procedure of preparing the design charts is explained below. It is worth mentioning that the values of (Pu/fckbD) and (Mu/fckbD2) are determined considering different locations of the neutral axis for the four cases mentioned in sec.10.24.2. Step 1: When the neutral axis is at infinity The governing equation is Eq.10.40. The strain profile EF and the corresponding stress block are in Fig.10.23.2b and c of Lesson 23, respectively. Step 2: When the column is subjected to axial load considering minimum

eccentricity

Lesson 22 presents the design of short columns subjected to axial load only considering minimum eccentricity as stipulated in cl.29.3 of IS 456, employing Eq.10.4, which is as follows:

Pu = 0.4 fck b D + (pbD/100) (0.67 fy - 0.4 fck) …. (10.4) Dividing both sides of Eq.10.4 by fck bD, we have


(Pu/fck bD) = 0.4 + (p/100 fck) (0.67 fy - 0.4 fck) (10.51)

The Pu obtained from Eq.10.51 can also resist Mu as per cl.39.3 of IS 456.

From the stipulation of cl. 39.3 of IS 456 and considering the maximum value of the minimum eccentricity as 0.05D, we have

Mu = (Pu) (0.05)D = 0.02 fck bD2 + (0.05 pbD2/100) (0.67 fy - 0.4 fck) Dividing both sides of the above equation by fckbD2, we have (Mu/fck bD2) = 0.02 + (0.05p/100 fck) (0.67 fy - 0.4 fck)

(10.52) Equations 10.51 and 10.52 are the two equations to be considered in this

case. Step 3: When the neutral axis is outside the section Figures 10.23.3b and c of Lesson 23 present one strain profile JK and the corresponding stress block, respectively, out of a large number of values of k from 1 to infinity, only values up to about 1.2 are good enough to consider, as explained in sec.10.23.5 of Lesson 23. Accordingly, we shall consider only one point, where k = 1.1, in this case. With the help of Eqs.10.42 and 10.44, Table 10.4 for the values of C1 and C2, Table 10.5 for the values of fsi and Eq.10.23 or Eq.10.27 for the values of fci, the non-dimensional parameters Pu/fck bD and Mu/fck bD2 are determined. Step 4: When the neutral axis is within the section One representative strain profile IU and the corresponding stress block are presented in Fig.10.23.4b and c, respectively, of Lesson 23. The following six points of the interaction diagram are considered satisfactory for preparing the design charts:

(a) Where the tensile stress of longitudinal steel is zero i.e., kD = D - d', (b) Where the tensile stress of longitudinal steel is 0.4fyd = 0.4(0.87 fy), (c) Where the tensile stress of longitudinal steel is) 0.8fyd = 0.8(0.87 fy) , (d) Where the tensile stress of longitudinal steel is fyd = 0.87fy and strain

= 0.87fy/Es, i.e., the initial yield point, (e) Where the tensile stress of longitudinal steel is fyd = 0.87fy and strain

= 0.87fy/Es + 0.002, i.e., the final yield point,


(f) When the depth of the neutral axis is 0.25D. For all six points, the respective strain profile and the corresponding stress blocks can be drawn. Therefore, values of (Pu/fck bD) and (Mu/fck bD2) are determined from Eqs.10.46 and 10.48, using Table 10.5 for fsc and Eq.10.34 for fci.

Step 5: When the column behaves like a steel beam As explained in sec.10.24.6, Eq.10.50 is used to compute Mu/fck bD2 in this case. Step 6: Preparation of design chart The ten pairs of (Pu/fck bD) and (Mu/fck bD2) (one set each in steps 1, 2, 3 and 5 and six sets in step 4) can be plotted to prepare the desired design chart. One illustrative example is taken up in the next section. 10.24.8 Illustrative Example Problem 1: Prepare a design chart for a rectangular column with 3 per cent longitudinal steel distributed equally on two faces using M 25 and Fe 415, and considering d'/D = 0.15. Solution 1: The solution of this problem is explained in six steps of the earlier section. Step 1: When the neutral axis is at infinity Figures 10.23.2b and c present the strain profile EF and the corresponding stress block, respectively. Using the values of p = 3 per cent, fck = 25 N/mm2 and determining the value of fsc = 327.7388 N/mm2 (using linear interpolation from the values of Table 10.5 of Lesson 23), we get the value of (Pu/fck bD) from Eq.10.40 as (Pu/fck bD) = 0.8259. Step 2: When the column is subjected to axial load considering minimum

eccentricity Using the value of p = 3 per cent, fck = 25 N/mm2 and fy = 415 N/mm2 in Eqs.10.51 and 10.52 of sec.10.24.6, we have


(Pu/fck bD) = 0.7217 (Mu/fck bD2 ) = 0.0361

Step 3: When the neutral axis depth = 1.1 D

Figures 10.24.1a, b and c show the section of the column, strain profile JK and the corresponding stress block, respectively, for this case. We use Eqs.10.42 and 10.44 for determining the value of (Pu/fck bD) and (Mu/fck bD2 ) for


this case using k = 1.1, fck = 25 N/mm2, p1 = p2 = 1.5 , y1/D = 0.35 and y2/D = - 0.35. Values of C1, C2, fs1 and fs2, fc1 and fc2 are obtained from equations mentioned in Step 3 of sec.10.24.6. The values of all the quantities are presented in Table 10.6A, mentioning the source equation no., table no. etc. to get the two non-dimensional parameters as given below:

(Pu/fck bD) = 0.67405 (Mu/fck bD2 ) = 0.06370


Step 4: When the neutral axis is within the section

In Step 4 of section 10.24.6, six different locations of neutral axis are mentioned; five of them (a to e) are specified by the magnitude of fs2 (tensile) of longitudinal steel and one of them is specified by the value of k = 0.25. The values of all the quantities are presented in Tables 10.6A and B, mentioning the source equation no., table no. etc.


Figures 10.24.2 to 10.24.7 present the respective strain profiles and the

corresponding stress block separately for all six different locations of the neutral axis.

Table 10.6A Parameters and results of Problem 1 of Section 10.24.8 Given data: fck = 25 N/mm2, fy = 415 N/mm2, p = 3 per cent, p1 = p2 = 1.5 per cent, d’/D = 0.15 Note: Units of fsi, fsc and fci are in N/mm2, (-) minus sign indicates tensile strain

or stress.

Given Sl.No. Description

k = 1.1 fs2 = 0 fs2 = -0.4 fyd fs2 = 0.8 fyd

1 Sec. No. 10.24.7 10.24.7 10.24.7 10.24.7 2 Step No. 3 4 4 4 3 Fig. No. 10.24.1 10.24.2 10.24.3 10.24.4 4 εs1 = εc1 0.002829 0.00288 0.00275 0.00263 5 εs2 = εc2 0.000744 0.0 -0.00072 -0.00144 6 Table No.

of fsi and fsc

10.5 10.5 10.5 10.5

7 fs1 352.407 352.871 351.669 348.392 8 fs2 148.914 0.0 -144.42 -288.84 9 fsc NA NA NA NA 10 Eq.Nos. of

fci

10.23 and 10.27

10.34 10.34 10.34

11 fc1 11.15 11.15 11.15 11.15 12 fc2 6.757 0.0 0.0 0.0 13 Table No.

of C1 and C2

10.4 NA NA NA

14 C1 0.384 NA NA NA 15 C2 0.443 NA NA NA 16 y1/D +0.35 +0.35 +0.35 +0.35 17 y2/D -0.35 -0.35 -0.35 -0.35 18 k 1.1 0.85 0.7046 0.6017 19 Eq.No. of

Pu/fck bD 10.42 10.46 10.46 10.46

20 Pu/fck bD 0.6740 0.5110 0.3713 0.2457 21 Eq.No. of

Mu/fck bD210.44 10.48 10.48 10.48

22 Mu/fck bD2 0.0643 0.1155 0.1536 0.1850


Table 10.6B Parameters and results of Problem 1 of Section 10.24.8 Given data: fck = 25 N/mm2, fy = 415 N/mm2, p = 3 per cent, p1 = p2 = 1.5 per cent, d’/D = 0.15 Note: Units of fsi, fsc and fci are in N/mm2, (-) minus sign indicates tensile strain

or stress.


fs2 = - fyd (Initial yield)

fs2 = - fyd (Final yield)

k = 0.25

1 Sec. No. 10.24.7 10.24.7 10.24.7 2 Step No. 4 4 4 3 Fig. No. 10.24.5 10.24.6 10.24.7 4 εs1 = εc1 0.00256 0.00221 0.0014 5 εs2 = εc2 -0.00180 -0.00380 -0.0084 6 Table No.

of fsi and fsc

10.5 10.5 10.5

7 fs1 346.754 335.484 281.0 8 fs2 -361.05 -361.05 -361.05 9 fsc NA NA NA 10 Eq.Nos. of

fci

10.34 10.34 10.34

11 fc1 11.15 11.15 10.146 12 fc2 0.0 0.0 0.0 13 Table No.

of C1 and C2

NA NA NA

14 C1 NA NA NA 15 C2 NA NA NA 16 y1/D +0.35 +0.35 +0.35 17 y2/D -0.35 -0.35 -0.35 18 k 0.5607 0.4072 0.25 19 Eq.No. of

Pu/fck bD 10.46 10.46 10.46

20 Pu/fck bD 0.1866 0.1246 0.0353 21 Eq.No. of

Mu/fck bD210.48 10.48 10.48

22 Mu/fck bD2 0.1997 0.1921 0.1680 Step 5: When the column behaves like a steel beam

For this case, the parameter (Mu/fck bD2) is determined from Eq.10.50 using p1 = p2 = 1.5 per cent, fck = 25 N/mm2, fy = 415 N/mm2, y1/D = 0.35 and y2/D = -0.35. Thus, we get


(Mu/fck bD2 ) = 0.15164

Step 6: Final results of design chart

The values of ten pairs of (Pu/fck bD) and (Mu/fck bD2) as obtained in steps 1 to 5 are presented in Sl. Nos. 1 to 10 of Table 10.6C. The design chart can be prepared by plotting these values.

Table 10.6C Final values of Pu/fck bD and Mu/fck bD2 of Problem 1 of Section

10.24.8

Sl. No. Particulars about the point Pu/fck bD Mu/fck bD2

1 k = α 0.8259 0.0 2 Minimum eccentricity 0.7217 0.0361 3 k = 1.1 0.6740 0.0643 4 fs2 = 0 0.5110 0.1155 5 fs2 = (-)0.4 fyd 0.3713 0.1536 6 fs2 = (-)0.8 fyd 0.2457 0.1850 7 fs2 = (-) fyd

(Initial yield) 0.1866 0.1997

8 fs2 = (-) fyd (Final yield)

0.1246 0.1921

9 k = 0.25 0.0353 0.1680 10 Steel Beam 0.0 0.1516

10.24.9 Practice Questions and Problems with Answers Q.1: Why do we need to have non-dimensional design chart? A.1: See sec. 10.24.1 Q.2: Name the different design parameters while designing a column. Mention

which one is the most important parameter. A.2: See sec. 10.24.2. Q.3: Prepare a design chart for a rectangular column within three per cent

longitudinal steel, equally distributed on two faces, using M 25 and Fe 250 and considering d'/D = 0.15.

.A.3: The solution of this problem is obtained following the same six steps of

Problem 1 of sec.10.24.8, except that the grade of steel here is Fe 250. Therefore, the final results and all the parameters are presented in Table 10.7 avoiding explaining step by step again.


Table 10.7 Final values of Pu/fck bD and Mu/fck bD2 of Q.3 of Section 10.24.9

Sl. No. Particulars about the point Pu/fck bD Mu/fck bD2

1 k = α 0.6936 0.0 2 Minimum eccentricity 0.5890 0.0295 3 k = 1.1 0.5931 0.0354 4 fs2 = 0 0.4298 0.0871 5 fs2 = (-)0.4 fyd 0.3438 0.1113 6 fs2 = (-)0.8 fyd 0.2645 0.1323 7 fs2 = (-) fyd

(Initial yield) 0.2268 0.1421

8 fs2 = (-) fyd (Final yield)

0.1559 0.1395

9 k = 0.25 0.0839 0.1248 10 Steel Beam 0.0 0.0914

10.24.10 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 10.24.11 Test 24 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Determine the parameters including the two non-dimensional

parameters, Pu and Mu of a rectangular reinforced concrete short column of b = 370 mm, D = 530 mm, d'/D = 0.1 and having 8-25 mm diameter bars as longitudinal steel distributed equally on two sides using M 20 and Fe 415 for each of the following three cases:

(a) when fs2 = - 0.4 fyd (b) when fs2 = - 0.8 fyd (c) when fs2 = - fyd (at final yield)

(16 + 17 + 17 = 50) A.TQ.1: This problem can be solved following the same procedure of explained

in Step 4b, c and d of sec.10.24.7. The step by step calculations are not shown here and the final results are presented in Table 10.8.

Table 10.8 Parameters and results of TQ.1 of Section 10.24.11 Given data: fck = 20 N/mm2, fy = 415 N/mm2, b = 370 mm, D = 530 mm, Longitudinal steel = 8-25 mm diameter equally distributed on

two sides, d'/D = 0.15


fs2 = -0.4 fyd fs2 = 0.8 fyd fs2 = - fyd (Final yield)

1 Sec. No. 10.24.7 10.24.7 10.24.7 2 Step No. 4 4 4 3 Fig. No. 10.24.3 10.24.4 10.24.6 4 εs1 = εc1 0.0030 0.0029 0.0027 5 εs2 = εc2 -0.00072 -0.00144 -0.0038 6 Table No.

of fsi and fsc

10.5 10.5 10.5

7 fs1 354.1702 353.468 349.956


8 fs2 -144.42 -288.84 -361.05 9 fsc NA NA NA 10 Eq.Nos. of

fci

10.34 10.34 10.34

11 fc1 11.15 11.15 11.15 12 fc2 0.0 0.0 0.0 13 Table No.

of C1 and C2

NA NA NA

14 C1 NA NA NA 15 C2 NA NA NA 16 y1/D +0.4 +0.4 +0.4 17 y2/D -0.4 -0.4 -0.4 18 k 0.7461 0.6371 0.4311 19 Eq.No. of

Pu/fck bD 10.46 10.46 10.46

20 Pu/fck bD 0.3690 0.2572 0.1452 21 Pu (kN) 1447.225 1008.792 569.568 22 Eq.No. of

Mu/fck bD210.48 10.48 10.48

23 Mu/fck bD2 0.1481 0.1799 0.1899 24 Mu (kNm) 307.777 374.125 394.779

10.24.12 Summary of this Lesson

This lesson explains the procedure of the preparation of design charts of rectangular reinforced concrete short columns subjected to axial load and uniaxial moment. Different positions of the neutral axis due to different pairs of Pu and Mu give rise to different strain profiles and stress blocks. Accordingly, the column may collapse when subjected to any pair of axial load and moment exceeding the capacities of the column. Design charts are very much useful to design the column avoiding lengthy numerical computations. Illustrative example, practice and test problems will help in understanding each step of the procedure to prepare the design chart.


Module 10


Lesson 25

Design of Short Columns under Axial Load with

Uniaxial Bending




• state the two types of problems that can be solved using the design charts of SP-16,

• mention the three sets of design charts specifying their parameters,

• state the approximations, limitations and usefulness of the design charts

of SP-16,

• mention the different steps of solving the analysis type of problems using the design charts of SP-16,

• mention the different steps of solving the design type of problems using

the design charts of SP-16,

• apply the methods in solving both types of problems using the design charts of SP-16.


Lesson 24 explains the procedure of preparing the design charts of short rectangular reinforced concrete columns under axial load with uniaxial bending. It is also mentioned that similar design charts can be prepared for circular and other types of cross-sections of columns by dividing the cross-section into several strips. This lesson explains the design of rectangular and circular short columns with the help of design charts. It is known that the design of columns by direct computations involves several trials and hence, time taking. On the other hand, design charts are very useful in getting several alternative solutions quickly. Further, design charts are also used for the analysis of columns for safety etc. However, there are limitations of using the design charts, which are mentioned in this lesson. Several numerical problems are solved in this lesson for the purpose of illustration covering both analysis and design types of problems using the design charts of SP-16.


10.25.2 Design Charts of SP-16 SP-16 has three sets of design charts prepared by following the procedure explained in Lesson 24 for rectangular and circular types of cross-sections of columns. The three sets are as follows: (i) Charts 27 to 38 are the first set of twelve charts for rectangular columns having symmetrical longitudinal steel bars in two rows (Fig.10.25.1) for three grades of steel (Fe 250, Fe 415 and Fe 500) and each of them has four values of d’/D ratios (0.05, 0.10, 0.15 and 0.20).

(ii) Charts 39 to 50 are the second set of twelve charts for rectangular columns having symmetrical longitudinal steel bars (twenty numbers) distributed equally on four sides (in six rows, Fig.10.25.2) for three grades of steel (Fe 250, Fe 415 and Fe 500) and each of them has four values of d’/D ratios (0.05, 0.10, 0.15 and 0.20).


(iii) The third set of twelve charts, numbering from 51 to 62, are for circular columns having eight longitudinal steel bars of equal diameter and uniformly spaced circumferentially (Fig.10.25.3) for three grades of steel (Fe 250, Fe 415 and Fe 500) and each of them has four values of d’/D ratios (0.05, 0.10, 0.15 and 0.20). All the thirty-six charts are prepared for M 20 grade of concrete only. This is a justified approximation as it is not worthwhile to have separate design charts for each grade of concrete. 10.25.3 Approximations and Limitations of Design Charts of SP-16 (i) Approximations The following are the approximations of the design charts of SP-16: (a) Grade of concrete As mentioned in the earlier section, all the design chars of SP-16 assume the constant grade M 20 of concrete. However, each chart has fourteen plots having different values of the parameter p/fck ranging from zero to 0.26 at an interval of 0.02. The designer, thus, can make use of the actual grade of concrete by multiplying the p/fck obtained from the plot with the actual fck for the particular grade of concrete to partially compensate the approximation.


(b) The d’/D ratio The three sets of charts have four fixed values of d’/D ratios (0.05, 0.10, 0.15 and 0.20). However, in the practical design, the d’/D ratio may be different from those values. In such situations intermediate values are determined by making linear interpolations. (c) Equal distribution of twenty longitudinal steel bars on four sides of rectangular

columns In spite of the above consideration, the design charts may be used without significant error for any number of bars greater than eight provided the bars are distributed equally on four sides. (d) Longitudinal bars in circular columns Though the design charts are prepared considering eight bars uniformly placed circumferentially, they may generally be used for any number of bars greater than six, uniformly placed circumferentially. (ii) Limitations

The following are the limitations of the design chars of SP-16: (a) Longitudinal bars equally distributed on four sides of rectangular columns Twenty bars, when equally placed on four sides, are placed in six rows avoiding any bar on the two axes. However, there will be bars on the axes for odd number of rows. A very common type is the 6-bar arrangement (Fig.10.25.4). Such arrangements, though symmetrical, are not covered in the design charts of SP-16. In such cases, the designer has to make his own assumptions judiciously in order to use the available charts of SP-16. Alternatively, he has to prepare the actual design chart depending on the bar arrangement to get accurate results.


(b) Unsymmetrical arrangement of longitudinal bars in rectangular cross-sections It is not covered in the charts. (c) Non-uniform placing of longitudinal bars in circular cross-sections It is not covered in the charts. (d) Cross-sections other than rectangular or circular like, I, T, H, X etc. These are not covered in the charts. The items under b, c and d, though rare, should be taken care of by preparing the respective design chars as and when needed. (e) Concluding remarks In spite of the above approximations and limitations, use of SP-16 has several advantages even by making some more approximations if the charts are not directly applicable. In the note of cl.39.5 of IS 456, the following is recommended, which is worth reproducing: “The design of members subject to combined axial load and uniaxial bending will involve lengthy calculation by trial and error. In order to overcome these difficulties interaction diagrams may be used. These have been prepared and published by BIS in “SP-16 Design aids for reinforced concrete to IS 456’.” Accordingly, the use of SP-16 is explained in the following sections for the solutions of both analysis and design types of problems. 10.25.4 Use of Design Charts in the Analysis Type of Problems

In many situations, it becomes necessary to assess the safety of a column with known cross-section dimensions, and longitudinal and transverse steel reinforcing bars. The objective is to examine if the column can resist some critical values of Pu or Mu or pairs of Pu and Mu, as may be expected to be applied on the column. This is done by comparing if the given values of pair of Pu and Mu are less than the respective strength capacities pair of Pu and Mu. The word “given” shall be used in the suffix of pairs of Pu and Mu to indicate that they are the given values for which the column has to be examined. The strength capacities of the column, either Pu or Mu alone or pair of Pu and Mu, will not have any suffix. Thus, the designer shall assess


(pair of Pu and Mu)given < pair of Pu and Mu, as strength capacities (10.53) This type of problem is known as analysis type of problem. The three steps are given below while using design charts of SP-16 for solving such problems. Step 1: Selection of the design chart The designer has to select a particular design chart, specified by the chart number, from the known value of d’/D and the grade of steel for circular columns; and considering also the distribution of longitudinal steel bars equally on two or four sides for the rectangular columns. Step 2: Selection of the particular curve The designer shall select the particular curve out of the family of fourteen curves in the chart selected in Step 1. The selection of the curve shall be made from the value of p/fck parameter which is known. Step 3: Assessment of the column This can be done in any of the three methods selecting two of the three parameters as known and comparing the third parameter to satisfy Eq.10.53. The parameters are Pu/fck bD, Mu/fck bD2 and p/fck for rectangular columns. For circular columns the breadth b shall be replaced by D (the diameter of the column). 10.25.5 Use of Design Charts in the Design Type of Problems It is explained in sec.10.24.2 of Lesson 24 that the design of columns mainly involves with the determination of percentage of longitudinal steel p, either assuming or knowing the dimensions b and D, grades of concrete and steel, distribution of longitudinal bars in two or multiple rows and d’/D from the analysis or elsewhere. However, the designer has to confirm the assumed data or revise them, if needed. The use of design charts of SP-16 is explained below in four steps while designing columns: Step 1: Selection of the design chart As in step 1 of sec.10.25.4, the design chart is selected from the assumed values of the parameter as explained in step 1 of sec.10.25.4. The only difference is that, here the assumed parameter may be revised, if required.


Step 2: Determination of the percentage of longitudinal steel

The two parameters (Pu/fck bD) and (Mu/fck bD2) are known and the point A is located on the design chart with these two coordinates (Fig.10.25.5). The point may be like A1, on a particular curve of specified p/fck, or like A2, in between two such curves having two values of p/fck, the difference between the two values of p/fck is 0.02. In the first case, the corresponding p/fck is obtained directly as specified on the curve. While, in the second case, liner interpolation is to be done by drawing a line KL perpendicular to the two curves and passing through the point A2. The percentage of longitudinal steel is obtained by multiplying the p/fck, so obtained, by the actual grade of concrete (which may be different from M 20 though the chart is prepared assuming M 20 only). Thus, percentage of longitudinal steel, p = (p/fck) (Actual fck) (10.54) This percentage of longitudinal steel (obtained from Eq.10.54) is a tentative value and shall be confirmed after finalizing the assumed data, i.e., d’/D, b, D etc. Step 3: Design of transverse reinforcement This should be done before confirming d’/D as the diameter of the lateral tie has a role in finalizing d’. The design of transverse reinforcement shall be done following the procedure explained in secs.10.21.8 and 10.21.9 of Lesson 21.


Step 4: Revision of the design, if required If the value of d’/D changes in step 3 requiring any change of other dimension etc., the repetition of steps 1 to 3 are needed. Otherwise, the design is complete. 10.25.6 Illustrative Examples

Problem 1: Figure 10.25.6 shows a rectangular short reinforced concrete column using M 25 and Fe 415. Analyse the safety of the column when subjected to Pu = 1620 kN and Mu = 170 kNm. Solution 1: This is an analysis type of problems. The data given are: b = 300 mm, D = 450 mm, d’ = 56 mm, Asc = 4021 mm2 (20 bars of 16 mm diameter), fck = 25 N/mm2, fy = 415 N/mm2, Pu = 1620 kN and Mu = 170 kNm. So, we have d’/D = 56/450 = 0.1244, Pu/fckbD = 0.48, Mu/fckbD2 = 0.111934 and p/fck = 0.11914. Step 1: Selection of design chart From the given data: d’/D = 0.1244, fy = 415 N/mm2 and longitudinal steel bars are equally distributed on four sides, the charts selected are 44 (for d’/D =


0.1) and 45 (for d’/D = 0.15). Linear interpolation has to done with the values obtained from these two charts. Step 2: Selection of the particular curve From the given value of p/fck = 0.11914, the two curves having p/fck = 0.1 and 0.12 are selected from both the charts (No. 44 and 45). Here also, linear interpolation has to be done. Step 3: Assessment of the column In order to assess the column, we select the two given parameters p/fck and Pu/fckbd2 to determine the third parameter Mu/fckbD2 to compare its value with the given parameter Mu/fckbD2. However, the value of Mu/fckbD2 is obtained by doing linear interpolation two times: once with respect to p/fck and the second time with respect to d’/D. The results are furnished in Table 10.9 below: Table 10.9: Values of Mu/fckbD2 when (Pu/fckbD2)given = 0.48 and (p/fck)given =

0.11914; and d’/D = 0.1244

d’/D Sl. No. p/fck0.1 0.15 0.1244

1 0.1 0.1* 0.09** 0.09512*** 2 0.12 0.12* 0.107** 0.113656*** 3 0.11914 0.1194*** 0.10649*** 0.1130941***

Note: * Values obtained from chart 44 ** Values obtained from chart 45 *** Linearly interpolated values Thus, the moment capacity of the column is obtained from the final value of Mu/fckbD2 = 0.1130941 as Mu = (0.1130941)(25)(300)(450)(450) Nmm = 171.762 kNm, which is higher than the given Mu = 170 kNm. Hence, the column can be subjected to the pair of given Pu and Mu as 1620 kN and 170 kNm, respectively. Problem 2: Design a short spiral column subjected to Pu = 2100 kN and Mu = 187.5 kNm using M 25 and Fe 415. The preliminary diameter of the column may be taken as 500 mm. Solution 2:


Step 1: Selection of design chart With the given fy = 415 N/mm2 and assuming d’/D = 0.1, the chart selected for this problem is Chart 56. Step 2: Determination of the percentage of longitudinal steel With the given fck = 25 N/mm2 and assuming the given D = 500 mm, we have: Pu/fckD2 = 2100000/25(500)(500) = 0.336, and Mu/fckD3 = 187.5(106 )/25(500)(500)(500) = 0.06 The particular point A (Fig.10.25.5) having coordinates of Pu/fckD2 = 0.336 and Mu/fckD3 = 0.06 in Chart 56 gives: p/fck = 0.08. Hence, p = 0.08(25) = 2 per cent (see Eq.10.54). Asc = 0.02(π )(500)(500)/4 = 3928.57 mm2

Provide 8-25 mm diameter bars to have Asc actually provided = 3927 mm2. Marginally less amount of steel than required will be checked considering the enhancement of strength for spiral columns as stipulated in cl.39.4 of IS 456. Step 3: Design of transverse reinforcement


The diameter of the helical reinforcement is taken as 8 mm (> 25 mm/4). The pitch p of the spiral is determined from Eq.10.11 of Lesson 22, which satisfies the stipulation in cl.39.4.1 of IS 456. From Eq.10.11, we have the pitch of the spiral p as: p ≤ 11.1(Dc - spφ ) asp fy/(D2 - f)2

cD ck …. (10.11) where, Dc = 500 – 40 – 40 = 420 mm, D = 500 mm, fck = 25 N/mm2, fy = 415 N/mm2, spφ = 8 mm and asp = 50 mm2. Using the above values in Eq.10.11, we have p ≤ 25.716 mm. As per cl.26.5.3.2d1, regarding the pitch of spiral: p >/ 420/6 (= 70 mm), p </ 25 mm and p </ 24 mm. So, pitch of the spiral = 25 mm is o.k. Figure 10.25.7 presents the cross-section with reinforcing bars of the column. Step 4: Revision of the design, if required Providing 25 mm diameter longitudinal steel bars and 8 mm diameter spirals, we have d’ = 40 + 8 + 12.5 = 60.5 mm. This gives d’/D = 60.5/500 = 0.121. In step 1, d’/D is assumed as 0.1. So, the revision of the design is needed. However, as mentioned in step 2, the area of steel required is not provided and this may be offset considering the enhanced strength of the spiral column, as stipulated in cl.39.4 of IS 456. We, therefore, assess the strength of the designed column, when d’/D = 0.121 and Asc = 3927 mm2, if it can be subjected to Pu = 2100 kN and Mu = 187.5 kNm. For the purpose of assessment, we determine the capacity Pu of the column when Mu = 187.5 kNm. Further, the revised d’/D = 0.121 needs to interpolate the values from Charts 56 (for d’/D = 0.1) and 57 (for d’/D = 0.15). The value of p/fck = 0.08 and Mu/fckbD3 = 0.06. Table 10.10 presents the results. Table 10.10: Value of Pu/fckbD2 when Mu/fckD3 = 0.06 and p/fck = 0.08

Sl.No. d’/D Pu/fckD2

1 0.1 0.336 (from Chart 56) 2 0.15 0.30 (from Chart 57) 3 0.121 0.32088 (Interpolated value)

From Table 10.10, thus, we get,


Pu/fckD2 = 0.32088, which gives Pu = (0.32088)(25)(500)(500) = 2005.5 kN. Considering the enhanced strength as 1.05 times as per cl.39.4 of IS 456, the actual capacity of this column is (1.05)(2005.5) = 2105 kN > 2100 kN. Thus, the design is safe to carry Pu = 2100 kN and Mu = 187.5 kNm. 10.25.7 Practice Questions and Problems with Answers Q.1: Name the two types of problems that can be solved using the design

charts of SP-16. A.1: See sec. 10.25.1. Q.2: Mention the three different sets of design charts available in SP-16

mentioning the number of charts and the parameters for their identification. A.2: See sec. 10.25.2. Q.3: State the approximations, limitations and usefulness of the design charts of

SP-16 in solving the analysis and design type of problems of short columns. A.3: See sec. 10.25.3. Q.4:


Assess the safety of the spiral column shown in Fig.10.25.8 using M 20 and Fe 415 when subjected to Pu = 1200 kN and Mu = 64 kNm, considering the enhanced strength of the spiral column.

A.4: In this problem, the given data are: D = 400 mm, d’ = 40 + 6 + 10 = 56 mm,

Asc = 2513 mm2 (8-20 mm diameter bars), fck = 20 N/mm2, fy = 415 N/mm2, Pu = 1200 kN and Mu = 64 kNm.

Step 1: Selection of the design charts With fy = 415 N/mm2 and d’/D = 56/400 = 0.14, we select two charts nos. 56 (for d’/D = 0.1) and 57 (for d’/D = 0.15). We have to interpolate the values obtained from these two charts. Step 2: Selection of the particular curve From the given data we have p/fck = 0.0999488 ≅ 0.1. So, we select the curve for p/fck = 0.1 in the two charts (Nos. 56 and 57). Step 3: Assessment of the column For the purpose of assessment, we select the two parameters p/fck and Mu/fckD3 and determine the values of Pu/fckD2 from the two charts for interpolation. The results are presented in Table 10.11 below. Table 10.11: Values of p/fckD2 and Mu/fckD3 and p/fck = 0.1

Sl.No. d’/D Pu/fckD2


From Table 10.11, thus, we get Pu/fckD2 = 0.4264, which gives Pu = 1364.48 kN. It may be noted that for more accuracy another set of values of d’/D = 0.08 is required. The interpolated value, thus obtained, shall be strictly applicable when p/fck = 0.0999488. However, for all practical designs, such accuracy is not required. Further, as per cl.39.4 of IS 456, the enhanced capacity of the spiral column = 1.05(1364.48) = 1432.704 kN, which is more than 1200 kN. It is also seen that the column is safe even without considering the enhanced capacity as the Pu = 1364.48 kN > 1200 kN. Q.5:


Design a short square tied column to carry Pu = 2240 kN and Mu = 112

kNm using M 25 and Fe 415, and assuming the dimension b = D = 400 mm, as shown in Fig.10.25.9.

A.5: The data given are: b = D = 400 mm, Pu = 2240 kN, Mu = 112 kNm, fck =

25 N/mm2 and fy = 415 N/mm2. Step 1: Selection of the design chart With the given data of fy = 415 N/mm2 and assuming d’/D = 0.15, we have to refer to Chart 45. Step 2: Determination of percentage of longitudinal steel Using the values of fck = 25 N/mm2 and assuming b = D = 400 mm as given, we have Pu/fckD2 = 0.56 and Mu/fckD3 = 0.07. From Chart 45, we get p/fck = 0.1, giving p = 2.5 per cent. Accordingly, Asc = 2.5(400)(400)/100 = 4000 mm2. Provide 20 bars of 16 mm diameter (Asc(provide) = 4021 mm2).


Step 3: Design of lateral tie The arrangement of lateral tie shall be like Fig.18 of IS 456 as the longitudinal bars are not spaced more than 75 mm on either side (cl.26.5.3.2b1 of IS 456). The pitch of the lateral tie of diameter 8 mm is kept at 250 mm c/c satisfying the stipulation in cl.26.5.3.2c1 of IS 456. Figure 10.25.9 presents the cross-section with reinforcing bars of the column. Step 4: Revision of the design, if required The value of d’ is now 56 mm which gives d’/D = 0.14. Accordingly, the assumed value of d’/D in step 1 as 0.15 is not valid. So, we have to check the capacity of the column interpolating the values when d’/D = 0.1 and 0.15 from Charts 44 and 45, respectively. Further, the longitudinal steel provided gives p/fck = 0.100525, which also is different from 0.1 as obtained in step 2 of this problem. Though the difference is marginal, both the interpolations are done for the academic interest and results are presented in Table 10.12 below. In assessing the capacity of this column, we keep p/fck = 0.100125 and Pu/fckD2 = 0.56 as constants and determine the value of Mu/fckD3 by two linear interpolations. Table 10.12: Values of Mu/fckD3 when Pu/fckbD2 = 0.56 and p/fck = 0.10025

d’/D Sl. No. p/fck0.1 0.15 0.14

1 0.1 0.1* 0.07** 0.072*** 2 0.12 0.08* 0.09** 0.092*** 3 0.100525 0.080525*** 0.070525*** 0.072525***

Note: * Values obtained from Chart 44 ** Values obtained from Chart 45 *** Linearly interpolated values So, the capacity of the column Mu = (0.072525)(25)(400)(400)(400) Nmm = 116 kNm > 112 kNm. Hence, the design of the column is o.k. 10.25.8 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 10.25.9 Test 25 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Mention the three different sets of design charts available in SP-16

mentioning the number of charts and the parameters for their identification. (10 marks)

A.TQ.1: See sec. 10.25.2. TQ.2: State the approximations, limitations and usefulness of the design charts

of SP-16 in solving the analysis and design type of problems of short columns. (10 marks)


A.TQ.2: See sec. 10.25.3. TQ.3:

Check the short square column of Fig.10.25.10 to carry Pu = 3250 kN and

Mu = 250 kNm using M 25 and Fe 415. (30 marks)

A.TQ.3: Given data are: b = D = = 500 mm, Asc = 6283 mm2 (20 bars of 20 mm

diameter), fck = 25 N/mm2, fy = 415 N/mm2, Pu = 3250 kN and Mu = 250 kNm.

Step 1: Selection of design chart From Fig.10.25.10, we get d’ = 65 mm giving d’/D = 0.13, and given fy = 415 N/mm2, we select Charts 44 (for d’/D = 0.1) and 45 (for d’/D = 0.15). We have to interpolate the values to get the result when d’/D = 0.13. Step 2: Selection of the particular curve With p = 628300/(500)(500) = 2.5132 per cent, we get p/fck = 0.100528 ≅ 0.1. Accordingly, the curve for p/fck = 0.1 is to be used in Charts 44 and 45.


Step 3: Assessment of the column For the assessment, we keep Pu/fckD2 = 3250/25(500)(500) = 0.52 and p/fck = 0.1 as constants to determine Mu/fckD3 from two charts. The results are given in Table 10.13 below. Table 10.13: Values of Mu/fckD3 when Pu/fckD2 = 0.52 and p/fck = 0.1

Sl.No. d’/D Mu/fckD3


So, we get Mu/fckD3 = 0.084, giving Mu = (0.084)(25)(500)(500)(500) =

262.5 kNm > 250 kNm. Hence, the column is safe to carry Pu = 3250 kN and Mu = 250 kNm. 10.25.10 Summary of this Lesson This lesson explains the approximations, limitations and usefulness of the three sets of design charts available in SP-16 for the purpose of solving analysis and design types of reinforced concrete columns. The use of design charts has been illustrated in several steps for the solution of both analysis and design types of problems. Several numerical problems in illustrative examples, practice problem and test will help in understanding the use of design charts to solve the two types of problems.


Module 10


Lesson 26

Short Compression Members under Axial Load with Biaxial Bending




• understand the behaviour of short columns under axial load and biaxial bending,

• understand the concept of interaction surface,

• identify the load contour and interaction curves of Pu-Mu in a interaction

surface,

• mention the limitation of direct application of the interaction surface in solving the problems,

• explain the simplified method of design and analysis of short columns

under axial load and biaxial bending,

• apply the IS code method in designing and analysing the reinforced concrete short columns under axial load and biaxial bending.


Beams and girders transfer their end moments into the corner columns of a building frame in two perpendicular planes. Interior columns may also have biaxial moments if the layout of the columns is irregular. Accordingly, such columns are designed considering axial load with biaxial bending. This lesson presents a brief theoretical analysis of these columns and explains the difficulties to apply the theory for the design. Thereafter, simplified method, as recommended by IS 456, has been explained with the help of illustrative examples in this lesson.


10.26.2 Biaxial Bending

Figures 10.26.1a and b present column section under axial load and uniaxial bending about the principal axes x and y, respectively. Figure 10.26.1c


presents the column section under axial load and biaxial bending. The eccentricities ex and ey of Fig.10.26.1c are the same as those of Fig.10.26.1a (for ex) and Fig.10.26.1b (for ey), respectively. Thus, the biaxial bending case (case c) is the resultant of two uniaxial bending cases a and b. The resultant eccentricity e, therefore, can be written as (see Fig.10.26.1c): (10.55)

2/122 ) ( yx eee +=

Designating the moments of cases a, b and c by Mux, Muy and Mu, respectively, we can write: (10.56)

2/122 ) ( uyuxu MMM +=

and the resultant Mu is acting about an inclined axis, so that tanθ = ex/ey = Muy/Mux (10.57) the angle of inclination θ is measured from y axis. This inclined resultant axis shall also be the principal axis if the column section including the reinforcing bars is axisymmetric. In such a situation, the biaxial bending can be simplified to a uniaxial bending with the neutral axis parallel to the resultant axis of bending. The reinforced concrete column cross-sections are, in general, non-axisymmetric with reference to the longitudinal axis and, therefore, the neutral axis is not parallel to the resultant axis of bending (θ is not equal to λ in Fig.10.26.1c). Moreover, it is extremely laborious to find the location of the neutral axis with successive trials. However, failure strain profile and stress block can be drawn for a given location of the neutral axis. Figs.10.25.1d and e present the strain profile and stress block, respectively, of the section shown in Fig.10.25.1c.


10.26.3 Interaction Surface

Figure 10.26.2 can be visualised as a three-dimensional plot of Pu-Mux-Muy, wherein two two-dimensional plots of Pu-Muy and Pu-Mus are marked as case (a) and case (b), respectively. These two plots are the interaction curves for the columns of Figs.10.26.1a and b, respectively. The envelope of several interaction curves for different axes will generate the surface, known as interaction surface. The interaction curve marked as case (c) in Fig.10.26.2, is for the column under biaxial bending shown in Fig.10.26.1c. The corresponding axis of bending is making an angle θ with the y axis and satisfies Eq.10.57. It has been explained in Lesson 24 that a column subjected to a pair of P and M will be safe if their respective values are less than Pu and Mu, given by its interaction curve. Extending the same in the three-dimensional figure of interaction surface, it is also acceptable that a column subjected to a set of Pu, Muy and Mux is safe if the set of values lies within the surface. Since Pu is changing in the direction of z, let us designate the moments and axial loads as mentioned below: Muxz = design flexural strength with respect to major axis xx under biaxial

loading, when Pu = Puz,


Muyz = design flexural strength with respect to minor axis yy under biaxial

loading, when Pu = Puz, Mux1 = design flexural strength with respect to major axis xx under uniaxial

loading, when Pu = Puz, and Muy1 = design flexural strength with respect to minor axis yy under uniaxial

loading, when Pu = Puz. The above notations are also shown in Fig.10.26.2. All the interaction curves, mentioned above, are in planes perpendicular to xy plane. However, the interaction surface has several curves parallel to xy plane, which are planes of constant Pu. These curves are known as load contour, one such load contour is shown in Fig.10.26.2, when Pu = Puz. Needless to mention that the load is constant at all points of a load contour. These load contour curves are also interaction curves depicting the interaction between the biaxial bending capacities. 10.26.4 Limitation of Interaction Surface The main difficulty in preparing an exact interaction surface is that the neutral axis for the case (c) of Fig.10.26.1c will not, in general, be perpendicular to the line joining the loading point Pu and the centre of the column (Fig.10.26.1c). This will require several trials with c and λ , where c is the distance of the neutral axis and λ angle made by the neutral axis with the x axis, as shown in Fig.10.26.1c. Each trial will give a set of Pu, Mux and Muy. Only for a particular case, the neutral axis will be perpendicular to the line joining the load point Pu to the centre of the column. This search makes the process laborious. Moreover, several trials with c and λ , giving different values of h (see Fig.10.26.1c), may result in a failure surface with wide deviations, particularly as the value of Pu will be increasing. Accordingly, the design of columns under axial load with biaxial bending is done by making approximations of the interaction surface. Different countries adopted different approximate methods. Clause 39.6 of IS 456 recommends one method based on Bresler's formulation, also known as "Load Contour Method", which is taken up in the following section. (For more information, please refer to: "Design Criteria for Reinforced Columns under Axial Load and Biaxial Bending", by B. Bresler, J. ACI, Vol.32, No.5, 1960, pp.481-490). 10.26.5 IS Code Method for Design of Columns under Axial Load and Biaxial Bending


IS 456 recommends the following simplified method, based on Bresler's formulation, for the design of biaxially loaded columns. The relationship between Muxz and Muyz for a particular value of Pu = Puz, expressed in non-dimensional form is: (10.58)

1 )/( )/( 11 ≤+ nnuyuyuxux MMMM αα

where Mux and Muy = moments about x and y axes due to design loads, and nα is related to Pu/Puz, (Fig.10.26.3), where Puz = 0.45 fck Ac + 0.75 fy Asc = 0.45 Ag + (0.75 fy - 0.45 fck) Asc (10.59) where Ag = gross area of the section, and Asc = total area of steel in the section Muxz, Muyz, Mux1 and Muy1 are explained in sec.10.26.3 earlier. It is worth mentioning that the quantities Mux, Muy and Pu are due to external loadings applied on the structure and are available from the analysis, whereas Mux1, Muy1 and Puz are the capacities of the column section to be considered for the design. Equation 10.58 defines the shape of the load contour, as explained earlier (Fig.10.26.2). That is why the method is also known as "Load Contour Method". The exponent nα of Eq.10.58 is a constant which defines the shape of the load


contour and depends on the value of Pu. For low value of the axial load, the load contour is approximated as a straight line and, in that case, nα = 1. On the other hand, for high values of axial load, the load contour is approximated as a quadrant of a circle, when nα = 2. For intermediate load values, the value of nα lies between 1 and 2. Chart 64 of SP-16 presents the load contour and Fig.10.26.3 presents the relationship between nα and Pu/Puz. The mathematical relationship between nα and Pu/Puz is as follows: nα = 1.0, when Pu/Puz ≤ 0.2 nα = 0.67 + 1.67 Pu/Puz, when 0.2 < (Pu/Puz) < 0.8 nα = 2.0, when (Pu/Puz) 0.8 (10.60)

≥

10.26.6 Solution of Problems using IS Code Method The IS code method, as discussed in sec.10.26.5, can be employed to solve both the design and analysis types of problems. The only difference between the design and analysis type of problems is that a trial section has to be assumed including the percentage of longitudinal reinforcement in the design problems. However, these data are available in the analysis type of problems. Therefore, a guide line is given in this section for assuming the percentage of longitudinal reinforcement for the design problem. Further, for both types of problems, the eccentricities of loads are to be verified if they are more than the corresponding minimum eccentricities, as stipulated in cl.25.4 of IS 456. Thereafter, the relevant steps are given for the solution of the two types of problems. (a) Selection of trial section for the design type of problems As mentioned in sec.10.24.2(i) of Lesson 24, the preliminary dimensions are already assumed during the analysis of structure (mostly statically indeterminate). Thus, the percentage of longitudinal steel is the one parameter to be assumed from the given Pu, Mux, Muy, fck and fy. Pillai and Menon (Ref. No. 4) suggested a simple way of considering a moment of approximately 15 per cent in excess (lower percentage up to 5 per cent if Pu/Puz is relatively high) of the resultant moment

2/122 ) ( (1.15) uyuxu MMM += (10.61) as the uniaxial moment for the trial section with respect to the major principal axis xx, if Mux M≥ uy; otherwise, it should be with respect to the minor principal axis.


The reinforcement should be assumed to be distributed equally on four sides of the section. (b) Checking the eccentricities ex and ey for the minimum eccentricities Clause 25.4 of IS 256 stipulates the amounts of the minimum eccentricities and are given in Eq.10.3 of sec.10.21.11 of Lesson 21. However, they are given below as a ready reference. exmin ≥ greater of (l/500 + b/30) or 20 mm …. (10.3) eymin ≥ greater of (l/500 + D/30) or 20 mm where l, b and D are the unsupported length, least lateral dimension and larger lateral dimension, respectively. The clause further stipulates that for the biaxial bending, it is sufficient to ensure that the eccentricity exceeding the minimum value about one axis at a time. (c) Steps for the solution of problems The following are the steps for the solution of both analysis and design types of problems while employing the method recommended by IS 456. (i) Verification of eccentricities It is to be done determining ex = Mux/Pu and ey = Muy/Pu from the given data of Pu, Mux and Muy; and exmin and eymin from Eq.10.3 from the assumed b and D and given l. (ii) Assuming a trial section including longitudinal reinforcement This step is needed only for the design type of problem, which is to be done as explained in (a) above. (iii) Determination of Mux1 and Muy1 Use of design charts should be made for this. Mux1 and Muy1, corresponding to the given Pu, should be significantly greater than Mux and Muy, respectively. Redesign of the section should be done if the above are not satisfied for the design type of problem only. (iv) Determination of Puz and nα The values of Puz and nα can be determined from Eqs.10.59 and 10.60, respectively. Alternatively, Puz can be obtained from Chart 63 of SP-16.


(v) Checking the adequacy of the section This is done either using Eq.10.58 or using Chart 64 of SP-16. 10.26.7 Illustrative Example

Problem 1: Design the reinforcement to be provided in the short column of Fig.10.26.4 is subjected to Pu = 2000 kN, Mux = 130 kNm (about the major principal axis) and Muy = 120 kNm (about the minor principal axis). The unsupported length of the column is 3.2 m, width b = 400 mm and depth D = 500 mm. Use M 25 and Fe 415 for the design. Solution 1: Step 1: Verification of the eccentricities Given: l = 3200 mm, b = 400 mm and D = 500 mm, we have from Eq.10.3 of sec.10.26.6b, the minimum eccentricities are: exmin = greater of (3200/500 + 400/30) and 20 mm = 19.73 mm or 20 mm = 20 mm


eymin = greater of (3200/500 + 500/30) and 20 mm = 23.07 mm or 20 mm = 23.07 mm Again from Pu = 2000 kN, Mux = 130 kNm and Muy = 120 kNm, we have ex = Mux/Pu = 130(106)/2000(103) = 65 mm and ey = Muy/Pu = 120(106)/2000(103) = 60 mm. Both ex and ey are greater than exmin and eymin, respectively. Step 2: Assuming a trial section including the reinforcement We have b = 400 mm and D = 500 mm. For the reinforcement,

, from Eq.10.61 becomes 203.456 kNm. Accordingly, we get

2/122 ) ( 1.15 uyuxu MMM +=

Pu/fckbD = 2000(103)/(25)(400)(500) = 0.4

Mu/fckbD2 = 203.456(106)/(25)(400)(500)(500) = 0.0814

Assuming d' = 60 mm, we have d'/D = 0.12. From Charts 44 and 45, the value of p/fck is interpolated as 0.06. Thus, p = 0.06(25) = 1.5 per cent, giving Asc = 3000 mm2. Provide 12-20 mm diameter bars of area 3769 mm2, actual p provided = 1.8845 per cent. So, p/fck = 0.07538. Step 3: Determination of Mux1 and Muy1 We have Pu/fckbD = 0.4 and p/fck = 0.07538 in step 2. Now, we get Mux1/fckbD2 from chart corresponding to d' = 58 mm (Fig.10.26.4) i.e., d'/D = 0.116. We interpolate the values of Charts 44 and 45, and get Mux1/fckbD2 = 0.09044. So, Mux1 = 0.0944(25)(400)(500(500)(10-6) = 226.1 kNm. For Mux1, d'/b = 58/400 = 0.145. In a similar manner, we get Muy1 = 0.0858(25)(400)(400)(500)(10-6) = 171.6 kNm. As Mux1 and Muy1 are significantly greater than Mux and Muy, respectively, redesign of the section is not needed. Step 4: Determination of Puz and nα From Eq.10.59, we have Puz = 0.45(25)(400)(500) + {0.75(415) - 0.45(25)}(3769) = 3380.7 kN.


Alternatively, Chart 63 may be used to find Puz as explained. From the upper section of Chart 63, a horizontal line AB is drawn at p = 1.8845, to meet the Fe 415 line B (Fig.10.26.5). A vertical line BC is drawn from B to meet M 25 line at C. Finally, a horizontal line CD is drawn from C to meet Puz/Ag at 17. This gives Puz = 17(400)(500) = 3400 kN. The difference between the two values, 19.3 kN is hardly 0.57 per cent, which is due to the error in reading the value from the chart. However, any one of the two may be employed. Now, the value of nα is obtained from Eq.10.60 for Pu/Puz = 2000/3380.7 = 0.5916, i.e., 0.2 < Pu/Puz < 0.8, which gives, nα = 0.67 + 1.67 (Pu/Puz) = 1.658. Alternatively, nα may be obtained from Fig.10.26.3, drawn to scale. Step 5: Checking the adequacy of the section Using the values of Mux, Mux1, Muy, Muy1 and nα in Eq.10.58, we have (130/226.1)1.658 + (120/171.6)1.658 = 0.9521 < 1.0. Hence, the design is safe.


Alternatively, Chart 64 may be used to determine the point (Mux/Mux1), (Muy/Muy1) is within the curve of Pu/Puz = 0.5916 or not. Here, Mux/Mux1 = 0.5749 and Muy/Muy1 = 0.6993. It may be seen that the point is within the curve of Pu/Puz = 0.5916 of Chart 64 of SP-16. Step 6: Design of transverse reinforcement As per cl.26.5.3.2c of IS 456, the diameter of lateral tie should be > (20/4) mm diameter. Provide 8 mm diameter bars following the arrangement shown in Fig.10.26.4. The spacing of lateral tie is the least of : (a) 400 mm = least lateral dimension of column, (b) 320 mm = sixteen times the diameter of longitudinal reinforcement (20 mm), (c) 300 mm Accordingly, provide 8 mm lateral tie alternately @ 250 c/c (Fig.10.26.4). 10.26.8 Practice Questions and Problems with Answers Q.1: Explain the behaviour of a short column under biaxial bending as the

resultant of two uniaxial bending. A.1: See sec. 10.26.2 Q.2: Draw one interaction surface for a short column under biaxial bending and

show typical interaction curves and load contour curve. Explain the safety of a column with reference to the interaction surface when the column is under biaxial bending.

A.2: See sec.10.26.3 and Fig.10.26.2. Q.3: Discuss the limitation of the interaction curve. A.3: See sec.10.26.4. Q.4: Illustrate the IS code method of design of columns under biaxial bending. A.4: See sec.10.26.5.


Q.5:

Analyse the safety of the short column of unsupported length 3.2 m, b =

450 mm, D = 500 mm, as shown in Fig.10.26.6, having 12-16 mm diameter bars as longitudinal reinforcement and 8 mm diameter bars as lateral tie @ 250 mm c/c, when subjected to Pu = 1600 kN, Mux = 120 kNm and Muy = 100 kNm. Use M 25 and Fe 415.

A.5: Step 1: Verification of the eccentricities From the given data: l = 3200 mm, b = 450 mm and D = 500 mm, exmin = 3200/500 + 450/30 = 21.4 > 20 mm, so, 21.4 mm eymin = 3200/500 + 5000/30 = 23.06 > 20 mm, so, 23.06 mm ex = Mux/Pu = 120(103)/1600 = 75 mm ey = Muy/Pu = 100(103)/1600 = 62.5 mm So, the eccentricities ex and ey are >> exmin and eymin. Step 2: Determination of Mux1 and Muy1


Given data are: b = 450 mm, D = 500 mm, fck = 25 N/mm2, fy = 415 N/mm2, Pu = 1600 kN, Mux = 120 kNm, Muy = 100 kNm and Asc = 2412 mm2 (12-16 mm diameter bars). We have p = (100)(2412)/(450)(500) = 1.072 per cent, and d'/D = 56/500 = 0.112, d'/b = 56/450 = 0.124, Pu/fckbD = 1600/(25)(450)(500) = 0.2844 and p/fck = 1.072/25 = 0.043. We get Mux1/fckbD2 from Charts 44 and 45 as 0.09 and 0.08, respectively. Linear interpolation gives Mux1/fckbD2 for d'/D = 0.112 as 0.0876. Thus, Mux1 = (0.0876)(25)(450)(500)(500) = 246.376 kNm Similarly, interpolation of values (0.09 and 0.08) from Charts 44 and 45, we get Muy1/fckdb2 = 0.085 for d'/b = 0.124. Thus Muy1 = (0.085)(25)(500)(450)(450) = 215.156 kNm Step 3: Determination of Puz and nα From Eq.10.59, Puz = 0.45(25)(450)(500) + {0.75(415) - 0.45(25)}(2412) = 3254.85 kN. This gives Pu/Puz = 1600/3254.85 = 0.491574. From Eq.10.60, nα = 0.67 + 1.67(Pu/Puz) = 0.67 + 1.67(0.491574) = 1.4909. Step 4: Checking the adequacy of the section From Eq.10.58, we have: (120/246.376)1.4909 + (100/215.156)1.4909 = 0.6612 < 1. Hence, the section is safe to carry Pu = 1600 kN, Mux = 120 kNm and Muy = 100 kNm. 10.26.9 References






TQ.1:

Analyse the safety of the short square column of unsupported length =

3.5 m, b = D = 500 mm, as shown in Fig.10.26.7, with 12-16 mm diameter bars as longitudinal reinforcement and 8 mm diameter bars as lateral tie @ 250 mm c/c, when subjected to Pu = 1800 kN, Mux = 160 kNm and Muy = 150 kNm.

A.TQ.1: Step 1: Verification of the eccentricities From the given data: l = 3500 mm, b = D = 500 mm, we have emin in both directions (square column) = (3500/500) + (500/30) = 23.67 mm ex = 160(103)/1800 = 88.88 mm and ey = 150(103)/1800 = 83.34 mm Therefore, ex and ey >> emin.


Step 2: Determination of Mux1 and Muy1 We have the given data: b = D = 500 mm, fck = 25 N/mm2, fy = 415 N/mm2, Pu = 1800 kN, Mux = 160 kNm, Muy = 150 kNm and Asc = 2412 mm2 (12-16 mm diameter bars). The percentage of longitudinal reinforcement p = 241200/(500)(500) = 0.9648 per cent, and d'/D = 56/500 = 0.112 and p/fck = 0.9648/25 = 0.03859. Linear interpolation of values of Mux1/fckbD2 from Charts 44 and 45 for d'/D = 0.112 is obtained as 0.08. Thus, Mux1 = (0.08)(25)(500)(500)(500) = 250 kNm Muy1 = Mux1 = 250 kNm (square column) Step 3: Determination of Puz and nα From Eq.10.59,

Puz = 0.45(25)(500)(500) + {0.75(415) - 0.45(25)}(2415) = 3536.1 kN.

Pu/Puz = 1800/3536.1 = 0.509. From Eq.10.60, nα = 0.67 + 1.67(0.509) = 1.52. Step 4: Checking the adequacy of the section From Eq.10.58, we have: (160/250)1.52 + (150/250)1.52 = 0.967 < 1. Hence, the section can carry Pu = 1800 kN, Mux = 160 kNm and Muy = 150 kNm. 10.26.11 Summary of this Lesson

This lesson explains the behaviour of short columns under axial load and biaxial bending with the help of interaction surface, visualised as a three-dimensional plot of Pu-Mux-Muy. The interaction surface has a set of interaction curves of Pu-Mu and another set of interaction curves of Muxz-Muyz at constant Puz, also known as load contour. The design and analysis of short columns are also explained with the help of derived equations and design charts of SP-16. Numerical examples in the illustrative example, practice problems and test will help in understanding the application of the theory in solving the analysis and design types of problems of short columns under axial load and biaxial bending.


Module 10


Lesson 27

Slender Columns




• define a slender column, • give three reasons for its increasing importance and popularity,

• explain the behaviour of slender columns loaded concentrically,

• explain the behaviour of braced and unbraced single column or a part of

rigid frame, bent in single or double curvatures,

• roles and importance of additional moments due to P- effect and moments due to minimum eccentricities in slender columns,

Δ

• identify a column if sway or nonsway type,

• understand the additional moment method for the design of slender

columns,

• apply the equations or use the appropriate tables or charts of SP-16 for the complete design of slender columns as recommended by IS 456.


Slender and short are the two types of columns classified on the basis of slenderness ratios as mentioned in sec.10.21.5 of Lesson 21. Columns having both lex/D and ley/b less than twelve are designated as short and otherwise, they are slender, where lex and ley are the effective lengths with respect to major and minor axes, respectively; and D and b are the depth and width of rectangular columns, respectively. Short columns are frequently used in concrete structures, the design of such columns has been explained in Lessons 22 to 26, loaded concentrically or eccentrically about one or both axes. However, slender columns are also becoming increasingly important and popular because of the following reasons: (i) the development of high strength materials (concrete and steel),

(ii) improved methods of dimensioning and designing with rational and reliable design procedures,


(iii) innovative structural concepts – specially, the architect’s expectations for creative structures.

Accordingly, this lesson explains first, the behaviour of slender elastic columns loaded concentrically. Thereafter, reinforced concrete slender columns loaded concentrically or eccentrically about one or both axes are taken up. The design of slender columns has been explained and illustrated with numerical examples for easy understanding. 10.27.2 Concentrically Loaded Columns It has been explained in Lessons 22 to 26 that short columns fail by reaching the respective stresses indicating their maximum carrying capacities. On the other hand, the slender or long columns may fail at a much lower value of the load when sudden lateral displacement of the member takes place between the ends. Thus, short columns undergo material failure, while long columns may fail by buckling (geometric failure) at a critical load or Euler’s load, which is much less in comparison to that of short columns having equal area of cross-section. The buckling load is termed as Euler’s load as Euler in 1744 first obtained the value of critical load for various support conditions. For more information, please refer to Additamentum, “De Curvis elasticis”, in the “Methodus inveiendi Lineas Curvas maximi minimive proprietate gaudentes” Lausanne and Geneva, 1744. An English translation of this work is given in Isis No.58, Vol.20, p.1, November 1933. The general expression of the critical load Pcr at which a member will fail by buckling is as follows: Pcr = π2EI /(kl)2

where E is the Young’s modulus I is the moment of inertia about the axis of bending, l is the unsupported length of the column and k is the coefficient whose value depends on the degree of restraints at the supports. Expressing moment of inertia I = Ar2, where A is the area of cross-section of the column and r is the radius of gyration, the above equations can be written as, Pcr = π2EA /(kl/r)2 (10.62) Thus, Pcr of a particular column depends upon kl/r or slenderness ratio. It is worth mentioning that kl is termed as effective length le of the column.


Figures 10.27.1 and 2 show two elastic slender columns having hinge supports at both ends and fixed supports against rotation at both ends,


respectively. Figure 10.27.3 presents a column of real structure whose end supports are not either hinged or fixed. It has supports partially restrained against rotation by the top and bottom beams. Each of the three figures shows the respective buckled shape, points of inflection PIs (points of zero moment), the distance between the PIs and the value of k. All the three columns, having supports at both ends, have the k values less than one or at most one. By providing supports at both ends, one end of the column is prevented from undergoing lateral movement or sidesway with respect to the other end.


However, cantilever columns are entirely free at one end, as shown in Fig.10.27.4. Figure 10.27.5 shows another type of column, rotationally fixed at both ends but one end can move laterally with respect to the other. Like that of Fig.10.27.3, a real column, not hinged, fixed or entirely free but restrained by top and bottom beams, where sideway can also take place. Each of these three figures, like those of Figs.10.27.1 to 3, presents the respective buckled shape, points of inflection (PIs), if any, the distance between the PIs and the value of k. All these columns have the respective k values greater than one or at least one.


Figures 10.27.7 and 8 present two reinforced concrete portal frames, a typical reinforced concrete rigid frame. Columns of Fig.10.27.7 are prevented from sidesway and those of Fig.10.27.8 are not prevented from sidesway, respectively, when subjected to concentric loadings. The buckled configuration of the frame, prevented from sidesway (Fig.10.27.7) is similar to that of Fig.10.27.3,


except that the lower ends of the portal frame are hinged. One of the two points of inflection (PIs) is at the lower end of the column, while the other PI is slightly below the upper end of the column, depending on the degree of restraint. The value of k for such a frame is thus less than 1. The critical load is, therefore, slightly more than Pcr of the hinge-hinge column of Fig.10.27.1. The buckled configuration of the other portal frame of Fig.10.27.8, where sidesway is not prevented, is similar to the column of Fig.10.27.4 when it is made upside down, except that the upper end is not fixed but partially restrained by the supporting beam. In this case, the value of k exceeds 2, depending on the degree of restraint. One of the two PIs is at the bottom of the column. The critical load of the column of Fig.10.27.8 is much less than that of the column of Fig.10.27.1. Table 10.14: Critical loads in terms of Pcr of hinge-hinge column and effective lengths le = kl of elastic and reinforced concrete columns with different boundary conditions and for a constant unsupported length l Sl. No.

Support conditions Critical load Pcr

Effective length le = kl

Fig. No.

(A) Elastic single columns 1. Hinged at both ends, no

sidesway Pcr l 10.27.1

2. Fixed against rotation at both ends – no sidesway

4Pcr 0.5 l 10.27.2

3. Partially restrained against rotation by top and bottom cross-beams, no sidesway

Between Pcr and 4Pcr

l > kl > l/2 10.27.3

4. Fixed at one end and entirely free at other end – sidesway not prevented

0.25 Pcr 2 l, one PI is on imaginary extension

10.27.4

5. Rotationally fixed at both ends – sidesway not prevented

Pcr l, one PI is on imaginary extension

10.27.5

6. Partially restrained against rotation at both ends – sidesway not prevented

Between zero and slightly

less than Pcr *

l < kl < α 10.27.6

(B) Reinforced concrete columns 7. Hinged portal frame – no

sidesway > Pcr kl < l 10.27.7

8. Hinged portal frame – sidesway not prevented

<< Pcr kl > 2 l 10.27.8

Notes: 1. Buckled shapes are half sine wave between two points of inflection (PIs).


2. * The critical load is slightly less than Pcr of hinge-hinge column

(Sl.No.1), when cross-beams are very rigid compared to columns, i.e., the case under Sl.No.6 approaches the case under Sl.No.1.

The critical load is zero when cross-beams are very much

flexible compared to columns, i.e., the case under Sl.No.6 approaches to hinge-hinge column of Sl.No.1, allowing sidesway. In that case, it becomes unstable and hence, carries zero load.

Table 10.14 presents the critical load in terms of that of hinge-hinge column Pcr and effective lengths le (equal to the distance between two points of inflection PIs = kl) of elastic and reinforced concrete columns for a constant value of the unsupported length l. The stress-strain curve of concrete, as shown in Fig.1.2.1 of Lesson 2, reveals that the initial tangent modulus of concrete Ec is much higher than Et (tangent modulus at higher stress level). Taking this into account in Eq.10.62, Fig.10.27.9 presents a plot of buckling load Pcr versus kl/r. It is evident from the plot that the critical load is reducing with increasing slenderness ratio. For very short columns, the limiting factored concentric load estimated from Eq.10.39 of Lesson 24 will be found to be less than the critical load, determined from Eq.10.62. The column, therefore, will fail by direct crushing and not by buckling. We can also find out the limiting value of kl/r when the crushing load and the buckling load are the same. The (kl/r)lim is shown in Fig.10.27.9. The limiting value of kl/r also indicates that a column having kl/r more than (kl/r)lim will fail by


buckling, while columns having any value of kl/r less than (kl/r)lim will fail by crushing of concrete. The following are the observations of the discussions about the concentrically loaded columns: 1. As the slenderness ratio kl/r increases, the strength of concentrically loaded column decreases. 2. The effective length of columns either in single members or parts of rigid frames is between 0.5l and l, if the columns are prevented from sidesway by bracing or otherwise. The actual value depends on the degree of end restraints. 3. The effective length of columns either in single members or parts of rigid frames is always greater than one, if the columns are not prevented from sidesway. The actual value depends on the degree of end restraints. 4. The critical load of braced frame against sidesway is always significantly larger than that of the unbraced frame.


10.27.3 Slender Columns under Axial Load and Uniaxial Moment (A) Columns bent in single curvature Figure 10.27.10a shows a column bent in single curvature under axial load P less than its critical load Pcr with constant moment Pe. The deflection profile marked by dotted line is due to the constant moment. However, there will be additional moment of Py at a distance z from the origin (at the bottom of column) which will deflect the column further, as shown by the solid line. The constant moment Pe and additional moment Py are shown in Fig.10.27.10b. Thus, the total moment becomes M = Mo + Py = P(e + y) (10.63) The maximum moment is P(e + Δ ) at the mid-height of the column. This, we can write Mmax = Mo + P = P(e + Δ Δ ) (10.64) This is known as P - Δ effect.


Figure 10.27.11a shows another column whose bending is caused by a

transverse load H. The bending moment at a distance z from the origin (bottom of the column) is Hz/2 causing deflection of the column marked by dotted line in the figure. The axial load P, less than its critical load Pcr, causes additional moment resulting in further deflection, marked by solid line in the figure. This additional deflection produces additional moment of Py at a section z from the origin. The two bending moment diagrams are shown in Fig.10.27.11b. Here again, the total moment is M = Mo + Py = Hz/2 + Py (10.65) The maximum moment at the mid-height of the column is Mmax = Mo,max + P = Hl/4 + PΔ Δ (10.66) The total moment in Eqs.10.63 and 10.65 consists of the moment Mo that acts in the presence of P and the additional moment caused by P (= Py). The deflections y can be computed from yo, the deflections without the axial load from the expression y = yo[1/{1 – (P/Pcr)}] (10.67) From Eq.10.64, we have Mmax = Mo + P = MΔ o + PΔ o[1/{1 – (P/Pcr)}] (10.68) Equation 10.68 can be written as

1 ( 1 - (

crmax o

cr

P / P )M MP / P )ψ+

=

(10.69) where ψ depends on the type of loading and generally varies between 0.20. Since P/P

±cr is always less than one, we can ignore ψ (P/Pcr) term of Eq.10.69, to

have Mmax = Mo/{1 – (P/Pcr)} (10.70) where 1/{1 – (P/Pcr)} is the moment magnification factor. In both the cases above (Figs.10.27.10 and 11), a direct addition of the maximum moment caused by


transverse load or otherwise, to the maximum moment caused by P gives the total maximum moment as that is the most unfavourable situation. However, this is not the case for situation taken up in the following. (B) Columns bent in double curvature

Figure 10.27.12a shows a column subjected to equal end moment of opposite signs. From the moment diagrams Mo and Py (Figs.10.27.12b and c), it is clear that though Mo moments are maximum at the ends, the Py moments are maximum at some distance from the ends. The total moment can be either as shown in d or in e of Fig.10.27.12. In case of Fig.10.27.12d, the maximum moment remains at the ends and in Fig.10.27.12e, the maximum moment is at some distance from the ends, where Mo is comparatively smaller than Mo max at the ends. Accordingly, the total maximum moment is moderately higher than Mo

max. From the above, it is evident that the moment Mo will be magnified most strongly if the section of Mo max coincides with the section of maximum value of y, as in the case of column bent in single curvature of Figs.10.27.10 and 11. Similarly, if the two moments are unequal but of same sign as in Fig.10.27.10, the moment Mo will be magnified but not so much as in Fig.10.27.10. On the other hand, if the unequal end moments are of opposite signs and cause bending in double curvature, there will be little or no magnification of Mo moment. This dependence of moment magnification on the relative magnitudes of the two moments can be expressed by modifying the earlier Eq.10.70 as


Mmax = Mo Cm/{1 – (P/Pcr)} (10.71) where Cm = 0.6 + 0.4(M1/M2) ≥ 0.4 (10.72) The moment M1 is smaller than M2 and M1/M2 is positive if the moments produce single curvature and negative if they produce double curvature. It is further seen from Eq.10.72 that Cm = 1, when M1 = M2 and in that case, Eq.10.71 becomes the same as Eq.10.70. For the column of Fig.10.27.12a, the deflections caused by Mo are magnified when axial load P is applied. The deflection can be obtained from y = yo [1/{1 – (P/4Pcr)}] (10.73)

(C) Portal frame laterally unbraced and braced Here, the sidesway can occur only for the entire frame simultaneously. A fixed portal frame, shown in Fig.10.27.13a, is under horizontal load H and compression force P. The moments due to H and P and the total moment diagrams are shown in Fig.10.27.13b, c and d, respectively. The deformations of the frame due to H are shown in Fig.10.27.13a by dotted curves, while the solid curves are the magnified deformations. It is observed that the maximum values of positive and negative Mo are at the ends of the column where the maximum


values of positive and negative moments due to P also occur. Thus, the total moment shall be at the ends as the two effects are fully additive.

Figure 10.27.14a shows a fixed portal frame, laterally braced so that no sidesway can occur. Figures 10.27.14b and c show the moments Mo and due to P. It is seen that the maximum values of the two different moments do not occur at the same location. As a result, the magnification of the moment either may not be true or shall be small. (D) Columns with different slenderness ratios


Figure 10.27.15 shows the interaction diagram of P and M at the mid-height section of the column shown in Fig.10.27.10. Three loading paths OA, OB and OC are also shown in the figure for three columns having the same cross-sectional area and the eccentricity of loads but with different slenderness ratios. The three columns are loaded with increasing P and M (at constant eccentricity) up to failure. The loading path OA is linear indicating Δ = 0, i.e., for a very short column. It should be noted that Δ should be theoretically zero only when either the effective length or the eccentricity is zero. In a practical short column, however, some lateral deflection shall be there, which, in turn will cause additional moment not more than five per cent of the primary moment and may be neglected. The loading path OA terminates at point A of the interaction diagram, which shows the failure load Psc of the short column with moment Msc = Psc e. The short column fails by crushing of concrete at the mid-height section. This type of failure is designated as material failure, either a tension failure or a compression failure depending on the location of the point A on the interaction curve. The load path OB is for a long column, where the deflection caused by increasing value of P is significant. Finally, the long column fails at load P

Δlc and

moment Mlc = Plc(e + ). The loading path OB further reveals that the secondary moment P

ΔlcΔ is comparable to the primary moment Plc e. Moreover, the failure

load and the primary moment of the long column Plc and Plc e, respectively, are less than those of the short column (Psc and Psc e, respectively), though both the columns have the same cross-sectional areas and eccentricities but different slenderness ratios. Here also, the mid-height section of the column undergoes material failure, either a compression failure or a tension failure, depending on the location of the point B on the interaction diagram.


The loading path OC, on the other hand, is for a very long column when the lateral deflection is so high that the slope of the path dP/dM at C is zero. The column is so slender that the failure is due to buckling (instability) at a comparatively much low value of the load P

Δ

cr, though this column has the same cross-sectional area and the eccentricity of load as of the other two columns. Such instability failure occurs for very slender columns, specially when they are not braced. The following points are summarised from the discussion made in sec.10.27.3. 1. Additional deflections and moments are caused by the axial compression force P in columns. The additional moments increase with the increase of kl/r, when other parameters are equal. 2. Laterally braced compression members and bent in single curvature have the same or nearby locations of the maxima of both Mo and Py. Thus, being fully additive, they have large moment magnification. 3. Laterally braced compression members and bent in double curvature have different locations of the maxima of both Mo and Py. As a result, the moment magnification is either less or zero. 4. Members of frames not braced laterally, the maxima of Mo and Py mostly occur at the ends of column and cause the maximum total moment at the ends of columns only. Additional moments and additional deflections increase with the increase of kl/r. 10.27.4 Effective Length of Columns Annex E of IS 456 presents two figures (Figs.26 and 27) and a table (Table 26) to estimate the effective length of columns in frame structures based on a research paper, “Effective length of column in multistoreyed building” by R.H. Wood in The Structural Engineer Journal, No.7, Vol.52, July 1974. Figure 26 is for columns in a frame with no sway, while Fig.27 is for columns in a frame with sway. These two figures give the values of k (i.e., le/l) from two parameters

21 and ββ which are obtained from the following expression: (10.74)

/ ∑ ∑ ∑+= bcc KKKβ

where Kc and Kb are flexural stiffnesses of columns and beams, respectively. The quantities 21 and ββ at the top and bottom joints A and B, respectively, are determined by summing up the K values of members framing into a joint at top


and bottom, respectively. Thus 21 and ββ for the frame shown in Fig.10.27.16 are as follows:

1β = (Kc + Kct)/(Kc + Kct + Kb1 + Kb2) (10.75) 2β = (Kc + Kcb)/(Kc + Kcb + Kb3 + Kb4) (10.76) However, assuming idealised conditions, the effective length in a given plane may be assessed from Table 28 in Annex E of IS 456, for normal use. 10.27.5 Determination of Sway or No Sway Column Clause E-2 of IS 456 recommends the stability index Q to determine if a column is a no sway or sway type. The stability index Q is expressed as: Q = P∑ u /HuΔ u hz (10.77) where ∑ Pu = sum of axial loads on all columns in the storey, = elastically computed first-order lateral deflection, uΔ Hu = total lateral force acting within the storey, and


hz = height of the storey. The column may be taken as no sway type if the value of Q is 0.4, otherwise, the column is considered as sway type.

≤

10.27.6 Design of Slender Columns The design of slender columns, in principle, is to be done following the same procedure as those of short columns. However, it is essential to estimate the total moment i.e., primary and secondary moments considering P- effects. These secondary moments and axial forces can be determined by second-order rigorous structural analysis – particularly for unbraced frames. Further, the problem becomes more involved and laborious as the principle of superposition is not applicable in second-order analysis.

Δ

However, cl.39.7 of IS 456 recommends an alternative simplified method of determining additional moments to avoid the laborious and involved second-order analysis. The basic principle of additional moment method for estimating the secondary moments is explained in the next section. 10.27.7 Additional Moment Method In this method, slender columns should be designed for biaxial eccentricities which include secondary moments (Py of Eq.10.63 and 10.65) about major and minor axes. We first consider braced columns which are bent symmetrically in single curvature and cause balanced failure i.e., Pu = Pub. (A) Braced columns bent symmetrically in single curvature and undergoing balanced failure For braced columns bent symmetrically in single curvature, we have from Eqs.10.63 and 10.65, M = Mo + Py = Mo + P ea = Mo + Ma (10.78) where P is the factored design load Pu, M are the total factored design moments Mux and Muy about the major and minor axes, respectively; Mo are the primary factored moments Moux and Mouy about the major and minor axes, respectively; Ma are the additional moments Max and May about the major and minor axes, respectively and ea are the additional eccentricities eax and eay along the minor and major axes, respectively. The quantities Mo and P of Eq.10.78 are known and hence, it is required to determine the respective values of ea, the additional eccentricities only.


Let us consider the columns of Figs.10.27.10 and 11 showing as the maximum deflection at the mid-height section of the columns. The column of Fig.10.27.10, having a constant primary moment M

Δ

o, causes constant curvature φ , while the column of Fig.10.27.11, having a linearly varying primary moment with a maximum value of Mo max at the mid-height section of the column, has a linearly varying curvature with the maximum curvature of φ max at the mid-height section the column. The two maximum curvatures can be expressed in terms of their respective maximum deflection Δ as follows: The constant curvature (Fig.10.27.10) (10.79)

2max /8 elΔ=φ

The linearly varying curvature (Fig.10.27.11) (10.80)

2max /12 elΔ=φ

where le are the respective effective lengths kl of the columns. We, therefore, consider the maximum φ as the average value lying in between the two values of Eqs.10.79 and 80 as (10.81)

2max /10 elΔ=φ

Accordingly, the maximum additional eccentricities ea, which are equal to the maximum deflections Δ , can be written as ea = = (10.82)

Δ /10 2elφ


Assuming the column undergoes a balanced failure when Pu = Pub, the maximum curvature at the mid-height section of the column, shown in Figs.10.27.17a and b, can be expressed as given below, assuming (i) the values of cε = 0.0035, stε = 0.002 and Dd /′ = 0.1, and (ii) the additional moment capacities are about eighty per cent of the total moment. φ = eighty per cent of {(0.0035 + 0.002)/0.9D} (see Fig.10.27.17c) or φ = 1/200D (10.83) Substituting the value of φ in Eq.10.82, ea = D(le/D)2/2000 (10.84) Therefore, the additional moment Ma can be written as,


Ma = Py = PΔ = Pea = (PD/2000) (le/D)2 (10.85) Thus, the additional moments Max and May about the major and minor axes, respectively, are: Max = (PuD/2000) (lex/D)2 (10.86) May = (Pub/2000) (ley/b)2 (10.87) where Pu = axial load on the member, lex = effective length in respect of the major axis, ley = effective length in respect of the minor axis, D = depth of the cross-section at right angles to the major axis, and b = width of the member. Clause 39.7.1 of IS 456 recommends the expressions of Eqs.10.86 and 87 for estimating the additional moments Max and May for the design. These two expressions of the additional moments are derived considering the columns to be braced and bent symmetrically undergoing balanced failure. Therefore, proper modifications are necessary for different situations like braced columns with unequal end moments with the same or different signs, unbraced columns and columns causing compression failure i.e., when Pu > Pub. (B) Braced columns subjected to unequal primary moments at the two ends For braced columns without any transverse loads occurring in the height, the primary maximum moment (Mo max of Eq.10.64), with which the additional moments of Eqs.10.86 and 87 are to be added, is to be taken as: Mo max = 0.4 M1 + 0.6 M2 (10.88) and further Mo max ≥ 0.4 M2 (10.89) where M2 is the larger end moment and M1 is the smaller end moment, assumed to be negative, if the column is bent in double curvature.


To eliminate the possibility of total moment Mu max becoming less than M2 for columns bent in double curvature (see Fig.10.27.12) with M1 and M2 having opposite signs, another condition has been imposed as Mu max M≥ 2 (10.90) The above recommendations are given in notes of cl.39.7.1 of IS 456. (C) Unbraced columns Unbraced frames undergo considerable deflection due to P- effect. The additional moments determined from Eqs.10.86 and 87 are to be added with the maximum primary moment M

Δ

o max at the ends of the column. Accordingly, we have Mo max = M2 + Ma (10.91) The above recommendation is given in the notes of cl.39.7.1 of IS 456. (D) Columns undergoing compression failure (Pu > Pub) It has been mentioned in part A of this section that the expressions of additional moments given by Eqs.10.86 and 10.87 are for columns undergoing balanced failure (Fig.10.27.17). However, when the column causes compression failure, the e/D ratio is much less than that of balanced failure at relatively high axial loads. The entire section may be under compression causing much less curvatures. Accordingly, additional moments of Eqs.10.86 and 10.87 are to be modified by multiplying with the reduction factor k as given below: (i) For Pu > Pubx: kax = (Puz – Pu)/(Puz – Pubx) (10.92) (ii) For Pu > Puby: kay = (Puz – Pu)/(Puz – Puby) (10.93) with a condition that kax and kay should be ≤ 1 (10.94) where Pu = axial load on compression member Puz is given in Eq.10.59 of Lesson 26 and is, Puz = 0.45 fck Ac + 0.75 fy Ast … (10.59)


Pubx, Puby = axial loads with respect to major and minor axes, respectively, corresponding to the condition of maximum compressive strain of 0.0035 in concrete and tensile strain of 0.002 in outermost layer of tension steel. It is seen from Eqs.10.92 and 10.93 that the values of k (kax and kay) vary linearly from zero (when Pu = Puz) to one (when Pu = Pub). Since Eqs.10.92 and 10.93 are not applicable for Pu < Pub, another condition has been imposed as given in Eq.10.94. The above recommendations are given in cl.39.7.1.1 of IS 456. The following discussion is very important for the design of slender columns. Additional moment method is one of the methods of designing slender columns as discussed in A to D of this section. This method is recommended in cl.39.7 of IS 456 also. The basic concept here is to enhance the primary moments by adding the respective additional moments estimated in a simple way avoiding laborious and involved calculations of second-order structural analysis. However, these primary moments under eccentric loadings should not be less than the moments corresponding to the respective minimum eccentricity, as stipulated in the code. Hence, the primary moments in such cases are to be replaced by the minimum eccentricity moments. Moreover, all slender columns, including those under axial concentric loadings, are also to be designed for biaxial bending, where the primary moments are zero. In such cases, the total moment consisting of the additional moment multiplied with the modification factor, if any, in each direction should be equal to or greater than the respective moments under minimum eccentricity conditions. As mentioned earlier, the minimum eccentricity consideration is given in cl.25.4 of IS 456. 10.27.8 Illustrative Example The following illustrative example is taken up to explain the design of slender columns. The example has been solved in step by step using (i) the equations of Lessons 21 to 27 and (ii) employing design charts and tables of SP-16, to compare the results.


Problem 1: Determine the reinforcement required for a braced column against sidesway with the following data: size of the column = 350 x 450 mm (Fig.10.27.18); concrete and steel grades = M 30 and Fe 415, respectively; effective lengths lex and ley = 7.0 and 6.0 m, respectively; unsupported length l = 8 m; factored load Pu = 1700 kN; factored moments in the direction of larger dimension = 70 kNm at top and 30 kNm at bottom; factored moments in the direction of shorter dimension = 60 kNm at top and 30 kNm at bottom. The column is bent in double curvature. Reinforcement will be distributed equally on four sides. Solution 1: Step 1: Checking of slenderness ratios lex/D = 7000/450 = 15.56 > 12, ley/b = 6000/350 = 17.14 > 12. Hence, the column is slender with respect to both the axes. Step 2: Minimum eccentricities and moments due to minimum

eccentricities (Eq.10.3 of Lesson21) ex min = l/500 + D/30 = 8000/500 + 450/30 = 31.0 > 20 mm ey min = l/500 + b/30 = 8000/500 + 350/30 = 27.67 > 20 mm


Mox (Min. ecc.) = Pu(ex min) = (1700) (31) (10-3) = 52.7 kNm Moy (Min. ecc.) = Pu(ey min) = (1700) (27.67) (10-3) = 47.04 kNm Step 3: Additional eccentricities and additional moments Method 1: Using Eq. 10.84 eax = D(lex/D)2/2000 = (450) (7000/450)2/2000 = 54.44 mm eay = b(lex/b)2/2000 = (350) (6000/350)2/2000 = 51.43 mm Max = Pu(eax) = (1700) (54.44) (10-3) = 92.548 kNm May = Pu(eay) = (1700) (51.43) (10-3) = 87.43 kNm Method 2: Table I of SP-16 For lex/D = 15.56, Table I of SP-16 gives: eax/D = 0.1214, which gives eax = (0.1214) (450) = 54.63 mm For ley/D = 17.14, Table I of SP-16 gives: eay/b = 0.14738, which gives eay = (0.14738) (350) = 51.583 mm It is seen that values obtained from Table I of SP-16 are comparable with those obtained by Eq. 10.84 in Method 1. Step 4: Primary moments and primary eccentricities (Eqs.10.88 and 89) Mox = 0.6M2 – 0.4M1 = 0.6(70) – 0.4(30) = 30 kNm, which should be ≥ 0.4 M2 (= 28 kNm). Hence, o.k.

Moy = 0.6M2 – 0.4M1 = 0.6(60) – 0.4(30) = 24 kNm, which should be ≥ 0.4 M2 (= 24 kNm). Hence, o.k. Primary eccentricities: ex = Mox/Pu = (30/1700) (103) = 17.65 mm ey = Moy/Pu = (24/1700) (103) = 14.12 mm


Since, both primary eccentricities are less than the respective minimum eccentricities (see Step 2), the primary moments are revised to those of Step 2. So, Mox = 52.7 kNm and Moy = 47.04 kNm. Step 5: Modification factors To determine the actual modification factors, the percentage of longitudinal reinforcement should be known. So, either the percentage of longitudinal reinforcement may be assumed or the modification factor may be assumed which should be verified subsequently. So, we assume the modification factors of 0.55 in both directions. Step 6: Total factored moments

Mux = Mox + (Modification factor) (Max) = 52.7 + (0.55) (92.548) = 52.7 + 50.9 = 103.6 kNm

Muy = Moy + (Modification factor) (May) = 47.04 + (0.55) (87.43) = 47.04 + 48.09 = 95.13 kNm

Step 7: Trial section (Eq.10.61 of Lesson 26) The trial section is determined from the design of uniaxial bending with Pu = 1700 kN and Mu = 1.15 . So, we have M2/122 ) ( uyux MM + u = (1.15){(103.6)2 + (95.13)2}1/2 = 161.75 kNm. With these values of Pu (= 1700 kN) and Mu (= 161.75 kNm), we use chart of SP-16 for the Dd /′ = 0.134. We assume the diameters of longitudinal bar as 25 mm, diameter of lateral tie = 8 mm and cover = 40 mm, to get = 40 + 8 + 12.5 = 60.5 mm. Accordingly, d ′ Dd /′ = 60.5/450 = 0.134 and

= 60.5/350 = 0.173. We have: bd /′ Pu/fck bD = 1700(103)/(30)(350)(450) = 0.3598 Mu/fck bD2 = 161.75(106)/(30)(350)(450)(450) = 0.076 We have to interpolate the values of p/fck for Dd /′ = 0.134 obtained from Charts 44 (for = 0.1) and 45 (Dd /′ Dd /′ = 0.15). The values of p/fck are 0.05 and 0.06 from Charts 44 and 45, respectively. The corresponding values of p are 1.5 and 1.8 per cent, respectively. The interpolated value of p for = 0.134 is 1.704 per cent, which gives A

Dd /′sc = (1.704)(350)(450)/100 = 2683.8 mm2. We

use 4-25 + 4-20 (1963 + 1256 = 3219 mm2), to have p provided = 2.044 per cent giving p/fck = 0.068. Step 8: Calculation of balanced loads Pb


The values of Pbx and Pby are determined using Table 60 of SP-16. For this purpose, two parameters k1 and k2 are to be determined first from the table. We have p/fck = 0.068, = 0.134 and Dd /′ bd /′ = 0.173. From Table 60, k1 = 0.19952 and k2 = 0.243 (interpolated for Dd /′ = 0.134) for Pbx. So, we have: Pbx/fckbD = k1 + k2 (p/fck) = 0.19952 + 0.243(0.068) = 0.216044, which gives Pbx = 0.216044(30)(350)(450)(10-3) = 1020.81 kN. Similarly, for Pby: = 0.173, p/fbd /′ ck = 0.068. From Table 60 of SP-16, k1 = 0.19048 and k2 = 0.1225 (interpolated for bd /′ = 0.173). This gives Pby/fckbD = 0.19048 + 0.1225(0.068) = 0.19881, which gives Pby = (0.19881)(30)(350)(450)(10-3) = 939.38 kN. Since, the values of Pbx and Pby are less than Pu, the modification factors are to be used. Step 9: Determination of Puz Method 1: From Eq.10.59 of Lesson 26 Puz = 0.45 fck Ag + (0.75 fy – 0.45 fck) Asc = 0.45(30)(350)(450) + {0.75(415) – 0.45(30)}(3219) = 3084.71 kN Method 2: Using Chart 63 of SP-16 We get Puz/Ag = 19.4 N/mm2 from Chart 63 of SP-16 using p = 2.044 per cent. Therefore, Puz = (19.4)(350)(450)(10-3) = 3055.5 kN, which is in good agreement with that of Method 1. Step 10: Determination of modification factors Method 1: From Eqs.10.92 and 10.93 kax = (Puz – Pu)/(Puz – Pubx) … (10.92) or kax = (3084.71 – 1700)/(3084.71 – 1020.81) = 0.671 and

kay = (Puz – Pu)/(Puz – Puby) … (10.93) or kay = (3084.71 – 1700)/(3084.71 – 939.39) = 0.645 The values of the two modification factors are different from the assumed value of 0.55 in Step 5. However, the moments are changed and the section is checked for safety.


Method 2: From Chart 65 of SP-16 From Chart 65 of SP-16, for the two parameters, Pbx/Puz = 1020.81/3084.71 = 0.331 and Pu/Puz = 1700/3084.71 = 0.551, we get kax = 0.66. Similarly, for the two parameters, Pby/Puz = 939.38/3084.71 = 0.3045 and Pu/Puz = 0.551, we have kay = 0.65. Values of kax and kay are comparable with those of Method 1. Step 11: Total moments incorporating modification factors Mux = Mox (from Step 4) + (kax) Max (from Step 3) = 52.7 + 0.671(92.548) = 114.8 kNm Muy = Moy (from Step 4) + kay (May) (from Step 3) = 47.04 + (0.645)(87.43) = 103.43 kNm. Step 12: Uniaxial moment capacities The two uniaxial moment capacities Mux1 and Muy1 are determined as stated: (i) For Mux1, by interpolating the values obtained from Charts 44 and 45, knowing the values of Pu/fckbD = 0.3598 (see Step 7), p/fck = 0.068 (see Step 7),

= 0.134 (see Step 7), (ii) for MDd /′ uy1, by interpolating the values obtained from Charts 45 and 46, knowing the same values of Pu/fckbD and p/fck as those of (i) and = 0.173 (see Step 7). The results are given below: Dd /′ (i) Mux1/fckbD2 = 0.0882 (interpolated between 0.095 and 0.085) (ii) Muy1/fckbb2 = 0.0827 (interpolated between 0.085 and 0.08) So, we have, Mux1 = 187.54 kNm and Muy1 = 136.76 kNm. Step 13: Value of nα Method 1: From Eq.10.60 of Lesson 26 We have Pu/Puz = 1700/3084.71 = 0.5511. From Eq.10.60 of Lesson 26, we have nα = 0.67 + 1.67 (Pu/Puz) = 1.59. Method 2: Interpolating the values between (Pu/Puz) = 0.2 and 0.6 The interpolated value of nα = 1.0 + (0.5511 – 0.2)/0.6 = 1.5852. Both the values are comparable. We use nα = 1.5852.


Step 14: Checking of column for safety Method 1: From Eq.10.58 of Lesson 26 We have in Lesson 26: 1 )/( )/( 11 ≤+ nn

uyuyuxux MMMM αα … (10.58) Here, putting the values of Mux, Mux1, Muy, Muy1 and nα , we get: (114.8/187.54)1.5452 + (103.43/136.76)1.5852 = 0.4593 + 0.6422 = 1.1015. Hence, the section or the reinforcement has to be revised. Method 2: Chart 64 of SP-16 The point having the values of (Mux/Mux1) = 114.8/187.54 = 0.612 and (Muy/Muy1) = 103.43/136.76 = 0.756 gives the value of Pu/Pz more than 0.7. The value of Pu/Puz here is 0.5511 (see Step 13). So, the section needs revision. We revise from Step 7 by providing 8-25 mm diameter bars (= 3927 mm2, p = 2.493 per cent and p/fck = 0.0831) as the longitudinal reinforcement keeping the values of b and D unchanged. The revised section is checked furnishing the repeated calculations from Step 8 onwards. The letter R is used before the number of step to indicate this step as revised one. Step R8: Calculation of balanced loads Pb Table 60 of SP-16 gives k1 = 0.19952, and k2 = 0.243. We have p/fck = 0.0831 now. So, Pbx = {0.19952 + (0.243)(0.0831)} (30)(350)(450)(10-3) = 1038.145 kN. Similarly, k1 = 0.19048, k2 = 0.1225 and p/fck = 0.0831 give Pby = {0.19048 + (0.1225)(0.0831)} (30)(350)(450)(10-3) = 948.12 kN. The values of Pbx and Pby are less than Pu (= 1700 kN). So, modification factors are to be incorporated. Step R9: Determination of Puz (Eq. 10.59 of Lesson 26) Puz = 0.45(30)(350)(450) + {0.75(415) – 0.45(30)}(3927) = 3295.514 kN. Step R10: Determination of modification factors (Eqs.10.92 and 10.93) kax = (3295.514 – 1700)/(3295.514 – 1038.145) = 0.707 kay = (3295.514 – 1700)/(3295.514 – 948.12) = 0.68 Step R11: Total moments incorporating modification factors


Mux = 52.70 + 0.707(92.548) = 118.13 kNm Muy = 47.04 + 0.68(87.43) = 106.49 kNm Step R12: Uniaxial moment capacities Using Charts 44 and 45 for Mux1 and Charts 45 and 46 for Muy1, we get (i) the coefficient 0.1032 (interpolating 0.11 and 0.10) and (ii) the coefficient 0.0954 (interpolating 0.1 and 0.09) for Mux1 and Muy1, respectively. Mux1 = (0.1032)(30)(350)(450)(450)(10-6) = 219.429 kNm Muy1 = (0.0954)(30)(450)(350)(350)(10-6) = 157.77 kNm Step R13: Value of nα (Eq.10.60 of Lesson 26) Pu/Puz = 1700/3295.514 = 0.5158 which gives nα = 1 + (0.5158 – 0.2)/0.6 = 1.5263 Step R14: Checking of column for safety (Eq.10.58 of Lesson 26) (118.13/219.424)1.5263 + (106.49/157.77)1.5263 = 0.3886 + 0.5488 = 0.9374 < 1.0 Hence, the revised reinforcement is safe. The section is shown in Fig.10.27.18. 10.27.9 Practice Questions and Problems with Answers Q.1: Define a slender column. Give three reasons for its increasing importance

and popularity. A.1: See sec. 10.27.1. Q.2: Explain the behaviour of a slender column subjected to concentric loading.

Explain Euler’s load. A.2: See sec.10.27.3. Q.3: Choose the correct answer.

(A) As the slenderness ratio increases, the strength of concentrically loaded column:

(i) increases (ii) decreases


(B) For braced columns, the effective length is between (i) l and 2l (ii) 0.5l and 2l (iii) 0.5l and l (C) The critical load of a braced frame is (i) always larger than that of an unbraced column (ii) always smaller than that of an unbraced column (iii) sometimes larger and sometimes smaller than that of an unbraced column A.3: A. (ii), B. (iii), C. (i) Q.4: Explain the behaviour of slender columns under axial load and uniaxial

bending, bent in single curvature. A.4: Part (A) of sec. 10.27.3. Q.5: Explain the behaviour of slender columns under axial load and uniaxial

bending, bent in double curvature. A.5: Part (B) of sec. 10.27.3. Q.6: Explain the behaviour of columns in portal frame both braced and unbraced. A.6: Part (C) of sec. 10.27.3.


Q.7: Check the column of Fig.10.27.19, if subjected to an axial factored load of

Pu = 1500 kN only when the unsupported length of the column = l = 8.0 m, lex = ley = 6.0 m, D = 400 mm, b = 300 mm, using concrete of M 20 and steel grade in Fe 415.

A.7: Solution: Step 1: Slenderness ratios Lex/D = 6000/400 = 15 > 12 Ley/b = 6000/300 = 20 > 12 The column is slender about both the axes. Step 2: Minimum eccentricities and moments due to minimum

eccentricities (Eq.10.3 of Lesson 21) ex min = l/500 + D/30 = 8000/500 + 400/30 = 29.33 mm > 20 mm ey min = 8000/500 + 300/30 = 26 mm > 20 mm Mx due to min. ecc. = Pu (ex min) = 1500(29.33) = 43.995 kNm My due to min. ecc. = Pu (ey min) = 1500(26.0) = 39.0 kNm Step 3: Primary moments Since the column is concentrically loaded, the primary moments are zero. Therefore, the additional moments must be greater than the respective moments due to minimum eccentricity. Step 4: Additional eccentricities and moments (Eq.10.84) eax = D(lex/D)2/2000 = 400(6000/400)2/2000 = 45 mm > ex min (= 29.23 mm) eay = b(ley/b)2/2000 = 300(6000/300)2/2000 = 60 mm > ey min (= 26 mm) Step 5: Calculation of balance loads Pbx and Pby Given Asc = 3927 mm2 (8 bars of 25 mm diameter give p = 3.2725 per cent. So, p/fck = 0.1636. Using 8 mm diameter lateral tie, d ′ = 40 + 8 + 12.5 =


60.5 mm giving /D = 60.5/400 = 0.15125 d ′ ≅ 0.15 and d ′ /b = 60.5/300 = 0.2017 0.20. ≅

From Table 60 of SP-16, we get k1 = 0.196 and k2 = 0.061. Thus, we have: Pbx = {0.196 + (0.061)(0.1636)}(20)(300)(400)(10-3) = 494.35 kN Similarly, for Pby: k1 = 0.184 and k2 = -0.011, we get Pby = {0.184 - (0.011)(0.1636)}(20)(300)(400)(10-3) = 437.281 kN Since, Pbx and Pby are less than Pu (= 1500 kN), modification factors are to be incorporated. Step 6: Determination of Puz (Eq.10.59 of Lesson 26) Puz = 0.45(20)(300)(400) + {0.75(415) – 0.45(20)}(3927)(10-3) = 2266.94 kN Step 7: Determination of modification factors kax = (2266.94 – 1500)/(2266.94 – 494.35) = 0.433 and kay = (2266.94 – 1500)/(2266.94 – 437.281) = 0.419 Step 8: Additional moments and total moments Max = 1500(0.433)(45) = 29.2275 kNm May = 1500(0.419)(60) = 37.71 kNm Since, primary moments are zero as the column is concentrically loaded, the total moment shall consist of the additional moments. But, as both the additional moments are less than the respective moment due to minimum eccentricity, the revised additional moments are: Max = 43.995 kNm and May = 39.0 kNm, which are the total moments also. Thus, we have: Mux = 43.995 kNm, Muy = 39.0 kNm and Pu = 1500 kN. Step 9: Uniaxial moment capacities


We have, Pu/fck bD = {1500/(20)(300)(400)}(1000) = 0.625, p/fck = 0.1636 and /D = 0.15 for Md ′ ux1; and d ′ /b = 0.2 for Muy1. The coefficients are 0.11 (from Chart 45) and 0.1 (from Chart 46) for Mux1 and Muy1, respectively. So, we get, Mux1 = 0.11(20)(300)(400)(400)(10-6) = 225.28 kNm, and Muy1 = 0.1(20)(300)(300)(400)(10-6) = 72.0 kNm Step 10: Value of nα (Eq.10.60 of Lesson 26) Here, Pu/Puz = 1500/2266.94 = 0.6617. So, we get nα = 1.0 + (0.4617/0.6) = 1.7695 Step 11: Checking the column for safety (Eq.10.58 of Lesson 26) 1 )/( )/( 11 ≤+ nn

uyuyuxux MMMM αα

Here, (43.995/225.28)1.7695 + (39.0/72.0)1.7695 = 0.0556 + 0.3379 = 0.3935 < 1 Hence, the column is safe to carry Pu = 1500 kN. 11.27.10 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Reinforced Concrete Limit State Design, 5th Edition, by Ashok K. Jain, Nem Chand & Bros, Roorkee, 1999.

11.27.11 Test 27 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions.


TQ.1: Determine the primary, additional and total moments of the column shown in Fig.10.27.20 for the three different cases:

(i) When the column is braced against sidesway and is bent in single curvature. (ii) When the column is braced against sidesway and is bent in double curvature. (iii) When the column is unbraced. Use the following data: Pu = 2000 kN, concrete grade = M 20, steel grade = Fe 415, unsupported length l = 8.0 m, lex = 7.0 m, ley = 6.0 m, Asc = 6381 mm2 (12-25 mm diameter bars), lateral tie = 8 mm diameter @ 250 mm c/c, d = 60.5 mm, D = 500 mm and b = 400 mm. The factored moments are: 70 kNm at top and 40 kNm at bottom in the direction of larger dimension and 60 kNm at top and 30 kNm at bottom in the direction of shorter dimension.

′

A.TQ.1: Solution The following are the common steps for all three cases. Step 1: Slenderness ratios lex/D = 7000/500 = 14 > 12 and ley/b = 6000/400 = 15 > 12 The column is slender about both axes. Step 2: Minimum eccentricities and moments due to minimum

eccentricities (Eq.10.3 of Lesson 21) ex min = l/500 + D/30 = 8000/500 + 500/30 = 32.67 mm > 20 mm, and ey min = l/500 + b/30 = 8000/500 + 400/30 = 29.34 mm > 20 mm

Mx (min. ecc.) = 2000(32.67)(10-3) = 65.34 kNm, and My (min. ecc.) = 2000(29.34)(10-3) = 58.68 kNm Step 3: Additional eccentricities and moments due to additional eccentricities (Eq.10.84) eax = D(lex/D)2/2000 = 500(7000/500)2/2000 = 49 mm > ex min (= 32.67 mm)


eay = b(ley/b)2/2000 = 400(6000/400)2/2000 = 45 mm > ey min (= 29.34 mm) Max = Pu(eax) = (2000)(49)(10-3) = 98 kNm, and May = Pu(eay) = (2000)(45)(10-3) = 90 kNm Step 4: Calculation of balanced loads Using d ′ /D = 0.121 and p/fck = 3.1905/20 = 0.159525 in Table 60 of SP-16, we have k1 = 0.20238 and k2 = 0.2755 (by linear interpolation). This gives Pbx = {0.20238 + 0.2755(0.159525)}(20)(400)(500)(10-3) = 983.32 kN Similarly, d /b = 0.15125 and p/f′ ck = 0.159525 in Table 60 of SP-16 gives k1 = 0.1957 and k2 = 0.198625 (by linear interpolation). So, we get Pby = {0.1957 + 0.198625(0.159525)}(20)(400)(500)(10-3) = 909.54 kN Both Pbx and Pby are smaller than Pu (= 2000 kN). Hence, modification factors are to be incorporated. Step 5: Calculation of Puz (Eq.10.59 of Lesson 26) Puz = 0.45 fck Ag + (0.75 fy – 0.45 fck) Asc

= 0.45(20)(400)(500) + {0.75(415) – 0.45(20)}(6381) = 3728.66 kN Step 6: Modification factors and revised additional moments (Eqs.10.92 and 10.93) kax = (3728.66 - 2000)/(3728.66 – 983.32) = 0.6297, and kay = (3728.66 - 2000)/(3728.66 – 909.54) = 0.6132 The revised additional moments are: Max = 98(0.6297) = 61.71 kNm, and May = 90(0.6132) = 55.19 kNm Now, the different cases are explained. Case (i): Braced column in single curvature


Primary moments = 0.4 M1 + 0.6 M2, but should be equal to or greater than 0.4 M2 and moment due to minimum eccentricities. So, we get, Mox = largest of 58 kNm, 28 kNm and 65.34 kNm = 65.34 kNm Moy = largest of 48 kNm, 24 kNm and 58.68 kNm = 58.68 kNm Additional moments are Max = 61.71 kNm and May = 55.19 kNm (incorporating the respective modification factors). Total moments = Mux = Mox + Max = 65.34 + 61.71 = 127.05 kNm > 65.34 kNm (moment due to minimum eccentricity), and Muy = Moy + May = 58.68 + 55.19 = 113.87 kNm > 58.68 kNm (moment due to minimum eccentricity). Case (ii): Braced column in double curvature

Primary moments = - 0.4 M1 + 0.6 M2, but should be equal to or greater than 0.4M2 and the moment due to minimum eccentricity. So, we get, Mox = largest of 26 kNm, 28 kNm and 65.34 kNm = 65.34 kNm Moy = largest of 24 kNm, 24 kNm and 58.68 kNm = 58.68 kNm Additional moments are Max = 61.71 kNm and May = 55.19 kNm Final moments = Mux = Mox + Max = 65.34 + 61.71 = 127.05 kNm > 65.34 kNm (moment due to minimum eccentricity), and Muy = 58.68 + 55.19 = 113.87 kNm > 58.68 kNm (moment due to minimum eccentricity). Case (iii): Unbraced column

Primary moments = M2 and should be greater than or equal to moment due to minimum eccentricity. Mox = 70 kNm > 65.34 kNm (moment due to minimum eccentricity), and Moy = 60 kNm > 58.68 kNm (moment due to minimum eccentricity). Additional moments are Max = 61.71 kNm and May = 55.19 kNm Final moments = Mux = Mox + Max = 70.0 + 61.71 = 131.71 kNm > 65.34 kNm (moment due to minimum eccentricity), and


Muy = Moy + Max = 60.0 + 55.19 = 115.19 kNm > 58.68 kNm (moment due to minimum eccentricity). 10.27.12 Summary of this Lesson

This lesson mentions the reasons of increasing importance and popularity of slender columns and explains the behaviour of slender columns loaded concentrically or eccentrically. The role of minimum eccentricity that cannot be avoided in any practical column is explained for slender columns. The moments due to minimum eccentricities in both directions should be taken into account for a slender column loaded concentrically as it should be designed under biaxial bending. On the other hand, the given primary moments are also to be checked so that they are equal to or greater than the respective moments due to minimum eccentricity for all slender columns. Both braced and unbraced columns, bent in single or double curvatures, are explained. The importance of modification factors of the additional moments due to P-Δ effect is explained. Effective lengths and important parameter to determine the slenderness ratios are illustrated for different types of support conditions either in single column or when the column is a part of rigid frames. Additional moment method, a simple method for the design of slender columns, is explained, which is recommended in IS 456. Numerical problems in illustrative example, practice problem and test questions will help in understanding and applying the method for the design of slender columns, as stipulated in IS 456. Direct computations from the given equations as well as use of design charts and tables of SP-16 are illustrated for the design.


Module 11

Foundations - Theory and Design


Lesson 28

Foundations - Theory




• explain the two major and other requirements of the design of foundation, • identify five points indicating the differences between the design of

foundation and the design of other elements of the superstructure,

• differentiate between footing and foundation,

• differentiate between shallow and deep foundations,

• identify the situations when a combined footing shall be used,

• explain the safe bearing capacity of soil mentioning the difference between gross and net safe bearing capacities,

• determine the minimum depth of foundation,

• determine the critical sections of bending moment and shear in isolated

footings,

• draw the distributions of pressure of soil below the footing for concentric and eccentric loads with e ≤ L/6 and e > L/6,

• determine the soil pressure in a foundation which is unsymmetrical.


Till now we discussed the different structural elements viz. beams, slabs, staircases and columns, which are placed above the ground level and are known as superstructure. The superstructure is placed on the top of the foundation structure, designated as substructure as they are placed below the ground level. The elements of the superstructure transfer the loads and moments to its adjacent element below it and finally all loads and moments come to the foundation structure, which in turn, transfers them to the underlying soil or rock. Thus, the foundation structure effectively supports the superstructure. However, all types of soil get compressed significantly and cause the structure to settle. Accordingly, the major requirements of the design of foundation structures are the two as given below (see cl.34.1 of IS 456):


1. Foundation structures should be able to sustain the applied loads, moments, forces and induced reactions without exceeding the safe bearing capacity of the soil. 2. The settlement of the structure should be as uniform as possible and it should be within the tolerable limits. It is well known from the structural analysis that differential settlement of supports causes additional moments in statically indeterminate structures. Therefore, avoiding the differential settlement is considered as more important than maintaining uniform overall settlement of the structure. In addition to the two major requirements mentioned above, the foundation structure should provide adequate safety for maintaining the stability of structure due to either overturning and/or sliding (see cl.20 of IS 456). It is to be noted that this part of the structure is constructed at the first stage before other components (columns / beams etc.) are taken up. So, in a project, foundation design and details are completed before designs of other components are undertaken. However, it is worth mentioning that the design of foundation structures is somewhat different from the design of other elements of superstructure due to the reasons given below. Therefore, foundation structures need special attention of the designers. 1. Foundation structures undergo soil-structure interaction. Therefore, the behaviour of foundation structures depends on the properties of structural materials and soil. Determination of properties of soil of different types itself is a specialized topic of geotechnical engineering. Understanding the interacting behaviour is also difficult. Hence, the different assumptions and simplifications adopted for the design need scrutiny. In fact, for the design of foundations of important structures and for difficult soil conditions, geotechnical experts should be consulted for the proper soil investigation to determine the properties of soil, strata wise and its settlement criteria. 2. Accurate estimations of all types of loads, moments and forces are needed for the present as well as for future expansion, if applicable. It is very important as the foundation structure, once completed, is difficult to strengthen in future. 3. Foundation structures, though remain underground involving very little architectural aesthetics, have to be housed within the property line which may cause additional forces and moments due to the eccentricity of foundation. 4. Foundation structures are in direct contact with the soil and may be affected due to harmful chemicals and minerals present in the soil and fluctuations of water table when it is very near to the foundation. Moreover,


periodic inspection and maintenance are practically impossible for the foundation structures. 5. Foundation structures, while constructing, may affect the adjoining structure forming cracks to total collapse, particularly during the driving of piles etc. However, wide ranges of types of foundation structures are available. It is very important to select the appropriate type depending on the type of structure, condition of the soil at the location of construction, other surrounding structures and several other practical aspects as mentioned above. 11.28.2 Types of Foundation Structures Foundations are mainly of two types: (i) shallow and (ii) deep foundations. The two different types are explained below: (A) Shallow foundations Shallow foundations are used when the soil has sufficient strength within a short depth below the ground level. They need sufficient plan area to transfer the heavy loads to the base soil. These heavy loads are sustained by the reinforced concrete columns or walls (either of bricks or reinforced concrete) of much less areas of cross-section due to high strength of bricks or reinforced concrete when compared to that of soil. The strength of the soil, expressed as the safe bearing capacity of the soil as discussed in sec.11.28.3, is normally supplied by the geotechnical experts to the structural engineer. Shallow foundations are also designated as footings. The different types of shallow foundations or footings are discussed below.


1. Plain concrete pedestal footings

Plain concrete pedestal footings (Fig.11.28.1) are very economical for columns of small loads or pedestals without any longitudinal tension steel (see cls.34.1.2 and 34.1.3 of IS 456). In Fig.11.28.1, the angle α between the plane passing through the bottom edge of the pedestal and the corresponding junction edge of the column with pedestal and the horizontal plane shall be determined from Eq. 11.3.


2. Isolated footings


These footings are for individual columns having the same plan forms of square, rectangular or circular as that of the column, preferably maintaining the proportions and symmetry so that the resultants of the applied forces and reactions coincide. These footings, shown in Figs.11.27.2 to 11.27.4, consist of a slab of uniform thickness, stepped or sloped. Though sloped footings are economical in respect of the material, the additional cost of formwork does not offset the cost of the saved material. Therefore, stepped footings are more economical than the sloped ones. The adjoining soil below footings generates upward pressure which bends the slab due to cantilever action. Hence, adequate tensile reinforcement should be provided at the bottom of the slab (tension face). Clause 34.1.1 of IS 456 stipulates that the sloped or stepped footings, designed as a unit, should be constructed to ensure the integrated action. Moreover, the effective cross-section in compression of sloped and stepped footings shall be limited by the area above the neutral plane. Though symmetrical footings are desirable, sometimes situation compels for unsymmetrical isolated footings (Eccentric footings or footings with cut outs) either about one or both the axes (Figs.11.28.5 and 6).


3. Combined footings


When the spacing of the adjacent columns is so close that separate isolated footings are not possible due to the overlapping areas of the footings or inadequate clear space between the two areas of the footings, combined footings are the solution combining two or more columns. Combined footing normally means a footing combining two columns. Such footings are either rectangular or trapezoidal in plan forms with or without a beam joining the two columns, as shown in Figs.11.28.7 and 11.28.8.


4. Strap footings

When two isolated footings are combined by a beam with a view to sharing the loads of both the columns by the footings, the footing is known as strap footing (Fig.11.28.9). The connecting beam is designated as strap beam. These footings are required if the loads are heavy on columns and the areas of foundation are not overlapping with each other.


5. Strip foundation or wall footings

These are in long strips especially for load bearing masonry walls or reinforced concrete walls (Figs.11.28.10). However, for load bearing masonry walls, it is common to have stepped masonry foundations. The strip footings distribute the loads from the wall to a wider area and usually bend in transverse direction. Accordingly, they are reinforced in the transverse direction mainly, while nominal distribution steel is provided along the longitudinal direction.


6. Raft or mat foundation

These are special cases of combined footing where all the columns of the building are having a common foundation (Fig.11.28.11). Normally, for buildings with heavy loads or when the soil condition is poor, raft foundations are very much useful to control differential settlement and transfer the loads not exceeding the bearing capacity of the soil due to integral action of the raft foundation. This is a threshold situation for shallow footing beyond which deep foundations have to be adopted.


(B) Deep foundations


As mentioned earlier, the shallow foundations need more plan areas due to the low strength of soil compared to that of masonry or reinforced concrete. However, shallow foundations are selected when the soil has moderately good strength, except the raft foundation which is good in poor condition of soil also. Raft foundations are under the category of shallow foundation as they have comparatively shallow depth than that of deep foundation. It is worth mentioning that the depth of raft foundation is much larger than those of other types of shallow foundations. However, for poor condition of soil near to the surface, the bearing capacity is very less and foundation needed in such situation is the pile foundation (Figs.11.28.12). Piles are, in fact, small diameter columns which are driven or cast into the ground by suitable means. Precast piles are driven and cast-in-situ are cast. These piles support the structure by the skin friction between the pile surface and the surrounding soil and end bearing force, if such resistance is available to provide the bearing force. Accordingly, they are designated as frictional and end bearing piles. They are normally provided in a group with a pile cap at the top through which the loads of the superstructure are transferred to the piles. Piles are very useful in marshy land where other types of foundation are impossible to construct. The length of the pile which is driven into the ground depends on the availability of hard soil/rock or the actual load test. Another advantage of the pile foundations is that they can resist uplift also in the same manner as they take the compression forces just by the skin friction in the opposite direction. However, driving of pile is not an easy job and needs equipment and specially trained persons or agencies. Moreover, one has to select pile foundation in such a situation where the adjacent buildings are not likely to be damaged due to the driving of piles. The choice of driven or bored piles, in this regard, is critical. Exhaustive designs of all types of foundations mentioned above are beyond the scope of this course. Accordingly, this module is restricted to the design of some of the shallow footings, frequently used for normal low rise buildings only. 11.28.3 Safe Bearing Capacity of Soil The safe bearing capacity qc of soil is the permissible soil pressure considering safety factors in the range of 2 to 6 depending on the type of soil, approximations and assumptions and uncertainties. This is applicable under service load condition and, therefore, the partial safety factors fλ for different load combinations are to be taken from those under limit state of serviceability


(vide Table 18 of IS 456 or Table 2.1 of Lesson 3). Normally, the acceptable value of qc is supplied by the geotechnical consultant to the structural engineer after proper soil investigations. The safe bearing stress on soil is also related to corresponding permissible displacement / settlement. Gross and net bearing capacities are the two terms used in the design. Gross bearing capacity is the total safe bearing pressure just below the footing due to the load of the superstructure, self weight of the footing and the weight of earth lying over the footing. On the other hand, net bearing capacity is the net pressure in excess of the existing overburden pressure. Thus, we can write Net bearing capacity = Gross bearing capacity - Pressure due to overburden soil (11.1) While calculating the maximum soil pressure q, we should consider all the loads of superstructure along with the weight of foundation and the weight of the backfill. During preliminary calculations, however, the weight of the foundation and backfill may be taken as 10 to 15 per cent of the total axial load on the footing, subjected to verification afterwards. 11.28.4 Depth of Foundation All types of foundation should have a minimum depth of 50 cm as per IS 1080-1962. This minimum depth is required to ensure the availability of soil having the safe bearing capacity assumed in the design. Moreover, the foundation should be placed well below the level which will not be affected by seasonal change of weather to cause swelling and shrinking of the soil. Further, frost also may endanger the foundation if placed at a very shallow depth. Rankine formula gives a preliminary estimate of the minimum depth of foundation and is expressed as d = (qc/λ ){(1 - sinφ )/(1 + sinφ )}2 (11.2) where d = minimum depth of foundation qc = gross bearing capacity of soil λ = density of soil φ = angle of repose of soil Though Rankine formula considers three major soil properties qc, λ and φ , it does not consider the load applied to the foundation. However, this may be a guideline for an initial estimate of the minimum depth which shall be checked subsequently for other requirements of the design.


11.28.5 Design Considerations (a) Minimum nominal cover (cl. 26.4.2.2 of IS 456) The minimum nominal cover for the footings should be more than that of other structural elements of the superstructure as the footings are in direct contact with the soil. Clause 26.4.2.2 of IS 456 prescribes a minimum cover of 50 mm for footings. However, the actual cover may be even more depending on the presence of harmful chemicals or minerals, water table etc. (b) Thickness at the edge of footings (cls. 34.1.2 and 34.1.3 of IS 456) The minimum thickness at the edge of reinforced and plain concrete footings shall be at least 150 mm for footings on soils and at least 300 mm above the top of piles for footings on piles, as per the stipulation in cl.34.1.2 of IS 456. For plain concrete pedestals, the angle α (see Fig.11.28.1) between the plane passing through the bottom edge of the pedestal and the corresponding junction edge of the column with pedestal and the horizontal plane shall be determined from the following expression (cl.34.1.3 of IS 456) tanα 0.9{(100 q≤ a/fck) + 1}1/2 (11.3) where qa = calculated maximum bearing pressure at the base of pedestal in N/mm2, and fck = characteristic strength of concrete at 28 days in N/mm2. (c) Bending moments (cl. 34.2 of IS 456) 1. It may be necessary to compute the bending moment at several sections of the footing depending on the type of footing, nature of loads and the distribution of pressure at the base of the footing. However, bending moment at any section shall be determined taking all forces acting over the entire area on one side of the section of the footing, which is obtained by passing a vertical plane at that section extending across the footing (cl.34.2.3.1 of IS 456). 2. The critical section of maximum bending moment for the purpose of designing an isolated concrete footing which supports a column, pedestal or wall shall be:

(i) at the face of the column, pedestal or wall for footing supporting a concrete column, pedestal or reinforced concrete wall, (Figs.11.28.2, 3 and 10), and


(ii) halfway between the centre-line and the edge of the wall, for footing under masonry wall (Fig.11.28.10). This is stipulated in cl.34.2.3.2 of IS 456.

The maximum moment at the critical section shall be determined as mentioned in 1 above.

For round or octagonal concrete column or pedestal, the face of the column or pedestal shall be taken as the side of a square inscribed within the perimeter of the round or octagonal column or pedestal (see cl.34.2.2 of IS 456 and Figs.11.28.13a and b). (d) Shear force (cl. 31.6 and 34.2.4 of IS 456) Footing slabs shall be checked in one-way or two-way shears depending on the nature of bending. If the slab bends primarily in one-way, the footing slab shall be checked in one-way vertical shear. On the other hand, when the bending is primarily two-way, the footing slab shall be checked in two-way shear or punching shear. The respective critical sections and design shear strengths are given below: 1. One-way shear (cl. 34.2.4 of IS 456) One-way shear has to be checked across the full width of the base slab on a vertical section located from the face of the column, pedestal or wall at a distance equal to (Figs.11.28.2, 3 and 10): (i) effective depth of the footing slab in case of footing slab on soil, and


(ii) half the effective depth of the footing slab if the footing slab is on piles (Fig.11.28.12).

The design shear strength of concrete without shear reinforcement is given in Table 19 of cl.40.2 of IS 456. 2. Two-way or punching shear (cls.31.6 and 34.2.4) Two-way or punching shear shall be checked around the column on a perimeter half the effective depth of the footing slab away from the face of the column or pedestal (Figs.11.28.2 and 3). The permissible shear stress, when shear reinforcement is not provided, shall not exceed ks cτ , where ks = (0.5 + cβ ), but not greater than one, cβ being the ratio of short side to long side of the column, and cτ = 0.25(fck)1/2 in limit state method of design, as stipulated in cl.31.6.3 of IS 456. Normally, the thickness of the base slab is governed by shear. Hence, the necessary thickness of the slab has to be provided to avoid shear reinforcement. (e) Bond (cl.34.2.4.3 of IS 456) The critical section for checking the development length in a footing slab shall be the same planes as those of bending moments in part (c) of this section. Moreover, development length shall be checked at all other sections where they change abruptly. The critical sections for checking the development length are given in cl.34.2.4.3 of IS 456, which further recommends to check the anchorage requirements if the reinforcement is curtailed, which shall be done in accordance with cl.26.2.3 of IS 456. (f) Tensile reinforcement (cl.34.3 of IS 456) The distribution of the total tensile reinforcement, calculated in accordance with the moment at critical sections, as specified in part (c) of this section, shall be done as given below for one-way and two-way footing slabs separately. (i) In one-way reinforced footing slabs like wall footings, the reinforcement shall be distributed uniformly across the full width of the footing i.e., perpendicular to the direction of wall. Nominal distribution reinforcement shall be provided as per cl. 34.5 of IS 456 along the length of the wall to take care of the secondary moment, differential settlement, shrinkage and temperature effects. (ii) In two-way reinforced square footing slabs, the reinforcement extending in each direction shall be distributed uniformly across the full width/length of the footing.


(iii) In two-way reinforced rectangular footing slabs, the reinforcement in the long direction shall be distributed uniformly across the full width of the footing slab. In the short direction, a central band equal to the width of the footing shall be marked along the length of the footing, where the portion of the reinforcement shall be determined as given in the equation below. This portion of the reinforcement shall be distributed across the central band:

Reinforcement in the central band = {2/( β +1)} (Total reinforcement in the short direction) (11.4) where β is the ratio of longer dimension to shorter dimension of the footing slab (Fig.11.28.14). Each of the two end bands shall be provided with half of the remaining reinforcement, distributed uniformly across the respective end band.


(g) Transfer of load at the base of column (cl.34.4 of IS 456)

All forces and moments acting at the base of the column must be transferred to the pedestal, if any, and then from the base of the pedestal to the footing, (or directly from the base of the column to the footing if there is no


pedestal) by compression in concrete and steel and tension in steel. Compression forces are transferred through direct bearing while tension forces are transferred through developed reinforcement. The permissible bearing stresses on full area of concrete shall be taken as given below from cl.34.4 of IS 456: brσ = 0.25fck, in working stress method, and (11.5) brσ = 0.45fck, in limit state method (11.6) It has been mentioned in sec. 10.26.5 of Lesson 26 that the stress of concrete is taken as 0.45fck while designing the column. Since the area of footing is much larger, this bearing stress of concrete in column may be increased considering the dispersion of the concentrated load of column to footing. Accordingly, the permissible bearing stress of concrete in footing is given by (cl.34.4 of IS 456): brσ = 0.45fck (A1/A2)1/2 (11.7) with a condition that (A1/A2)1/2 2.0 (11.8)

≤

where A1 = maximum supporting area of footing for bearing which is

geometrically similar to and concentric with the loaded area A2, as shown in Fig.11.28.15

A2 = loaded area at the base of the column. The above clause further stipulates that in sloped or stepped footings, A1 may be taken as the area of the lower base of the largest frustum of a pyramid or cone contained wholly within the footing and having for its upper base, the area actually loaded and having side slope of one vertical to two horizontal, as shown in Fig.11.28.15. If the permissible bearing stress on concrete in column or in footing is exceeded, reinforcement shall be provided for developing the excess force (cl.34.4.1 of IS 456), either by extending the longitudinal bars of columns into the footing (cl.34.4.2 of IS 456) or by providing dowels as stipulated in cl.34.4.3 of IS 456 and given below: (i) Sufficient development length of the reinforcement shall be provided to transfer the compression or tension to the supporting member in accordance with


cl.26.2 of IS 456, when transfer of force is accomplished by reinforcement of column (cl.34.4.2 of IS 456). (ii) Minimum area of extended longitudinal bars or dowels shall be 0.5 per cent of the cross-sectional area of the supported column or pedestal (cl.34.4.3 of IS 456). (iii) A minimum of four bars shall be provided (cl.34.4.3 of IS 456). (iv) The diameter of dowels shall not exceed the diameter of column bars by more than 3 mm.

(v) Column bars of diameter larger than 36 mm, in compression only can be doweled at the footings with bars of smaller size of the necessary area. The dowel shall extend into the column, a distance equal to the development length of the column bar and into the footing, a distance equal to the development length of the dowel, as stipulated in cl.34.4.4 of IS 456 and as shown in Fig.11.28.16. (h) Nominal reinforcement (cl. 34.5 of IS 456) 1. Clause 34.5.1 of IS 456 stipulates the minimum reinforcement and spacing of the bars in footing slabs as per the requirements of solid slab (cls.26.5.2.1 and 26.3.3b(2) of IS 456, respectively).


2. The nominal reinforcement for concrete sections of thickness greater than 1 m shall be 360 mm2 per metre length in each direction on each face, as stipulated in cl.34.5.2 of IS 456. The clause further specifies that this provision does not supersede the requirement of minimum tensile reinforcement based on the depth of section. 11.28.6 Distribution of Base Pressure


The foundation, assumed to act as a rigid body, is in equilibrium under the action of applied forces and moments from the superstructure and the reactions from the stresses in the soil. The distribution of base pressure is different for different types of soil. Typical distributions of pressure, for actual foundations, in sandy and clayey soils are shown in Figs.11.28.17 and 18, respectively. However, for the sake of simplicity the footing is assumed to be a perfectly rigid body, the soil is assumed to behave elastically and the distributions of stress and stain are linear in the soil just below the base of the foundation, as shown in Fig.11.28.19. Accordingly, the foundation shall be designed for the applied loads, moments and induced reactions keeping in mind that the safe bearing capacity of the soil is within the prescribed limit. It is worth mentioning that the soil bearing capacity is in the serviceable limit state and the foundation structure shall be designed as per the limit state of collapse, checking for other limit states as well to ensure an adequate degree of safety and serviceability. In the following, the distributions of base pressure are explained for (i) concentrically loaded footings, (ii) eccentrically loaded footings and (iii) unsymmetrical (about both the axes) footings.


(i) Concentrically loaded footings

Figure 11.28.20 shows rectangular footing symmetrically loaded with service load P1 from the superstructure and P2 from the backfill including the weight of the footing. The assumed uniformly distributed soil pressure at the base of magnitude q is obtained from: q = (P1 + P2)/A (11.9) where A is the area of the base of the footing. In the design problem, however, A is to be determined from the condition that the actual gross intensity of soil pressure does not exceed qc, the bearing capacity of the soil, a known given data. Thus, we can write from Eq.11.9: A = (P1 + P2)/qc (11.10)


From the known value of A, the dimensions B and L are determined such that the maximum bending moment in each of the two adjacent projections is equal, i.e., the ratio of the dimensions B and L of the footing shall be in the same order of the ratio of width b and depth D of the column. (ii) Eccentrically loaded footings

In most of the practical situations, a column transfers axial load P and moment M to the footing, which can be represented as eccentrically loaded footing when a load P is subjected to an eccentricity e = M/P. This eccentricity may also be there, either alone or in combined mode, when • the column transfers a vertical load at a distance of e from the centroidal axis

of the footing, and


• the column or the pedestal transfers a lateral load above the level of foundation, in addition to vertical loads.

Accordingly, the distribution of pressure may be of any one of the three types, depending on the magnitude of the eccentricity of the load, as shown in Figs.11.28.21b to d. The general expression of qa, the intensity of soil pressure at a distance of y from the origin is: qa = P/A (Pe/I± x)y (11.11) We would consider a rectangular footing symmetric to the column. Substituting the values of A = BL, Ix = BL3/12 and y = L/2, we get the values of qa at the left edge. qa at the left edge = (P/BL) {1 - (6e/L)} (11.12) It is evident from Eq.11.12, that the three cases are possible: (A) when e < L/6, qa at the left edge is compression (+), (B) when e = L/6, qa at the left edge is zero, and (C) when e > L/6, qa at the left edge is tension (-). The three cases are shown in Figs.11.28.21b to d, respectively. It is to be noted that similar three cases are also possible when eccentricity of the load is negative resulting the values of qa at the right edge as compression, zero or tension. Evidently, these soil reactions, in compression and tension, should be permissible and attainable. Case (A): when | e | ≤ L/6 Figures 11.28.21b and c show these two cases, when |e| < L/6 or |e| = L/6, respectively. It is seen that the entire area of the footing is in compression having minimum and maximum values of q at the two edges with a linear and non-uniform variation. The values of q are obtained from Eq.11.11. In the limiting case i.e., when |e| = L/6, the value of qa is zero at one edge and the other edge is having qa = 2P/BL (compression) with a linear variation. Similarly, when e = 0, the footing is subjected to uniform constant pressure of P/BL. Thus, when |e| = L/6, the maximum pressure under one edge of the footing is twice of the uniform pressure when e = 0.


In a more general case, as in the case of footing for the corner column of a building, the load may have biaxial eccentricities. The general expression of qa at a location of (x,y) of the footing, when the load is having biaxial eccentricities of ex and ey is,

qa = P/A ± P exy/Ix ± P eyx/Iy (11.13)

Similarly, it can be shown that the rectangular footing of width B and length L will have no tension when the two eccentricities are such that 6ex/L + 6ey/B ≤ 1 (11.14) Case (B): when | e | > L/6


The eccentricity of the load more than L/6 results in development of tensile stresses in part of the soil. Stability, in such case, is ensured by either anchoring or weight of overburden preventing uplift. However, it is to ensure that maximum compressive pressure on the other face is within the limit and sufficient factor of safety is available against over turning. Accordingly, the maximum pressure in such a case can be determined considering the soil under compression part only. Further, assuming the line of action of the eccentric load coincides with that of resultant soil pressure (Fig.11.28.22) we have: qmax = P/L'B + 12P(0.5 C)(1.5 C)/BL' = 2P/L'B (11.15) where L' = 3C (11.16) (iii) Unsymmetrical footings It may be necessary to provide some cutouts in the foundation to reduce the uplift pressure or otherwise. The footing in such cases becomes unsymmetrical about both the axes. It is possible to determine the soil pressure distribution using the structural mechanics principle as given below. qa(x,y) = P/A {(M± yIx - MxIxy)(x)/(IxIy - )} + {(M2

xyI xIy - MyIxy)(y)/(IxIy - )} (11.17)

2xyI

where Mx = moment about x axis, My = moment about y axis, Ix = moment of inertia about x axis, Iy = moment of inertia about y axis, Ixy = product of inertia 11.28.7 Practice Questions and Problems with Answers Q.1: (A) What are the two essential requirements of the design of foundation? (B) Mention five points indicating the differences between the design of

foundation and the design of other elements of superstructure. A.1: See sec. 11.28.1. Q.2: Draw sketches of different shallow foundations.


A.2: Figure Nos. 11.28.1 to 11. Q.3: Explain the difference between gross and net safe bearing capacities of soil.

Which one is used for the design of foundation? A.3: See sec. 11.28.3. Q.4: How would you determine the minimum depth of foundation? A.4: See sec.11.28.4. Q.5: What are the critical sections of determining the bending moment in

isolated footing? A.5: See part (c)2 of sec.11.28.5. Q.6: Explain the one-way and two-way shears of foundation slabs. A.6: See part (d) of sec.11.28.5. Q.7: Draw the actual distributions of base pressures of soil below the footing in

sandy and clayey soils. Draw the assumed distribution of base pressure below the footing.

A.7: Figure Nos. 11.28.17 and 18. Q.8: Draw the distributions of pressure in a footing for concentric and

eccentric loadings (e ≤ L/6 and e > L/6). A.8: Figure Nos. 11.28.20 and 21. Q.9: How would you determine the pressure at any point (x,y) of a foundation

which is unsymmetrical? A.9: See part (iii) of sec.11.28.6. 11.28.8 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 11.28.9 Test 28 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: (A) What are the two essential requirements of the design of foundation?

(5 marks)

(B) Mention five points indicating the differences between the design of

foundation and the design of other elements of superstructure. (5 marks)

A.TQ.1: See sec. 11.28.1. TQ.2: How would you determine the minimum depth of foundation? (10 marks)


A.TQ.2: See sec.11.28.4. TQ.3: What are the critical sections of determining the bending moment in

isolated footing? (10 marks)

A.TQ.3: See part (c)2 of sec.11.28.5. TQ.4: Explain the one-way and two-way shears of foundation slabs. (10 marks) A.TQ.4: See part (d) of sec.11.28.5. TQ.5: Draw the distributions of pressure in a footing for concentric and

eccentric loadings (e ≤ L/6 and e > L/6). (10 marks)

A.TQ.5: Figure Nos. 11.28.20 and 21. 10.26.11 Summary of this Lesson

This lesson explains the two major and other requirements of the design of foundation structures. Various types of shallow foundations and pile foundation are discussed explaining the distribution of pressure in isolated footings loaded concentrically and eccentrically with e ≤ L/6 and e > L/6. The gross and net safe bearing capacities are explained. The equation for determining the minimum depth of the foundation is given. Various design considerations in respect of minimum nominal cover, thickness at the edge of footing, bending moment, shear force, bond, tensile reinforcement, transfer of load at the base of the column, and minimum distribution reinforcement are discussed, mentioning the codal requirements. The actual and the assumed distributions of base pressure are discussed. The distributions of base pressure for concentric and eccentric loads with eccentricity ≤ L/6 and > L/6 are explained. Determination of bearing pressure of soil for unsymmetrical footing is also discussed. All the discussions are relevant in understanding the load carrying mechanism of the foundation and the behaviour of soil. These understandings are essential in designing the foundation structures which is taken up in the next lesson.


Module 11

Foundations - Theory and Design


Lesson 29

Design of Foundations



• understand and apply the design considerations to satisfy the major and other requirements of the design of foundations,

• design the plain concrete footings, isolated footings for square and

rectangular columns subjected to axial loads with or without the moments, wall footings and combined footings, as per the stipulations of IS code.


The two major and some other requirements of foundation structures are explained in Lesson 28. Different types of shallow and deep foundations are illustrated in that lesson. The design considerations and different codal provisions of foundation structures are also explained. However, designs of all types of foundations are beyond the scope of this course. Only shallow footings are taken up for the design in this lesson. Several numerical problems are illustrated applying the theoretical considerations discussed in Lesson 28. Problems are solved explaining the different steps of the design. 11.29.2 Numerical Problems Problem 1: Design a plain concrete footing for a column of 400 mm x 400 mm carrying an axial load of 400 kN under service loads. Assume safe bearing capacity of soil as 300 kN/m2 at a depth of 1 m below the ground level. Use M 20 and Fe 415 for the design. Solution 1: Plain concrete footing is given in secs.11.28.2(A)1 and 11.28.5(b). Step 1: Transfer of axial force at the base of column It is essential that the total factored loads must be transferred at the base of column without any reinforcement. For that the bearing resistance should be greater than the total factored load Pu. Here, the factored load Pu = 400(1.5) = 600 kN.


The bearing stress, as per cl.34.4 of IS 456 and given in Eqs.11.7 and 8 of sec.11.28.5(g) of Lesson 28, is brσ = 0.45 fck (A1/A2)1/2 (11.7) with a condition that

(A1/A2)1/2 ≤ 2.0 (11.8) Since the bearing stress brσ at the column-footing interface will be governed by the column face, we have A1 = A2 = 400(400) = 160000 mm2. Using A1 = A2, in Eq.11.7, we have Pbr = Bearing force = 0.45 fck A1 = 0.45(20)(160000)(10-3) = 1440 kN > Pu (= 600 kN). Thus, the full transfer of load Pu is possible without any reinforcement. Step 2: Size of the footing


Let us assume the weight of footing and back fill soil as 15 per cent of Pu. Then, the base area required = 400(1.15)/300 = 1.533 m2. Provide 1250 x 1250 mm (= 1.5625 m2) as shown in Fig.11.29.1. The bearing pressure qa = 400(1.15)/(1.25)(1.25) = 294.4 kN/m2. Step 3: Thickness of footing The thickness of the footing h is governed by Eq.11.3 of sec.11.28.5 of Lesson 28. From Eq.11.3, we have tan ≤α 0.9{(100qa/fck) + 1}1/2 …. (11.3) ≤ 0.9[{100(0.2944)/20} + 1]1/2

≤ 1.415 We have from Fig.11.29.1a: h = {(1250 - 400)/2}(tanα ) = 601.375 mm Provide 1250 x 1250 x 670 mm block of plain concrete. Step 4: Minimum reinforcement The plain concrete block 1250 x 1250 x 670 shall be provided with the minimum reinforcement 0.12 per cent for temperature, shrinkage and tie action. Minimum Ast = 0.0012(1250)(670) = 1005.0 mm2. Provide 9 bars of 12 mm diameter (= 1018 mm2) both ways as shown in Fig.11.29.1b. The spacing of bars = (1250 - 50 - 12)/8 = 148.5 mm c/c. Provide the bars @ 140 mm c/c. Step 5: Check for the gross base pressure Assuming unit weights of concrete and soil as 24 kN/m3 and 20 kN/m3

Service load = 400.00 kN Weight of footing = (0.67)(1.25)(1.25)(24) = 25.125 kN Weight of soil = (0.33)(1.25)(1.25)(20) = 10.3125 kN Total = 435.4375 kN qa = 435.4375/(1.25)(1.25) = 278.68 kN/m2 < 300 kN/m2


Hence, o.k. Problem 2: Design an isolated footing for a square column, 400 mm x 400 mm with 12-20 mm diameter longitudinal bars carrying service loads of 1500 kN with M 20 and Fe 415. The safe bearing capacity of soil is 250 kN/m2 at a depth of 1 m below the ground level. Use M 20 and Fe 415. Solution 2: Step 1: Size of the footing


Given P = 1500 kN, qc = 250 kN/m2 at a depth of 1 m below the ground level. Assuming the weight of the footing and backfill as 10 per cent of the load, the base area required = 1500(1.1)/250 = 6.6 m2. Provide 2.6 m x 2.6 m, area = 6.76 m2 (Fig.11.29.2b). Step 2: Thickness of footing slab based on one-way shear Factored soil pressure = 1500(1.5)/(2.6)(2.6) = 0.3328 N/mm2, say, 0.333 N/mm2. Assuming p = 0.25% in the footing slab, for M 20 concrete cτ = 0.36 N/mm2 (Table 19 of IS 456). Vu = 0.36(2600)d and Vu (actual) = 0.333(2600)(1100 - d). From the condition that Vu should be more than or equal to the actual Vu, we have (see Fig.11.29.2a, sec.11): 0.36(2600)d ≥ 0.333(2600) (1100 – d) So, d 528.57 mm. ≥ Provide d = 536 mm. The total depth becomes 536 + 50 + 16 + 8 (with 50 mm cover and diameter of reinforcing bars = 16 mm) = 610 mm. Step 3: Checking for two-way shear The critical section is at a distance of d/2 from the periphery of the column. The factored shear force (Fig.11.29.2b, sec. 2222) = 0.333{(2600)2 – (400 + d)2}(10)-3 = 1959.34 kN. Shear resistance is calculated with the shear strength = ks cτ = ks(0.25) (fck)1/2; where ks = 0.5 + cβ (cl. 31.6.3 of IS 456). Here cβ = 1.0, ks = 1.5 >/ 1; so ks = 1.0. This gives shear strength of concrete = 0.25 (fck)1/2 = 1.118 N/mm2. So, the shear resistance = (1.118) 4 (936)(536) = 2243.58 kN > 1959.34 kN. Hence, ok. Thus, the depth of the footing is governed by one-way shear. Step 4: Gross bearing capacity Assuming unit weights of concrete and soil as 24 kN/m3 and 20 kN/m3, respectively: Given, the service load = 1500 kN Weight of the footing = 2.6(2.6)(0.61)(24) = 98.967 kN Weight of soil = 2.6(2.6)(1.0-0.61)(20) = 52.728 kN (Assuming the depth of the footing as 1.0 m). Total = 1635.2 kN


Gross bearing pressure = 1635.2/(2.6)(2.6) = 241.893 kN/m2 < 250 kN/m2. Hence, ok. Step 5: Bending moment The critical section (Fig.11.29.2b, sec.3.3), is at the face of the column. Mu = 0.333(2600)(1100)(550) Nmm = 523.809 kNm Moment of resistance of the footing = Rbd2 where R = 2.76 (Table 3.3 of Lesson 5). Moment of resistance = 2.76(2600)(536)(536) = 2061.636 kNm > 523.809 kNm. Area of steel shall be determined from Eq.3.23 of Lesson 5, which is: Mu = 0.87 fy Ast d {1 – (Ast fy/fck bd)} …. (3.23) Substituting Mu = 523.809 kNm, fy = 415 N/mm2, fck = 20 N/mm2, d = 536 mm, b = 2600 mm, we have: - 67161.44578 A2

stA st + 181.7861758 (105) = 6 Solving, we get Ast = 2825.5805 mm2. Alternatively, we can use Table 2 of SP-16 to get the Ast as explained below: Mu/bd2 = 523.809(106)/(2600)(536)(536) = 0.7013 N/mm2. Table 2 of SP-16 gives p = 0.2034. Ast = 0.2034(2600)(536)/100 = 2834.58 mm2. This area is close to the other value = 2825.5805 mm2. However, one-way shear has been checked assuming p = 0.25%. So, use p = 0.25%. Accordingly, Ast = 0.0025(2600)(536) = 3484 mm2. Provide 18 bars of 16 mm diameter (= 3619 mm2) both ways. The spacing of bars = {2600 – 2(50) – 16}/17 = 146.117 mm. The spacing is 140 mm c/c (Fig.11.29.2). The bending moment in the other direction is also the same as it is a square footing. The effective depth, however, is 16 mm more than 536 mm. But, the area of steel is not needed to be determined with d = 552 mm as we are providing 0.25 per cent reinforcement based on one-way shear checking.


Step 6: Development length Ld = fs φ /4( bdτ ) = 0.87(415)(16)/4(1.6)(1.2) = 47(16) = 752 mm (cl.26.2.1 of IS 456). Length available = 1100 – 50 = 1050 mm > 752 mm. Step 7: Transfer of force at the base of the column Pu = 1500(1.5) = 2250 kN Compressive bearing resistance = 0.45 fck(A1/A2)1/2. For the column face A1/A2 = 1 and for the other face A1/A2 > 2 but should be taken as 2. In any case, the column face governs. Force transferred to the base through column at the interface = 0.45(20)(400)(400) = 1440 kN < 2250 kN. The balance force 2250 – 1440 = 810 kN has to be transferred by the longitudinal reinforcements, dowels or mechanical connectors. As it is convenient, we propose to continue the longitudinal bars (12-20 mm diameter) into the footing. The required development length of 12-20 mm diameter bars, assuming a stress level of 0.87fy(810/2250) = 129.978 N/mm2, is 129.978(20)/4(1.6)(1.2)(1.25) = 270.8 mm. Here bdτ for M 20 = 1.2 N/mm2, increased factor of 1.6 is due to deformed bars and increased factor of 1.25 is for the compression. Length available = 610 – 50 – 16 – 16 – 16 = 512 mm > 270.8 mm (Fig.11.29.2a). Hence, o.k. The arrangement is shown in Fig.11.29.2c. Alternatively: Design of dowels


For the balance force 810 kN, the area of dowels = 810000/0.67(415) = 2913.15 mm2. Minimum area = 0.5(400)(400)/100 = 800 mm2 < 2913.15 mm2 (cl.34.4.3 of IS 456). Therefore, number of 16 mm dowels = 2913.15/201 = 15. The development length of 16 mm dowels in compression = 0.87(415)(16)/4(1.6)(1.2)(1.25) = 601.76 mm. Available vertical embedment length = 610 – 50 – 16 – 16 – 16 = 512 mm. So, the dowels will be extended by another 100 mm horizontally, as shown in Fig.11.29.2c. Problem 3: Design a sloped footing for a square column of 400 mm x 400 mm with 16 longitudinal bars of 16 mm diameter carrying a service load of 1400 kN. Use M 20 and Fe 415 both for column and footing slab. The safe bearing capacity of soil is 150 kN/m2. Solution 3:


Step 1: Size of the footing

Given P = 1400 kN and qc = 150 kN/m2. Assuming the weight of the footing and the back file as 10 per cent of the load, the required base area is: 1400(1.1)/150 = 10.27 m2. Provide 3400 x 3400 mm giving 11.56 m2 (Fig.11.29.3b). Step 2: Thickness of footing slab based on one-way shear


Factored bearing pressure = 1400(1.5)/(3.4)(3.4) = 181.66 kN/m2 = 0.18166 N/mm2. Assuming 0.15 per cent reinforcement in the footing slab, Table 19 of IS 456 gives cτ for M 20 = 0.28 N/mm2. From the condition that the one-way shear resistance one-way shear force, we have at a distance d from the face of the column (sec.1-1of Figs.11.29.3a and b).

≥

0.28(3400)d ≥ 0.18166(1500 – d)(3400) or d ≥ 590.24 mm. Provide total depth of footing as 670 mm, so that the effective depth = 670 – 50 – 16 – 8 = 596 mm. (The total depth is, however, increased to 750 mm in Step 7.) Step 3: Checking for two-way shear (See sec.11.28.5.d-2) At the critical section 2222 (Figs.11.29.3a and b), the shear resistance = 4(400 + 596)(596)(0.25)(fck)1/2 = 2654.73 kN. The shear force = {(3.4)(3.4) – (0.996)(0.996)}0.18166 = 1919.78 kN < 2654.73 kN. Hence, o.k. Step 4: Gross bearing capacity Assuming unit weights of concrete and soil as 25 kN/m3 and 18 kN/m3, respectively, we have: Load on footing = 1400.00 kN Weight of footing = (3.4)(3.4)(0.67)(25) = 193.63 kN Weight of soil = (3.4)(3.4)(1.25 - 0.67)(18) = 120.69 kN (Assuming the depth of the footing as 1.25 m). Total = 1714.32 kN Gross bearing capacity = 1714.32 / (3.4)(3.4) = 148.30 kN/m2 < 150 kN/m2. Hence o.k. Step 5: Bending moment We have to determine the area of steel in one direction as it is a square footing. So, we consider the lower effective depth which is 596 mm. The critical section is sec.33 (Figs.11.29.3a and b), where we have Mu = 3400(1500)(0.18166)(1500)/2 = 694.8495 kNm


Mu/bd2 = 694.8495(106)/(3400)(596)(596) = 0.575 N/mm2

Table 2 of SP-16 gives, p = 0.165%. Accordingly, area of steel = 0.165(3400)(596)/100 = 3343.56 mm2. Provide 30 bars of 12 mm diameter (= 3393 mm2), both ways. Step 6: Development length Development length of 12 mm diameter bars = 0.87(415)(12)/4(1.6)(1.2) = 564.14 mm. Hence, o.k. Step 7: Providing slope in the footing slab Since the three critical sections (.i.e., of bending moment, two-way shear and one-way shear) are within a distance of 596 mm from the face of the column, the full depth of the footing slab is provided up to a distance of 700 mm from the face of the column. However, by providing slope the available section now is a truncated rectangle giving some less area for the one-way shear. Accordingly, the depth of the footing is increased from 670 mm to 750 mm. With a cover of 50 mm and bar diameter of 12 mm in both directions, the revised effective depth = 750 – 50 – 12 – 6 = 682 mm. Providing the minimum depth of 350 mm at the edge, as shown in Figs.11.29.3a and b, we check the one-way shear again, taking into account of the truncated rectangular cross-section at a distance of 682 mm from the face of the column.

One-way shear force = 0.18166(1500 – 682)(3400) = 505232.792 N

Area of truncated rectangle = 1800(682) + 1600(282) + 1600(682 – 282)/2 = 1998800 mm2

The shear stress = 505232.792/1998800 = 0.2527 N/mm2 < 0.28 N/mm2. Hence, o.k. Step 8: Revised area of steel The bending moment in step 5 is 694.8495 kNm at the face of the column. With d = 682 mm now, we have Mu/bd2 = 694.8495(106)/(3400)(682)(682) = 0.4394 N/mm2

Table 2 of SP-16 gives, p is less than 0.15 per cent. Provide p = 0.15 per cent due to the one-way shear. So, Ast = 0.15(3400)(682)/100 = 3478.2 mm2. Provide 31 bars of 12 mm (Ast = 3506 mm2), both ways. Effectively, the number of bars has increased from 30 to 31 now.


Step 9: Transfer of force at the base of the column Pu = 1400(1.5) = 2100 kN. Compressive bearing resistance = 0.45 fck(A1/A2)1/2 = 0.45(20)(1) = 9 N/mm2. Force transferred at the base through the column = 9(400)(400)(10-3) = 1440 kN < 2100 kN. Provide dowels for the excess (2100 – 1440) = 660 kN. The area of dowels = 660(103)/(0.67)(415) = 2373.67 mm2. Minimum area of dowels = 0.5(400)(400)/100 = 800 mm2. Provide 12 dowels of 16 mm diameter (area = 2412 mm2). The development length of 16 mm dowels = 0.87(415)(16)/4(1.6)(1.2)(1.25) = 601.76 mm. The vertical length available 750 – 50 – 12 – 12 – 16 = 660 mm > 601.76 mm. Hence, o.k. The arrangement of reinforcement and dowels is shown in Fig.11.29.3a and b. Problem 4: Design one isolated footing for a column 300 mm x 450 mm, having 20 bars of 20 mm diameter (Ast = 4021 mm2) of Problem 1 of sec.10.25.6 of Lesson 25 carrying Pu = 1620 kN and Mu = 170 kNm using M 25 and Fe 415. Assume that the moment is reversible. The safe bearing capacity of the soil is 200 kN/m2 at a depth of 1 metre from ground level. Use M 25 and Fe 415 for the footing. Solution 4: Step 1: Size of the footing


Given Pu = 1620 kN and Mu = 170 kNm. The footing should be symmetric with respect to the column as the moment is reversible. Assuming the weights of footing and backfill as 15 per cent of Pu, the eccentricity of load Pu at the base is e = Mu/P(1.15) = 170(106)/1620(1.15)(103) = 91.25 mm. This eccentricity may be taken as < L/6 of the footing.


The factored bearing pressure is 200(1.5) = 300 kN/m2. For the footing of length L and width B, we, therefore, have: Pu/BL + 6M/BL2 300 ≤ or, 1620(1.15)/BL + 6(170)/BL2 ≤ 300 or, BL2 – 6.21L – 3.4 ≤ 0 (1) For the economic proportion, let us keep equal projection beyond the face of the column in the two directions. This gives (L – 0.45)/2 = (B – 0.3)/2 or, B = L – 0.15 (2) Using Eq.(2) in Eq.(1), we have (L – 0.15) L2 – 6.212 – 3.4 ≤ 0 or L3 – 0.15 L2 – 6.21 L – 3.4 ≤ 0 We have L = 2.8 m and B = 2.65 m. Let us provide L = 2.85 m and B = 2.70 m (Fig.11.29.4b). We get the maximum and minimum pressures as 1620(1.15)/(2.85)(2.70) 170(6)/(2.7)(2.85)(2.85) = 242.105 ± ± 46.51 = 288.615 kN/m2 and 195.595 kN/m2, respectively (Fig.11.29.4c). Both the values are less than 300 kN/m2. Hence, o.k. Step 2: Thickness of footing slab based on one-way shear The critical section (sec.11 of Figs.11.29.4a and b) is at a distance d from the face of the column. The average soil pressure at sec.11 is {288.615 – (288.615 – 195.595)(1200 – d)/2850} = 249.449 + 0.0326d. The one-way shear force at sec.11 = (2.7)(1.2 – 0.001d((249.449 + 0.0326d) kN. Assuming 0.15 per cent reinforcement in the footing slab, the shear strength of M 25 concrete = 0.29 N/mm2. Hence, the shear strength of the section = 2700(d)(0.29)(103) kN. From the condition that shear strength has to be

shear force, we have ≥ 2700(d)(0.29)(10-3) = (2.7)(1.2 – 0.001d)(249.449 + 0.0326d) This gives,


d2 + 15347.51534d – 9182171.779 = 0 Solving, we get d = 576.6198. Let us assume d = 600 mm Step 3: Checking for two-way shear At the critical section 2222 (Figs.11.29.4a and b), the shear resistance is obtained cl.31.6.31 of IS 456, which gives cτ = (0.5 + 450/300)(0.25)(25)1/2 but the multiplying factor (0.5 + 450/300) >/ 1.0. So, we have cτ = 0.25(25)1/2 = 1.25 N/mm2. Hence, the shear resistance = (1.25)(2){(300 + 600) + (450 + 600)}(600) = 2925 kN. Actual shear force is determined on the basis of average soil pressure at the centre line of the cross-section which is (195.595 + 288.615)/2 = 242.105 kN/m2 (Fig.11.29.4c). So, the actual shear force = Vu = (242.105){(2.7)(2.85) – (0.3 + 0.6)(0.45 + 0.6)} = 1634.209 kN < shear resistance (= 2925 kN). Hence, the depth of the footing is governed by one-way shear. With effective depth = 600 mm, the total depth of footing = 600 + 50 (cover) + 16 (bar dia) + 8 (half bar dia) = 674 mm. Step 4: Gross bearing capacity Assuming the unit weights of concrete and soil as 25 kN/m3 and 18 kN/m3, respectively, we have the bearing pressure for (i) Pu = 1620 kN, (ii) Mu = 170 kNm and (iii) self weight of footing and backfill soil. (i) Due to Pu = 1620 kN: pressure = 1620/(2.7)(2.85) = 210.53 kN/m2

(ii) Due to Mu = 170 kNm: pressure = ± 170(6)/(2.7)(2.85)(2.85) = 46.51 kN/m

±2

(iii) Self weight of footing of depth 674 mm and soil of (1000 – 674) = 326 mm: pressure = 0.674(25) + 0.326(18) = 22.718 kN/m2

Thus, the maximum and minimum pressures are = 210.53 + 22.718 46.51 = 279.758 kN/m

±2 and 186.738 kN/m2 < 300 kN/m2. Hence, o.k.

Step 5: Bending moment (i) In the long direction (along the length = 2850 mm) Bending moment at the face of column (sec.33 of Figs.11.29.4a and b) is determined where the soil pressure = 288.615 – (288.615 – 195.595)(1200)/2850 = 249.45 kN/m2. So, the bending moment = 249.45(2.7)(1.2)(0.6) + (288.615 – 249.45)(2.7)(1.2)(2)/(2)(3) = 527.23 kNm.


M/Bd2 = 527.23(106)/(2700)(616)(616) = 0.515 N/mm2 < 3.45 N/mm2 for M 25 concrete. Table 3 of SP-16 gives p = 0.1462 < 0.15 per cent as required for one-way shear. Thus, Ast = 0.15(2700)(616)/100 = 2494.8 mm2. Provide 13 bars of 16 mm diameter (area = 2613 mm2), spacing = (2700 – 100 – 16)/12 = 215.33 mm, say 210 mm c/c. (ii) In the short direction (B = 2700 mm) The average pressure on soil between the edge and centre of the footing = (288.615 + 242.105)/2 = 265.36 kN/m2. The bending moment is determined with this pressure as an approximation.

Bending moment = (265.36)(1.2)(0.6)(2.85) kNm = 544.519 kNm

M/Ld2 = 544.519(106)/(2850)(600)(600) = 0.531 Table 3 of SP-16 gives p = 0.15068, which gives area of steel = 0.15068(2850)(600)/100 = 2576.628 mm2. Provide 13 bars of 16 mm diameter (area = 2613 mm2) @ 210 mm c/c; i.e. the same arrangement in both directions. Step 6: Development length Development length of 16 mm diameter bars (M 25 concrete) = 0.87(415)(16)/4(1.6)(1.4) = 644.73 mm. Length available = 1200 – 50 – 8 = 1142 mm > 644.73 mm. Hence, o.k. Step 7: Transfer of force at the base of the column Since the column is having moment along with the axial force, some of the bars are in tension. The transfer of tensile force is not possible through the column-footing interface. So, the longitudinal bars of columns are to be extended to the footing. The required development length of 20 mm bars = 0.87(415)/4(1.4)(1.6) = 805.92 mm. Length available = 600 mm < 805.92 mm. The bars shall be given 90o bend and then shall be extended by 200 mm horizontally to give a total length of 600 + 8(20) (bend value) + 200 = 960 mm > 805.92 mm (Figs.11.29.4 a and b). The arrangement of reinforcement is shown in Figs.11.29.4a and b.


Problem 5:

Design a combined footing for two columns C1, 400 mm x 400 mm with 8 bars of 16 mm diameter carrying a service load of 800 kN and C2, 300 mm x 500 mm with 8 bars of 20 mm diameter carrying a service load of 1200 kN (Figs.11.29.5a and b). The column C1 is flushed with the property line. The columns are at 3.0 m c/c distance. The safe bearing capacity of soil is 200 kN/m2 at a depth of 1.5 m below the ground level. Use M 20 and Fe 415 for columns and footing.


Solution : Step 1: Size of the footing Assuming the weight of combined footing and backfill as 15 per cent of the total loads of the columns, we have the required base area, considering qc = 200 kN/m2, Area of the base = (800 + 1200)(1.15)/200 = 11.5 m2. It is necessary that the resultant of the loads of two columns and the centroid of the footing coincide so that a uniform distribution of soil pressure is obtained. Thus, the distance of the centroid of the footing y from column C1 (Fig.11.29.5b) is: y = 800(0) + 1200(3)/2000 = 1.8 m (Fig.11.29.5b). Since y is greater than half the c/c distance of column, a rectangular footing has to be designed. Let us provide 4 m x 3 m and the dimensions are shown in Fig.11.29.5b coinciding the centroid of the footing and the resultant line of action of the two loads, i.e. at a distance of 2 m from the left edge.


Step 2: Thickness of footing slab based on one-way shear Considering the footing as a wide beam of B = 3 m in the longitudinal direction, the uniformly distributed factored load = (800 + 1200)(1.5)/4 = 750 kN/m. Figures 11.29.6a, b and c present the column loads, soil pressure, shear force and bending moment diagrams. The critical section of one-way shear is sec.11 (at point K) of Figs.11.29.5a and 11.29.6a, at a distance of d + 250 mm from G (the location of column C2). The one-way shear force is Shear force = (1600 – d – 250)1200/1600 = (1012.5 – 0.75d) kN Assuming p = 0.15 per cent reinforcement in the footing slab, the shear strength of M 20 concrete = 0.28 N/mm2. Hence, the shear strength of section 11 = (3000)d(0.28)(10-3) kN. From the condition that shear strength shear force, we have

≥

(3000)d(0.28)(10-3) 1012.5 – 0.75d, which gives d 636.79 mm. Provide d = 650 mm and the total depth = 650 + 50 + 16 + 8 = 724 mm (assuming cover = 50 mm and the diameter of bars = 16 mm).

≥ ≥

Step 3: Checking for two-way shear (i) Around column C2 The effective depth along 4.0 m is 650 + 16 = 666 mm. The critical section for the two-way shear around column C2 is at a distance of 666/2 = 333 mm from the face of the column and marked by 2222 line in Fig.11.29.5b. The two-way punching shear force, considering the soil pressure = 750/3 = 250 kN/m2, is Vu = 1800 – (1.166)(0.966)(250) = 1518.411 kN As per cl.31.6.3.1 of IS 456, here ks = 0.5 + (500/300) but >/ 1.0; so, ks = 1.0. Therefore, shear strength of concrete = 0.25(20)1/2 (2){(300 + 666) + (500 + 666)}(666) = 3174.92 kN > 1518.411 kN. Hence, o.k. (ii) Around column C1 The effective depth of footing is 666 mm. The critical section is marked by 3333 in Fig.11.29.5b. The two-way punching shear = 1200 – (1.066)(0.733)(250) = 1004.65 kN. The resistance to two-way shear = 0.25(20)1/2 (2){(1066 + 733)}(666) = 2679.1 kN > 1004.65 kN. Hence, o.k. Thus, the depth of the footing is governed by one-way shear.


Step 4: Gross bearing capacity Assuming unit weights of concrete and soil as 25 kN/m3 and 18 kN/m3, respectively, the gross bearing capacity under service load is determined below. (i) Due to two loads: (800 + 1200)/(3)(4) = 166.67 kN/m2

(ii) Due to weight of the footing: With a total depth of the footing = 724 mm, the pressure = 0.724(25) = 18.1 kN/m2. (iii) Due to backfill of 1500 – 724 = 776 mm, the pressure = 0.776(18) = 13.968 kN/m2. The total pressure = 166.67 + 18.1 + 13.968 = 198.738 kN/m2 < 200 kN/m2. Hence, o.k. Step 5: Bending moments (longitudinal direction) (i) Maximum positive moment Figure 11.29.6c shows the maximum positive bending moment = 720 kNm at a distance of 1.4 m from the column C1 (at point J). With effective depth d = 666 mm, we have M/Bd2 = 720(106)/(3000)(666)(666) = 0.541 N/mm2

Table 2 of SP-16 gives p = 0.1553 per cent. Ast = 0.1553(3000)(666)/100 = 3102.894 mm2

Provide 16 bars of 16 mm diameter (area = 3217 mm2), spacing = (3000 – 50 – 16)/15 = 195.6 mm c/c, say 190 mm c/c. Development length of 16 mm bars = 47.01(16) = 752.16 mm Length available = 1600 – 50 – 16 = 1534 mm > 752.16 mm Hence, o.k. (ii) Maximum negative moment Figure 11.29.6c shows the maximum negative moment = 240 kNm at a distance of 800 mm from the right edge. With the effective depth = 666 mm, we have M/Bd2 = 240(106)/(3000)(666)(666) = 0.018 N/mm2


It is very nominal. So, provide 0.15 per cent steel, which gives Ast = 0.15(3000)(666)/100 = 2997 mm2. Provide 27 bars of 12 mm (area = 3053 mm2) at spacing = (3000 – 50 – 12)/26 = 113 mm c/c; say 110 mm c/c. Development length = 47.01(12) = 564 mm Length available = 800 – 50 – 12 = 738 mm > 564 mm Hence, o.k. Step 6: Design of column strip as transverse beam

Figure 11.29.7 shows the two column strips under columns C1 and C2.


(i) Transverse beam under column C1 The width of the transverse beam is 0.75d from the face of column C1. The effective depth is 666 – 6 – 8 = 652 mm, as the effective depth in the longitudinal direction = 666 mm, bottom bar diameter in longitudinal direction = 12 mm and assuming the bar diameter in the transverse direction as 16 mm. We have to check the depth and reinforcement in the transverse direction considering one-way shear and bending moment. (A) One-way shear The factored load for this transverse strip = 1200/3 = 400 kN/m. The section of the one-way shear in sec.44 (Fig.11.29.7) at a distance of d = 652 mm from the face of column C1. The width of the transverse strip = 400 + 0.75(652) = 889 mm. One-way shear force in sec.44 = (1500 – 652 – 200)(400)(10-3) = 259.2 kN Shear stress developed = 259.2(103)/(889)(652) = 0.447 N/mm2

To have the shear strength of concrete 0.447 N/mm≥ 2, the percentage of reinforcement is determined by linear interpolation from Table 19 of IS 456, so that the depth of footing may remain unchanged. Table 19 of IS 456 gives p = 0.43125. Accordingly, Ast = 0.43125(889)(652)/100 = 2499.65 mm2. Provide 13 bars of 16 mm (area = 2613 mm2), spacing = 889/12 = 74.08 mm c/c, say 70 mm c/c. However, this area of steel shall be checked for bending moment consideration also. (B) Bending moment at the face of column C1 in the transverse strip under column Bending moment = (1.3)(1.3)(400)/2 = 338 kNm. We, therefore, have M/(width of strip)d2 = 338(106)/(889)(652)(652) = 0.89 N/mm2

Table 2 of SP-16 gives p = 0.2608 < 0.43125. Hence, the area of steel as determine for one-way shear consideration is to be provided. Provide 13 bars of 16 mm @ 70 mm c/c in the column strip of width 889 mm under the column C1. Development length of 16 mm bars = 47.01(16) = 752.16 mm Length available = 1300 – 50 – 16 = 1234 mm > 752.16 mm Hence, o.k. (ii) Transverse beam under column C2


Figure 11.29.7 shows the strip of width = 500 + 0.75d + 0.75 d = 500 + 1.5(652) = 1478 mm, considering the effective depth of footing = 652 mm. (A) One-way shear The factored load for this transverse strip = 1800/3 = 600 kN/m. The one-way shear section is marked by sec.5.5 in Fig.11.29.7 at a distance of d = 652 mm from the face of the column C2. One-way shear in sec.55 (of width = 1478 mm) = (1500 – 652 – 150)(600)(10-3) = 418.8 kN The shear stress developed = 418.8(103)/(1478)(652) = 0.434 N/mm2

The corresponding percentage of area of steel, as obtained from Table 19 of IS 456 is, p = 0.404 per cent. Accordingly, Ast = 0.404(1478)(652)/100 = 3893.17 mm2. Provide 20 bars of 16 mm (area = 4021 mm2), spacing = 1478/19 = 77.78 mm c/c, say 75 mm c/c. However, this area of steel shall be checked for bending moment consideration also. (B) Bending moment at the face of column C2 in the transverse strip under column C2

The bending moment = (1.35)(1.35)(600)/2 = 546.75 kNm. We, therefore, have M/(width of strip)d2 = 546.75(106)/(1478)(652)(652) = 0.87 N/mm2

Table 2 of SP-16 gives: p = 0.2544 < 0.404 per cent as required for one-way shear. So, provide 20 bars of 16 mm diameter @ 75 mm c/c in the column strip of width 1478 under column C2. Development length as calculated in Step (B) of col strip C1 = 752.16 mm The length available = 1350 – 50 – 16 = 1284 mm > 752.16 mm Hence, o.k. Step 7: Transfer of forces at the base of the columns (A) For column C1 The limiting bearing stress at the column face governs where the bearing stress = 0.45fck = 9 N/mm2, since the column C1 is at the edge of the footing.


The force that can be transferred = 9(400)(400)(10-3) = 1440 kN > 1200 kN, load of column C1. Hence full transfer of force is possible without the need of any dowels. However, four 16 mm nominal dowels are provided as shown in Figs.11.29.5a and b. (B) For column C2

Figure 11.29.8 shows the dimensions to determine A1 and A2 for the column C2 and the footing. Accordingly, A1 = (1600)(1400) mm2, and A2 = (300)(500) mm2

(A1/A2)1/2 = {(16)(14)/(3)(5)}1/2 = 3.86 but limited to 2


The bearing stress at the column face = 0.45fck = 9 N/mm2, and the bearing stress at the footing face = 0.45fck(2) = 18 N/mm2. However, the bearing stress of 9 N/mm2 governs. The force that can be transferred through the column C2 = 9(300)(500) = 1350 kN < 1800 kN. For the excess force (1800 – 1350) = 450 kN, dowels shall be provided. The area of dowels = 450(103)/0.67(415) = 1618.414 mm2. The minimum area of dowels = 0.5(300)(500)/100 = 750 mm2. Provide 6 dowels of 20 mm diameter (area = 1885 mm2). The development length required in compression

= 0.97(415)(20)/4(1.6)(1.2)(1.25) = 752.2 mm. The length available = 652 – 20 = 632 mm Therefore, the dowels shall be given a 90o bend and shall be extended horizontally by 100 mm to have a total length of 632 + 8(20) + 100 = 892 mm > 752.2 mm (Figs.11.29.5a and b). Step 8: Distribution reinforcement Nominal distribution reinforcement shall be provided at top and bottom where the main reinforcement bars are not provided. The amount @ 0.12 per cent comes to 0.12(1000)(652)/100 = 782.4 mm2/metre. Provide 12 mm diameter bars @ 140 mm c/c (area = 808 mm2/m). All the reinforcing bars and dowels are shown in Figs.11.29.5a and b.


11.29.3 Practice Questions and Problems with Answers


Q.1: Design an isolated footing for a rectangular column, 300 mm x 500 mm

with 16-20 mm diameter longitudinal bars carrying a service load of 3000 kN (Figs.11.29.9a and b). Assume safe bearing capacity of soil as 260 kN/m2 at a depth of 1 m below ground level. Use M 20 and Fe 415.

A.1: Step 1: Size of the footing Given P = 3000 kN and qc = 260 kN/m2. Assuming the weight of the footing and backfill as 25 per cent of the load, the base area required = 3000(1.25)/260 = 14.42 m2. Provide 3 m x 5 m giving 15 m2 area. Step 2: Thickness of footing slab based on one-way shear (i) Along 5 m direction The critical section of one-way shear is marked by sec.11 in Fig.11.29.9b.


Factored soil pressure = 3000(1.5)/15 = 300 kN/m2 = 0.3 N/mm2

Assuming 0.2 per cent reinforcement in the footing slab, Table 19 of IS 456 gives cτ for M 20 = 0.32 N/mm2. From the condition that the shear resistance shear force, we have: ≥ 0.32(3000)d 0.3(2250 – d)(3000) ≥ or d ≥ 1088.71 mm Using 16 mm diameter bars with 50 mm cover the total depth = 1088.71 + 50 + 16 + 8 = 1162.71 mm. Provide total depth of 1165 mm, so that d = 1091 mm. (ii) Along 3 m direction The critical section of one-way shear is marked by sec.2-2 in Fig.11.29.9b. Resistance shear force = 0.32(5000)(1091) = 1745.6 kN Actual shear force = 0.3(259)(5000) = 388.5 kN < 1745.6 kN Hence, o.k. Step 3: Checking for two-way shear The critical section is marked by 3-3-3-3 in Fig. 11.29.9b. The permissible shear strength of concrete M 20

= (0.5 + β ) 0.25(fck)1/2 = 0.25(20)1/2 = 1.118 N/mm2

Total perimeter of section 3-3-3-3 = 2(500 + 1091 + 300 + 1091) = 5964 mm giving area = 5964(1091) = 6506724 mm2. Actual shear force = {15 – (1.591)(1.391)}(300) = 3836.0757 kN Shear stress = 3836.0757(103)/6506724 = 0.5896 N/mm2 < 1.118 N/mm2

Hence, o.k. The depth of the footing slab is governed by one-way shear here.


Step 4: Gross bearing capacity Assuming unit weights of concrete and soil as 25 kN/m3 and 20 kN/m3, respectively, we have Service load = 3000 kN Weight of footing slab = 3(5)(1.165)(25) = 436.875 kN Weight of backfill = 3(5)(1.165)(20) = 349.5 Total loads = 3786.375 kN Gross bearing capacity = 3786.375/15 = 252.425 kN/m2 < 260 kN/m2

Hence, o.k. Step 5: Bending moments (i) In the long direction The critical section of the bending moment in the long direction is marked by sec.4-4, (at the face of column) in Fig.11.29.9b. With factored soil pressure q = 0.3 N/mm2, the bending moment is 0.3(3000)(2250)(2250)/2 = 2278.125 kNm. Moment of resistance of the section = RBd2 = 2.76(3000)(1091)(1091)(10-

6) = 9855.53 kNm > 2278.125 kNm. Hence, o.k. The area of steel is determined from Eq.3.23 of Lesson 5, which is, Mu = 0.87 fy Ast d {1 – Ast fy/fck b d} or 2278.125(106) = 0.87(415)(Ast)(1091) – 0.87(Ast

2)(415)2/(20)(3000) or Ast

2 – 157734.9398 Ast + 912.249313(106) = 0, which gives Ast = 6012.6245, i.e., p = 0.1837 per cent. However, we have to provide 0.2 per cent, as required for the one-way shear. So, Ast = 0.2(3000)(1091)/100 = 6546 mm2. Provide 33 bars of 16 mm diameter (area = 6634 mm2, shown in Fig. 11.29.9c). The uniform spacing = (3000 – 50 – 16)/32 = 91.69 mm c/c, say @ 90 mm c/c. Alternatively, Mu/Bd2 = 2278.125(106)/(3000)(1091)(1091) = 0.638


Table 2 of SP-16 gives p = 0.1834 per cent, which is very close to the computed p = 0.1837 per cent. So, provide 33 bars of 16 mm diameter @ 90 mm c/c in the long direction. The development length of 16 mm bar is 47.01(16) = 752.16 mm Length available = 2250 – 50 – 16 = 2184 mm > 752.16 mm Hence, o.k. (ii) In the short direction The critical section is marked by sec.5-5 in Fig.11.29.9b, (at the face of column) where the bending moment is: 0.3(5000)(1350)(1350)/2 = 1366.875 kNm We get Mu/Ld2 = 1366.875(106)/(5000)(1091)(1091) = 0.2296 This is a low value, which is not included in SP-16. So, provide 0.15 per cent to have Ast = 0.15(5000)(1091)/100 = 8182.5 mm2

However, we have to check for one-way shear as the p provided 0.15 per cent is less than 0.2 per cent, required for the one-way shear and for which cτ = 0.28 N/mm2 (Table 19 of IS 456). The one-way shear force, as calculated in step 2(ii) is 388.5 kN. The resistance force of shear at this section = 0.28(5000)(1091) = 1527.4 kN > 388.5 kN. Hence, o.k. Further, the total area of steel 8182.5 mm2 has to be distributed in the central band of width 3 m and two outer bands of width 0.75 m each as stipulated in cl.34.3.1c of IS 456. The reinforcement in the central band is obtained as given below: Ast for central band = (2/ β + 1) (total area of steel) where β = 5/3 = 1.67. Thus,

Area of steel in the central band = (2/2.67)(8182.5) = 6129.21 mm2

Provide 55 bars of 12 mm (area = 6220 mm2) at the spacing of 3000/54 = 55.55 mm c/c, say, @ 55 mm c/c. Each of two outer band needs (8182.5 – 6129.21)/2 = 1026.65 mm2. Provide 10 bars of 12 mm diameter (area = 1131 mm2). The


spacing = (1000 – 50)/10 = 95 mm c/c, say @ 90 mm c/c. Thus, the total area of steel provided = 6220 + 1131 + 1131 = 8482 mm2 > 8182 mm2 (Fig. 11.29.9c). The development length required for 12 mm bars = (47.01)(12) = 564.12 mm Length available = 1350 – 50 – 16 = 1284 mm > 564.12 mm Hence, o.k. Step 6: Transfer of force at the base of the column Factored load = 3000(1.5) = 4500 kN. From the limiting bearing stress at the column-footing interface 0.45 fck(A1/A2)1/2, we have (i) At the column face, where A1 = A2; bearing stress = 0.45 fck = 9.0 N/mm2

(ii) At the footing face, A1 = 15 m

2 and A2 = (0.3)(0.5) = 0.15 m2. But, (A1/A2)1/2 >/ 2. So, the bearing stress = 2(9) = 18 N/mm2. Therefore, the column face governs. The force that can be transferred through bearing = 9(500)(300) = 1350 kN. The excess force to be transferred by dowels = 4500 – 1350 = 3150 kN. Area of dowels = 3150(103)/0.67(415) = 11328.9 mm2. Minimum area of dowels = 0.5(300)(500)/100 = 750 mm2. So, the number of 20 mm dowels = 36. Instead of providing 36 dowels, it is convenient to extend the sixteen column bars of 20 mm diameter to the footing. The development length required = 0.87(415)(20)/4(1.6)(1.25)(1.2) = 752.2 mm Length available = 1091 – 74 = 1017 mm > 752.2 mm Step 7: Nominal reinforcement Clause 34.5.2 stipulates to provide @ 360 mm2 per metre length in each direction on each force, when the thickness of footing slab is greater than one metre. So, the minimum reinforcement is 8 mm bars @ 130 mm c/c (area = 387 mm2/m), wherever there is no other reinforcement. The reinforcement bars are shown in Figs.11.29.9a and c.


Q.2: Design a reinforced concrete footing for a wall of 400 mm thickness transmitting a load of 200 kN/m (Fig.11.29.10) under service condition. Assume the safe bearing capacity of soil as 100 kN/m2 at a depth of 1 m below the ground level. Use M 20 and Fe 415. A.2: Step 1: Size of the footing Given axial load = 200 kN/m and safe bearing capacity of soil qc = 100 kN/m2 at a depth of 1 m below the ground level. Assuming the self weight of footing and backfill as 10 per cent, the area of the base required = 200(1.1)/100 = 2.2 m2. Provide width of 2.2 m for every one metre to get the required area of 2.2 m2. Step 2: Thickness of footing slab based on shear The critical section of shear is marked as sec.1-1 in Fig.11.29.10, at a distance of effective depth d of the footing.


Factored soil pressure = 200(1.5)/2.2 = 136.36 kN/m2 = 0.13636 N/mm2. Assuming 0.15 per cent reinforcement in the footing slab, the shear strength of M 20 cτ = 0.28 N/mm2. From the condition that the shear strength of the section actual shear force in sec.11, we have: ≥ 0.28(1000)d ≥ (0.13636)(1000)(900 – d) This gives d 294.755 mm. Using cover of 50 mm and diameter of reinforcing bar as 10 mm, the total depth of footing = 295 + 50 + 10 + 5 = 360 mm. This gives effective depth = 295 mm.

≥

Step 3: Checking for the moment The critical section for the bending moment is marked marked as sec.2-2 in Fig.11.29.10, where the bending moment (factored) is, Mu = (1)(0.13636)(1000)(1000)/2 = 68.18 kNm/m The capacity of the section = (2.76)(1000)(295)(295) = 240.189 kNm/m > 68.18 kNm/m. Hence, o.k. M/bd2 = 68.18(106)/1000(295)(295) = 0.784 N/mm2

Table 2 of SP-16 gives p = 0.2282 per cent. Accordingly, Ast = 0.2282(1000)(295)/100 = 673.19 mm2/m. Provide 10 mm diameter bars @ 110 mm c/c which gives 714 mm2/m > 673.19 mm2/m. Step 4: Development length The development length of 10 mm bars = 47.01(10) = 470.1 mm Length available = 900 – 50 = 850 > 542.88 Step 5: Distribution reinforcement Minimum reinforcement @ 0.12 per cent should be provided longitudinally. Ast = 0.12(2200)(360)/100 950.4 mm2

Provide 13 bars of 10 mm diameter (area = 1021 mm2). The spacing = (2200 – 50 – 10)/12 = 178.33 mm c/c, say @ 175 mm c/c. Step 6: Transfer of loads at wall-footing base


Bearing stress = 0.45 fck (A1/A2)1/2 = 0.45(20)(1) = 9 N/mm2

The forces that can be transferred = 9(1000)(2200)(10-3) kN = 19800 kN >> factored load of 200(1.5) = 300 kN. Hence, o.k. 11.29.4 References
















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Reinforced Concrete Limit State Design, 5th Edition, by Ashok K. Jain, Nem Chand & Bros, Roorkee, 1999.


11.29.5 Test 29 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions.


TQ.1: Redesign the footing of Problem 4 of section 11.29.2 (Fig.11.29.4) to have uniform base pressure considering that moments are due to dead load and hence irreversible (Fig.11.29.11). Other data of Prolem 4 are: column size = 300 mm x 450 mm, longitudinal bars of column = 20 bars of 20 mm (Ast = 4021 mm2), Pu = 1620 kN, Mu = 170 kNm, safe bearing capacity of soil = 200 kN/m2 at a depth of 1 metre below the ground level, grade of concrete = M 25 and grade of steel = Fe 415.

(50 marks)

A.TQ.1: Step 1: Size of the footing The required eccentricity between the centroids of column and footing = Mu/Pu = 170(106)/1620(103) = 104.938 mm, say 105 mm. Assuming the weight of the footing and backfill as 15 per cent, the required base area = 1620(1.15)/200(1.5) = 6.21 m2. It is desirable that the cantilever projections in the two direction from the respective column face should be equal. Accordingly, the footing is selected as 2.7 m x 2.64 m where the centre line of column is located at a distance of 105 mm left from the centre line of footing, as shown in Fig.11.29.11c. The arrangement shows that the cantilever projections are equal (1200 mm). Step 2: Thickness of footing based on one-way shear Factored soil pressure = 1620/(2.7)(2.64) = 227.27 kN/m2 (Fig.11.29.11b). The critical section of one-way shear is marked by sec.1-1 in Figs.11.29.11a and c, at a distance of d from the face of the column, where the factored shear Vu is, Vu = 0.22727(2700)(1200 – d) = (736354.8 – 613.629 d) N Assuming 0.15 per cent reinforcement in the footing slab, Table 19 of IS 456 gives cτ = 0.29 for M 25 concrete. Accordingly, the resistance shear of the section is 0.29(2700)d = 783d. From the condition that the resistance shear ≥ actual shear, we have 783 d ≥ 736354.8 – 613.629 d, which gives d 527.237. Let us use full depth = 527.237 + 50 + 16 + 8 = 601.237 mm, say 605 mm, so that d = 605 – 50 – 8 = 547 mm in the long direction and 605 – 50 – 16 – 8 = 531 mm in the short direction.

≥

Step 3: Checking for two-way shear


The shear strength of concrete for two-way shear = 0.25(fck)1/2 (cl.31.6.3 of IS 456) = 1.25 N/mm2. The resistance shear of the section is (sec.2222 of Fig.11.29.11c): (1.25)(2){(450 + 547) + (300 + 547)}(547) = 2521.67 kN The actual shear force in sec.2222 (Fig. 11.29.11c) is {(2.7)(2.64) – (0.997)(0.847)}(227.27) = 1428.06 kN < 2521.67 kN Hence, o.k. Step 4: Gross bearing capacity Assuming unit weights of concrete and soil as 25 kN/m3 and 18 kN/m3, we have the factored gross bearing capacity as q = (1620)/(2.7)(2.64) + {0.605(25) + 0.395(18)} = 249.509 kN/m2 < 300 kN/m2

Step 5: Bending moment (i) In the short direction The critical section is marked by sec.33 in Fig.11.29.11c, at the face of column, where the bending moment is: Mu = (2.7){(1.2)(1.2)/2}227.27 = 441.82 kNm Mu/Ld2 = 441.82(106)/(2.7)(103)(547)(547) = 0.546 N/mm2 < 3.45 N/mm2 for M 25 and Fe 415. Hence, the section has the capacity to resist Mu = 441.82 kNm. Table 3 of SP-16 gives p = 0.15488 per cent. Accordingly, Ast = 0.15488(2700)(547)/100 = 2287.42 mm2

Provide 12 bars of 16 mm diameter (area = 2412 mm2). The spacing = (2700 – 50 – 16)/11 = 239.45 mm c/c, say 220 mm c/c. Provide 12 bars of 16 mm diameter @ 220 mm c/c. Development length of 16 mm diameter bars

= 0.87(415)(16)/4(1.4)(1.6) = 644.73 mm


Length available = 1200 – 50 – 16 = 1134 mm > 644.73 mm Hence, o.k. (ii) In the long direction The critical section is marked by sec.4-4 in Fig.11.29.11c, at the face of column, where the moment Mu is, Mu = (2.64){(1.2)(1.2)/2}(227.27) = 431.994 kNm Mu/Bd2 = 431.994(106)/(2640)(531)(531) = 0.5803 N/mm2 < 3.45 N/mm2

Hence, the section can carry this moment. Table 3 of SP-16 gives p = 0.16509 per cent. Accordingly, Ast = 0.16509(2640)(531)/100 = 2314.2976 mm2

Provide 12 bars of 16 mm diameter (area = 2412 mm2). The spacing = (2640 – 50 – 16)/11 = 234 mm c/c, say 220 mm c/c. Provide 12 bars of 16 mm diameter @ 220 mm c/c. Development length as calculated in (i) for 16 mm diameter = 644.73 mm Length available = 1200 – 50 – 16 = 1134 mm > 644.73 mm Hence, o.k. Step 6: Transfer of force at the base of column Since the column is having moment along with the axial force, some of the bars are in tension. The transfer of tensile force is not possible through the column-footing interface. So, the longitudinal bars of column are to extended to the footing. The required development length of 20 mm bars = 0.87(415)/4(1.4)(1.6) = 805.92 mm. Length available = 531 mm < 805.92 mm. The bars shall be given 90o bend and then shall be extended by 200 mm horizontally to give a total length of 531 + 8(20) (bend value) + 200 = 891 mm > 805.92 mm. The arrangement of reinforcing bars is shown in Fig.11.29.11a.



This lesson illustrates the applications of the requirements for the design of foundations, explained in Lesson 28. Five illustrative examples, two practice problems and one test problem include plain concrete footings, isolated footings of square and rectangular columns transferring axial loads with or without moments to the footings, wall footings and combined footings. The solutions of illustrative problems and practice of the practice and test problems will give a clear understanding of the design of footings under different circumstances and soil conditions. All the designs are done following the stipulations of IS codes. It is, however, worth mentioning that all the footings explained in this lesson are shallow foundations. The design of deep foundations is beyond the scope of this course.


Module 12

Yield Line Analysis for Slabs


Lesson 30

Basic Principles, Theory and One-way Slabs




• state the limitations of elastic analysis of reinforced concrete slabs,

• explain the meaning of yield lines,

• explain the basic principle of yield line theory,

• state the assumptions of yield line analysis,

• state the rules for predicting yield patterns and locating the axes of rotation

of slabs with different plan forms and boundaries,

• state the upper and lower bound theorems,

• explain the two methods i.e., (i) method of segmental equilibrium and (ii)

method of virtual work,

• explain if the yield line analysis is a lower or upper bound method,

• analyse one-way slab problems to determine the location of yield lines and

determine the collapse load applying the theoretical formulations of (i)

method of segmental equilibrium and (ii) method of virtual work.


The limit state of collapse method of design of beams and slabs considers

the actual inelastic behaviour of slabs when subjected to the factored loads. Accordingly, it is desirable that the structural analysis of beams and slabs has to be done taking the inelastic behaviour into account. Though the coefficients in Annex D-1 of IS 456 to determine the bending moments for the design of two-way slabs are based on inelastic analysis, the code also recommends the use of linear elastic theory for the structural analysis in cl. 22.1. Moreover, IS 456 further stipulates the use of coefficients of moments and shears for continuous beams given in Tables 12 and 13 of cl. 22.5 in lieu of rigorous elastic analysis. These coefficients of beams are also applicable for the design of one-way slabs, based on linear elastic theory. Thus, there are inconsistencies between the methods of analysis and design.


The above discussion clearly indicates the need of adopting the inelastic analysis or collapse limit state analysis for all structures. However, there are sufficient justifications for adopting the inelastic analysis for slabs as evident from the following limitations of the elastic analysis of slabs:

(i) Slab panels are square or rectangular. (ii) One-way slab panels must be supported along two opposite sides

only; the other two edges remain unsupported.

(iii) Two-way slab panels must be supported along two pairs of opposite sides, supports remaining unyielding.

(iv) Applied loads must be uniformly distributed.

(v) Slab panels must not have large opening.


Therefore, for slabs of triangular, circular and other plan forms, for loads other than uniformly distributed, for support conditions other than those specified above and for slabs with large openings (Figs.12.30.1a to d); the collapse limit state analysis has been found to be a powerful and versatile method. Inelastic or limit analysis is similar to the plastic analysis of continuous steel beams which is based on formation of plastic hinges to form a mechanism of collapse. However, full plastic analysis of reinforced concrete beams and frames is tedious and time consuming. One important advantage of the reinforced concrete slabs over the reinforced concrete beams and frames is that the slabs are mostly under-reinforced. This gives a large rotation capacity of slabs, which may be taken as the presence of sufficient ductility. 12.30.2 Yield Line Theory The yield line theory, thus, is an ultimate or factored load method of analysis based on bending moment on the verge of collapse. At collapse loads, the slab begins to crack as they are mostly under-reinforced, with the yielding of reinforcement at points of high bending moment. With the propagation of cracks the yield lines are developed gradually. Finally, a mechanism is formed when the slab collapses due to uncontrolled rotation of members. Yield lines, therefore, are lines of maximum yielding moments of the reinforcement of slab. The essence is to find out the locations of the appropriate yield lines. Yield line analysis, though first proposed by Ingerslev in 1923 (vide, “The strength of rectangular slabs”, by A. Ingerslev, J. of Institution of Structural Engineering, London, Vol. 1, No.1, 1923, pp. 3-14), Johansen is more known for his large extension of the analysis (please refer to (i) Brutlinieteorier, Jul. Gjellerups Forlag, Copenhagen, 1943, by K.W. Johansen, English translation, Cement and Concrete Association London 1962; and (ii) Pladeformler, 2d ed., Polyteknisk Forening, Copenhagen, 1949, by K.W. Johansen, English translation,” Yield Line formulas for slabs, Cement and Concrete Association, London, 1972). Its importance is reflected in the recommendation of the use of this method of analysis of slabs in the Note of cl. 24.4 of IS 456. It is to note that only under-reinforced bending failure is considered in this theory ignoring the effects due to shear, bond and deflection. Effect of in-plane forces developed is also ignored.


12.30.3 Basic Principles


As mentioned earlier, reinforced concrete slabs are mostly under-reinforced and they fail in flexure. Figures 12.30.2a and b show such a reinforced concrete simply supported one-way slab subjected to short term loading. The maximum bending moment is along CC, at a distance of L/2 from the supports, where the deflection and curvature will be the maximum. The curvature φ is expressed by (Fig. 12.30.2c): φ = ∈max/x (12.1)

where ∈max is the compressive strain in concrete in the outermost fibre and x is the depth of the neutral axis. Figure 12.30.2b shows the plastic curvature φp distributed over a short length DE adjacent to the failure section and the ignored elastic curvatures φe elsewhere on the slab. Figure 12.30.3 shows the schematic moment-curvature diagrams of the slab of Fig.12.30.2a. It is evident from Fig.12.30.3 that up to the point D, when the crack first appears anywhere along the line CC of Figs.12.30.2a, the line OD is elastic. Thereafter, the slope of the line changes gradually with the progress of cracks. Accordingly, the stiffness of the slab is reduced. The reinforcement starts yielding at point F, when the ultimate strength is almost reached. The two paths FH for the mild steel reinforcing bars and FI for the deformed bars show marginal increase in moment capacities. Two points H and I are the respective failure points showing larger disproportionate deformations and curvatures when the maximum moment capacity i.e., the resistance of the cross-section Mp is reached. It may be noted that at these points (H and I), the plastic curvatures φp


are much larger than the elastic curvature φe. Neglecting these marginal moment capacities in the regions beyond F, the moment curvature diagram is idealised as OEG. The extended zone of increasing curvature at nearly constant moment and considering φp as much larger than φe, it is justified to neglect the elastic curvature φe totally. This amounts to shifting the point E to E′ in Fig. 12.30.3. The first crack starting anywhere along CC of the one-way slab of Fig.12.30.2a, after reaching the maximum moment capacity Mp, proceeds forming plastic hinges. In the process, the crack line or yield line CC is formed when the slab collapses forming a mechanism. It is worth mentioning that even at collapse; the two segments AC and BC remain elastic and plane.


It us thus seen that only one yield line is needed to form a mechanism with two real hinges for the collapse of the statically determinate one-way slab of Fig.12.30.2a. In case of statically indeterminate slab, the clamped-clamped slab of Fig.12.30.4a, however, can sustain the loads without collapse even after the formation of one or more yield lines. Figures 12.30.4c to f show the bending moment and deflection diagrams of the slab with increasing loads w1 < w2 < w3 < w4. It is known that during the elastic range, the negative moment at the clamped ends (w1l2/12) is twice of the positive moment at mid-span (w1l2/24), as shown in Fig.12.30.4c. Assuming that the slab has equal reinforcement for positive and negative moments, the highly stressed clamped ends start yielding first. Though the support line hinges rotate, the restraining moments continue to act till the mid-span moment becomes equal to the moment at the supports. Accordingly, a third hinge is formed (Fig.12.30.4f). Now, the three hinges form the mechanism and the slab collapses showing large deflection and curvature. The moment diagram just before the collapse is shown in Fig.12.30.4f. It is clear from the above discussion that such mechanism is possible with the bending moment diagram of Fig.12.30.4f, if the slab is having adequate reinforcement to resist equal moments at the support and mid-span. The elastic bending moment ratio of 1:2 between the mid-span and support could be increased to 1:1 by the redistribution of moment, which depends on the reinforcement provided in the supports and mid-span sections and not on the elastic bending moment diagram shown in Fig.12.30 4c. 12.30.4 Assumptions The following are the assumptions of the yield line analysis of reinforced concrete slabs.

1. The steel reinforcement is fully yielded along the yield lines at collapse. Rotation following yield is at constant moment.

2. The slab deforms plastically at collapse and is separated into

segments by the yield lines. The individual segments of the slab behave elastically.

3. The elastic deformations are neglected and plastic deformations are

only considered. The entire deformations, therefore, take place only along the yield lines. The individual segments of the slab remain plane even in the collapse condition.

4. The bending and twisting moments are uniformly distributed along the

yield lines. The maximum values of the moments depend on the capacities of the section based on the amount of reinforcement provided in the section.


5. The yield lines are straight lines as they are the lines of intersection

between two planes. 12.30.5 Rules for Yield Lines The first requirement of the yield line analysis is to assume possible yield patterns and locate the axes of rotation. It has been observed that assuming the possible yield patterns and locating the axes of rotation are simple to establish for statically determinate or indeterminate (simply supported or clamped) one-way slabs. For other cases, however, suitable guidelines are needed for drawing the yield lines and locating the axes of rotation. It is worth mentioning that other cases of two-way slabs will have sufficient number of real or plastic hinges to form a mechanism while they will be on the verge of collapse. The yield lines will divide the slabs into a number of segments, which will rotate as rigid bodies about the respective axes of rotation. The axes of rotations will be located along the lines of support or over columns, if provided as point supports. The yield line between two adjacent slab segments is a straight line, as the intersection of two-plane surfaces is always a straight line. The yield line should contain the point of intersection, if any, of the two axes of rotation of two adjacent segments as such point of intersection is common to the two planes. The two terms, positive and negative yield lines, are used in the analysis to designate the yield lines for positive bending moments having tension at the bottom and negative bending moments having tension at the top of the slab, respectively. The following are the guidelines for predicting the yield lines and axes of rotation:

1. Yield lines between two intersecting planes are straight lines. 2. Positive yield line will be at the mid-span of one-way simply supported

slabs.

3. Negative yield lines will occur at the supports in addition to the positive yield lines at the mid-span of one-way continuous slabs.


4. Yield lines will occur under point loads and they will be radiating outward from the point of application of the point loads.

5. Yield line between two slab segments should pass through the point of

intersection of the axes of rotation of the adjacent slab segments.

6. Yield lines should end at the boundary of the slab or at another yield line.

7. Yield lines represent the axes of rotation.

8. Supported edges of the slab will also act as axes of rotation. However,

the fixed supports provide constant resistance to rotation having negative yield lines at the supported edges. On the other hand, axes of rotation at the simply supported edges will not provide any resistance to rotation of the segment.

9. Axis of rotation will pass over any column support, if provided, whose

orientation will depend on other considerations.


Some examples are taken up to illustrate the applications of the guidelines for predicting the possible yield patterns in Figs.12.30.5a to g. 12.30.6 Upper and Lower Bound Theorems According to the general theory of structural plasticity, the collapse load of a structure lies in between the upper bound and lower bound of the true collapse load. Therefore, the solution employing the theory of plasticity should ensure that lower and upper bounds converge to the unique and correct values of the collapse load. The statements of the two theorems applied to slabs are given below:

(A) Lower bound theorem: The lower bound of the true collapse load is that external load for which a distribution of moments can be found satisfying the requirements of equilibrium and boundary conditions so that the moments at any location do not exceed the yield moment.

(B) Upper bound theorem: The upper bound of the true collapse load is that

external load for which the internal work done by the slab for a small increment of displacement, assuming that moment at every plastic hinge is equal to the yield moment and satisfying the boundary conditions, is equal to the external work done by that external load for the same amount of small increment of displacement.


Thus, the collapse load satisfying the lower bound theorem is always lower than or equal to the true collapse load. On the other hand, the collapse load satisfying the upper bound theorem is always higher than or equal to the true collapse load.

The yield line analysis is an upper bound method in which the predicted failure load of a slab for given moment of resistance (capacity) may be higher than the true value. Thus, the solution of the upper bound method (yield line analysis) may result into unsafe design if the lowest mechanism could not be chosen. However, it has been observed that the prediction of the most probable true mechanism in slab is not difficult. Thus, the solution is safe and adequate in most of the cases. However, it is always desirable to employ a lower bound method, which is totally safe from the design point of view.

12.30.7 Methods of Analysis After predicting the general yield pattern and locating the axes of rotation, the specific pattern and locations of axes of rotation and the collapse load for the slab can be determined by one of the two methods given below:

(1) Method of segmental equilibrium (2) Method of virtual work.

The two methods are briefly explained below.

(1) Method of segmental equilibrium

In this method, equilibrium of the individual slab segments causing the collapse forming the required mechanism is considered to arrive at a set of simultaneous equations. The solutions of the simultaneous equations give the values of geometrical parameters for finalising the yield pattern and the relation between the load capacity and resisting moment. Thus, in segmental equilibrium method, each segment of the slab is studied as a free body (Fig.12.30.5b) which is in equilibrium at incipient failure under the action of the applied loads, moment along the yield lines, and reactions or shear along the support lines. Since, yield moments are principal moments, twisting moments are zero along the yield lines. Further, in most of the cases, shear forces are also zero. Thus, with reference to Fig.12.30.5b, the vector sum of moments along yield lines AO


and OB (segment AOB) is equal to moments of the loads on the segment AOB about the axis of rotation 1-1 just before the collapse. It should be specially mentioned that equilibrium of a slab segment should not be confused with the general equilibrium equation of the true equilibrium method, which is lower bound. Strip method of slab design, developed by A. Hillerborg, (vide “Equilibrium theory for reinforced concrete slabs” (in Swedish), Betong, vol.4 No. 4, 1956, pp. 171-182) is a true equilibrium method resulting in a lower bound solution of the collapse load, which is safe and preferable too. The governing equilibrium equation for a small slab element of lengths dx and dy is

2 2 2

2 22Mx Mxy My wx yx y

∂ ∂ ∂ + + = −∂ ∂∂ ∂

(12.2)

where w is the external load per unit area; Mx and My are the bending

moments per unit width in x and y directions, respectively; and Mxy is the twisting moment. However, strip method of analysis is beyond the scope of this course. For more information about strip method, the reader may refer to Chapter 15 of “Design of concrete structures” by A.H. Nilson, Tata-McGraw – Hill Publishing Company Limited, New Delhi. (2) Method of virtual work

This method is based on the principle of virtual work. After predicting the

possible yield pattern and the axes of rotation, the slab, which is in equilibrium with the moments and loads on the structure, is given an infinitesimal increase in load to cause the structure further deflection. The principle of virtual work method is that the external work done by the loads to cause a small virtual deflection should be equal to the internal work done by the yield moments to cause the rotation in accommodating the virtual deflection. The relation between the applied loads and the ultimate resisting moments of the slab is obtained by equating the internal and external works. As the elastic deflections and rotations are small compared to the plastic deformations and rotations, they are neglected in the governing equation. Further, the compatibility of deflection must be maintained. The work equation is written as follows:

∑ w Δ = ∑ M θ l (12.3) where w = collapse load, Δ = vertical deflection through which the collapse load w moves,


M = moment capacity of the section per unit length, θ = rotation of the slab segment satisfying the compatibility of deflection, and l = length of the yield line. As mentioned earlier, both the methods of segmental equilibrium and virtual work are upper bound methods. Therefore, the collapse load obtained by either method of yield line analysis will be at the higher end of the true collapse load. Accordingly, each of the two methods should be developed to get the correct solution for predicted mechanism. However, the true collapse load will be obtained only if the correct mechanism has been predicted. Thus, the solution of any of the two upper bound methods has two essential parts: (1) predicting the correct yield pattern, and (2) determining the geometric parameters that define the exact location and orientation of the yield pattern and solving for the relation between applied load and resisting moments. 12.30.8 Analysis of One-Way Slab


Figures12.30.6a and b show one-way continuous slab whose one span of width one metre is considered. The factored moments of resistance are MA and MB at the continuous supports A and B where negative yield lines have formed and MC at C where the positive yield line has formed. All the moments of resistance are for one metre width. Figure 12.30.6c shows the free body diagrams of the two segments of slab, which are in equilibrium. We now apply the method of segmental equilibrium in the free body diagrams of Fig.12.30.6c. (1) Method of segmental equilibrium The line CC, where the positive yield line has formed is at a distance of x from AA. The shear V = 0 at C as the bending moment is the maximum (positive) there. Here there are two unknowns x, the locations of the positive yield line and w, the collapse load, which are determined from the two equations of equilibrium ∑V = 0 and ∑M = 0. From the statical analysis, we know that the vertical reactions VA and VB at A and B, respectively are, (assuming MA > MB): VA = (wL/2) + (MA – MB) / L (12.4) VB = (wL/2) – (MA – MB) / L (12.5) Now, ∑V = 0 gives: VA – wx = 0 (12.6) Substituting the expression of VA from Eq. 12.4 in Eq. 12.6, we have, wL/2 + (MA – MB) / L – wx = 0, which gives: w = -2 (MA – MB) / L (L-2x) (12.7) ∑M = 0 gives: VA x – MA – wx2/2 – Mc = 0 (12.8) Substituting the expression of VA from Eq. 12.4 in Eq. 12.8, we have, {wL/2 + (MA – MB) / L}x – MA – wx2/2 – MC = 0 (12.9) Substituting the expression of w from Eq. 12.7 into Eq. 12.9, we have,

(MB – MA) x2 + 2 (MA + MC) Lx – (MA + MC) L2 = 0 (12.10)

Equation 12.10 will give the values of x for known values of MA, MB and MC. Equation12.7 will give the value of w after getting the value of x from Eq. 12.10.


(2) Method of virtual work

The work equation (Eq.12.3) is written equating the total work done by the collapse loads during the rotation of slab segments, maintaining the deflection compatibility to the total internal work done by bending and twisting moments on all the yield lines. Figure12.30.6d presents the rigid body rotations of the two slab segments. The left segment AACC rotates by θA in the clockwise direction and the right segment BBCC rotates by θB in the anticlockwise direction, while maintaining the deflection Δ compatible, as shown in Fig.12.30.6d.

External work done = wx (Δ/2) + w (L-x) Δ/2 = wLΔ/2 (12.11)

Internal work done = + MA θA + MB θB + MC (θA + θB) (12.12)

From Fig.12.30.6d:

θA = Δ/x and θB = Δ/(L-x) (12.13)

Substituting the values of θA and θB from Eq.12.13 into Eq.12.12, we have:

Internal work done = MA (Δ/x) + MB (Δ /L-x) + MC (Δ/x + Δ/L-x) (12.14)

Equating the works, we have form Eqs.12.11 and 12.14, wL Δ/2 = MA (Δ/x) + MB (Δ/L-x) + MC (Δ/x + Δ/L-x)

or, w = 2MA / Lx + 2MB / L (L-x) + 2 MC / x (L-x) (12.15) In the method of segmental equilibrium, we have two equations (Eqs.12.7 and 12.9) for the two unknowns x and w. In the method virtual work, however, we have only one equation (Eq.12.15) to determine the two unknowns x and w. Accordingly; we generate another equation with the help of differential calculus. Since the method of analysis is upper bound, we have to consider the minimum value of w satisfying Eq.12.15. This can be obtained by differentiating w of Eq.12.15 with respect to x and putting dw/dx = 0. Hence, we have: dw/dx = (2MA / L) (-1 /x2) + (2 MB /L){1/(L-x)2}+ 2 MC [(1/L-x) (-1/x2) + (1/x){1/(L-x)2}] = 0 (12.16) After arranging the coefficients of x2, x and constant terms, we have:


(MB – MA) x2 + 2 (MA + MC) Lx – (MA + MC) L2 = 0 (12.10) This is the same equation (Eq.12.10), as we obtained in the method of segmental equilibrium. In the method of virtual work, therefore, we get the values of x from the same equation, Eq.12.10, and then we get w from Eq.12.15. We consider two special cases of simply supported and clamped slabs from the above equations. Case (i): Simply Supported Slab Here, we have MA = MB = 0. We get the values of x and w from Eqs.12.10 and 12.7, respectively, by the method of segmental equilibrium and from Eqs.12.10 and 12.15, respectively, by the method of virtual work. Equation 12.10 becomes 2 MC Lx – MC L2 = 0, when MA = MB = 0, which gives x = L/2. However, Eq. 12.7 gives division by zero for the simply supported case, when MA = MB = 0 and x = L/2. So, we use Eq.12.8 ( ∑M = 0) when MA = MB = 0 and x = L/2. This gives w = 8MC / L2. In the method of virtual work, Eq.12.10 (the same as in the method of segmental equilibrium) gives x = L/2. Thereafter, Eq.12.15 is used for determining w, which gives (when MA = MB = 0 and x = L/2), w = 8MC / L2

Thus, we get the following values of x and w for the simply supported slab x = L/2 and w = 8MC / L2 (12.17) Case (ii) Clamped Slab with MA = MB = 2MC Here, we get x = L/2 from Eq.12.10. Then, we use Eq.12.8 as Eq.12.7 involves zero by zero case, as explained in case (i) above. This gives w = 24 MC / L2 (when MA = MB = 2MC and x = L/2). These values are by the method of segmental equilibrium. Similarly, by the method of virtual work, Eq.12.10 gives x = L/2. Then, Eq.12.15 gives, w = 24 MC / L2.


Thus, we get for the clamped slab, when MA = MB = 2 MB C, x = L/2 and w = 24 MC /L2 (12.18) It is clear from the two cases that there will be positive yield line at the centre of the slab when it is simply supported and there will be two negative yield lines at the two supports in addition to the positive yield line at the centre of the slab when the slab is clamped at both ends. The collapse loads are: w = 8 MC / L2 and 24 MC / L2 for the simply supported and clamped slabs, respectively. However, it is worth mentioning that the formation of those specific yield lines and the respective collapse loads are possible only if the slab is designed with adequate positive and negative reinforcement, as assumed to get the solution. In the practical cases of continuous one-way slabs, the negative moments over the supports may be anywhere in the range of wl2/8 and wl2/12. Moreover, it is not essential that the negative moment of resistances at the two ends should be equal. Depending on the values of MA, MB and MC as per the reinforcement provided, the values of x and w shall be determined using the respective equations employing either the method of segmental analysis or the method of virtual work. We now take up numerical problems in the next section for the purpose of illustration. 12.30.9 Numerical Problems


Problem 1. The factored moment capacities of the one-way continuous

reinforce concrete slab of Figs. 12.30.7 are MA = 32 kNm, MB = 36 kNm and MC = 30 kNm. The span of the slab is 5 m. Determine the location of plastic hinges and the collapse load employing methods of segmental equilibrium and virtual work.

Solution 1:

(A) Method of segmental equilibrium

We have two equations (Eqs.12.10 and 12.7) to determine the location of positive yield line (distance x) and the collapse load (Figs.12.30.6a to d).


(MB – MA) x2 + 2 (MA + MC) Lx – (MA + MC) L2=0 (12.10) Using the given values of MA = 32 kNm, MB = 36 kNm and MC = 30

kNm in Eq.12.10, we have 4 x2 + 2(62) (5) x – (62) (25) = 0 or, x2 + 155 x – 387.5 = 0 gives x = 2.4609 m. w = - 2 (MA – MB) / L (L-2x) (12.7)

Using the values of MA = 32 kNm, MB = 36 kNm and x = 2.4609 m in Eq.12.7, we have, w = 20.4604 kN/m.

Therefore, the locations of the positive yield line is at a distance of 2.4609 m from the left support at A, the negative yield lines are at the supports A and B, and the collapse load = 20.4604 kN/m.

(B) Method of virtual work In this method, we have Eqs.12.10 and 12.15 to determine x and w

(Figs.12.30.6a to d). In the method of segmental equilibrium (A) above, Eq.12.10 gives: x = 2.4609 m.

w = 2 MA/Lx + 2 MB / L(L-x) + 2 MC /x (L-x) (12.15)

Using MA = 32 kNm, MB = 36 kNm, MC = 30 kNm and x = 2.4609 m, in Eq.12.15 above, we get w = 20.4749 kN/m. Thus, we get the same location of positive yield line i.e., at x = 2.4609 m from the left support A, negative yield lines are at the support A and B and the collapse load w = 20.4749 kN/m .

The marginal difference of the two collapse loads is due to the truncation error in the calculation. Otherwise, both the values should be the same. The bending moment and shear force diagrams are shown in Figs.12.30.7b and c, respectively.

12.30.10 Practice Questions and Problems with Answers Q.1: Name five limitations of elastic analysis of reinforced concrete slabs.


A.1: Second paragraph of sec.12.30.1 Q.2: What are yield lines and what is yield line theory? A.2: First paragraph of sec.12.30.2 Q.3: Explain the basic principles of yield line theory. A.3: Sec.12.30.3 Q.4: State the assumptions of yield line theory. A.4: Sec.12.30.4 Q.5: What are the guidelines to draw the possible yield patterns and locate the

axes of rotations? A.5: The last paragraph of sec.12.30.5 Q.6: State upper and lower bound theorems. A.6: Sec.12.30.6 Q.7: Explain the two methods of yield line analysis. A.7: Sec.12.30.7 Q.8: Compare the locations of positive and negative yield lines and the values

of the respective collapse loads of one 3 m × 6 m slab supported along 6 m direction (Figs.12.30.6a) carrying a total factored load of 20 kN/m for the three cases: (i) the slab is simply supported, (ii) the slab is clamped and (iii) the positive and negative reinforcements are identical to have the equal resistance for the continuous slab. Discuss the results.

A.8: The moments MC at the mid-span for one simply supported slab is wl2/8 =

90 kNm and for another clamped supported slab MC = wl2 /24= 30 kNm. The support moments for clamped slab MA = MB = -wl2/12 = - 60 kNm. For the third case the magnitude of MA, MB and MC are equal for a continuous slab. So, MA = MB = - 45 kNm and MC = + 45 kNm. Method of virtual work is employed.


Case (i): From Eq. 12.17, we have the distance of positive yield line x = 6/2 = 3 m and w = 8 MC /L2 = 8 (90) / 6(6) = 20 kN/m. There are no negative yield lines as the slab is simply supported.

Case (ii): From Eq. 12.18, we have the distance of positive yield line x =

6/2 = 3 m and w = 24 MC /L2 = 24 (30) / 6(6) = 20 kN/m. The negative yield lines of the clamped slab are at the two sides AA and BB.

Case (iii): Employing method of virtual work (MB – MA)x2 + 2(MA + MC)Lx – (MA + MC)L2 = 0

(12.10) w = 2MA / Lx + 2MB/L (L-x) + 2MC / x (L-x)

(12.15) Using MA = MB = MC = 45 kNm in Eq.12.10, we have, 2x – L = 0 or x = L/2 = 6/2 = 3 m. Using x = 3 m and MA = MB = MC = 45 kNm in Eq. 12.15, we have w = 90{1/6(3) + 1/6(3) + 1/3(3)} = 20 kN/m. The negative yield lines are over the supports of continuous slab i.e., along AA and BB. Discussion of results: All the three slabs have x =3 m and w = 20 kN/m. However, the third case is the most economic as the depth of the slab may be provided for M = 45 kNm and the areas of positive and negative steel are the same. The simply supported slab has no negative moment and the clamped slab has large difference between the positive and negative moments and accordingly, wide variation is seen in the magnitudes of positive and negative steel. 12.30.11 References


2. Limit State Design of Reinforced Concrete, 2nd Edition, by P.C. Varghese, Prentice-Hall of India Pvt. Ltd., New Delhi, 2002.

3. Advanced Reinforced Concrete Design, by P.C. Varghese, Prentice-Hall of India Pvt. Ltd., New Delhi, 2001.

4. Reinforced Concrete Design, 2nd Edition, by S. Unnikrishna Pillai and Devdas Menon, Tata McGraw-Hill Publishing Company Limited, New Delhi, 2003.


5. Limit State Design of Reinforced Concrete Structures, by P. Dayaratnam, Oxford & I.B.H. Publishing Company Pvt. Ltd., New Delhi, 2004.

6. Reinforced Concrete Design, 1st Revised Edition, by S.N. Sinha, Tata McGraw-Hill Publishing Company. New Delhi, 1990.

7. Reinforced Concrete, 6th Edition, by S.K. Mallick and A.P. Gupta, Oxford & IBH Publishing Co. Pvt. Ltd. New Delhi, 1996.







14. Reinforced Concrete Designer’s Handbook, 10th Edition, by C.E.Reynolds and J.C. Steedman, E & FN SPON, London, 1997.


16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 12.30.12 Test 30 with Solutions Maximum Marks = 50 Maximum Time = 30

minutes Answer all questions. TQ.1: Explain the basic principles of yield line theory. A.TQ.1: Sec.12.30.3

[10 Marks] TQ.2: State the assumptions of yield line theory. A.TQ.2: Sec.12.30.4

[10 Marks] TQ.3: What are the guidelines to draw the possible yield patterns and locate the

axes of rotations? A.TQ.3: The last paragraph of sec.12.30.5


[10 Marks] TQ.4: Determine the location of plastic hinges and collapse load of the one-way

slab supported at 4 m c/c (Fig.12.30.6a) having the positive and negative reinforcements to have the factored moments of resistance MA = 25 kNm, MB = 35 kNm and MC = 30 kNm. Use method of virtual work.

[20 Marks]

A.TQ.4: (MB – MA)x2 + 2(MA + MC)Lx – (MA + MC)L2 = 0 (12.10)

w = 2MA / Lx + 2MB/L (L-x) + 2MC / x (L-x) (12.15) Using MA = 25 kNm, MB = 35 kNm and MC = 30 kNm in Eq. 12.10, we

have, x = 1.9165 m and using the values of MA, MB, MC and x in Eq. 12.15, we get w = 29.9478 kN/m2.

12.30.13 Summary of this Lesson This lesson explains the basic principle of yield line analysis which is required to remove the inconsistency between the elastic analysis and the design by limit state method considering inelastic behaviour. Moreover, the limitations of elastic analysis of slab are mentioned. The upper and lower bound theorems are explained to show that the two methods of yield line analysis viz. (i) method of segmental equilibrium and (ii) method of virtual work are upper bound methods. The governing equations of both the methods are derived. The first requirement of the yield line analysis is to predict the possible yield pattern and locate the axes of rotation. Suitable guidelines are given for the same as the correctness of the solution depends on the prediction of true yield line pattern and location of the axes of rotation. Numerical problems are solved by both methods of analysis.


Module 12



Lesson 31

Nodal Forces and Two-way Slabs




• derive the expression for determining the work done by bending and

twisting moments when the yield lines are at angles with the directions of

reinforcing bars,

• state the need for considering the nodal forces and to estimate their values

when one yield line meets another yield line or a free edge,

• to select the possible yield pattern of a two-way slab supported at four sides

either by simple supports or fixed supports,

• to finalise the yield patterns and to evaluate the collapse loads of two-way

slabs, either simply supported or clamped at four sides,

• apply the theory in solving numerical problems of slabs to finalise the yield

pattern and to determine the collapse load employing (i) the method of

segmental equilibrium and (ii) the method of virtual work.


Lesson 30 introduces the yield line analysis, which is an upper bound method of analysis for slabs. The different rules for predicting the yield lines are stated. The two methods i.e., (i) method of segmental equilibrium and (ii) method of virtual work are explained. Applications of both the methods are illustrated through numerical problems of one-way slabs – either simply supported or continuous.

This lesson presents the derivations of the expressions for determining

bending and torsional moments when yield lines are at angles with the directions of reinforcement. The need for the nodal forces and their determinations are explained when yield line meets another yield line or the free edge. Thereafter, different possible yield patterns of two-way slabs are explained. Numerical illustrative problems of two-way slabs with or without nodal forces are illustrated.


12.31.2 Work Done by Yield Line Moments

Normally, the reinforcing bars are placed in two mutually perpendicular

directions parallel to the sides of rectangular and square slabs. However, the yield lines may be at an angle with the direction of reinforcing bars as shown in Fig. 12.31.1, in which the yield line AB of length L has bending moment Mb and twisting moment Mt per unit length of the yield line. The slab segment is undergoing rigid body rotation whose components are θx and θy. The horizontal and vertical projections of the yield line are having moment capacities of Mx and My per unit length, respectively. All moments and rotations are shown using the right hand thumb rule. The following expression is derived for obtaining the absolute values of the work done by Mb and Mt on the yield line AB.

With reference to Fig. 12.31.1, the total work done by the bending and

twisting moments Mb and Mt is,

W = Mb L (θx cos θ + θy sin θ) + Mt L (- θx sin θ + θy cos θ) (12.19)


Equating the work done by bending and twisting moments Mb and Mt along the yield line AB with the respective work done by their components along the projections of the yield line, we have the following two equations: Mb L = Mx L sin θ sin θ + My L cosθ cosθ , which gives Mb = Mx sin2θ + My cos2θ (12.20) and Mt L = Mx sin θ cos θ - My L cos θ sin θ , which gives Mt = (Mx – My) sinθ cosθ (12.21) Substituting the expressions of Mb and Mt from Eqs.12.20 and 12.21 in Eq.12.19, we have, W = (Mx sin2θ + My cos2θ) (θx L cosθ + θy L sinθ) + (Mx – My) sinθ cosθ (-θx L sinθ + θy L cosθ), which gives W = Mx L sinθ θy + My L cosθ θxor W = Mx Ly θy + My Lx θx (12.22) The two terms on the right hand side of Eq. 12.22 give the external work done by the moments MxLy and MyLx acting on the horizontal and vertical projections of the yield line. It is thus seen that the expression of Eq.12.19 involving bending and twisting moments Mb and Mt may be replaced by Mx and My of Eq.12.22 to get the same work done by Eq. 12.22 as that of Eq. 12.19. 12.31.3 Special Conditions at Edges and Corners


Figures 12.30.5d, e and f of Lesson 30 present yield patterns of slabs

when positive yield lines intersect the free edges at angles. In actual case, however, both bending and twisting moments are zero at the free or simply supported edges. The directions of the principal stresses are, therefore, parallel and perpendicular to the respective edge. Accordingly, the yield lines should enter the edge perpendicular to it, which is confirmed by experimental tests also. In Fig. 12.31.2a, it is shown that the yield line ADC normally turns quite close to the edge, say at D, which is at a distance s. The yield line is approximated by extending it in a straight line AB to the edge introducing a pair of concentrated shear force, (+) V and (-) V, as shown in Fig. 12.31.2b. These shear forces, which are parallel, equal and opposite forces for the equilibrium, are designated as nodal forces acting upward at an obtuse corner, marked by (-)ve sign on the left of the yield line AB and acting downward at an acute corner, marked by (+)ve sign on the right of the yield line AB, as shown in Fig. 12.31.2b.

They are, in fact, the static equivalent of twisting moments and shear

forces near the edge. We now establish the required expression for determining the magnitude of these nodal forces V.

Figure 12.31.3a shows an element ACB of a slab where the yield line AB

is making an angle α with the free edge CB. The angle between the yield line AB and the element side AC (i.e., angle CAB) is dα. The bending and twisting moments Mb and Mt on the yield lines AB and AC are shown in Fig. 12.31.3a, neglecting their differential increments. The free body diagram is shown in Fig. 12.31.3b. The distances BD, AD and BC are Lx, Ly and dLx, respectively. Taking moments about the line AC and equating it to zero, we get My (Lx +dLx) cos (α -


dα) – My Lx cos (α - dα) + Mx Ly sin (α - dα) Mx Ly sin (α - dα) – V dLx sin (α - dα) = 0

or V sin (α - dα) = My cos (α - dα)

or V (sin α cos dα - cos α sin dα) = My (cos α cos dα + sin α sin dα) For small values of dα, sin dα = 0 and cos dα = 1. So, we have V sin α = My cos α or V = My cot α (12.23) Equation 12.23 gives the magnitude of the nodal force V where α is measured anticlockwise. When α = 90 degree, V = 0, i.e., the nodal force is zero if the yield line intersects the free edge at an angle of 90 degree.

The nodal force V shall be used in the method of segmental equilibrium as

we consider the equilibrium of individual segment. In the method of virtual work, however, the work done is determined for the entire slab involving all the elements, when the total work done by the positive and negative nodal forces is zero. Hence, it is not needed to consider the nodal force in the method of virtual work.


Figures 12.31.4a and b present two typical yield patterns of two-way slabs

clamped at four sides and Fig. 12.31.4c and d show two typical yield patterns of a slab simply supported on three sides and free at one side. In all the yield patterns, it is assumed that the yield lines enter the corners between the two intersecting sides. It may not be the case always. Sometimes, yield lines fork before they reach the corner as shown in Fig. 12.31.4e and thus form corner lever. The triangular element EFB in Fig. 12.31.4e will pivot about the axis EF and lift off the supports if the corner is not held down. It has been observed that such yield patterns with corner levers are more critical than those without them. However, these patterns are generally neglected. It should be mentioned that the introduction of corner levers makes the analysis more complicated and does not produce much error by neglecting them. This is illustrated in Lesson 32.

12.31.4 Two-way Slabs of Yield Pattern 1 of Figure 12.31.4a


We consider the two-way slab of Fig. 12.31.4a subjected to uniformly distributed collapse load of intensity w kN/m2. Yield pattern 1 divides the slab into four segments marked by 1, 2, 3 and 4 in Fig. 12.31.5a, where the positive and negative yield lines are shown along with the positive and negative nodal forces. The slab undergoes a displacement Δ at the center point O. The free body diagrams of segments 1 and 2 are shown in Figs. 12.31.5b and c, respectively. The positive and negative moments along x and y directions are designated by Mpx, Mpy, Mnx and Mny in these two figures. We are employing both (A) the method of segmental equilibrium and (B) the method of virtual work to determine the magnitude of the collapse load w of the slab.

(A) Method of segmental equilibrium At the equilibrium, the moment of all the forces and moments of segment 1 about the left edge AB is zero. This gives (Fig. 12.31.5b): (1/2) (Lx /2) (Ly) w(Lx/6) – V(Lx/2) – Mnx Ly – Mpx Ly= 0 or wLxLy – 12 V = 24 (Mpx + Mnx) (Ly / Lx) (12.24) Similarly, at the equilibrium, the moment of all forces and moments of segment 2 about the bottom edge BC is zero. This gives (Fig. 12.31.5c): (1/2) Lx (Ly/2) w (Ly/6) + V (Ly/2) – Mny Lx – Mpy Lx = 0 or, wLxLy + 12 V = 24 (Mpy + Mny) (Lx / Ly) (12.25) Eliminating V from Eqs. 12.24 and 12.25 by adding the two equations, we have 2w Lx Ly = 24 {(Mpx + Mnx)(Ly / Lx)+ (Mpy + Mny) (Lx / Ly)} or w = 12 {(Mpx + Mnx) / Lx

2 + (Mpy + Mny) / Ly2}

(12.26) The collapse load w is determined from Eq. 12.26 from known values of Mpx, Mnx Mpy and Mny. (B) Method of virtual work The total external work done by the load in causing the segments to undergo deflection and the total internal work done by the moments in rotating all the four segments are computed. As mentioned in sec. 12.31.3, the effect of all the nodal forces is zero. Accordingly, the total external work done (TEW) is,

TEW = 4 (1/2) Lx (Ly/2) w (Δ/3) = wLx Ly (Δ/3) (12.27)


The total internal work done (TIW) is,

TIW = 2 (Mpx + Mnx) Ly θx + 2 (Mpy + Mny) Lx θy

= 4Δ {(Mpx + Mnx) (Ly /Lx) + (Mpy + Mny) (Lx / Ly)} (12.28) Equating the two works from Eqs. 12.27 and 12.28

w Lx Ly (Δ/3) = 4Δ{(Mpx + Mnx) (Ly /Lx) + (Mpy + Mny) (Lx/Ly)}

or w = 12 {(Mpx + Mnx) / Lx2 + (Mpy+Mny) / Ly

2} (12.26) The above equation is the same as obtained by the method of segmental equilibrium to get the value of the collapse load w. We now consider the two cases below, first the square and simply supported at all edges and secondly the square and clamped at all edges. Case (i) Square and simply supported slab For square and simply supported slab, Lx = Ly = L, Mnx = Mny = 0 and let us assume Mpx = Mpy = Mp. Using the conditions mentioned above in Eq. 12.26, we get w = 24 Mp / L2 (12.29) or, Mp = w L2/24 (12.30) Case (ii) Square and clamped slab For square and clamped slab, let us assume Mpx =Mpy = Mnx = Mny = Mp. Using these conditions in Eq. 12.26, we get w = 48 Mp / L2 (12.31) or Mp = wL2/48 (12.32) The values of w and Mp in the two cases above reveal that the factored load intensity of a clamped slab is twice of that of simply supported slab having the same moment carrying capacity Mp. Further, we observe that for the same factored load intensity w, a clamped slab would have half the factored moment of simply supported slab and, therefore, would be economic.


12.31.5 Two-way Slabs of Yield Pattern 2 of Figure 12.31.4b


We now consider the two-way slab having the yield pattern 2 as shown in Fig. 12.31.4b and subjected to uniformly distributed collapse load of intensity w kN/m2. Yield pattern 2 divides the slab into four segments marked by 1, 2, 3 and 4 in Fig. 12.31.6a, where the positive and negative yield lines are shown. The slab undergoes a displacement of Δ along the yield line EF. The free body diagrams of segments 1 and 2 are shown in Figs. 12.31.6b and c, respectively. The positive and negative moments along x and y directions are designated by Mpx, Mpy, Mnx and Mny in these two figures. We are employing both (A) method of segmental equilibrium and (B) method of virtual work to determine the distance x and the magnitude of the collapse load w of the slab. (A) Method of segmental equilibrium At the equilibrium, the moment of all the forces and moments of segment 1 about the left edge AB is zero. This gives (Fig. 12.31.6b): (Ly/2) x w (x/3) – (Mpx + Mnx) = 0 or w = 6 (Mpx + Mnx) / x2 (12.33) Similarly, at the equilibrium, the moment of all forces and moments of segment 2 about the bottom edge BC is zero. This gives (Fig. 12.31.6c): (1/2) x (Ly/2) (Ly/6) 2w + w(Lx – 2x) (Ly/2) (Ly/4) – (Mpy + Mny)Lx = 0 or, w = 24 (Mpy + Mny) Lx /{2x Ly

2 + 3Ly2 (Ly –2x)}

(12.34) Equating the two expressions of w from Eqs. 12.33 and 12.34, we have

)2(32

)(24)(6222 xLxLxL

LMMx

MM

yy

xnypynxpx

−+

+=

+

or, 4 (Mpy + Mny) Lxx2 + 4 (Mpx + Mnx) Ly

2 x – 3 (Mpx + Mnx) Lx Ly2 = 0

(12.35) Equation 12.35 is used to determine the values of x when Mpx, Mpy, Mnx and Mny are known. Thereafter, Eq. 12.34 is used to determine the magnitude of the collapse load w. (B) Method of virtual work The total external work done by the load in causing deflection of the four segments of Fig. 12.31.6a, TEW is:


TEW = 2W1 + 2 (W21 + W22 + W23), where W1 is the work done for the segment 1 and the work done of segment 2 is subdivided into three parts as W21, W22 and W23. Noting that W21 = W23, we get the TEW as, TEW = 2w{(1/2)x Ly (Δ/3)} + 2w {(1/2)x (Ly/2) (Δ/3) + 2w {(Lx-2x) (Ly/2) (Δ/2)} = (wΔ/6) (3 Lx Ly – 2x Ly) (12.36) The total internal work done by the yield moments (TIW) is, TIW = 2 (Mpx + Mnx) θx Ly + 2 (Mpy + Mny) θy Lx

= 2 (Mpx + Mnx) (Δ/x) Ly + 2 (Mpy + Mny) (2Δ/Ly) Lx (12.37) Equating the two works from Eqs. 12.36 and 12.37, we have, (w Δ/6) (3Lx Ly - 2x Ly) = 2 (Mpx + Mnx) (Δ/x) + 2 (Mpy + Mny) (2Δ/Ly) Lx

or )23(

)( 24)(1222

2

xxLL

xLMMLMMw

xy

xnypyynxpx

−

+++=

(12.38) To get the minimum collapse load, we put dw/dx = 0, which gives Ly

2 (3x Lx – 2x2) {24 Lx (Mpy + Mny)} – 12 [(Mpx + Mnx) Ly2 + 24 x Lx (Mpy + Mny)]

(3Lx – 4x) Ly2 = 0

or 4 (Mpy + Mny) Lx x2 + 4 (Mpx + Mnx) Ly

2 x – 3 (Mpx + Mnx) LxLy2 = 0

(12.35) The above equation is the same as obtained by the method of segmental equilibrium to determine the values of x. Thereafter, Eq. 12.38 is used to get the value of w, the collapse load. We observe two points from secs. 12.31.4 and 12.31.5. They are as follows: (i) Only one equation (Eq. 12.26) is needed for determining the value of the collapse load w for the yield pattern 1 of Fig. 12.31.5a. On the other hand, two equations are need for the yield pattern 2. This is because the yield pattern 1 is already determined but the yield pattern 2 is determined only after finding the values of x. Therefore, two equations are needed for determining the two unknowns x and w in the case of yield pattern 2.


(ii) The one equation needed for the yield pattern 1 is the same equation (Eq. 12.26) by the two methods, viz. method of segmental equilibrium and method of virtual work. Out of the two equations needed for the yield pattern 2, only one equation (Eq. 12.35) is the same by both the methods. After getting the values of x from Eq. 12.35, the values of collapse load w is determined from Eq. 12.33 by the method of segmental equilibrium and Eq. 12.38 by the method of virtual work. Let us now take up Eq. 12.35, which is used to determine the values of x in the case of yield pattern 2. Substituting the value of x = Lx/2 in Eq. 12.35, we get 4 (Mpy + Mny) (Lx

3 )(1/4) + 4 (Mpx + Mnx) Ly2 (Lx/2) – 3 (Mpx + Mnx) Lx Ly

2 = 0

or, 2

2

LyLx

MMMM

nypy

nxpx =+

+

(12.39) Noting that, x = L/2 gives the yield pattern 1 of Fig.12.31.5a, Eq.12.39 is used to check if the slab has the yield pattern 1 from the known values of Mpx, Mnx, Mpy, Mny, Lx and Ly of the slab. Similarly, it can be shown that for the yield pattern 2, i.e., when x < Lx/2, the required condition is:

2

⎟⎟⎠

⎞⎜⎜⎝

⎛<

+

+

y

x

nypy

nxpx

LL

MMMM

(12.40) We now take up numerical problems in the next section for the purpose of illustration. 12.31.6 Numerical Problems


Problem 1. Determine the yield pattern and collapse load of the trapezoidal slab

of Fig.12.31.7a having clamped edges along BA, AF and FE while the edge BE is free. Given that Mn = Mp = 70 kNm/m. Employ the method of segmental equilibrium.

Solution 1. Here, in this problem the yield pattern drawn in Fig.12.31.7a shows

that there is one additional unknown x to finalise the yield pattern. Thus, we have two unknowns x and the collapse load w.

The yield pattern divides the slab into three segments marked by 1, 2

and 3. Yield lines AC and FD meet the free edge BE. So, we have to


consider the nodal forces (+)V and (-)V, as shown on the left and right of C. Since the problem is symmetrical, segments 1 and 3 are identical. We first consider the equilibrium of segment 1 having positive moment Mp along yield line AC and negative moment Mn along yield line AB (Fig. 12.31.7b). The nodal force V at C of segment 1 is negative, i.e., acting downward.

Equation 12.23 gives the magnitude of the nodal force V = Mn cot α,

where α is the angle ACB. Geometric properties:

In triangle ABK, the side AB = (BK2 + AK2)1/2 = 6 m. The angle ABC = 600. Assuming the angle BAK = θ, we have sinθ = BK/BA = 0.5 giving θ = 300. The line CJ is perpendicular to AB. The angle JCB = 300 gives BJ = BC cos 60 = 0.5x. Therefore, AJ = 6 - 0.5x. The distance CJ = BC sin 60 =31/2 (x/2). From triangle ACK, we have cotα = CK/AK = (x-3)/3(3)1/2. Area of the triangle ABC = (1/2) (BC) (AK) = (x/2) 3(3)1/2. The load of segment 1 is acting at a distance of CJ/3 from the side AB, which is equal to 05x/(3)1/2. Taking moments of load of the segment 1, nodal force V and considering Mn and Mp about the edge AB, we have:

Mp (AJ) + Mn (AB) – V (CJ) – w (area of segment ABC) (CJ/3) = 0 Substituting the values of Mp, Mn, V, AJ, AB, CJ and the area of

triangle ABC, we have 70 (6 –0.5x) + 70 (6) – {70 (x – 3) / 3 (3)1/2} (3)1/2 (x/2) – (x/2) 3(3)1/2 w

(3)1/2 (x/2) / 3 = 0 or w = 4(2520 – 35x2) / 9x2 (i)

Now, we take up segment 2 for writing the equilibrium equation. The segment is subdivided into three parts marked by 21, 22 and 23. The sub-parts 21 and 23 are identical. The area of the triangle ACH = (1/2) (x - 3) 3 (3)1/2 and the load is considered at a distance of CH/3 = (3)1/2 m from the edge AF. The area of rectangle CDGH = (12 - 2x) 3(3)1/2 and the load is considered at a distance of (3/2) (3)1/2 from the edge AF. The equilibrium equation of segment 2 is obtained by taking moments of loads on this segment, two nodal forces of (+) V at C and D about the edge AF and considering Mp along AC and FD and Mn along AF (Fig.12.31.7c). This gives 2Mp (AH) + Mn(AF) + 2V (CH) – 2w(area of sub segment AHC) (HC/3) – w (area of rectangle CDGH) (HC/2) = 0. Substituting the values of Mp, Mn, V, AH, AF, CH and the areas of triangle AHC and rectangle CDGH, we have,


2 (70) (x – 3) + 70 (6) + 2{(70) (x-3) / 3 (3)1/2} 3(3)1/2 – 2 w (1/2) (x –3) (3) (3)1/2 (3)1/2 + w [{6 – 2(x – 3)} 3 (3)1/2 (3/2) (3)1/2 ] = 0 This gives w = (280 x – 420) / (135 – 18 x) (ii) Equating the two expressions of w from Eqs. (i) and (ii), we have: 4 (2520 – 35x2) / 9x2 = (280x – 420) / (135 – 18x) This gives x2 + 12x – 90 = 0 (iii) Solving we get x = 5.2249 m From Eq. (i) w = 4 (2520 – 35x2) / 9x2 = 25.465 kN/m2

From Eq. (ii) w = (280x – 420) / (135 – 18x) = 25.4695 kN/m2

So, we get the same value of w from Eqs. (i) and (ii). Thus, the yield pattern is finalised when x = 5.2249 m and the collapse load of the slab is 25.4695 kN/m2.

Problem 2. Determine the yield pattern and the collapse load of the two-way slab

of Fig.12.31.8 which is having clamped edges along four sides. Assume that Mpx = Mnx = 50 kNm/m and Mpy = Mny = 70 kNm/m.


Solution 2. In this problem, we have to find out whether yield pattern 1 or 2 will be the governing from Eqs.12.39 and 12.40 of sec.12.31.5.

From the given data, we have: (Mpx + Mnx) / (Mpy + Mny) = 100/140 = 0.714 and Lx

2 /Ly2 = 64/36 = 1.77.

Therefore, yield pattern 2 will govern (see Eq. 12.40).

Thus, we have two unknowns: (i) the value of x for finalising the yield pattern and (ii) the value of the collapse load w. Since, the governing equations of yield pattern 2 are derived in sec.12.31.5, the problem is solved by direct application of the equations. We have Eq.12.35 to determine the value of x. Using the values of Mpx, Mnx, Mpy, Mny, Lx and Ly in Eq. 12.35, we have:

4 (Mpy + Mny) Lx x2 + 4 (Mpx + Mnx) Ly2 x – 3 (Mpx + Mnx) Lx Ly

2 = 0 (12.35) or 4 (140) (8) x2 + 4 (100) (36) x – 3 (100) (8) (36) = 0 or 14 x2 + 45 x – 270 = 0, which gives x = 3.069 m

Now, we use the three equations (Eqs.12.33 and 12.34 by the method of segmental equilibrium and Eq. 12.38 by the method of virtual work). Using the values of Mpx, Mpy, Mnx, Mny, Lx, Ly and x in those three equations, we determine the value of w for comparing them.

(i) Eq.12.33 (i.e., w = 6 (Mpx + Mnx) x2 ) gives w = 63.69 kN/m2

(ii) Eq. 12.34 (i.e., )2(32

)(2422 xLLxL

LMMw

xyy

xnypy

−+

+= ) gives w = 63.69 kN/m2

(iii) Eq.12.38 (i.e., )23(

)(24)(1222

2

xxLL

xLMMLMMw

xy

xnypyynxpx

−

+++= ) gives

w = 63.69 kN/m2. Thus, we observe that the collapse load is the same from the three equations. Usually, they may differ marginally depending on the truncation of the value of x. 12.31.7 Practice Questions and Problems with Answers. Q.1: Establish the work done by the yield line moments Mb and Mt when the

yield line is at an angle with the two orthogonal directions of the reinforcement.


A.1: See sec. 12.31.2 Q.2: Explain the nodal force and derive the expression to determine its value. A.2: See sec. 12.31.3 Q.3: Draw the possible two yield patterns of a two-way slab clamped at four

sides. Derive the equation to find out which one will govern in a particular case given the values of Mpx, Mnx, Mpy, Mny, Lx and Ly.

A.3: Figures 12.31.4a and b are the two possible yield patterns. For finding the

governing yield pattern, the equations (Eqs. 12.39 and 12.40) are established in sec. 12.31.5.


Q.4: (a) Establish the general equations for determining the yield pattern and

uniformly distributed collapse load of an isosceles triangular slab, shown in Fig. 12.31.9 having the positive and negative moment capacity of Mp and Mn, respectively. Use the method of virtual work.


(b) Determine the specific yield pattern and uniformly distributed collapse load of such a slab when B = 6 m, 2L = 12 m, Mp = 9 kNm/m and Mn = 12 kNm/m, as shown in Fig.12.31.10. Use method of virtual work.

A.4: (a): The possible yield pattern is drawn in Fig.12.31.9 having unknown y.

Thus, we have two unknowns y and the collapse load w of the slab. The yield pattern divides the slab into three segments of which segments 2 and 3 are symmetrical.

Geometric properties. The angle DEJ = θ = tan-1 (L/B). The perpendicular distance from G to ED

is GH = EG sinθ = (B – y) {sin tan-1 (L/B)}. The length of side DE = (B2 + L2)1/2.

Let us assume the displacement of the slab at point G = Δ. The rotation of

the segment 1 = θ1 = Δ/y and the rotation of the segments 2 and 3 = θ2 = Δ/GH = Δ / [(B-y){sin tan-1 (L/B)}].

The total external work (TEW) done by the loads of three segments is

obtained considering the displacement of the centroid of all three segments as Δ/3.

Thus, TEW = (1/2) (2L) (B) w (Δ/3) = B L w Δ/3

(1) The internal work done by the negative yield lines DF, DE and EF is: Mn {(DF) (θ1) + 2(DE) (θ2)}

(2A) For the positive yield moment along DG, let us project DG along DE and

DF. The projected lengths are DH and DJ, respectively. Similarly, projecting the moments of positive yield lines of EG and FG along the sides of the triangle DEF, we have the total internal works done by the three positive yield lines = Mp {(DF) (θ1) + 2(DE)(θ2)} (2B)

Therefore, the total internal work done (TEW) is obtained adding the two

expressions of 2A and 2B as: TIW = (Mp + Mn) {(DF) (θ1) + 2 (DE) (θ2)}

(2) Equating TEW and TIW from Eqs. 1 and 2, we have


(B)(L) w Δ/3 = (Mp + Mn) {(DF) (θ1) + 2 (DE) (θ2)} (3)

Substituting the values of DF, DE, θ1 and θ2, we have from Eq. 3 (B)(L) w Δ/3 = (Mp + Mn) [(2L) (Δ/y) + 2 (B2 + L2)1/2 Δ / [(B-y) {sin tan-1

(L/B)}]]

or, ⎥⎦

⎤⎢⎣

⎡−

++

+= − )}/(tan){sin(

)(22))((

)(31

2/122

BLyBLB

yL

LBMM

w np

(4) Equation 4 is the only one equation from the method of virtual work to

determine y and w. Therefore, we differentiate w with respect to y to get the lowest value of the load and determine y. Thereafter, Eq. 4 shall be used to find the value of w.

:gives 0=dydw

)}/( tan){sin()(22

1-

2/122

2 BLyBLB

yL

−+−

=−

or [L{sin tan-1 (L/B)} – (B2 + L2)1/2] y2 – [2 (L) (B) {sin tan-1 (L/B)}] y + LB2

{sin tan-1 (L/B)} = 0 (5)

Thus, Eqs. 4 and 5 are the general equations to determine y and w of the

slab. A 4. (b): For the specific case when 2L = 12 m, B = 6 m, Mp = 9 kNm/m and Mn

= 12 kNm/m, we have angle DEJ = θ = 450. The distance GH = EG sinθ = (6-y) / (2)1/2. The length of side DE = EF = (36 + 36)1/2 = 6(2)1/2 m. The rotations θ1 = Δ/y and θ2 = Δ/[(B-y) {sin tan-1 (L/B)}] = Δ(2)1/2 / (6-y).

Equating the TEW and TIW from Eqs. 1 and 2 of A.4a, we have:

12 2412 6

p n( M M )w

y y+ ⎧ ⎫

= +⎨ ⎬−⎩ ⎭

(6)

:gives 0=dydw y2 + 12y – 36 = 0

(7)


Solution of Eq. 7 gives y = 2.485 m. Using the value of y in Eq. 6, we get w = 20.399 kN/m2.

12.31.8 References

















12.31.9 Test 31 with Solutions


Maximum Marks = 50 Maximum Time = 30 minutes

Answer all questions.


TQ.1: Determine the yield pattern and the uniformly distributed collapse load w

kN/m2 of the triangular slab shown in Fig. 12.31.11a having simple supports along AB and AC and the edge BC is free. The reinforcing bars


along x and y directions have the moment capacities Mx = 50 kNm/m and My = 60 kNm/m, respectively. Use both the methods i.e., (i) method of segmental equilibrium and (ii) method of virtual work.

[25×2 = 50] A.TQ.1: (i) Method of segmental equilibrium. The yield pattern of the slab is

drawn in Fig.12.31.11a involving x as unknown. Thus, we have two unknowns x and w here.

The yield pattern divides the slab into two segments 1 and 2, whose free

body diagrams are shown in Figs. 12.31.11b and c, respectively. The nodal force V has the magnitude (90 – 15x)/ 2 obtained from Eq.12.23 ( V = My cotα).

Taking moment of all forces and moments of segment 1 about AB and

equating it to zero gives

(1/2) x (8) w (DE)/3 – V (DE) – 8 Mx sin β- My (6-x) cosβ = 0, which gives w = (-18 x2 + 1608) / (3.2) x2 (1)

Similarly, taking moment of all forces and moments about AC of segment 2 and equating it to zero gives: (1/2) (6-x) w (8) (6-x) / 3 + V (CD) – 8 Mx = 0, which gives,

w = (-22.5 x2 + 270 x + 390) / (144 + 4x2 – 48x) (2) Equating Eqs. 1 and 2

xx

xxx

x484144

3902705.222.3

1608182

2

2

2

−+++−

=+−

or, 27 x2 – 804 x + 2412 = 0 (3) which gives x = 3.3847 m.

From Eq. 1: w = (1608 – 18x2) / 3.2 x2 = 38.2369 kN/m2 and From Eq.2: w = (-22.5 x2+ 270 x + 390) / (144 + 4 x2 – 48 x) = 38.2369 kN/m2.

(ii) Method of virtual work


Total external work TEW = {w (Δ) / 3 (2)} {8x + (6 – x) 8} = 8 wΔ (4)

Total internal work TIW = 8Mx (θ1) sin β + (6-x) My θ1 cos β + (6-x) My θ2 cos 90 + 8 Mx θ2

= 8 (50) (Δ / DE) (0.8) + (6 – x) (60) (Δ/DE) (0.6) + 0 + 8 (50) Δ/(6-x)}

= ⎥⎦

⎤⎢⎣

⎡−

+−

+)x(x

)x(x

Δ6400645400

(5) Equating Eqs. 4 and 5

400 45 6 4008 Δ Δ6

( x )wx x x

−⎡ ⎤= + +⎢ ⎥−⎣ ⎦

or, ⎥⎦⎤

⎢⎣⎡

−+−=

xxw

640045670

81

(6)

Equation 6 is the only equation to determine x and w. Differentiating w with respect to x and equating it to zero will give the lowest value of w. Thus, dw/dx = 0 gives:

(-) 670/(x2) + 400 / (6 –x)2 = 0 or, 27 x2 – 804 x + 2412 = 0 (7)

Equation 7 is the same as Eq. 3, obtained by the method of segmental equilibrium. The solution of Eq.7 is the same as that of Eq.3 and so, we get x = 3.3847 m.

Using the value of x in Eq.6, we get

2369.38640045670

81

=⎥⎦⎤

⎢⎣⎡

−+−=

xxw kN/m2

Thus, we get the same values of x and w by both methods.



This lesson presents the derivations of the expressions for determining the work done by bending and twisting moments when yield lines are at angles with the two orthogonal directions of the reinforcing bars. The need for the nodal forces and their determination are explained when one yield line meets another yield line or the free edge. Different possible yield patterns of two-way slabs are explained. Numerical problems are solved for the purpose of illustration taking examples with or without nodal forces employing both (i) method of segmental equilibrium and (ii) method of virtual work. Illustrative examples, practice problems and problem of test will give a clear understanding of analysing the slabs by the two methods.


Module 12



Lesson 32

Two-way Rectangular, Square, Triangular and

Circular Slabs Version 2 CE IIT, Kharagpur


At the end of this lesson, the student should be able to: • analyse rectangular slabs simply supported at three edges and free at the

other edge considering the two possible yield patterns, employing (i) the method of segmental equilibrium and (ii) the method of virtual work,

• analyse square slab with forking yield patterns when the corners are having inadequate reinforcement,

• predict yield lines of fan pattern for slabs in case this may be a possibility, • analyse the fan pattern of yield lines to determine the collapse loads of

triangular and circular slabs with different support conditions, • analyse the fan pattern of yield lines to determine the collapse loads of

circular slabs clamped along the circumference and having a column support at the centre.

12.32.1 Introduction Rectangular / square slabs may have different yield patterns depending on the support conditions and type of loads. Simply supported slabs at three edges and free at the other edge may have two types of yield patterns depending on the ratio of moment resisting capacities and the aspect ratio. This lesson first takes up such slabs to determine the condition for selecting a particular one out of the two possible yield patterns. The case of a square slab having forking yield pattern is explained when the corner reinforcement is inadequate. Several cases of triangular and circular slabs with or without a central column support are taken up to explain the yield lines of fan pattern. All the expressions are derived by employing the method of either segmental equilibrium or virtual work. In some cases both the methods are taken up to compare the values. 12.32.2 Rectangular Slabs Simply Supported at Three Edges and Free at the Other Edge Considering Yield Pattern 1


Figure 12.31.4c of Lesson 31 shows the yield pattern 1 of such slabs involving one additional unknown y. So, there are two unknowns – y and w to be determined. The yield pattern divides the slab into three segments marked by 1, 2 and 3 (Fig.12.32.1a). The slab carrying uniformly distributed load of w kN/m2, undergoes deflection of Δ at point E. The free body diagrams of segments 1 and 2 are shown in Figs.12.32.1b and c, respectively. Due to symmetry, segments 1 and 3 are identical. Segment 1 is subdivided into two parts as 11 and 12. We employ the method of segmental equilibrium first. (A) Method of segmental equilibrium Here, the nodal forces are zero since the moment capacities of three intersecting yield lines are identical. The equilibrium equation of segment 1 is developed taking the moments of loads and moments of segment 1 about AB,


and equating the same to zero. Thus, we get: (Lx/2) (Ly – y) w (Lx / 4) + (Lx /4)y w (Lx / 6) – Mx Ly = 0, or w = (24 Mx Ly) / {3Lx2 (Ly – y) + Lx

2 y} (12.41) Similarly, the equilibrium equation of segment 2 is developed by taking moment of loads and moments of segment 2 about the base BC and equating the same to zero, which gives: (Lx / 2) y w (y/3) – My Lx = 0 or w = 6 My / y2 (12.42) Equating the expression of w from Eqs.12.41 and 12.42, we get: 24 Mx Ly / {3Lx

2 (Ly – y) + Lx2 y} = 6My / y2

or 4 Mx Ly y2 + 2My Lx

2 y – 3 My Lx2 Ly = 0

(12.43) The solution of Eq. 12.43 is

y = [-2 My Lx2 + {4 My

2 Lx4 + 48 Mx My Lx

2 Ly2}1/2] / 8Mx Ly

(12.44) After getting the value of y from Eq. 12.44, the value of the collapse load w is obtained from either Eq. 12.41 or Eq. 12.42. The condition that y < Ly gives: (- 2)(My/Mx) Lx

2 + {(4) (My/Mx)2 Lx4 + 48(My / Mx) Lx

2 Ly2} < 8Ly

2

or My / Mx < 4 (Ly/Lx)2 (12.45) (B) The method of virtual work Referring to Figs.12.32.1a, b and c, total external work done by the loads of three segments TEW is as follows: TEW = 2w [(Lx/2) (Ly-y) (Δ/2) + (Lx/2)y (Δ/3)] or TEW = w Lx (3Ly – y) (Δ/6) (12.46) Total internal work done by the yield moments TIW is: TIW = 2 Mx (Ly) θx + My (y) θy.

Using θx = 2Δ / Lx and θy = Δ/y, we have


TIW = Δ {4Mx (Ly/Lx) + My (Lx/y)} (12.47)

From the two equations of TEW and TIW (Eqs. 12.46 and 12.47), we have w = (24 Mx Ly y + (My Lx

2) / {Lx2 (3y Ly – y2)}

(12.48) This is the only equation in the method of virtual work to determine y and the collapse load w. Differentiating w with respect to y and equating that to zero to get the lowest w, we have: Lx

2 (3y Ly – y2) (24 Mx Ly) – (24 Mx Ly y + 6My Lx2) Lx

2 (3Ly – 2y) = 0

or 4 Mx Ly y2 + 2 My Lx

2 y – 3My Lx2 Ly = 0

(12.43) Thus, the method of virtual work gives the same equation (Eq. 12.43) as that obtained by the method of segmental equilibrium. After getting the value of y from Eq. 12.44, the value of the collapse load w is obtained from Eq. 12.48. 12.32.3 Rectangular Slabs Simply Supported at Three Edges and Free at the Other Edge Considering Yield Pattern 2


Figure 12.31.4d of Lesson 31 shows that yield pattern 2 of such slabs

involving one additional unknown x. So, there are two unknowns – x and w to determine. The yield pattern divides the slab into three segments marked by 1, 2 and 3. The slab, carrying the uniformly distributed load of w kN/m2, undergoes deflection of Δ along EF (Fig.12.32.2a). Free body diagrams of segments 1 and 2 are shown in Figs.12.32.2b and c, respectively. Due to symmetry, segments 1 and 3 are identical. Segment 2 is further subdivided into three parts 21, 22 and 23, of which sub-segments 21 and 23 are also symmetrical. We consider the method of segmental equilibrium first.

(A) Method of segmental equilibrium The nodal forces on the left and right of point E of the yield line BE are (+) V, acting downward and (-) V, acting upward, respectively. Similarly, nodal forces on the left and right of point F of yield line CF are (-) V, acting upward and (+) V, acting downward, respectively. The magnitude of the nodal force, as given in Eq.12.23 of Lesson 31, is V = My cot α = My (AE/AB) = My (x/Ly) (12.49)


We now develop the equilibrium equation of segment 1 taking moments of loads and moments of segment 1 about AB and equating the same to zero. w (x/2) Ly (x/3) + Vx – Mx Ly = 0, which gives, w = 6 (MxLy

2 – My x2) / (x2 Ly2)

(12.50) Similarly, the equilibrium equation of segment 2 is developed by taking moment of loads and forces of segment 2 about BC and equating the same to zero. 2 w (x/2) (Ly) (Ly/3) + w (Lx – 2x) Ly (Ly/2) – 2V Ly – 2My x = 0 , which gives, w = (24 My x) / { Ly

2 (3Lx – 4x)} (12.51) Equating the two expressions of w from Eqs.12.50 and 12.51, we have,

)xL(L

xM

Lx

)xMLM(6

xy

y

y

yyx

43

24222

22

−=

−

or 3(My / Mx) Lx x2 + 4 Ly

2 x – 3Ly2 Lx = 0

(12.52) The solution of the above equations is:

x = [- 4Ly2 + {16 Ly

4 + 36 (My / Mx) Lx2 Ly

2}1/2] / {6 Lx (My / Mx)} (12.53) From Fig. 12.32.2a, it is evident that x < Lx / 2. So, we get from Eq. 12.53 for the condition that x < Lx/2,

- 4 Ly2 + {16 Ly

4 + 36 (My / Mx) Lx2 Ly

2}1/2 ⟨ (Lx / 2) (6 Lx) (My /Mx), which finally gives

(My / Mx) > (4/3) (Ly / Lx)2 (12.54) Therefore, Eq. 12.54 shall be used to confirm if yield pattern 2 is possible or not. After getting the value of x from Eq. 12.53 (the solution of Eq. 12.52), the collapse load w is determined either from Eq. 12.50 or from Eq. 12.51.


(B) Method of virtual work Referring to Figs. 12.32.2a, b and c, the total external work done by the loads of three segments TEW is as follows: TEW = 2 (W1) + 2 (W21) + W22 = 2{(x/2) Ly w (Δ/3)} + 2 {(x/2) Ly w (Δ/3)}+ (Lx – 2x) Ly w (Δ/2) or TEW = w Ly (Δ/6) (3 Lx – 2x) (12.55) The total internal work done by the yield moments TIW is: TIW = 2 Mx Ly θx + 2 My x θy. Using θx = Δ/x and θy = Δ/Ly, we have: TIW = 2Mx Ly (Δ/x) + 2 Myx (Δ/Ly), or TIW = Δ{2Mx Ly/x + 2 My x/Ly) (12.56) From the two expressions of TEW and TIW (Eqs. 12.55 and 12.56), we have: w (Ly/6) (3Lx – 2x) = 2Mx (Ly/x) + 2My (x/Ly), or w = {12 Mx Ly

2 + 12 My x2} / {Ly2 (3x Lx – 2x2)

(12.57)

The above is the only equation to determine x and the collapse load w in the method of virtual work. Differentiating w with respect to x and setting that to zero shall give the lowest load. Hence, we have: Ly

2 (3 x Lx – 2x2) (24 My x) – (12 Mx Ly

2 + 12 My x2) Ly2 (3Lx – 4x) = 0,

or 3 (My / Mx) Lx x2 + 4 Ly x2 – 3Ly

2 Lx = 0 (12.52)

Thus, the method of virtual work gives the same equation (Eq.12.52) as that obtained by the method of segmental equilibrium. After getting the value of y from Eq.12.52 (or Eq.12.53, the solution of Eq.12.52), the collapse load w is determined from Eq. 12.57.

12.32.4 Special Cases for Predicting Yield Patterns

Let us examine the two conditions of Eqs.12.45 and 12.54 for determining the correct yield pattern.

Equation 12.45 specifies that for the yield pattern 1, (My / Mx) < 4 (Ly /Lx)2,

while Eq. 12.54 specifies that for the yield pattern 2, My/Mx > 4/3 (Ly/Lx)2. However, for slabs when both the conditions are satisfied, it appears that both the yield patterns are possible. In such cases, both the yield patterns should be considered and the one giving the lowest load (either 1 or 2) is the correct yield pattern. We explain the above taking two specific cases of square slabs having


simply supported edges on three sides and free at the other side, as shown in Fig.12.32.1a, with two different ratios of My / Mx. (i) Case 1: Ly / Lx = 1 and My / Mx = 1.5 Here, (My/Mx) is less than 4 (Ly /Lx)2 (= 4). So, yield pattern 1 is possible. Similarly, (My/Mx) is greater than (4/3) (Ly/Lx)2 (= 4/3). So, yield pattern 2 is also possible. Therefore, we should consider both the yield patterns and select the one giving the lowest value of the collapse load. For the yield pattern 1, the value of the collapse load is obtained from any of the three equations (Eqs.12.41, 12.42 and 12.48), after determining the value of y from Eq.12.43. The results are as follows.

4 Mx Ly y2 + 2My Lx2 y – 3 My Lx

2 Ly = 0 (12.43) Using My = 1.5 Mx and Ly = Lx, we have 4y2 + 3Lx y – 4.5 Lx

2 = 0 (12.58) from which y = 0.75 Lx (12.59) Using the value of y in the three equations, we get the value of w as:

From Eq.12.41: 22 2

2416

3x y

x xx y x

M Lw (

L ( L y ) ( L y )= =

− +M / L )

(12.60) From Eq. 12.42: w = 6My / y2 = 16 (Mx / Lx

2) (12.60)

From Eq. 12.48: )L/M()yyL(L

LMyLMw xx

yx

xyyx 222

216

3

624=

−

+=

(12.60) We now consider yield pattern 2 in which the unknown distance x is obtained from Eq. 12.52, which is 3 (My / Mx) Lx x2 + 4 Ly2 x – 3Ly2 Lx = 0 (12.52)


When My = 1.5 Mx and Ly = Lx, the above equation becomes 4.5 x2 + 4 Lx x – 3 Lx

2 = 0 (12.61) which gives x = 0.4852 Lx (12.62) Using the value of x in Eqs. 12.50, 12.51 and 12.57, we get the value of w as given below.

From Eq. 12.50: 2 2

22 2

616 488x y y

x xy

( M L M x )w .

x L−

= = ( M / L )

(12.63)

From Eq. 12.51: )L/M(.)xL(L

x)M(w xx

xy

y 22 48816

43

24=

−=

(12.63)

From Eq. 12.57: )L/M(.)xxL(L

xMLMw xx

xy

yyx 222

2248816

23

1212=

−

+=

(12.63) Computing the values of w from Eqs. 12.60 and 12.63, it is evident that yield pattern 1 is the correct yield pattern. (ii) Case 2: Ly / Lx =1 and My/Mx = 3.5 Here also both the conditions of Eqs. 12.45 and 12.54 are satisfied. So, we consider them separately. Proceeding in the same manner, we determine the values of y and w for yield pattern 1, and x and w for yield pattern 2. Only the final results are given below. (a) For the yield pattern 1, y = 0.966 Lx (12.64) and w = 22.487 (Mx /Lx

2) (12.65) (b) For the yield pattern 2, x = 0.377 Lx (12.66)


and w = 21.222 (Mx / Lx2)

(12.67) Thus, in this case the yield pattern 2 is the correct yield pattern. 12.32.5 Square Slabs with Forking Yield Pattern


Yield lines of two-way slabs having inadequate corner reinforcement fork out before they reach the corners, as shown in Fig.12.31.4e of Lesson 31. We analyse a square slab having simply supported edges in all four sides subjected to uniformly distributed load w kN/m2 and having yield pattern as shown in Fig.12.32.3a. Yield lines divide the slab into symmetrical segments, of which we consider segments 1 and 2 employing the method of segmental equilibrium.

The yield lines have two unknown parameters x and r (Fig.12.32.3a),

where x is the distance AE and r is the perpendicular distance GV from point G to the yield line EF. In segment 1, EF is the negative yield line, and EG and FG are positive yield lines. Figure 12.32.3b shows the free body diagram of segment 1. The negative and positive moments are represented by Mnb and Mpb, respectively. These moments are resolved into their respective components Mnx, Mny, Mpx and Mpy along the sides of the slab, as shown in the figure. Segment 2 is shown in Fig.12.32.3c which is bounded by yield lines FG, GO, OH and HI, and the side FI. All the yield lines of segment 2 are having positive moment Mpb which are resolved as Mpx and Mpy along the sides of the square slab. Segment 2 is further subdivided into four sub-segments marked by 21, 22, 23 and 24 of which sub-segments 21 and 23 are symmetrical.


Before we take up the equilibrium of the two segments, let us determine the dimensions of different lengths of the two segments. With reference to Figs.12.32.3a, b and c,

AE = AF = BI = BU = x (assumed) GV = r (assumed) AK = KG = GJ = AJ = {(x /√2) + r}/ √2 = r/√2 + x/ 2 FK = EJ = AJ – AE = r/√2 – x/2 We further assume that Mnx = Mny = Mn and Mpx = Mpy = Mp.

The area of triangle EFG = (1/2) (EF) (FG) = x r /√2. Taking moments of the loads and moments of segment 1 about EF and equating the same to zero, we get: w (area EFG) (r /3) + (- Mnx AF – Mny AE + Mpy EJ – Mpx JG – Mpy GK + Mpx KF) cos 450 = 0 or r = {6(Mn + Mp) / w}½ (12.68) We now consider the equilibrium of segment 2 taking moment of loads and moment about FI. For easy understanding, the contributions of different sub-segments are computed separately and then added with the contribution of the moments. The contributions of different sub-segments and intermediate addition are:

(i) For the sub-segment 21:

)3/(222

)2/1()3/KG()KG)(FK)(2/1(2

wx2

r xrw ⎟⎠

⎞⎜⎝

⎛+⎟

⎠

⎞⎜⎝

⎛−=

(ii) For the sub-segment 22:

( )2/222

2L)2/()KG)(KS(2

2 wx2

r xrw ⎟⎠

⎞⎜⎝

⎛+

⎭⎬⎫

⎩⎨⎧

⎟⎠

⎞⎜⎝

⎛+−=

(iii) For the sub-segment 24: (1/2) (GH) {(L/2) – KG} {KG + {(L/2 – KG)/3} w =

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

⎟⎠

⎞⎜⎝

⎛+−+⎟

⎠

⎞⎜⎝

⎛+

⎭⎬⎫

⎩⎨⎧

⎟⎠

⎞⎜⎝

⎛+−

⎭⎬⎫

⎩⎨⎧

⎟⎠

⎞⎜⎝

⎛+− 3/

22222r

222

222

2xrLxxrLxrLw

=

⎥⎦

⎤⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

−−+⎟⎠

⎞⎜⎝

⎛+

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠

⎞⎜⎝

⎛++⎟

⎠

⎞⎜⎝

⎛+−⎟

⎠

⎞⎜⎝

⎛+− 3/

22222r

222

222222

22 xrLxxrxrLxrLLw


= ⎥⎦

⎤⎢⎣

⎡−−++

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

623622r

22224

22 xrLxxrwxrwLwL

=

⎭⎬⎫

⎩⎨⎧

++⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

6332

22224

22 LxrxrwxrwLwL

=

⎭⎬⎫

⎩⎨⎧

++⎟⎟⎠

⎞⎜⎜⎝

⎛++

⎭⎬⎫

⎩⎨⎧

++⎭⎬⎫

⎩⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

6332

226332

224

22 LxrxrwLxrxrwLwL

=

⎭⎬⎫

⎩⎨⎧

++⎟⎟⎠

⎞⎜⎜⎝

⎛++

⎭⎬⎫

⎩⎨⎧

++⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+

63322633224332

4

22 Lxr2xrwLxr2 x2

rwL- wLxrwL 3

(iv) The addition of {2 (sub-segment 21) + (sub-segment 22)} = ⎟

⎠⎞

⎜⎝⎛⎟

⎠

⎞⎜⎝

⎛+

⎭⎬⎫

⎩⎨⎧

⎟⎠

⎞⎜⎝

⎛+−+⎟

⎠⎞

⎜⎝⎛⎟

⎠

⎞⎜⎝

⎛+⎟

⎠

⎞⎜⎝

⎛−

222222

32222

22wxrxrLwxrxr

= ⎟

⎠

⎞⎜⎝

⎛−−+−⎟

⎠

⎞⎜⎝

⎛+

22262322

2xrLxrxrw

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−⎟

⎠

⎞⎜⎝

⎛+

2L

3x2

3r2

2x

2rw

2

(v) Therefore, {2 (sub-segment 21) + (sub-segment 22) + (sub-segment 24)}

⎭⎬⎫

⎩⎨⎧

++⎟⎠

⎞⎜⎝

⎛ +−+⎟⎟⎠

⎞⎜⎜⎝

⎛+

+⎟⎟⎠

⎞⎜⎜⎝

⎛++++−−⎟

⎠

⎞⎜⎝

⎛ +=

6332

2224332

4

6332

232

32

2232

2

LxrxrwLwLxrwL

LxrLxrxrw

⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟

⎠

⎞⎜⎝

⎛ ++++⎟⎠⎞

⎜⎝⎛ +−⎟

⎠

⎞⎜⎝

⎛ +=623

2222412263

2322

3222 Lxr xrwL- wLxwLrwLLxxrw

241226

6332

3232

22w

322322

22

wLxwLrwL

LxLrLxLrL xrxxrw

+++

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−+⎟

⎠

⎞⎜⎝

⎛ ++⎟⎠⎞

⎜⎝⎛ −⎟

⎠

⎞⎜⎝

⎛ +=

2423126

3232 wLwrxwxxwr+−−−=

(vi) Including the contributions of moment, the equilibrium equation of segment 2 becomes


0)2(2423126

3232

=−−+−−− xLMwLwrxwxxrw p

(12.69) Using the expression of r from Eq.12.68 in the above equation, we have: - w 6(Mn + Mp) x / 6w – wx3/12 – (w x2/3√2) {6(Mn + Mp)/w}1/2 + (w L3/24) – Mp (L-2x) = 0 or w x3/ 12 + x2{w (Mn + Mp) / 3}1/2 + (Mn – Mp)x + MpL – wL3/24 = 0 (12.70) Setting dw/dx = 0, Eq.12.70 finally gives: wx2/4 + 2x{ (Mn + Mp) w/3}1/2 + (Mn-Mp) = 0 (12.71) which has the solution of x as

x = 2{(Mn + 7Mp) / 3 w}1/2 – 4 {(Mn + Mp) / 3w}1/2 (12.72) Thus, the three unknowns x, r and w are determined from the three equations, Eqs. 12.68, 12.70 and 12.72. We now consider two cases below: (i) with adequate corner reinforcement and (ii) with inadequate corner reinforcement. Case (i): When the corner reinforcement is adequate The value of x = 0 when the corner reinforcement is adequate as there will be no forking of the yield line. Putting x = 0 in Eq. 12.70, we have: MpL – wL3 / 24 = 0 or w = 24 Mp / L2 (12.73) Using x = 0 in Eq. 12.72, we get: Mn + 7Mp = 4 (Mn + Mp), which gives: Mn = Mp (12.74) Using w = 24 Mp /L2 and Mn = Mp from Eqs 12.73 and 12.74 in Eq. 12.68, we have: r = L / √2 (12.75)


Case (ii): When the corner reinforcement is inadequate In this case, Mn = 0 and only Mp is present. Putting Mn = 0 in Eq. 12.71, we have: x = 2(7Mp / 3w)1/2 – 4 (Mp /3w)1/2 = 0.746 (Mp/w)1/2 (12.76) Using Mn = 0 and substituting the value of x from Eq.12.76 in Eq.12.70, we have: w x3/12 + x2 (w Mp/3)1/2 – Mpx + MpL – wL3 / 24 =0 or wL3 – 24 MpL + 9.3624 Mp (Mp/w)1/2 = 0 (12.77) The numerical solution of w of Eq.12.77 is: w = 22 M /L2 (12.78) The two values of w of the two cases (i) and (ii) reveal the reduction of w by about 8.33 per cent (= 24 – 22 / 0.24) when the corner reinforcement is such that Mn = 0. Therefore, it is justified to reduce the load carrying capacity by ten per cent rather than involving in the complicated analysis. Further, if the negative moment carrying capacity Mn is larger than Mp, the value of x becomes negative. In that case, the yield pattern is the same as that with adequate corner reinforcement. 12.32.6 Yield Lines of Fan Pattern

Triangular or circular slabs subjected to uniformly distributed loads or point loads may have yield lines of fan pattern. We explain below different cases of such slabs to estimate the collapse loads.


(A) Triangular slab subjected to a point load

Figure 12.32.4a shows a triangular slab clamped along the three edges and subjected to a point load P away from the edges and corners. Negative yield line of approximately circular pattern is formed with positive yield lines radiating outward from the point of application of the load. Assuming the resisting moment capacities of Mp and Mn per unit length for positive and negative moments, respectively, we consider one segment ODE, as shown in Fig.12.32.4b. The angle DOE is dθ and the resultant of the positive moments Mp along DO and OE, is Mp rdθ along DE, as shown in Fig.12.32.4c. The resultant Mp rdθ is in the same direction of the negative moment Mn rdθ along DE. The fractional part of the total load P acting on the segment is P dθ / 2π.


Taking moment of the load and moments of the segment DOE about DE and equating the same to zero, we have: (Mp + Mn) r dθ - P(dθ / 2π) r = 0, which gives: P = 2π (Mp + Mn) (12.79)

If we assume Mn = kMp, Eq.12.79 gives: P = 2π Mp (1 + k) (12.80) When Mn = Mp i.e., k =1, we have: P = 4π Mp (12.81)

Equations12.79 to 12.81 reveal that the collapse load P is independent of the radius of the fan patterns of yield lines. Thus, at the collapse load, the triangular slab clamped along three edges may have complete fan pattern of yield lines of any radius without any change of the collapse load. However, it is necessary that the boundaries must have the resisting moment capacities Mn at all points.

(B) Simply supported circular slabs subjected to uniformly distributed load w kN/m2


Figure 12.32.5a shows the positive yield lines of radiating type having the moment resistance capacity of Mp per unit length. Figure 12.32.5b shows the free body diagram of the segment DOE making an angle dθ at the center of the slab. Equating the moment of the load and moment about DE to zero, we have: MpR dθ = (1/2) (R dθ) R w (R/3), which gives: w = 6 Mp / R2 (12.82) (C) Clamped circular slab subjected to uniformly distributed load w kN/m2

Figure 12.32.6a shows the circular slab clamped at the periphery having

negative yield lines along the periphery of moment resisting capacity of Mn per unit length. Positive yield lines radiating from the centre of the slab are also shown in the figure having moment resisting capacity of Mp per unit length. Figure 12.32.6b shows the free body diagram of one segment DOE where dθ is


the angle made by the yield lines DO and OE at the centre. As explained earlier, the resultant of the two positive moments Mp of magnitude MpR dθ and the negative moment Mn are in the same direction as shown in the figure. Equating the moment of load and moments of segment DOE about DE to zero, we have: Mp r dθ + Mn r dθ - (1/2) (rdθ) r w (r/3) = 0. This gives: w = 6(Mn + Mp) / R2 (12.83) Assuming Mn = kMp, Eq.12.83 gives: w = 6 Mp (1 + k) / R2 (12.84) (D) Circular slabs clamped along the circumference and having a column

support at the centre


In many practical examples of circular slabs, a column support is provided at the centre. The yield pattern of a circular slab clamped along the periphery and having a central column support, subjected to uniformly distributed load of w kN/m2 is shown in Fig.12.32.7a. Negative yield lines of moment resisting capacity Mn are along the periphery and radiating from the centre of the slab. Positive yield line is shown in approximately circular pattern having a radius of r. Figure 12.32.7b shows the free body diagram of a segment OCD where dθ is the angle made by the yield lines OC and OD at the centre. Figure 12.32.7c shows the deflection profile of the segment OCD having a deflection of Δ along AB. The segment is further divided into four sub-segments 11, 12, 13 and 14, of which sub-segments 13 and 14 are symmetrical. The deflection of the centroid of the sub-segment 11 (OAB) is 2 Δ/3 at G and the deflection of the centroid of the sub-segments 13 and 14 is Δ/3. The deflection of the centroid of the sub-segment 12 is Δ/2. We employ the method of virtual work here. The external work done by the loads of the four sub-segments are presented in Table 12.1 giving the respective area of the sub-segment and the deflection at the centroid. By summing them considering that the work done by the load of segment ACE is the same as that of segment BDF, the total external work done TEW is as follows: TEW = (w Δdθ) {r2 / 3 + r (R –r) / 2 + (R – r)2 / 6} (12.85) The internal works done by different positive and negative yield lines of the segment OCD are furnished in Table 12.2 giving the respective moment and rotation. The total internal work (TIW) is then obtained by summing them considering that the work done by the moment of yield line AC is the same as that of yield line BD. Thus, the expression of TIW is, TIW = Δ{Mn (AB) / r + 2 Mn (CE) / (R – r) + Mn (CD) / (R – r) + Mp (AB) / r + Mp (AB) / (R – r)} (12.86) Table 12.1: Total External Work (TEW) by loads of segment OCD Sl. No.

Sub-Segment Area Deflection at the centroid

External work done

1. OAB (11) (r/2) r dθ 2 Δ/3 wΔ dθ (r2 / 3) 2. AEFB (12) r dθ (R – r) Δ/2 (wΔ dθ ) {r (R – r)/2} 3. ACE and BFD

(13 and 14) (R – r) (R – r) (dθ /2)

Δ/3 (wΔ dθ ) {(R – r)2 / 6}

Adding, we get: TEW = (wΔ dθ ) {(r2 / 3) + r (R – r) / 2 + (R – r)2 / 6} … Eq. (12.85)


Table 12.2: Total Internal Work (TIW) by positive and negative yield lines of segment OCD Sl. No.

Yield line Moment Rotation Internal work done

1. OA and OB

Mn (AB) Δ/r Mn (AB) (Δ/r)

2. AC and BD 2 Mn (CE)

Δ/(R – r) 2 Mn (CE) {Δ/(R – r)}

3. CD Mn (CD) Δ/(R – r) Mn (CD) {Δ/(R – r)} 4. AB Mp (AB) {Δ/r + Δ/(R – r)} Mp (AB) {Δ/r + Δ/(R – r)} Adding, we get: TIW = Δ{Mn (AB)/r + 2Mn (CE)/(R – r) + Mn(CD)/(R – r) + Mp(AB) /r + Mp (AB)/(R – r)} … Eq. (12.86) Equating the two works TEW and TIW form Eqs.12.85 and 12.86, and assuming that Mn = Mp = M, we get: w dθ { r2/3 + r(R-r) / 2 + (R – r)2/6} = M { 2(AB)/ r = (2 CE + CD + AB) / (R –r)} or w dθ { R (R + r) / 6} = M{2 (dθ) + 2R (dθ) / (R-r)} or w = {12M (2 R – r)} / {R (R2 – r2)} (12.87) Setting dw/dr = 0, we have the most critical layout of the mechanism and then r2 – 4 R r + R2 = 0 (12.88) which has a solution, r = 0.2679 R (12.89) Using the value of r from Eq. 12.89 in Eq. 12.87, we have w = 22.39 (M / R2) (12.90) or M = w R2 / 22.39 (12.91) 12.32.7 Practice Questions and Problems with Answers


Q.1: Determine the collapse load for the two possible yield patterns of a rectangular slab, simply supported at three edges and free at the other edge, when subjected to uniformly distributed loads employing (i) method of segmental equilibrium and (ii) method of virtual work.

A.1: secs. 12.32.2 and 3 Q.2: When do the yield lines of square slabs fork before reaching the corners?

What should be the approach to estimate the uniformly distributed collapse load in such cases?

A.2: sec. 12.32.5 Q.3: Determine the collapse point load in a triangular slab clamped along the

three edges. A.3: sec. 12.32.6 – part A Q.4: Determine the uniformly distributed collapse load of a circular slab clamped

along the periphery. A.4: sec. 12.32.6 – part C Q.5: Determine the uniformly distributed collapse load of a circular slab having

clamped support along the periphery and a column support at the centre of the slab.

A.5: sec.12.32.6 – part D 12.32.8 References


















12.32.9 Test 32 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Choose the correct answer for each of the following statements.

(4 × 5 = 20 Marks) (A) Rectangular slabs simply supported at three edges and free at the other

edge will have yield pattern 1 (Fig.12.32.1a),

(i) when (Mx/ My) < 4 (Ly / Lx) (ii) when (My / Mx) < 4 (Ly / Lx)2 (iii) when (Mx /My) < 4 (Ly/ Lx)2 (iv) when (My / Mx) > 4 (Ly/ Lx)2

A.TQ.1 (A): (ii). (B) Nodal forces are to be considered when

(i) the slab is subjected to vertical loads at the corners of the slab (ii) the slab is one-way (iii) the slab is polygonal having more than five corner node points


(iv) when the yield line meets a free edge A.TQ.1 (B): (iv) (C) Rectangular slabs simply supported at three edges and free at the other

edge will have yield pattern 2 (Fig.12.32.2a),

(i) when (My/ Mx) > (4/3) (Ly / Lx)2

(ii) when (Mx/ My) > (4/3) (Ly / Lx)2

(iii) when (Mx/ My) = (4/3) (Ly / Lx)2

(iv) when (My/ Mx) = (4/3) (Ly / Lx)2

A.TQ.1 (C): (i) (D) Yield lines of fan pattern will occur in

(i) any slab having a central cut out (ii) any slab subjected to unsymmetrical loading (iii) any slab having point load (iv) any slab having torsional moment

A.TQ.1 (D): (iii) TQ.2: Determine the uniformly distributed collapse load of a circular slab having

clamped support along the periphery and a column support at the centre of the slab.

(30 marks)

A.TQ.2: sec.12.32.6 – part D 12.32.10 Summary of this Lesson This lesson explains two different yield patterns of two-way slabs simply supported on three edges and free at the other edge. The expressions of determining the respective collapse loads are derived both by the method of segmental equilibrium and method of virtual work. The forking out type of yield pattern of square / rectangular slabs having inadequate corner reinforcement is explained. Slabs subjected to point loads or circular slabs having yield lines of fan pattern are taken up to determine the collapse loads. Practical circular slabs supported at the periphery and with a central column support is also taken up in this lesson.


Module 12



Lesson 33

Numerical Examples



At the end of this lesson, the student should be able to: • apply the theory of segmental equilibrium for the solution of different types of

slab problems, • apply the theory of virtual work for the solution of different types of slab

problems. 12.33.1 Introduction The theoretical formulations of the yield line analysis of slabs are explained in earlier Lessons 30 to 32, considering (i) the method of segmental equilibrium and (ii) the method of virtual work. Illustrative examples are solved in Lesson 30. This lesson includes different types of illustrative examples of slabs with different combinations of support conditions. Two types of loadings, viz., (i) point loads and (ii) uniformly distributed loads, are considered. In some cases, two or three possible yield patterns are examined to select the correct one and the corresponding loads are determined. The problems include different types of rectangular or square two-way slabs, triangular, quadrantal and circular slabs. Circular slabs also include one practical example where it is supported by a central column in addition to clamped support along the periphery. Some practice problems and test problem are also included. Understanding the numerical problems and solving practice and test problems will give a better understanding of the theories of yield line analysis and their applications. 12.33.2 Illustrative Examples


Problem 1. Determine the uniformly distributed collapse load w kN/m2 of a square slab (L m × L m), simply supported at three edges and the other edge is free. Assume My = 2Mx = 2M.

Solution 1. The slab is shown in Fig.12.33.1a. Given data are:

Lx = Ly = L and My = 2Mx = 2M. Step 1. To examine the possibility of yield patterns 1, 2 or both Equation12.45 of Lesson 32 stipulates that (My/Mx) ⟨ 4 (Ly /Lx)2 for

yield pattern 1 and Eq.12.54 of Lesson 32 gives the condition that (My/Mx) ⟩ (4/3) (Ly/Lx)2 for yield pattern 2.

Here, Ly/Lx = 1 and My/Mx = 2. So, (My/Mx) ⟨ 4 (Ly /Lx)2 and (My/Mx) ⟩

(4/3) (Ly /Lx)2. Thus, both the yield patterns are to be examined. Step 2. Value of y for yield pattern 1 The value of y is obtained from Eq.12.43 of Lesson 32, which is 4 Mx Ly y2 + 2 My Lx

2 y – 3 My Lx2 Ly = 0

(12.43) or 2y2 +2L y – 3L2 = 0

(1) which gives y = 0.823L

(2) Step 3. Determination of collapse load w kN/m2

Equations 12.41 and 12.42 of the method of segmental equilibrium

of Lesson 32 and Eq.12.48 of the method of virtual work of Lesson 32 shall be used to determine w. The results are given below.

(i) From Eq. 12.41: w = (24ML)/{3L2 (L– y) + L2y}= 17.722 (M /L2) (3) (ii) From Eq. 12.42: w = 12M/y2 = 17.722 (M/L2) (3) (iii) From Eq. 12.48: w = {12 M (2y + L)} / {L(3yL – y2)} = 17.722 (M/L2) (3)

Step 4. Value of x for yield pattern 2


The value of x is obtained from Eq. 12.52 of Lesson 32, which is 6x2 + 4Lx – 3L2 = 0 (4)

which gives: x = 0.448L (5)

Step 5. Determination of collapse load w kN/m2

Equations 12.50 and 12.51 of the method of segmental equilibrium

of Lesson 32 and Eq. 12.57 for the method of virtual work of Lesson 32 shall be used to determine w. The results are given below:

(i) From Eq. 12.50: w = {6M(L2 – 2x2)} / x2L2 = 17.841 (M /L2) (6) (ii) From Eq.12.51: w = 24 Mx / {L2 (3L-4x)}= 17.841 (M/L2) (6) (iii) From Eq.12.57: w = (M/L2) {12 (L2 + 2x2)}/ {x (3L – 2x)} = 17.841 (M/L2) (6) Step 6: Correct yield pattern and collapse load Comparison of results of steps 3 and 5 reveals that yield pattern 1 is

the correct one giving the lower value of w. For this yield pattern, y = 0.823 L and w = 17.722 (M/L2).

Problem 2: Determine the uniformly distributed collapse load w kN/m2 of a

rectangular slab whose Ly/Lx = 3/4, simply supported at three edges and free at the other edge. Assume My = 2Mx.

Solution 2: The slab is shown in Fig.12.33.1a. Given data are: Ly/Lx = 3/4 and

My/Mx = 2. Step 1: To examine the possibility of yield patterns 1, 2 or both Equation 12.45 of Lesson 32 stipulates that (My/Mx) ⟨ 4 (Ly/Ln)2 for

yield pattern 1 and Eq. 12.54 of Lesson 32 gives the condition that (My/Mx) ⟩ (4/3) (Ly/Lx)2 for yield pattern 2.

Here, Ly/Lx = 3/4 and My/Mx = 2 give (My/Mx) ⟨ 4 (Ly/Lx)2 and (My/Mx)

⟩ (4/3) (Ly/Lx)2. Thus, both the yield patterns are to be examined. Step 2: Value of y for yield pattern 1 The value of y is obtained from Eq.12.43 of Lesson 32, which gives:


6y2 + 8Lx – 9Lx2= 0 (7)

The solution of Eq. 7 is: y = 0.728 Lx (8) Step 3: Determination of collapse load w kN/m2 Equations 12.41 and 12.42 of the method of segmental equilibrium

of Lesson 32 and Eq. 12.48 of the method of virtual work of Lesson 32 shall be used to determine w. The results are given below:

(i) From Eq. 12.41: w = 24Mx Ly/{3Lx

2(Ly - y)+ Lx2y}= 22.657(Mx/Lx

2) (9)

(ii) From Eq. 12.42: w = 6My / y2 = 22.657 (Mx/Lx2)

(9) (iii) From Eq. 12.48: w = (24MxLy y + 6My Lx

2) / {Lx2 (3y Ly - y2)}

= 22.657 (Mx/Lx2) (9)

Step 4: Value of x for yield pattern 2

The value of x is obtained from Eq. 12.52 of Lesson 32, which is 6x2 + 2.25 Lx x – 1.6875 Lx

2 = 0 (10) The solution of the above equation is x = 0.375 Lx (11)

Step 5: Determination of collapse load w kN/m2

Equations 12.50 and 12.51 of the method of segmental equilibrium of Lesson 32 and Eq. 12.57 of the method of virtual work of Lesson 32 shall be used to determine w. The results are:

(i) From Eq. 12.50: w = 6(Mx Ly

2 – My x2) /(x2Ly2)= 21.334 (Mx/Lx

2) (12) (ii) From Eq. 12.51: w = 24 My x /{Ly

2 (3Lx – 4x)}= 21.334 (Mx/Lx2)

(12) (iii) From Eq. 12.57: w = 12{Mx Ly

2+Myx2}/{Ly2 x(3Lx - 2x)}

= 21.334 (Mx/Lx2) (12)

Step 6: Correct yield pattern and collapse load Comparison of results of steps 3 and 5 reveals that yield pattern 2 is

the correct one for which w is lower than that of yield pattern 1. For yield pattern 2, x = 0.375 Lx and w = 21.334 (Mx / Lx

2).


Problem 3: Determine the uniformly distributed collapse load w kN/m2 of a

square slab (L m × L m), simply supported at three edges and free at the other edge. Assume My = Mx = M.

Solution 3: The slab is shown in Fig. 12.33.1a. Given data are: Lx = Ly = L and My = Mx = M. Step 1. To examine the possibility of yield patterns 1, 2 or both From Eq. 12.45 of Lesson 32, it is seen that (My/Mx) < 4 (Ly/Ln)2. So,

yield pattern 1 has to be considered. But, (My/Mx) is not greater than (4/3) (Ly/Lx)2 (vide Eq. 12.54 of Lesson 32). So, yield pattern 2 is not to be considered.

Step 2. Value of y for yield pattern 1 Equation 12.43 of Lesson 32 gives: 4 y2 + 2Ly – 3L2 = 0 (13) The solution of Eq.13 is: y = 0.651 L (14) Step 3. Collapse load w kN/m2

(i) From Eq.12.41 of Lesson 32: w = (M/L2){(24)/(3 –2y)}=14.141(M/L2)

(15) (ii) From Eq. 12.42 of Lesson 32: w = 6M/y2 = 14.141 (M/L2)

(15) (iii) From Eq. 12.48 of Lesson 32: w = (M/ L2) {(24 y + 6)/{3y-y2)}=

14.141(M/L2) (15) Problem 4: Determine the uniformly distributed collapse load w kN/m2 of a

rectangular slab whose Ly/Lx = 0.4, simply supported at three edges and free at the other edge. Assume My/Mx = 1.

Solution 4: The slab is shown in Fig.12.33.1a. Given data are: Ly/Lx = 0.4 and

My/Mx = 1. Step 1: To examine the possibility of yield patterns 1, 2 or both


Here, (My/Mx) is not less than 4 (Ly/Lx)2. So, yield pattern 1 is not to be considered. However, (My/Mx) is ⟩ (4/3) (Ly/Lx)2. So, yield pattern 2 has to be considered (vide Eqs. 12.45 and 12.54 of Lesson 32).

Step 2: Value of x for yield pattern 2 Equation 12.52 (Lesson 32) gives: 3x2 + 0.64 Lx

2 – 0.48 Lx2 = 0 (16)

The solution of Eq. 16 is: x = 0.3073 Lx (17) Step 3: Collapse load w kN/m2

(i) Eq.12.50 gives: w = (Mx/Lx

2) {6(0.16-x2}/0.16x2} = 26.032 (Mx/Lx2)

(18) (ii) Eq.12.51 gives: w = (Mx/Lx

2){24x / 0.16(3-4x)}= 26.032 (Mx / Lx2)

(18) (iii) Eq. 12.57 gives: w = (Mx/Lx

2){12 (0.16+x2)/0.16(3x – 2x2)} = 26.032 (Mx / Lx

2) (18) (Equations 12.50, 12.51 and 12.57 are from Lesson 32.) Problem 5: Determine the correct yield pattern and the collapse point load P of

the isosceles triangular slab of Q.4b of sec.12.31.7 of Lesson 31 (Fig.12.31.10), if the load is applied at the centroid of the slab. The slab is also shown in Fig. 12.33.2a having B = 6 m and 2L = 12 m. Assume Mp = 9 kNm/m and Mn = 12 kNm/m. Use the method of virtual work.


Solution 5: The slab is shown in Fig. 12.32.2a. Given data are: 2L=12 m, B = 6 m, Mp = 9 kNm/m and Mn = 12 kNm/m. In this problem, two possible yield patterns, as shown in Figs.12.33.2b and c, are to be considered. The lower of the two loads shall be taken, as the correct collapse load and the corresponding yield pattern is the correct one.

Step 1: Yield pattern 1 Yield pattern 1, shown in Fig. 12.33.2b, divides the slab into three

segments. Assuming the deflection of the slab at the centroid G = Δ, the rotation θ1 of segment DFG = θ1 = Δ/2 (as the distance GJ = 2 m). The length of the side DE = 6√2 m and the perpendicular


distance from G to ED is GH = EG sin 450 = 2√2 m. The rotation of the segment DEG = rotation of the segment FEG = Δ/GH = Δ/2√2.

Step 2: TIW and TEW due to moments and loads TIW = (Mn + Mp) {DF (θ1) + 2DE (θ2)} = 21{12(Δ/2) + 2(6√2) (Δ/2√2)}

= 252 Δ So, TIW = 252 Δ (19) TIW = P Δ (20) Step 3: Collapse load P Equating TIW and TEW from Eqs. 19 and 20, we have, P = 252 kN (21) Thus, the collapse load for the yield pattern 1 is 252 kN. Step 4: Yield pattern 2 Yield pattern 2 divides the segment into a large number of sub-

segments as shown in Fig. 12.33.2c. The free body diagram of a typical segment GSR is shown in Fig. 12.33.2d. The internal work done by moments Mp and Mn of the segment GSR is given below.

Internal work done by moments = Mp r dθ (Δ/r) + Mn r dθ (Δ/r) = (Mn + Mp) dθ Δ (22) Total number of such segment = 2π/dθ (23) So, TIW = (Mn + Mp) dθ Δ (2π/dθ) = (Mn + Mp) 2π (Δ) (24) Total external work done by the load: TEW = PΔ (25) Equating TIW and TEW from Eqs. 24 and 25, we have (Mn + Mp) 2π (Δ) = PΔ which gives P = (Mn + Mp) 2π = 42 π = 119.428 kN (26) Step 5: Correct yield pattern and the collapse load


The correct yield pattern is the second one as it gives the lower collapse load P =119.428 kN.

Problem 6: A circular slab of 6 m radius is shown in Fig. 12.33.3 which is clamped along the circumference and has a column support at the centre. Determine the yield pattern and uniformly distributed collapse load of the slab if the positive and negative moments of resistance Mp and Mn are 30 kNm/m. Use the method of virtual work.

Solution 6: Figure 12.33.3 shows the slab. Given data are R = 6 m and Mn = Mp

= 30 kNm/m. The analysis of this type of slab is explained in sec. 12.32.6D of

Lesson 32 by the method of virtual work. We have Eqs.12.89 and 12.90 to determine the values of r and w (Fig. 12.32.7a to c). Using those equations with reference to Fig. 12.32.7 of Lesson 32, we have

(i) From Eq. 12.89: r = 0.2679 R = 1.6074 m (27) (ii) From Eq. 12.90: w = 22.39 (M/R2) = 18.658 kN/m2 (28)


Problem 7: Determine the yield pattern and uniformly distributed collapse load

of the quadrantal slab of radius R = 6 m as shown in Fig.12.33.4, which is clamped along the straight edges OA and OB, and free along the curved edge ACB. Assume Mn = Mp= 30 kNm/m. Use the method of segmental equilibrium for this problem.

Solution 7: Step 1: Yield pattern The slab and the yield pattern are shown in Fig. 12.33.4. Here, the

negative yield lines are along OA and OB, and the positive yield line OC divides the slab into two symmetrical segments. The point G is the centroid of the segment COA. The angle GOA = π/8. We know,

OG = (2R/3) (sin π/8 / π/8), GD = OG sin π/8 and area of the

segment COA = πR2/8 (29)

Step 2: Segmental equilibrium equation Taking moment of the load of segment COA about OA and

considering Mn and Mp, we have Mn (OA) + Mp (OC cos450) – w (π R2/8) (GD) = 0


or Mn (R) + Mp (R/√2) – w (πR2/8) (2R/3) (sin π/8 / π/8) sin π/8 = 0 or Mn +Mp/√2 – (2 w R2/3) sin 2 (π/8) = 0 (30) The value of the collapse load w can be determined from known

values of Mn , Mp and R from Eq.30. Let us assume: Mn = k Mp (31) Equation 30 then gives w = {3 (√2 k +1) Mp} /{R2(√2 –1)} (32) For this particular problem k = 1 and R = 6 m. Thus, from Eq. 32, we have: w = 14.571 kN/m2 (33) 12.33.3 Practice Questions and Problems with Answers Q.1: Determine the uniformly distributed collapse load w kN/m2 of a

circular slab of radius = 6 m simply supported along the periphery, if Mp = 30 kNm/m. Use the method of segmental equilibrium.

A.1: The analysis of such slabs is explained in sec. 12.32.6B of Lesson

32 employing the method of segmental equilibrium with reference to Fig. 12.32.5a and b. Equation 12.82 gives the value of the uniformly distributed load, which is

w = 6Mp/R2 = 6(30)/(36) = 5 kN/m2 (34) Therefore, the collapse load of this slab is 5 kN/m2. Q.2: Determine the uniformly distributed collapse load w kN/m2 of a

circular slab of radius = 6 m and clamped along the periphery, if Mp = Mn = 30 kNm/m. Use the method of segmental equilibrium.

A.2: The analysis of such slabs is explained in sec. 12.32.6C of Lesson

32 employing the method of segmental equilibrium with reference to Figs.12.32.6a and b. Equation 12.83 gives the value of the uniformly distributed load, which is


w = 6 (Mn + Mp) / R2 = 60(60)/36 = 10 kN/m2 (35) Therefore, the collapse load of this slab is 10 kN/m2. Q.3: Determine the uniformly distributed collapse load of a 6 m × 6 m

square slab clamped along four edges and has the moment of resistance Mpx = Mpy = Mnx = Mny = Mp = Mn = 30 kNm/m.


A.3: For this problem, we have to examine three yield patterns 1, 2 and

3, shown in Figs.12.33.5a, b and c, respectively. Step 1: Selection of yield patterns In this problem, (Ly / Lx)2 = 1 and (Mpy + Mny)/ (Mpx +Mnx) = 1. So,

yield pattern 1 has to be considered. Since, (Mpy + Mny) / (Mpx + Mnx) is not greater than (4/3) (Ly/Lx)2, yield pattern 2 is not possible. Yield pattern 3 is also possible. So, we have to examine yield patterns 1 and 3.

Step 2: Yield pattern 1 Equation 12.26 of Lesson 31 gives the value of w for yield pattern 1,

which is w = 12{(Mpx + Mnx)/(Lx)2 + (Mpy + Mny)/(Ly)2

(12.26) = 12 {60/36 + 60/36} = 40 kN/m2

Thus, w = 40 kN/m2 for yield pattern 1 Step 3: Yield pattern 3


Equation 12.83 of Lesson 32 gives the value of load w for yield pattern 3, which is:

w = 6(Mn + Mp) / R2 (12.83) Using Mn = Mp = 30 kNm/m as R = 3 m, we have w = 40 kN/m2. Therefore, the collapse load of this slab is 40 kN/m2 either for the

yield pattern 1 or for the yield pattern 3. 12.33.4 Reference

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 12.33.5 Test 33 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Determine the uniformly distributed collapse load of a rectangular

slab (Figs. 12.33.1a, b and c) having Ly/Lx = 0.5, simply supported at three edges and free at the other edge. Assume My = 0.5 Mx.

A.TQ.1: The slab is shown in Figs. 12.33.1a, b and c. Given data are: Ly/Lx =

0.5 and My/Mx = 0.5. Step 1: To examine the possibility of yield patterns 1, 2 or both Here, (My/Mx) ⟨ 4(Ly/Lx)2. So, yield pattern 1 is possible. Again

(My/Mx) ⟩ (4/3) (Ly/Lx)2 . So, yield pattern 2 is also possible. Thus, both yield patterns 1 and 2 are to be examined.

Step 2: Value of y for yield pattern 1 Equation 12.43 of Lesson 32 gives: 2y2 + Lx y – 0.75 Lx

2 = 0, which gives, y = 0.4114 Lx (36) Step 3: Determination of collapse load w kN/m2

(i) Eq. 12.41 gives: w = (Mx/Lx

2) {12/(1.5-2y)} = 17.722(Mx/Lx2)

(37) (ii) Eq. 12.42 gives: w = (Mx/Lx

2) (3/y2) = 17.722 (Mx / Lx2)

(37) (iii) Eq. 12.48 gives: w = (Mx/Lx

2) {(12y + 3) / y(1.5-y)} = 17.722 (Mx/Lx

2) (37) (Equations 12.41, 12.42 and 12.48 are from Lesson 32.) Step 4: Value of x for yield pattern 2 The value of x is obtained from Eq. 12.52 of Lesson 32, which is, 1.5 x2 + Lx x – 0.75 Lx

2 = 0 (38)


The solution of Eq. 38 is: x = 0.448 Lx (39) Step 5: Determination of collapse load w kN/m2 (i) Eq. 12.50 of Lesson 32 gives: w = 17.841 (Mx/Lx

2) (40) (ii) Eq. 12.51 of Lesson 32 gives: w = 17.841 (Mx/Lx

2) (40)

(iii) Eq. 12.57 of Lesson 32 gives: w = 17.841 (Mx/Lx2) (40)

Step 6: Correct yield pattern and collapse load Comparison of results of Steps 3 and 5 reveals that the yield

pattern 1 is the correct one giving lower value of w = 17.722 (Mx/Lx

2). 12.33.6 Summary of this Lesson

This lesson explains the different types of yield pattern of triangular, rectangular, quadrantal and circular slabs through several numerical problems. Both the methods of segmental equilibrium and virtual work are employed. The loadings considered are either point load or uniformly distributed load. All possible yield patterns of a particular problem are examined to select the correct one and the corresponding lower/lowest load for the slab problems. The moments of resistance of the slab in the two directions are also varied in the problems. Different types of support conditions are taken up in the numerical problems. In addition to illustrative examples, practice problems and test problem will help in understanding the theories and their applications in analysing slab problems with different types of support conditions and loadings.


Module 13

Working Stress Method Version 2 CE IIT, Kharagpur

Lesson 34

Rectangular Beams under Flexure



At the end of this lesson, the student should be able to: • explain the philosophy of the design by working stress method, • explain the concepts of permissible stresses in concrete and steel and the

factors of safety in this method, • state the limit of increase of permissible stresses under specific loading

considerations, • state the assumptions for the design of flexural members employing working

stress method, • explain the concept of modular ratio, • derive the governing equations of rectangular, balanced and under-reinforced

(singly-reinforced) sections, • name the two types of problems for the designer, • write down the steps separately for the two types of problems in the working

stress method, • justify the need for doubly-reinforced beams, and • derive the governing equations of the doubly-reinforced rectangular flexural

members. 13.34.1 Introduction Design of reinforced concrete structures started in the beginning of this century following purely empirical approach. Thereafter came the so called rigorous elastic theory where it is assumed that concrete is elastic and reinforcing steel bars and concrete act together elastically. The load-deflection relation is linear and both concrete and steel obey Hooke’s law. The method is designated as working stress method as the loads for the design of structures are the service loads or the working loads. The failure of the structure will occur at a much higher load. The ratio of the failure loads to the working loads is the factor of safety. Accordingly, the stresses of concrete and steel in a structure designed by the working stress method are not allowed to exceed some specified values of stresses known as permissible stresses. The


permissible stresses are determined dividing the characteristic strength fck of the material by the respective factor of safety. The values of the factor of safety depend on the grade of the material and the type of stress. Thus, for concrete in bending compression, the permissible stress of grade M 20 is 7 N/mm2, which is obtained by dividing the characteristic strength fck of M 20 concrete by a number 3 and then rationalising the value to 7. This permissible stress is designated by σcbc, the symbol σ stands for permissible stress and the letters c, b and c mean concrete in bending compression, respectively. 13.34.2 Permissible Stresses in Concrete The permissible stress of concrete in direct tension is denoted by σtd. The values of σtd for member in direct tension for different grades of concrete are given in cl. B-2.1.1 of IS 456. However, for members in tension, full tension is to be taken by the reinforcement alone. Though full tension is taken by the reinforcement only, the actual tensile stress of concrete ftd in such members shall not exceed the respective permissible values of σtd to prevent any crack. Table 13.1 presents the values of σtd for selective grades of concrete as a ready reference. It may be worth noting that the factor of safety of concrete in direct tension is from 8.5 to 9.5. The permissible stresses of concrete in bending compression σcbc, in direct compression σcc and the average bond for plain bars in tension τbd are given in Table 21 of IS 456 for different grades of concrete. However, Table 13.1 presents these values for selective grades of concrete, as a ready reference. The factors of safety of concrete in bending compression, direct compression and average bond for plain bars are 3, 4 and from 25 to 35, respectively. For plain bars in compression, the values of average bond stress are obtained by increasing the respective value in tension by 25 percent, as given in the note of Table 21 of IS 456. For deformed bars, the values of Table 21 are to be increased by sixty per cent, as stipulated in cl. B-2.1.2 of IS 456. Table 13.1 Permissible stresses in concrete

Grade of concrete

Direct tension σtd (N/mm2)

Bending compression σcbc (N/mm2)

Direct compression σcc (N/mm2)

Average bond τbd for plain

bars in tension (N/mm2)

M 20 2.8 7.0 5.0 0.8 M 25 3.2 8.5 6.0 0.9 M 30 3.6 10.0 8.0 1.0


M 35 4.0 11.5 9.0 1.1 M 40 4.4 13.0 10.0 1.2

From the values of the permissible stresses and the respective characteristic strengths for different grades of concrete, it may be seen that the factors of safety of concrete in direct tension, bending compression, direct compression and average bond for plain bars in tension are from 8.5 to 9.5, 3, 4 and from 25 to 35, respectively. 13.34.3 Permissible Stresses in Steel Reinforcement Permissible stresses in steel reinforcement for different grades of steel, diameters of bars and the types of stress in steel reinforcement are given in Table 22 of IS 456. Selective values of permissible stresses of steel of grade Fe 250 (mild steel) and Fe 415 (high yield strength deformed bars) in tension (σst and σsh) and compression in column (σsc) are furnished in Table 13.2 as a ready reference. It may be noted from the values of Table 13.2 that the factor of safety in steel for these stresses is about 1.8, much lower than concrete due to high quality control during the production of steel in the industry in comparison to preparing of concrete.

Table 13.2: Permissible stresses in steel reinforcement

Type of stress in steel reinforcement

Mild steel bars, Fe 250, (N/mm2)

High yield strength deformed bars, Fe 415,

(N/mm2) Tension σst or σss (a) up to and including

20 mm diameter (b) over 20 mm diameter

140

130

230

230

Compression in column bars σsc

130 190

13.34.4 Permissible Shear Stress in Concrete τc

Permissible shear stress in concrete in beams without any shear reinforcement depends on the grade of concrete and the percentage of main tensile reinforcement in beams. Table 23 of IS 456 furnishes the values of τc for wide range of percentage of tensile steel reinforcement for different grades concrete. Other relevant clauses regarding the permissible shear stress of concrete are given in cls.B-5.2.1.1, B-5.2.2 and B-5.2.3 of IS 456.


13.34.5 Increase in Permissible Stresses Clause B-2.3 of IS 456 recommends the increase of permissible stresses of concrete and steel given in Tables 21 to 23 up to a limit of 33.33 per cent, where stresses due to wind (or earthquake), temperature and shrinkage effects are combined with those due to dead, live and impact loads.

13.34.6 Assumptions for Design of Members by Working Stress Method As mentioned earlier, the working stress method is based on elastic theory, where the following assumptions are made, as specified in cl. B-1.3 of IS 456.

(a) Plane sections before bending remain plane after bending.

(b) Normally, concrete is not considered for taking the tensile stresses except otherwise specifically permitted. Therefore, all tensile stresses are taken up by reinforcement only.

(c) The stress-strain relationship of steel and concrete is a straight line under working loads.

(d) The modular ratio m has the value of 280/3σcbc, where σcbc is the permissible compressive stress in concrete due to bending in N/mm2. The values of σcbc are given in Table 21 of IS 456. The modular ratio is explained in the next section.

13.34.7 Modular Ratio m In the elastic theory, structures having different materials are made equivalent to one common material. In the reinforced concrete structure, concrete and reinforcing steel are, therefore, converted into one material. This is done by transformation using the modular ratio m which is the ratio of modulus of elasticity of steel and concrete. Thus, m = Es/Ec. where Es is the modulus of elasticity of steel which is 200000 N/mm2. However, concrete has different moduli, as it is not a perfectly elastic material.

The short-term modulus of concrete Ec = 5000 ckf in N/mm2, where fck is the characteristic strength of concrete. However, the short-term modulus does not take into account the effects of creep, shrinkage and other long-term effects. Accordingly, the modular ratio m is not computed as m = Es/Ec = 200000/(5000


ckf ). The value of m, as given in sec. 13.34.6 d, i.e., 280/3σcbc, partially takes into account long-term effects. This is also mentioned in the note of cl. B-1.3 of IS 456.

13.34.8 Flexural Members – Singly Reinforced Sections

A simply supported beam subjected to two point loads shall have pure moment and no shear in the middle-third zone, as shown in Fig. 1.1.1 of Lesson 1.The cross-sections of the beam in this zone are under pure flexure. Figures 13.34.1a, b and c show the cross-section of a singly-reinforced beam, strain profile and stress distribution across the depth of the beam, respectively due to the loads applied on the beam. Most of the symbols are used in earlier lessons. The new symbols are explained below.

x = kd = depth of the neutral axis, where k is a factor,

fcbc = actual stress of concrete in bending compression at the top fibre which should not exceed the respective permissible stress of concrete in bending compression σcbc,

fst = actual stress of steel at the level of centroid of steel which should not exceed the respective permissible stress of steel in tension σst,

jd = d(1-k/3) = lever arm i.e., the distance between lines of action of total compressive and tensile forces C and T , respectively.


Figures 13.34.1b and c show linear strain profile and stress distribution, respectively. However, the value of the stress at the level of centroid of steel of Fig. 13.34.1 c is fst /m due to the transformation of steel into equivalent concrete of area mAst.

13.34.9 Balanced Section – Singly-Reinforced

In a balanced cross-section both fcbc and fst reach their respective permissible values of σcbc and σst at the same time as shown in Fig.13.34.2c. The depth of neutral axis is xb = kbd. From the stress distribution of Fig.13.34.2c, we have

σst /m = σcbc (1-kb) / kb (13.1)

An expression of kb is obtained by substituting the expression of m as

m = 280 / 3σcbc (13.2)

into Eq. 13.1. This gives

kb = 93.33 / (σst + 93.33) (13.3)

Equation 13.3 shows that the value of kb for balanced section depends only on σst . It is independent of σcbc.

The lever arm, jb d = (1- kb/3) d (13.4)


The expressions of total compressive and tensile forces, C and T are: C = (1/2) σcbc b xb = (1/2) σcbc b kb d (13.5) T = Ast σst (13.6) The total compressive force is acting at a depth of xb/3 from the top fibre of the section. The moment of resistance of the balanced cross-section, Mb is obtained by taking moment of the total compressive force C about the centroid of steel or moment of the tensile force T about the line of action of the total compressive force C. Thus, 21 2 b b cbc b bM C( j d ) ( / ) k j ( bd )σ= = (13.7) or, 2 /100) b b st st b t ,bal st bM T( j d ) A j d ( p j (bd )σ σ= = = (13.8) as (13.9) )/bd(pA bal,tst 100= where pt,bal = balanced percentage of steel From Eqs.13.7 and 13.8, we can write Mb = Rb bd2 (13.10) where Rb = (1/2) σcbc kb jb = (pt,bal /100) σst jb (13.11) and jb = 1- (kb/3) (13.12) The expression of the balanced percentage of steel pt,bal is obtained by equating the total compressive force C to the tensile force T from Eqs. 13.5 and 13.6. This gives, Ast σst = (σcbc/2) b kb d, which gives: pt,bal (bd/100) σst = (σcbc/2) b kbd or pt,bal = 50 kb(σcbc/σst) (13.13) It is always desirable, though may not be possible in most cases, to design the beam as balanced since the actual stresses of concrete fcbc at the top compression fibre and steel at the centroid of steel should reach their respective permissible stresses σcbc and σst in this case only. The procedure of the design is given below.


Treating the design moment as the balanced moment of resistance Mb and assuming the width, b of the beam as 250 mm, 300 mm or 350 mm, the effective depth d is obtained from Eq.13.10. The required balanced area of steel Ast is then obtained from Eq. 13.9 getting the values of kb from Eq. 13.3 and then pt,bal from Eq. 13.13. The values of Rb, the moment of resistance factor Mb/bd2 are obtained from Eq. 13.11 for different values of σcbc and σst (different grades of concrete and steel) and are presented in Table 13.3. Similarly, the values of balanced percentage of tensile reinforcement, pt, bal obtained form Eq. 13.13 for different grades of concrete and steel are given in Table 13.4. Table 13.3 Moment of resistance factor Rb in N/mm2 for balanced rectangular section.

σst (N/mm2)

σcbc(N/mm2) 140 230 275

7.0 1.21 0.91 0.81 8.5 1.47 1.11 0.99 10.0 1.73 1.30 1.16

Table 13.4 Percentage of tensile reinforcement pt,bal for singly-reinforced balanced section.

σst (N/mm2)

σcbc(N/mm2) 140 230 275

7.0 1.0 0.44 0.32 8.5 1.21 0.53 0.39 10.0 1.43 0.63 0.46

Values of Rb and pt.bal from Tables 13.3 and 13.4 reveal the following: 1. For given values of width and effective depth, b and d of a rectangular

section, the balanced moment of resistance Mb increases with higher grade of concrete for a particular grade of steel. However, the balanced moment of resistance decreases with higher grade of steel for a particular grade of concrete.

2. The balanced percentage of steel, pt.bal increases with the increase of

grade of concrete for a particular grade of steel for given values of width and effective depth of a rectangular section. On the other hand, the


balanced percentage of steel, pt.bal decreases with the increased grade of steel for a particular grade of concrete.

As mentioned earlier, it may not be possible to design a balanced section since the area of steel required for the balanced condition is difficult to satisfy with available bar diameters. In such cases, it is essential that the beam should be provided with the steel less than the balanced steel so that the actual stress of steel in tension reaches the permissible value σst and the actual stress of concrete fcbc is less than the permissible value. Such sections are designated as under-reinforced sections and moment of resistance shall be governed by the tensile stress of steel σst, which is known. The depth of the neutral axis will be less than the balanced depth of the neutral axis, as shown in Fig. 13.34.3b. The relevant equations for the design of under-reinforced section are established in the next section.

13.34.10 Under-reinforced Section -- Singly Reinforced Figure 13.34.3a shows the cross-section where x (= kd) is the depth of the neutral axis. The depth of the neutral axis is determined by taking moment of the area of concrete in compression (= bx) and the transformed area of steel (= mAst) about the neutral axis, which gives b kd (kd/2) = m (pt bd/100) (d-kd)


or k2 + (pt mk/50) - (pt m/50) = 0 (13.14) Equation 13.14 has two roots of k given by k= - (pt m/100) ± {(pt m/100)2+ (pt m/50)}1/2 (13.15) Since k cannot be negative, we have the positive root to be considered as k= - (ptm /100) +{(pt m /100)2 + (pt m/50)}1/2 (13.16) The moment of resistance of the under-reinforced section is obtained from M = T (lever arm) = Ast σst d(1 - k/3) = (pt bd /100) σst d (1 - k/3) Therefore, we have: M = (pt/100)σst (1 - k/3) bd2 (13.17) which can also be expressed as M = R bd2 (13.18) where R = (pt/100) σst (1 – k/3) (13.19) is the moment of resistance factor M/bd2. The values of R are obtained for given values of pt for different grades of steel and concrete. Tables 68 to 71 of SP-16 furnish the values of R for four grades concrete and five values of σst. The actual stress of concrete at the top fibre fcbc shall not reach σcbc in under- reinforced sections. The actual stress fcbc is determined from the equation C=T as explained below: With reference to Fig. 13.34.3c, the compressive force C of concrete and tensile for T of steel are: C = (1/2) fcbc b kd (13.20) T = σst Ast (13.6)


For the tensile force, the actual stress of steel fst shall reach the value of σst. So, we are using Eq. 13.6, the same equation as for the balanced section. Equating C and T from Eqs. 13.20 and 13.6, we get (1/2)fcbc b kd = σst Ast or, fcbc = 2 σst Ast / b k d (13.21) Expressing Ast = pt bd /100 (13.22) and using Eq. 13.22 in Eq. 13.21, we get fcbc = pt σst / 50 k (13.23) The two types of problems: (i) Analysis type and (ii) Design type are now taken up in the following sections. 13.34.11 Analysis Type of Problems

For the purpose of analysing a singly-reinforced beam where the working loads, area of steel, b and d of the cross section are given, the actual stresses of concrete at the top fibre and steel at the centroid of steel are to be determined in the following manner. Step 1: To determine the depth of the neutral axis kd from Eq. 13.16. Step 2: The beam is under-reinforced, balanced or over-reinforced, if k is less than, equal to or greater than kb, to be obtained from Eq. 13.3. Step 3: The actual compressive stress of concrete fcbc and tensile stress of steel at the centroid of steel fst are determined in the following manner for the three cases of Step 2. Case (i) When k < kb (under-reinforced section) From the moment equation, we have M= Ast fst d(1 - k/3) or fst =M /{Ast d(1 - k/3)} (13.24)


where M is obtained from the given load, Ast and d are given, and k is determined in Step 1. Equating C = T, we have: (1/2) fcbc b(kd) = Ast fst or fcbc = (2 Ast fst) / (b kd) (13.25) where Ast, b and d are given and k and fst are determined in steps 1 and 2, respectively. Case (ii) When k = kb (balanced section) In the balanced section fcbc = σcbc and fst = σst. Case (iii) When k > kb (over-reinforced section) Such beams are not to be used as in this case fcbc shall reach σcbc while fst shall not reach σst. These sections are to be redesigned either by increasing the depth of the beam or by providing compression reinforcement. Beams with compression and tension reinforcement are known as doubly-reinforced beam and is taken up in sec. 13.34.13. 13.34.12 Design Type of Problems It has been explained earlier in sec.13.34.9 that it is difficult, though desirable, to design a balanced section, as the concrete and steel should attain their respective permissible values at the same time. Therefore, a practical design shall preferably be under-reinforced. However, the given data are the working load, span and support conditions of the beam. These data shall enable to get the design moment of the section. The breadth b, effective depth d and area of steel are to be determined in these problems. The steps to be followed are given below. Step 1. Selection of preliminary dimension b and D The width is normally taken as 250 mm, 300 mm or 350 mm so that the reinforcing bars can be accommodated without difficulty. The total depth of the beam may be supplied in architectural drawings or may be estimated from the span to depth ratios, as stiputed in cl. 23.2.1 of IS 456 for the control of deflection. However, these values should be revised, if needed. Step 2. Determination of d


The effective depth of a balanced section is first obtained from Eq. 13.10, where the value of Rb is taken from Table 13.3 after finalising the grades of concrete and steel. This balanced effective and total depths of beam, as given in the architectural drawings or may be needed for the control of deflection, give some idea while finalising the depth of the beam, higher than the balanced depth, so that the beam becomes under-reinforced. Step 3. Determination of Ast First, the balanced percentage of steel pt,bal is taken from Table 13.4 for the selected grades of concrete and steel. The amount of increase of the depth of the beam of step 2 may also depend on the amount pt, bal. In fact, there are sets of values of depth and the respective area of steel, to be selected judiciously on the basis of practical considerations. Step 4. Checking for the stresses After finalising the depth and area of steel (depth larger than the balanced depth and area of steel lower than the balanced area of steel), the section is to be analysed as an analysis type of problem, as explained in sec. 13.34.11. Alternatively, appropriate table from Tables 63 to 71 of SP-16 may be used for getting several possible combinations quickly. 13.34.13 Doubly-Reinforced Beams In some situations, it may not be possible to provide the required depth of the beam as singly-reinforced either due to restrictions given by the architects or due to functional requirements. In such cases, the designer has to satisfy the structural requirement with a depth less than that of a singly-reinforced section. This is done by providing steel reinforcing bars in the compression side along with the tensile steel in the tension side of the beam. These beams are designated as doubly-reinforced beams.


Figures 13.34.4a to c show the cross-section, strain profile and stress distribution of a doubly-reinforced section. Since, the design moment is more than the balanced moment of resistance of the section, we have

M = Mb + M′ (13.26)

The additional moment M′ is resisted by providing compression

reinforcement Asc (= pcbd/100) and additional tensile reinforcement Ast2. The modular ratio of the compression steel is taken as 1.5 m, where m is the modular ratio as explained in sec. 13.34.7.

Figure 13.34.4c shows that the stress of concrete at the level of compression steel is σcbc (kbd - d′) / kdd. Accordingly, the stress in the compression steel reinforcement is 1.5m σcbc (kbd - d′) / kbd.


Figure 13.34.4d and e present separate stress distribution for the balanced beam (shown in Fig. 13.34.2c) and the compressive and tensile forces of compressive and tensile reinforcing bars C2 and T2, respectively. The expression of the additional moment M′ is obtained by multiplying C2 and T2 with the lever arm (d - d′ ), where d′ is the distance of the centroid of compression steel from the top fibre. We have, therefore,

C2 = Asc (1.5m –1) σcbc (kd - d′) / kd

(13.27) T2 = (pt – pt, bal) (bd / 100) σst

(13.28) M′ = C2 (d - d′) = (pc bd /100) (1.5m – 1) σcbc (kd - d′ )/ kd (d- d′ )

or, M′ = (pc / 100)(1.5m – 1) σcbc (1- d′ / kd)(1- d′/d)bd2 (13.29) also, M′ = T2 (d - d′ ) d = (pt –pt, bal) (bd / 100)σst (d - d′ ) or, M′ = (pt – pt, bal) / 100 σst (1- d′/d) bd2 (13.30) Equating T2 = C2 from Eqs. 13.28 and 13.27, we have (pt – pt, bal) σst = pc (1.5m – 1) σcbc (1- d′ / kd) (13.31) The total moment M is obtained by adding Mbal and M′, as given below:

M = Mbal + (pt – pt, bal) / 100 σst (1- d′ / d) bd2 (13.32)

The total tensile reinforcement Ast has two components Ast1 + Ast2 for Mbal and M′, respectively. The equation of Ast is: Ast = Ast 1 + Ast 2 (13.33) where Ast1 = pt bal (bd / 100) (13.34) and Ast 2 = M′ / σst (d - d′) (13.35)


The compression reinforcement Asc is expressed as a ratio of additional tensile reinforcement Ast2, as given below:

(Asc / Ast2) = {pc /(pt – pt bal )} or, (Asc / Ast2) = σst / {σcbc (1.5m – 1) (1 - d′ / kd)} (13.36) Table M of SP-16 presents the values of Ast /Ast2 for different values of d′ / d and σcbc for two values of σst =140 N/mm2 and 230 N/mm2. Selective values are furnished in Table 13.5 as a ready reference. Tables 72 to 79 of SP-16 provide values of pt and pc for four values of d′ / d against M/bd2 for four grades of concrete and two grades of steel. Table 13.5: Values of Asc / Ast 2

d′ / d

σst(N/mm2)

σcbc(N/mm2) 0.05 0.10 0.15 0.20

7.0 1.20 1.40 1.68 2.11 8.5 1.22 1.42 1.70 2.13

140

10.0 1.23 1.44 1.72 2.15 7.0 2.09 2.65 3.60 5.54 8.5 2.12 2.68 3.64 5.63

230

10.0 2.14 2.71 3.68 5.76 13.34.14 Practice Questions and Problems with Answers Q.1: Justify the name working stress method of design. A.1: Paragraph 2 of Sec. 13.34.1 Q.2: How are the permissible stresses of concrete in direct tension, bending

compression, direct compression and average bond for plain bars in tension related to the factor of safety in the working stress method of design?

A.2: Sec. 13.14.2 Q.3: How is the permissible stress of steel in tension related to the factor of

safety in the working stress method of design?


A.3: Sec. 13.34.3

Q.4: Is it possible to increase the permissible stress? If yes, when is it done?

A.4: Sec. 13.34.5 Q.5: State the assumption for the design of members by working stress method.

A.5: Sec. 13.34.6 Q.6: Explain the concept of modular ratio.

A.6: Sec. 13.34.7 Q.7: Draw a cross-section of a singly-reinforced rectangular beam, the strain

and stress distributions along the depth of the section. A.7: Figs. 13.34.1a, b and c Q.8: What do you mean by balanced rectangular beam? Establish the

equations for determining the moment of resistance and percentage of tension steel in a balanced rectangular beam.

A.8: Sec. 13.34.9 Q.9: Establish the equations for determining the depth of neutral axis, moment

of resistance and area of tension steel of an under-reinforced rectangular beam.

A.9: Sec. 13.34.10 Q.10: Write down the steps for solving the analysis type of problems of singly-

reinforced rectangular beams. A.10: Sec. 13.34.11 Q.11: Write down the steps for solving the design type of problems of singly-

reinforced rectangular beams. A.11: Sec. 13.34.12 Q.12: When do we go for doubly-reinforced beams? Establish the equation for

determining areas of steel for the doubly reinforced rectangular beams. A.12: Sec. 13.34.13


13.34.15 References


















TQ.1: State the assumption for the design of members by working stress method. (15 marks)

A.TQ.1: Sec. 13.34.6 TQ.2: Establish the equations for determining the depth of neutral axis, moment

of resistance and area of tension steel of an under-reinforced rectangular beam.

(15 marks) A.TQ.2: Sec. 13.34.10

TQ.3: When do we go for doubly-reinforced beams? Establish the equation for determining areas of steel for the doubly-reinforced rectangular beams.

(20 marks) A.TQ.3: Sec. 13.34.13

13.34.17 Summary of this Lesson This lesson explains the philosophy of the analysis and design of singly-reinforced rectangular flexural members employing working stress method. The concepts of permissible stresses of concrete and reinforcement and the factors of safety are explained. The basic assumptions of the working stress method are mentioned. Balanced and under-reinforced sections are explained establishing the governing equations for the analysis and design of such members. The solution procedures for the two types of problems, viz., analysis and design types, are given in steps. Due to the architect’s restrictions or for the functional requirements, the doubly-reinforced rectangular beams are needed to be designed. The governing equations of doubly-reinforced rectangular beams are established.


Module 13

Working Stress Method Version 2 CE IIT, Kharagpur

Lesson 35

Numerical Problems



At the end of this lesson, the student should be able to: • employ the equations established for the analysis and design of singly and

doubly-reinforced rectangular beams, • use tables of SP-16 for the analysis and design of singly and doubly-

reinforced rectangular beams, • understand the economy in the design by limit state of collapse. 13.35.1 Introduction Lesson 34 explains the equations for the analysis and design of singly and doubly-reinforced rectangular beams. The applications of the equations are illustrated in this lesson through the solutions of several numerical problems. Direct computation method and use of tables of SP-16 are the two approaches for the analysis and design of singly and doubly-reinforced rectangular beams. In direct computation method, the derived equations are employed directly, while the use of tables of SP-16 gives the results quickly with several alternatives avoiding tedious calculations. Some of the numerical problems are solved using both methods. Problems having the same width and effective depth but of different grades of steel are solved to compare the results. Moreover, problems of singly and doubly-reinforced rectangular beams, solved earlier by limit state of collapse method are also taken up here to compare the results. It is shown that the beams designed by limit state of collapse method are economical. In addition to illustrative examples, practice and test problems are also given in this lesson. Understanding the solved numerical examples and solving the practice and test problems will help in following the applications of the equations and the use of tables of SP-16 in analysing and designing singly and doubly-reinforced rectangular beams.



Problem 1. (a) Determine the moment of resistance of the rectangular beam of Fig. 13.35.1 having b = 350 mm, d = 600 mm, D = 650 mm, Ast = 804 mm2 (4-16T), σcbc = 7 N/mm2 and σst = 230 N/mm2. (b) Determine the balanced moment of resistance of the beam and the balanced area of tension steel. (c) Determine the actual compressive stress of concrete fcbc and tensile stress of steel fst when 60 kNm is applied on the beam. Use direct computation method for all three parts. Solution 1. 1 (a): Given data are: b = 350 mm, d = 600 mm, Ast = 804 mm2, σcbc = 7 N/mm2 and σst = 230 N/mm2. We have: pt = Ast (100) / bd = 80400 / (350) (600) = 0.383 per cent and m = 93.33 /σcbc = 93.33 /7 = 13.33. We determine the value of k from Eq. 13.16.

k = - (pt m /100) + {(pt m/100)2 + (pt m/50)}1/2

(13.16) = - 0.051 + 0.323 = 0.272 j = 1- k/3 = 0.909

Equation 13.17 gives the moment of resistance of the beam as,

M = (pt/100) σst (1 – k/3) bd2 (13.17)


= (0.383/100) (230) (0.909) (350) (600) (600) = 100.89 kNm. 1 (b): The balanced moment of resistances and pt, bal of the beam is obtained

from Eqs. 13.10, 13.11 and 13.13. Mb = Rbbd2 (13.10) Rb = (1/2) σcbc kb jb = (pt, bal /100) σst jb (13.11) pt, bal = 50 kb (σcbc /σst ) (13.13)

where kb = 93.33 /(σst + 93.33) (13.3) jb = 1- kb /3

(13.12)

The values of Rb and pt, bal may also be taken from Tables 13.3 and 13.4, respectively of Lesson 34. Here, kb = 93.33 / (230 + 93.33) = 0.288 and jb = 1- 0.288/3 = 0.904.

pt, bal = 50 kb (σcbc /σst) = 50 (0.288) (7/230) = 0.438 pt, bal is 0.44 from Table 13.4.

Therefore, Mb = (1/2)σcbc kb jb (bd2) = 114.81 kNm, and Mb = (pt, bal /100) σst jb (bd2) = 114.75 kNm Taking Rb from Table 13.3, we get: Mb = Rb (bd2) = 0.91 (350) (600)2 = 114.66 kNm The balanced area of steel = pt,bal (bd /100) = 0.438 (350) (600) /100 = 919.8 mm2. The beam has 4-16 mm diameter bars having 804 mm2. So, additional amount of steel = 919.8 – 804 = 115.8 mm2 is needed to make the beam balanced. 1`(c): The actual stresses in steel and concrete fst and fcbc are obtained from

Eqs. 13.24 and 13.25, respectively. fst = M /{Ast d (1- k/3)} (13.24) fcbc = (2 Ast fst) / ( b k d) (13.25) where Ast = 804 mm2 and k = 0.272 (obtained in part a of the solution of this problem). Thus, we have:


fst = 60000000 / 804 (600) (0.909) = 136.83 N/mm2 ⟨ 230 N/mm2

fcbc = 2(804) (136.83) / (350) (0.272) (600) = 3.85 N/mm2 ⟨ 7 N/mm2

Problem 2. Solve part a of Problem 1 (Fig. 13.35.1) by employing table of SP-16. Solution 2. Given data are: pt = 0.383 per cent, σcbc = 7 N/mm2 and σst = 230 N/mm2. Table 69 of SP-16 gives R = 0.7956 (by linear interpolation). Accordingly, the moment of resistance of the beam is M = R bd2 = (0.7956) (350) (600) (600) = 100.246 kNm. The moment of resistance of the beam is 100.89 kNm by direct computation, as obtained in part a of Problem 1. Problem 3. Solve Problem 1 (Fig. 13.35.1) when σcbc = 7 N/mm2 and σst = 140 N/mm2 (mild steel) for all three parts. For part c, the applied moment is 40 kNm. Other values/data remain unchanged. Solution 3. 3 (a): Given data are: b = 350 mm, d = 600 mm, Ast= 804 mm2, σcbc 7 N/mm2 and σst = 140 N/mm2. We have from Problem 1: pt = 0.383 per cent, m = 13.33, k = 0.272, and j = 0.909. From Eq. 13.17, we get: M = (pt / 100) σst (1 – k/3) bd2 (13.17) = (0.383 / 100) (140) (0.909) (350) (600) (600) = 61.41 kNm 3 (b): Here kb = 93.33 / (140 + 93.33) = 0.4; jb = 0.87,

pt,bal = 50 kb (σcbc /σst) = 50 (0.4) (7/140) = 1.0 (Table 13.4 also gives pt,bal = 1). Therefore, Mb = (1/2) σcbc kb jb (bd2) (13.10) = (1/2) (7) (0.4) (0.87) (350) (600) (600) = 153.47 kNm Mb = (pt, bal/100) σst (jb) (bd2) (13.11) = (1/100) (140) (0.87) (350) (600) (600) = 153.47 kNm


and, taking Rb from Table 13.3, we have from Eq. 13.10, Mb = R bd2 = (1.21) (350) (600) (600) = 152.46 kNm. The balanced area of steel is obtained from taking pt, bal from Table 13.4. Accordingly, we get balanced area of steel = 1 (350) (600)/100 = 2100 mm2. The beam is having 4-16 mm diameter bars of area 804 mm2. The additional area of steel = 1296 mm2 is needed to make the beam balanced. 3 (c): fst = M / {Ast d (1- k/3)} (13.24) This gives fst = 40000000 / (804) (600) (0.909) = 91.22 N/mm2 ⟨ 140 N/mm2

fcbc = (2 Ast fst) / b kd (13.25) So, fcbc = 2 (804) (91.22) / (350) (0.272) (600) = 2.57 N/mm2 ⟨ 7 N/mm2

Problem 4. Solve part a of Problem 3 (Fig. 13.35.1) by using table of SP-16. Solution 4. Given data are: pt = 0.383 per cent, σcbc = 7 N/mm2 and σst = 140 N/mm2. Table 69 of SP-16 gives R = 0.48436 (by linear interpolation). Accordingly, the moment of resistance of the beam is

M = R b d2 = (0.48436) (350) (600) (600) = 61.03 kNm. The moment of resistance of this beam is 61.41 kNm by direct computation, as obtained in part a of Problem 3.


Problem 5. Design a singly-reinforced rectangular beam to carry the design moment = 100 kNm using M 25 concrete and Fe 415 grade of steel. Use b = 300 mm and d = 700 mm (Fig. 13.35.2) as preliminary dimensions. Solution 5. Steps 1 and 2 are not needed as the preliminary dimension of b = 300 mm and d = 700 mm are given. Step 3. The balanced moment of resistance of the beam = Mb = Rb d2= (1.11) (300) (700) (700) = 163.17 kNm (taking Rb = 1.11 from Table 13.3 for σcbc = 8.5 N/mm2 and σst = 230 N/mm2). The balanced area of steel = pt, bal (bd) / 100 = (0.53) (300) (700) / 100 = 1113 mm2 (taking pt, bal = 0.53 per cent from Table 13.4 for σcbc = 8.5 N/mm2 and σst = 230 N/mm2). We provide 4-16 mm diameter bars of Ast = 804 mm2 ⟨ 1113 mm2 to have pt = 804 (100) / (300) (700) = 0.383 per cent. This is more than the minimum tensile steel = 0.85 bd / fy = 430.12 mm2, as stipulated in cl. 26.5.1.1 of IS 456. Step 4. Now, the section has to be checked for the stresses in concrete and steel, fcbc and fst, respectively. For this problem m = 93.33 /σcbc = 10.98 and pt = 0.383 per cent, obtained in Step 3. Equation 13.16 gives: k = - (pt m /100) + {(pt m/100)2 + (pt m /50)}1/2 = 0.251. So, j = 1 – k/3 = 1 – 0.251/3 = 0.916. Equation 13.24 gives: fst = M / {Ast d (1-k/3)}= 100 (106) / (804) (700) (0.916) =

193.98 N/mm2 ⟨ 230 N/mm2. Equation 13.25 gives: fcbc = 2 Ast fst / b kd = 2(804) (193.98) / (300) (0.251) (700)

= 5.92 N/mm2 ⟨ 8.5 N/mm2

Therefore, the beam having b = 300 mm, d = 700 mm and Ast = 4-16 mm diameter bars is safe. Problem 6. Design the singly-reinforced rectangular beam of Problem 5 (Fig. 13.35.2) by using tables of SP-16. Solution 6. Given data are: M = 100 kNm, b = 300 mm and d = 700 mm. M/bd2 = 100 (106) / (300) (700) (700) = 0.6803. Using Table 70 of SP-16, we get pt = 0.321, which gives Ast = 0.321 (300) (700)/100 = 674 mm2. This area of steel is the minimum


amount of steel required to resist the design moment of 100 kNm. In fact, we have several alternatives of selecting area of steel from 674 mm2 to the balanced area of steel = 1113 mm2, as obtained in Step 3 of Problem 5. While solving Problem 5 by direct computation method, steel area of 804 mm2 is selected (4-16 mm diameter bars), as the amount is less than the balanced area of steel = 1113 mm2 . However, selecting area of 674 mm2 is difficult as 3-16 mm diameter bars have the area of steel 603 mm2. Only one 12 mm diameter is to be added to give additional area of 113 mm2, which satisfies the requirement. However, 12 mm bars are normally not preferred in beams. So, for the practical considerations, we prefer 4-16 mm diameter bars, as assumed in the direct computation method. The other alternatives are changing the depth of the beam. For studying all these options, use of tables of SP-16 is very much time-saving. Moreover, in the direct computation method, the selected area of steel has to be checked so that it is more than the minimum area of steel, as stipulated in cl. 26.5.1.1 of IS 456. On the other hand, the areas obtained from the table of SP-16 already take care so that they are more than the minimum area of steel. Problem 7. Determine the area of steel of the beam of Problem 3 to carry a design moment of 200 kNm. The relevant data of Problem 3 are: b = 350 mm, d = 600 mm, D = 650 mm (Fig. 13.35.1) σcbc = 7 N/mm2 and σst = 140 N/mm2. Solution 7. The moment of resistance of the balanced beam = 152.46 kNm obtained by using Rb = 1.21 in the equation Mb = Rb bd2, in part b of the solution of Problem 3. The balanced area of the steel = 2100 mm2, which is taken from the solution of part b of Problem 3. The values of m = 13.33, kb = 0.4, jb = 0.87 and pt, bal = 1 are also taken from the solution of part b of Problem 3. It is evident that the design moment (= 200 kNm) being greater than balanced moment (= 152.46 kNm), a doubly-reinforced beam has to be designed. The additional moment M′ is obtained from Eq.13.26, as

M′ = M - Mb = 200 – 152.46 = 47.54 kNm.

The additional tensile steel, obtained from Eq.13.35, is: Ast2 = M′ /σst (d-d′) = 47.54 (106)/140 (600–50) = 617.4 mm2 (assuming d′ = 50 mm). The amount of compression steel, obtained from Equation 13.36 is: Asc = (Ast2) / (σst)/{σcbc(1.5m -1)(1- d′/kd)} = (617.4)(140)/7(18.995)(0.792) = 820.79 mm2.


Total area of steel, obtained from Eq.13.33 is Ast = Ast1 + Ast2 = 2100 + 617.4 = 2717.4 mm2. Problem 8. Solve Problem 7 (Fig. 13.35.1) using tables of SP-16. Solution 8. The data of Problem 7 are: b = 350 mm, d = 600 mm, D = 650 mm, σcbc = 7 N/mm2 and σst = 140 N/mm2. We assume d′ = 50 mm, as in Problem 7. The design of doubly-reinforced beam is vary simple with the help of tables of SP-16. There are two ways. One is the direct solution using the appropriate table from Tables 72 to 79. The other option is to determine Ast1 and Ast2 from the equations and then to obtain Asc from Table M of SP-16. Both the methods are explained below. Method 1: Here, the appropriate table is Table 73. From the values of M/bd2 and d′ /d, we get the percentages of tensile (total) and compressive reinforcement Ast and Asc. Thereafter, the total tensile area of steel and compression area of steel are determined. M/bd2 = 200(106) / (350) (600) (600) = 1.59 and d′/d = 50/600 = 0.083. All tables from 72 to 79 of SP-16 have step values of M/bd2 and d′/d. So, the linear interpolations are to be done three times. The adjacent M/bd2 and d′/d are 1.55 and 1.60, and 0.05 and 0.1. First, we carry out linear interpolation for d′/d = 0.5. For M/bd2 = 1.59 (when d′/d = 0.5) pt = 1.253 + (0.038 / 0.05) 0.04 = 1.2834 and pc = 0.305 + (0.045 /0.05) 0.04 = 0.341 Again for M/bd2 = 1.59, when d′/d = 0.1 pt = 1.267 + (0.04/0.05) (0.04) = 1.299 and pc = 0.375 + (0.056 / 0.05) (0.04) = 0.4198. Now, the linear interpolation is done for M/bd2 = 1.59 and d′/d = 0.083. The results are given below: pt = 1.2834 + (0.033) (1.299 – 1.2834) / 0.05 = 1.2936 pc = 0.341 + (0.033) (0.4198 – 0.341) / 0.05 = 0.404


Thus, we have area of total tensile steel, Ast = (1.2936) (350) (600)/100 = 2716.56 mm2 and the area of compression steel, Asc = 0.404 (350) (600)/100 = 848.4 mm2. In the above calculations, more number of digits are taken to minimise the truncation error. We get almost the same values of Ast and Asc for Problem 7 by direct computation method (same as those of Problem 8). They are as follows: Ast = 2717.4 mm2 and Asc = 820.79 mm2. The values are reasonably comparable. Method 2: Here, in this method the values of Ast1 and Ast2 are obtained by direct computation as in Problem 7 giving Ast1 = 2100 mm2 and Ast2 = 617.4 mm2. Table M of SP-16 gives the ratio of Asc to Ast2 for different d′/d ratios. For d′/d = 0.083, by linear interpolation, we get Asc / Ast2 = 1.2 + (0.2) (0.333)/(0.05) = 1.332 So, Asc = (1.332) (617.4) = 822.37 mm2. This value is close to that of direct computation method as the ratio is depending on Ast2 obtained by direct computation method. 13.35.3 Practice Questions and Problems with Answers Q.1: Design the beam of Problem 5 of sec. 13.35.2 (Fig. 13.35.2) using σcbc = 8.5

N/mm2 and σst = 140 N/mm2. The given data of Problem 5 are: b = 300 mm, d = 700 mm and M = 100 kNm.

A.1: Steps 1 and 2 are not needed as preliminary dimensions of b and d are given. Step 3. Mb = Rb bd2 = (1.47) (300) (700) (700) = 216.09 kNm (Using Rb = 1.47 from Table 13.3). Ast, bal = pt, bal (bd/100) = (1.21) (300) (700) / 100 = 2541 mm2 (using pt, bal = 1.21 from Table 13.4). We provide 4-20 + 2-16 (=1256 + 402 = 1658 mm2) as Ast, giving pt = 1658 (100) / (300) (700) = 0.79 per cent. The minimum area of steel = 0.85 bd / fy = (0.85) (300) (700) / 250 = 714 mm2 ⟨1658 mm2. Hence, ok. Step 4. Check for stresses With the modular ratio m = 93.33 /8.5 = 10.98 and pt = 0.79; Eq. 13.16 gives: k = - (pt m/100) + {(pt m/100)2 + (pt m/50)}1/2 = 0.339 ; and j = 1- k/3 = 0.887.


Equation 13.24 gives: fst = M /{Ast d (1 – k/3)} = 100(106) / (1658) (700) (0.887) = 97.14 N/mm2 ⟨ 140 N/mm2

Equation 13.25 gives: fcbc = 2 Ast fst / bkd = 2(1658) (97.14) / (300) (0.339) (700)

= 4.53 N/mm2 ⟨ 8.5 N/mm2. Hence, this design is ok. In fact, there are several possible alternatives as explained in the solution of Problem 5. These alternatives are obtained quickly using tables of SP-16. Q.2: Design the beam of Q.1 (Fig. 13.35.2) using tables of SP-16. A.2: Steps 1 and 2 are not needed as the given data are: b = 300 mm, d = 700

mm and M = 100 kNm. Step 3. M/bd2 = 100(106) / (300) (700) (700) = 0.6803. Table 70 of SP-16 gives pt = 0.53775 by linear interpolation. Accordingly, Ast = (0.53775) (300) (700)/100 = 1129.275 mm2. Provide 4-20 mm diameter bars of area 1256 mm2, which has pt = 1256 (100) / (300) (700) = 0.598 Step 4. Check for stresses The modular ratio m = 10.98, k = 0.339 and j = 0.887 from A.1 of Q.1. Equation 13.24 gives: fst = M/Ast d(j) = 100 (106) / (1256) (700) (0.887) = 128.23 N/mm2 ⟨ 140 N/mm2. Equation 13.25 gives: fcbc = 2Ast fst / b kd = (2) (1256) (128.23) / (300) (0.339) (700) = 4.52 N/mm2 ⟨ 8.5 N/mm2 . Hence, the design is ok. Q.3: The problem of Q.4 of sec. 3.6.11 of Lesson 6 (Fig. 3.6.4) has the following data: b = 300 mm, d = 550 mm, D = 600 mm, Mu, lim = 220.45 kNm, Ast, lim = 1107.14 mm2, M 20 and Fe 500. (a) Determine the balanced moment of resistance and Ast,b in working stress

method. (b) Determine the areas of steel required to resist the design moment = Mu,lim

/1.5 = 146.97 kNm. A.3: (a) To determine Mb and Ast, b


Mb = Rb bd2 = (0.99) (300) (550) (550) = 89.84 kNm (using Rb = 0.99 from Table 13.3).

Using pt, bal = 0.39 from Table 13.4, Ast, b = (0.39) bd/100 = 643.5 mm2. (b) Design of doubly-reinforced beam From Eq.13.26: M′ = 146.97 – 89.84 = 57.13 kNm. From Eq.13.34: Ast1 = 643.5

mm2 for the balanced moment of 89.84 kNm (see sol.a of this question when pt, bal =

0.39). From Eq.13.35: Ast2 = M′/σst (d - d′ ) = (57.13) (106)/ (275) (500) = 415.49

mm2 (assuming d′ = 50 mm). Total Ast = Ast1 + Ast2 = 643.5 + 415.49 = 1058.99 mm2. From Eq.13.36, Asc = (Ast2) (σst) / {(σcbc) (1.5 m – 1) (1- d′ /kd)} =

(415.49)(275) / {(7) (18.995) (0.636)} = 1350.36 mm2. The beam requires Ast = 1058.99 mm2 and Asc = 1350.36 mm2. In the limit state method the values are: Ast = 1107.14 mm2 and Asc = 0, i.e., only singly-reinforced beam with almost the same amount of Ast is sufficient. This shows the economy in the design when done by employing limit state method. This problem, however, cannot be solved by using tables of SP-16 as the tables of SP-16 do not have the values for Fe 500 grade of steel. 13.35.4 Limitations of Working Stress Method The basis of the analysis by the working stress method is very simple. This method was used for the design of steel and timber structures also. However, due to the limitations of the method, now the limit state methods are being used. The limitations of working stress method are the following:

i) The assumptions of linear elastic behaviour and control of stresses within specially defined permissible stresses are unrealistic due to several reasons viz., creep, shrinkage and other long term effects, stress concentration and other secondary effects.

ii) The actual factor of safety is not known in this method of design. The

partial safety factors in the limit state method is more realistic than the


concept of permissible stresses in the working stress method to have factor of safety in the design.

iii) Different types of load acting simultaneously have different degrees of

uncertainties. This cannot be taken into account in the working stress method.

Accordingly, the working stress method is gradually replaced by the limit state method. The Indian code IS 456 has given working stress method in Annex B to give greater emphasis to limit state design. Moreover, cl. 18.2.1 of IS 456 specifically mentions of using limit state method normally for structures and structural elements. However, cl.18.2.2 recommends the use of working stress method where the limit state method cannot be conveniently adopted. Due to its simplicity in the concept and applications, better structural performance in service state and conservative design, working stress method is still being used for the design of reinforced concrete bridges, water tanks and chimneys. In fact, design of tension structures and liquid retaining structures are not included in IS 456 for the design guidelines in the limit state method of design. Accordingly, Lessons 34 and 35 include the basic concept of this method. The design of T-beam in flexure, rectangular and T-beams under shear, torsion and other topics of the limit state method, covered in different lessons, are not included adopting working stress method. However, the designs of tension structures and liquid retaining structures are taken up in the next lesson, as these are not included by the limit state method. 13.35.5 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 13.35.6 Test 35 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Problem 4.1 of sec. 4.9.2 of Lesson 4 has the following data: (Fig. 4.9.1) b

= 300 mm, d = 630 mm, d′ = 70 mm, D = 700 mm, M 20, Fe 415, Mu, lim = 492.96 kNm, Ast1 =1809.14 mm2, Ast2 = 783.621 mm2 (Ast = 2572.834 mm2), Asc = 806.517 mm2 (by direct computation method). Tension steel reinforcement = 4-25 mm diameter bars + 2-20 mm diameter bars (= 2591 mm2) and compression steel reinforcement = 2-20 mm diameter bars + 2-12 mm diameter bars (= 854 mm2).

Determine the areas of steel by working stress method when the design

moment = Mu, lim / 1.5 = 328.64 kNm. Use the same dimensions of the beam as mentioned above. Use direct computation method and also tables of SP-16.

(25+25 = 50 marks) A.TQ.1: (1) Direct computation method Given data are: b = 300 mm, d = 630 mm, σcbc = 7 N/mm2 and σst = 230 N/mm2.

Mb = Rb bd2 = 0.91 (300) (630) (630) = 108.35 kNm (using Rb = 0.91 from Table 13.3). Therefore, M′ = M – Mb = 328.64 – 108.35 = 220.29 kNm.


Ast b = pt, bal (bd/100) = 0.44 (300) (630) / 100 = 831.6 mm2 (using pt, bal = 0.44 from Table 13.4) and Ast2 = M′ / {σst (d-d′ )} = 220.29 (106) / {(230) (630 – 70)} = 1710.32 mm2. Therefore, Ast = 831.6 + 1710.32 = 2541.92 mm2. The modular ratio m = 93.33 / σcbc = 13.33 and kb = 93.33/(σst +93.33) = 0.289. Asc =(Ast2) (σst) / {σcbc (1.5 m –1) (1-d′/kd) = 4806.37 mm2 , giving pc = Asc (100)/ bd = (4806.37) (100) / (300) (630) = 2.54 per cent ⟨ maximum percentage allowed (4 per cent) as per cl.26.5.1.2 of IS 456.

So, Ast = 2541.92 mm2 = (4-25 + 2-20 giving 2591 mm2) and Asc =4806.37 mm2 = (8-25 + 3-20 giving 4869 mm2)

(2) Using tables of SP-16

M/bd2 = 328.64 (106) / (300) (630) (630) = 2.76 d′/d = 70/630 = 0.11 Table 77 has the values of pt and pc for different values of M/bd2 and d′/d. Therefore, the final values of pt and pc are to be determined by carrying out the linear interpolation three times, as explained in the solution of Problem 8. The results are given below:

(a) For Mu/bd2 = 2.75, d′/d = 0.11

pt = 1.327 + 0.052 (0.01) / 0.05 = 1.3374 pc = 2.348 + 1.03 (0.01) / 0.005 = 2.555

(b) For Mu/bd2 = 2.8 and d′/d = 0.11 pt = 1.351 + (0.053) (0.01) / 0.05 = 1.3616 pc = 2.412 + (1.063) (0.01) / 0.05 = 2.6246

(c) Therefore, for Mu/bd2 = 2.76 and d′/d = 0.11

pt = 1.3374 + (0.0342) (0.01) / (0.05) = 1.34424 pc = 2.555 + (0.0696) (0.01) / (0.05) = 2.5689

So, Ast = (1.34424) (300) (630) / 100 = 2540.6 mm2


and Asc = (2.5689) (300) (630) / 100 = 4855.2 mm2.

The values of Ast = 2541.92 mm2 and Asc = 4806.37 mm2, obtained by direct computation method in part (1) of this problem, are reasonably comparable with the values obtained by using tables of SP-16.

13.35.7 Summary of this Lesson This lesson illustrates the applications of the equations developed in Lesson 34 for the analysis and design of singly and doubly-reinforced rectangular beams by working stress method. For this purpose, several numerical problems are solved. Some of the problems have the same dimensions of width and effective depth but with different grades of steel so that the results are compared. Moreover, problems solved earlier on singly and doubly-reinforced beams are also taken up to compare the results. The method of direct computation i.e., using the appropriate equations directly and the use of tables of SP-16 are employed for some of the problems to illustrate the simplicity in using the tables of SP-16. Several practice problems, test problem and illustrative examples will help in understanding the applications for the analysis and design of singly and doubly-reinforced rectangular beams.


Module 14

Tension Members Version 2 CE IIT, Kharagpur

Lesson 36

Structural Requirements, Code

Stipulations and Governing Equations




• state the need to design the tension structures using reinforced concrete though concrete is very weak in tension,

• justify the use of working stress method for designing reinforced concrete tension members,

• name the two essential requirements for the design of reinforced concrete tension structures,

• explain why some amount of cracking may be allowed in such structures,

• explain why porous aggregates are not allowed in the reinforced concrete tension structures,

• state the basis of design of reinforced concrete tension members,

• state the different cases of reinforced concrete tension members, and

• establish the governing equations for all the cases of tension members to determine the stresses of concrete and steel following the working stress method of design.

14.36.1 Introduction Reinforced concrete is used in the design of several practical members of tension structures though it is not a potential material due to very inadequate tensile strength of concrete compared to its compression strength. In such cases, reinforced concrete develops tensile stresses either due to direct tension force or combined with bending. Tie members of trusses and arches, walls of rectangular tanks and bunkers, suspended roofs, cylindrical pipes, walls of liquid retaining structures are some of the examples where tension stresses also develop. Traditional elastic approach (working stress method) and the limit state method can be applied for the design of such structures. Based on elastic theory, the working stress method is simpler in concept and applications, while the limit state method, based on cracking behaviour of concrete, is not fully developed yet. Some of the structures or members mentioned above may not be in direct contact with any liquid. On the other hand, some may be in direct contact with liquid. Though control of cracking is important for all types of structures, it is an essential requirement for liquid retaining structures. Impermeability of structure is not only for preventing leakage but also for improving durability, resistance to leaching, chemical action, erosion, frost damage and protecting the reinforcement from corrosion.


IS 456 gives due considerations for compression either direct or associated with bending and shear when limit state method is employed. However, it is almost silent on tension members. IS 3370, dealing with liquid retaining structures, is still adopting working stress method. Thus, the applications of limit state method for the design of tension members are still in the development stage. Accordingly, the working stress method shall be employed in this lesson to explain the design of such members. 14.36.2 Resistance to Cracking and Strength

Figures 14.36.1a and b show one rectangular section under flexural tension and the distribution of stresses, respectively. The symmetric section of depth D is having the neutral axis at a depth of 0.5 D. Assuming that the actual tension stress of M 25 concrete (say) in bending ftb reaches the permissible value of σtb = 1.8 N/mm2 for resistance to cracking (see Table 14.1), the tensile stress of steel fst at the level of steel shall be modular ratio times 0.8 σtb or 15.81 N/mm2 {(10.98) (0.8) (1.8) N/mm2, where modular ratio of M 25 concrete m = 93.33 / 8.5 = 10.98 and d′ = 0.1 D}. The value of fst is very low when compared with the values of the permissible tension stress of steel (115 N/mm2 for Fe 250 and 150 N/mm2 for Fe 415 as given in Table 14.2. Therefore, some amount of cracking may be permitted when the structure is not in contact with any liquid at all. Moreover, the remote face of the liquid retaining structures, which is not in contact with the liquid, also is a possible place of allowing minor cracks. However, it is also important to note that sufficient depth is available to prevent the corrosion of the reinforcement.


Allowing minor cracks in such cases means increasing the values of permissible stresses of concrete and steel reinforcement specifying the particular cases. IS 3370 Part II recommends the permissible stresses of concrete and reinforcement for different types of situations, which are given in Tables 14.1 and 14.2. Hence, the design of tension members shall provide adequate control of cracking in addition to normal requirements of strength. There are many other issues involved in the design of tension structures. IS 3370 Parts I to IV deal with such issues in detail. The relevant important aspects are briefly mentioned in the following section. 14.36.3 IS Code Stipulations (a) Porous aggregates – (cl.2.1.a of IS 3370, Part I) Use of porous aggregates, such as burnt clay and broken bricks or tiles is not

allowed for parts of structure either in contact with the liquids on any face or enclosing the space above the liquid.

(b) Concrete mix - (cl.3.1 a of IS 3370, Part I) Concrete mix weaker than M 20 shall not be used except in thick sections

(thickness > 450 mm) and parts of structure neither in contact with the liquid on any face nor enclosing the space above the liquid.

(c) Basis of design - (cl.3.2 of IS 3370, Part II)

(i) The parts of the structure not in contact with the liquid shall be designed in accordance with the requirements of IS 456.

(ii) Design of members other than those in (i) above shall be based on

consideration of adequate resistance to cracking as well as adequate strength.

(iii) Calculation of stresses shall be based on the following assumptions in

addition to the general assumptions given in IS 456.

1. The whole section of concrete including the cover together with the reinforcement shall be taken into account in calculating both flexure and direct tension (or combination of both) relating to resistance to cracking.

2. Total shear stress = Q/(b jd), shall not exceed the permissible value given

in Table 14.1, where Q, b and jd are the total shear, breadth and lever arm, respectively.


3. Concrete has no tensile strength in strength calculations. (d) Permissible stresses in concrete - (cl.3.3 of IS 3370 Part II)

(i) For resistance to cracking

Table 14.1 furnishes the permissible stresses in tension σtd and σtb (direct and due to bending) and shear τsh for calculations relating to the resistance of members to cracking. The permissible tensile stresses due to bending apply to the face of the member in contact with the liquid. In members less than 225 mm thick and in contact with the liquid on one side, these permissible stresses in bending apply also to the face remote from the liquid.

Table 14.1: Permissible stresses in concrete in calculations relating to

resistance to cracking

Permissible stresses (N/mm2)

Grade of concrete Direct Tension,

σtd

Tension due to Bending σtb

Shear τsh (Q / b jd)

M 15 1.1 1.5 1.5 M 20 1.2 1.7 1.7 M 25 1.3 1.8 1.9 M 30 1.5 2.0 2.2 M 35 1.6 2.2 2.5 M 40 1.7 2.4 2.7

(ii) For strength calculations

In strength calculations, the permissible concrete stresses shall be in accordance with IS 456 (Table 13.1 of Lesson 34). Where the calculated shear stress in concrete alone exceeds the permissible value, reinforcement acting in conjunction with diagonal compression in the concrete shall be provided to take the whole of the shear.

(e) Permissible stresses in steel - (cl.3.4 of IS 3370, Part II)

(i) For resistance to cracking When steel and concrete are assumed to act together for checking the

tensile stress in concrete for avoidance of cracks, the tensile stress in the steel will be limited by the requirement that the permissible tensile stress in concrete is not exceeded, so that the tensile stress in steel shall be


equal to the product of modular ratio of steel and concrete and the corresponding allowable tensile stress in concrete.

(ii) For strength calculations For strength calculations the permissible stresses in steel reinforcement

are given in Table 14.2. Table 14.2: Permissible stresses in steel reinforcement for strength

calculations

Permissible stresses σst N/mm2 Sl. No

Types of stress in steel reinforcement

Fe 250 Fe 415

(i) Tensile stress in members under direct tension

115 150

(ii) Tensile stress in members in bending (a) On liquid retaining face of

members (b) On face away from liquid for

members less than 225 mm in thickness

(c) On face away from liquid for members 225 mm or more in thickness

115

115

125

150

150

190

(iii) Tensile stress in shear reinforcement: (a) For members less than 225

mm in thickness (b) For members 225 mm or

more in thickness

115

125

150

175

(iv) Compressive stress in columns subjected to direct load

125 175

Note: Stress limitations for liquid retaining faces shall also apply to the

following: a) Other faces within 225 mm of the liquid retaining face. b) Outside or external faces of structures away from the liquid but placed in

water logged soils up to the level of the highest subsoil water.


14.36.4 Types of Tension Structures

Figure 14.36.2 summarises the two types of tension structures made of reinforced concrete indicating the different cases. The two types are (i) structures having no contact with the liquid and (ii) structures in direct contact with the liquid or liquid retaining structures. The different cases are first due to the three sources of tension stresses: (i) due to direct tension, (ii) pure flexure and (iii) direct tension associated with the flexure. Secondly, for the liquid retaining structures if the tension is on the (i) liquid face or (ii) on the remote face. Further two cases are there for the structures having tension on the remote face depending on, if the depth of section is (i) < 225 mm or (ii) ≥ 225 mm. It is worth mentioning that understanding of the behaviour of tension structures and developing the governing equations of different cases are possible by taking up three main types: (i) when subjected to axial tension only, (ii) when subjected to bending moment only and (iii) when subjected to combined axial tension and bending moment. Accordingly, we take up these three main types in the following sections.


14.36.5 Members Subjected to Axial Tension only Based on the discussion made so far, the criteria of design of such tension members are:

(i) Reinforcement alone shall take the full tension force. (ii) The tension stress in steel shall not exceed the recommended

permissible stress values given in Table 22 of IS 456 and Table 13.2 of Lesson 34.

(iii) The tensile stress of concrete shall be determined transforming the

steel into equivalent concrete.

(iv) The tensile stress in the transformed section shall not exceed the stipulated permissible values given in cl. B-2.1.1 of IS 456 (Table 13.1 of Lesson 34). This will regulate the cracking of concrete.

Therefore, we can determine the actual tensile stress of steel fst and

concrete ftd from the following equations:

fst = Ft / Ast (14.1)

and ftd = Ft /(Ac + m Ast) (14.2)

where Ft = total tension force on the member minus pretension in steel, if any,

before concreting, Ast = cross-sectional area of reinforcing steel in tension, Ac = cross-sectional area of concrete excluding any finishing material and

reinforcing steel, and m = modular ratio

14.36.6 Members Subjected to Pure Flexure For this, we will consider two cases: (i) when the section is uncracked and

(ii) when the section is cracked. The first case is applicable when either the tension is on the liquid face or when the reinforcement is away from the liquid by a distance less than 225 mm. The second case is applicable when there is no direct contact of the structure with the liquid. The two cases are explained below.


(A) Uncracked Section

Figures 14.36.3a and b show the uncracked section and the stress distribution of such section subjected to bending moment alone. The maximum bending stresses in compression or tension are given by the expression. fcb = M x / Iyy (14.3) and ftb = M (D-x) / Iyy (14.4) where M = Applied moment on the member

x and (D - x) = maximum distances of the compression and tension fibres from the neutral axis, x is also the depth of the neutral axis.

Iyy = moment of inertia of the uncracked section about the neutral axis of the section transforming area of steel into equivalent concrete.

The depth of the neutral axis x is obtained by taking moment of the transformed section about the top face, which gives: bD2/2 + (m-1) pt b Dd/100 = {bD + (m-1) pt bD / 100}x or x = {b D2/2 + (m –1) pt b Dd/100} / {bD + (m-1) pt bD / 100} (14.5) The moment of inertia of the transformed section about the neutral axis is obtained from: :


Iyy = b x3 / 3 + b (D – x)3 / 3 + (m –1) pt bD (D-x)2/100 (14.6) The values of fcb and ftb are then obtained from Eqs. 14.3 and 14.4, respectively. The moment of resistance of the section is determined from Eq.14.4 as ftb is the governing stress of the section. This gives M = ftb Iyy / (D – x) (14.7) (B) Cracked Section

Figures 14.36.4a and b show the section and stress distribution of the cracked section when subjected to moment alone. This particular case has been discussed in secs. 13.34.8 and 13.34.9 when the section is balanced and under-reinforced, respectively. Therefore, the equations are not derived here and only the final equations are given for the balanced and under-reinforced sections separately. The only difference is the notation of compressive stress of concrete in bending compression, which is fcbc in secs. 13.34.8 and 9, while the same is denoted by fcb here.

(i) Equations for the under-reinforced sections

The depth of the neutral axis is determined from x = kd, where k = - (pt m / 100) + {(ptm / 100)2 + (pt m / 50)}1/2 (13.16) The compressive and tensile forces C and T are obtained from:


C = (1/2) fcb b kd

(13.20) T = fst Ast

(13.6) The moment of resistance with respect to steel is obtained from: M = Ast fst d(1 – k/3) (13.24) M = (pt /100)fst (1 – k/3) bd2 (13.19)

(ii) Equations for the balanced section are as follows: The depth of the balanced neutral axis is determined from: xb = kb d, where kb = 93.33 / (σst + 93.33) (13.3) The expressions of compressive and tensile forces C and T are: C = (1/2) (σcb) b kb d (13.5) T = Ast σst (13.6) The expressions of the balanced moment of resistance with respect to concrete and steel are: Mb = (1/2) σcb kb jb (b d 2) (13.7) Mb = (pt, bal / 100) σst jb (b d2) (13.8) 14.36.7 Members Subjected to Combined Axial Tension and Moment. Here also, we will consider two cases: (i) when the section is uncracked and (ii) when the section is cracked.


(A) Uncracked Section If tension is there on the liquid face, the tensile stress actually developed, i.e., ftd and ftb in direct tension and bending tension, respectively should satisfy the following condition: (ftd /σtd) + (ftb / σtb)≤ 1.0 (14.8) where σtd and σtb are the permissible stresses of concrete in direct tension and bending tension, respectively and are to be taken from Table 14.1, and ftd and ftb are the actual stresses of concrete in direct tension and bending tension, respectively and are determined from Eqs.14.2 and 14.4, respectively. (B) Cracked Section This case is classified under two groups: (i) when the eccentricity e of the tension force Ft is small; i.e., tensile force is large and (ii) when the eccentricity e of the tension force is large, i.e., tension force is small. We are explaining them separately. (i) When the eccentricity is small, i.e., Ft is large

Figures 14.36.5a and b show the cracked section subjected to tension force at a small eccentricity e within the depth of the section. As the eccentricity is small, measured from the center of gravity of the two areas of steel, As1 and As2, the whole section is in tension and hence the entire concrete is ineffective.


We have two equations of equilibrium: (i) equilibrium of forces and (ii) equilibrium of moments. They are expressed as follows: Ft = Ts1 + Ts2 = fst1 As1 + fst2 As2 (14.9) Moment of the forces about the bottom steel gives: Ft (d1 – e) – Ts2 (d1 + d2) = 0 or Ft (d1 – e) – fst2 As2 (d1 + d2) = 0 (14.10)

Substituting the expression of Ft from Eq.14.9 into Eq.14.10, fst2 can be computed. The value of fst1 can then be computed from Eq.14.9. (ii) When the eccentricity is large, i.e., Ft is small

Figure 14.36.6a and b show the section and stress distribution of cracked section subjected to tension force Ft having large value of the eccentricity e measured from the mid-depth of the section. The line of action of Ft is outside the section. The areas of compression and tension steel are denoted by Asc and Ast, respectively. Due to large eccentricity, the direct and bending stresses are equally dominant. The forces of compression and tension steel are represented by Csc and Ts, respectively, while the force of compressive concrete is represented by Cc. The distance of the neutral axis is x.


The expressions of these three forces are: Csc = fsc Asc (14.11) Ts = fst Ast (14.12) Cc = 0.5 fcb bx (14.13) The expressions of fsc and fst, obtained from the stress distribution of Fig. 14.36.6 b, are: fsc = (mc –1) fcb (x - d′)/x (14.14) fst = m fcb (d – x) / x (14.15) Using the expression of fsc and fst from Eqs. 14.14 and 14.15 in Eqs. 14.11 and 14.12, we have: Csc = (mc – 1) Asc fcb (x - d′) / x (14.16) Ts = m Ast fcb (d – x) / x (14.17) The equation of the equilibrium of forces is: Ft – Ts + Cc + Cse = 0 (14.18) Taking moment of all forces about the line of action of Ft, the equation of equilibrium of moment is: Ts (e + 0.5D - d) – Cc (e + 0.5D – x/3) – Csc (e + 0.5D - d′ ) = 0 (14.19) Substituting the expressions of Cc, Csc and Ts from Eqs. 14.13, 14.16 and 14.17, respectively into Eqs. 14.18 and 14.19, we have: Ft = fcb {m Ast (d – x) / x – 0.5 bx – (mc – 1)Asc (x -d′) / x} (14.20)


m Ast (d –x) (e + 0.5D –d) / x = 0.5 bx (e + 0.5D – x/3) + (mc – 1) Asc (x-d′) (e + 0.5D - d′) / x (14.21) Equations 14.20 and 14.21 have seven unknowns b, d, D, x, fcb, Ast and Asc of which b, d and D are assumed or taken from the preliminary dimensions. So, we have four unknowns x, fcb, Ast and Asc to be determined from two equations (Eqs. 14.20 and 14.21). Therefore, trial and error method has to be used. As a particular case, when Asc = 0, Eqs. 14.20 and 14.21 are reduced to: Ft = fcb {m Ast (d –x) / x – 0.5 bx} (14.22) and m Ast (d – x) (e + 0.5 D – d) / x = 0.5 bx (e + 0.5 D – x/3) (14.23)

Here also the three unknowns x, fcb and Ast are determined from Eqs. 14.22 and 14.23 by trial and error. However, for such problems of singly-reinforced section, the initial value of Ast can be determined as explained below. We use the equation of equilibrium of forces i.e., Eq. 14.24 as such and write the equation of equilibrium of moment by taking moment about Ts, which gives Cc (jd) – Ft (e + 0.5D – d) = 0 (14.24) or Cc = Ft (e + 0.5 D – d)/jd (14.25) Using the expression of Cc from Eq. 14.25 in Eq. 14.18, when Asc = 0, we get: Ft = Ts - Cc = Ts – Ft (e + 0.5 D – d)/ jd or Ts =Ft {1 + (e + 0.5 D –d)/ jd} (14.26)


Assuming Ts = Ast σst in Eq. 14.26, we have: Ast = (Ft / σst){1+ (e + 0.5 D – d)/ jd} (14.27)

Equation 14.27 is used to get the starting value of Ast assuming j as 0.87 (say). This value of Ast now can be used in Eq. 14.23 to get the value of x. Using that value of x in Eq. 14.22, the value of fcb can be determine. If fcb >σcb, either the value of Ast shall be increased keeping D as the same or the value of D may be increased keeping Ast as the same. The third option is to increase both Ast and D, as appropriate. Numerical problems are taken up in Lesson 37 for tension structures having no contact with liquid and for liquid retaining structures covering all the cases explained in Fig. 14.36.2. 14.36.8 Practice Questions and Problems with Answers Q.1: Why is it needed to design the reinforced concrete tension structures when

concrete is not good in tension? A.1: Sec. 14.36.1, Paragraph 1. Q.2: Name the two criteria of design of reinforced concrete tension structures. A.2: Sec. 14.36.2 Q.3: State the basis of design of reinforced concrete tension structures. A.3: Sec. 14.36.3, Part (c) Q.4 Draw a schematic diagram of different cases of tension structures made of

reinforced concrete. A.4: Fig. 14.36.2 Q.5: Establish the equations to determine the tensile stresses of concrete and

steel when the member is subjected to axial force Ft alone. A.5: Sec. 14.36.5 Q.6: Establish the equations to determine the stresses in concrete due to applied

moment only when the section is uncracked. A.6: Sec. 14.36.6, Part A


Q.7: Answer Q.6, if the section is cracked. A.7: Sec. 14.36.6, Part B Q.8: Answer Q.6. if the section is uncracked and subjected to combined axial

tension and moment. A.8: Sec. 14.36.7, Part A Q.9: Answer Q.6, if the section is cracked and subjected to combined axial

tension and moment. Assume the eccentricity of the tension force Ft is small.

A.9: Sec. 14.36.7, Part B (i) Q.10: Answer Q.6, if the section is cracked and subjected to combined axial

tension and moment. Assume the eccentricity of the tension force Ft is large.

A.10: Sec. 14.36.7, Part B, (ii) 14.36.9 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 14.36.10 Test 36 with Solutions Maximum Marks = 50, Maximum Time = 30 minutes Answer all questions. TQ.1: Draw a schematic diagram of different cases of tension structures made of

reinforced concrete. (20 marks) A.TQ.1: Fig. 14.36.2 TQ.2: Establish the equations to determine the stresses in concrete due to

applied moment only when the section is uncracked. (15 marks)

A.TQ.2: Sec. 14.36.6, Part A TQ.3: Answer TQ.2, if the section is cracked and subjected to combined axial

tension and moment. Assume the eccentricity of the tension force Ft is small.

(15 marks) A.TQ.3: Sec. 14.36.7, Part B (i)


14.36.11 Summary of this Lesson This lesson presents the analysis and design of tension structures following the working stress method as stipulated in IS 456 and IS 3370. The two requirements i.e., the resistance to cracking and strength must be considered for all types of tension structures whether they are in direct contact with the liquid or not. The respective permissible stresses of concrete in direct tension, bending tension, bending compression and shear are taken from the IS Codes for both cases: when no cracking is allowed and when some cracking may be allowed. The different cases of tension structures are classified depending on if they are in direct contact with the liquid or not, if the tension is on the liquid face or not and on the depth of the section. Understanding of the philosophy and concept of design and the different equations established for different cases will help to apply them in numerical problems. The numerical problems are taken up in Lesson 37.


Module 14

Tension Members Version 2 CE IIT, Kharagpur

Lesson 37

Numerical Problems




• select the category of the problem if it is a liquid retaining structure or a

structure having no contact with the liquid,

• identify the appropriate equations to be applied for the solution of a

particular problem belonging to specific category,

• select the permissible stresses in concrete and steel depending on the

category of the problems,

• apply the appropriate equations for the solution of the problems,

• solve the problems by establishing the equations from the first principle,

• revise the sections, if needed, to satisfy the requirements as per IS codes.


The requirements for the design of tension structures / members are discussed in Lesson 36. The stipulations of Indian Standard Codes are also mentioned. Tension structures / members are classified depending on (i) if they are liquid retaining or not having any contact with the liquid, (ii) if the tensile stresses are developed due to pure axial force, pure moment or combined effects of axial force and moment, (iii) if the tension is developed on the liquid face or on the remote face and (iv) if the depth of the member is less than, equal to or greater than 225 mm when the tension is developed in the remote face. Accordingly, the permissible stresses of concrete and steel are different. The equations are also different based on if the section is assumed to be cracked or uncracked. All these issues are explained in Lesson 36.

This lesson explains the applications of the established equations for the different cases as mentioned above through several numerical problems. The problems are solved in step-by-step mostly utilising the equations. However, a few problems are solved from the first principle for better understanding. Understanding the solutions of illustrative examples and solutions of the practice and test problems will give confidence in applying the appropriate equations and selecting the permissible stresses for analysing and designing tension structures or members using working stress method and following the stipulations of Indian Standard Codes.


14.37.2 Numerical Problems Problem 1. Determine the dimensions and area of tension steel of a

reinforced concrete rectangular tie member of a truss for carrying a direct tensile force Ft = 470 kN using M 25 and Fe 415.

Solution 1. The theory of such tensile members made of reinforced concrete is

discussed in sec.14.36.5 of Lesson 36. We have Eqs. 14.1 and 14.2 to determine the required parameters. The modular ratio m of M 25 concrete is 93.33/8.5 = 10.98, fst of Fe 415 is 230 N/mm2 and ftd ≤ 3.2 N/mm2 of M 25 concrete (Table 13.1 of Lesson 34).

Step 1. Determination of Ast

From Eq. 14.1 of Lesson 36, we have: Ast = Ft / fst assuming that fst, the actual stress will reach the value of σst = 230 N/mm2, which gives Ast = 470,000/230 = 2043.48 mm2. Provide 4 bars of 20 mm diameter and 4 bars of 16 mm diameter (= 2060 mm2).

Step 2. Determination of width and depth of the rectangular tie member From Eq. 14.2 of Lesson 36, we have: Ac = (Ft /ftd) – m Ast

or Ag = (Ft / ftd) – (m – 1)Ast, where Ag = gross area of the tie member. Substituting the values of Ft =470 kN, ftd = σtd = 3.2 N/mm2, m = 10.98 and Ast = 2060 mm2, we get Ac = 126316.2 mm2. Provide 300 mm x 425 mm (= 127500 mm2) as the width and depth of the tie member.

Step 3. Checking for the tensile stress in concrete From Eq. 14.2 of Lesson 36, we have: ftd = Ft / (Ac + m Ast) = Ft / {Ag + (m – 1)Ast} = 3.17 N/mm2 < 3.2 N/mm2. Hence, ok

Therefore, the tie member of 300 mm x 425 mm with 4 bars of 20 mm diameter and 4 bars of 16 mm diameter is the solution.

Problem 2. Design a bunker wall of 250 mm depth for resisting a moment of

33 kNm using M 20 and Fe 415. Solution 2. Let us assume the section as cracked. The necessary equations of a

cracked section subjected to pure flexure are given in sec.14.36.6B. For M 20 grade of concrete, m = 93.33/7 = 13.33 and kb = 93.33 / (σst + 93.33), as per Eq. 13.3 of Lesson 36. Here, kb = 0.288, using


σst = 230 N/mm2 and assume d = 225 mm from the given value of D = 250 mm.

Step 1. Determination of preliminary area of steel The lever arm = jd = (1 – kb / 3) d = 0.904 (225) = 203.4 mm.

Therefore, from Eq. 13.8 of Lesson 34, preliminary Ast = M/{σst (lever arm)} = 33 (106) / (230) (203.4) = 705.399 mm2. Let us provide 12 mm diameter bars @ 160 mm c/c to have 707 mm2/m.

Step 2. Determination of depth of neutral axis

The depth of neutral axis can be determined from Eq. 13.16 of

Lesson 34. Alternatively, we can determine x by taking moment of the area of concrete and steel, converting it into equivalent concrete. With reference to Fig. 14.37.1, taking moment of the two areas about the neutral axis, 1000 (x2/2) = 13.33 (707) (225-x) or x2 + 18.84862 x – 4240.9395 = 0, which gives x = 56.38 mm. This gives the lever arm = 225 – 56.38/3 = 206.21 mm.

Step 3. Determination of tensile stress of steel Tensile stress of steel fst = M/{(Ast) (lever arm) = 33 (106) / {(707)

(206.21)} = 226.35 N/mm2 < 230 N/mm2. Hence, the section is ok. Step 4. Checking of the maximum bending compression stress in

concrete


Figure 14.37.2a shows the section. The maximum bending

compression stress of concrete is determined either form the stress distribution diagram of Fig. 14.37.2b or after determining the moment of inertia of the section about the neutral axis. From the stress distribution diagram of Fig. 14.37.2b, we have: fcb = (fst / m) {x / (d-x)} = (226.35 / 13.33) (56.38 / 168.62) = 5.678 N/mm2 ⟨ 7 N/mm2.

The moment of inertia about the neutral axis is: Iyy = 1000 (56.38)3 / 3 + (13.33) (707) (168.62)2 = 0.3277 (109) mm4.

Then, fcb = 33 (106) (56.38) / {0.3277 (109)} = 5,678 N/mm2 ⟨ 7 N/mm2. Hence, the section, as shown in Fig. 14.37.2a, is ok.

Problem 3. Determine the area of tensile steel of a singly-reinforced bunker

wall of depth 250 mm subjected to Ft = 50 kN/m and M = 35 kNm/m at the horizontal level. Use M 20 and Fe 415 grade of steel.

Solution 3. In this problem, the eccentricity of the tension force Ft is M/Ft = 35(103) / 50 = 700 mm, i.e., the tension force is acting outside the section (Figs. 14.36.6a and b). The governing equations of such section are given in sec. 14.36.7.B(ii), assuming the section as cracked. Given D = 250 mm, so d = 225 mm. The value of the modular ratio m = 93.33 / 7 = 13.33. For M 20 concrete fcb ≤ 7 N/mm2. Let us assume j = 0.87.


Step 1. Preliminary area of tension steel From Eq. 14.27 of Lesson 36, we have: Ast =(Ft /σst) {1 + (e + 0.5 D – d) / jd}

(14.27) or Ast = (5000/230) {1 + (700 + 125 – 225) / (0.87) (225)} = 883.72 mm2.

Provide 12 mm diameter bars @ 120 mm c/c to have 942 mm2. Step 2. Determination of the depth of the neutral axis From Eq. 14.23 of Lesson 36, we get m(Ast) (d-x) (e+0.5 D – d) = 0.5 bx2(e + 0.5D – x/3) or 13.33 (942) (225 – x) (700+125-225) = 500 x2 (700 +125-x/3) or x3 – 2475 x2 – 45204.696 x + 1017105.6 = 0 By Newton’s method, we get x = 56.171 mm. Step 3. Determination of stresses of concrete and steel From Eq. 14.22 of Lesson 36, we have: fcb = Ft /{m Ast (d –x)/x – (0.5 b x)} = 50000 / {(13.33) (942) (225 – 56.171)/56.171 – 0.5 (1000) (56.171)} = 5.18 N/mm2 ⟨ 7.00 N/mm2

From Eq. 14.15 of Lesson 35, we have: fst = m fcb (d – x) / x = 13.33 (5.18) (225-56.171)/56.171 = 207.54 N/mm2 ⟨ 230 N/mm2. Hence, the design is ok. Problem 4. Design an interior slab panel of dimensions 4000 mm x 4000

mm of a flat bottom water tank to support 3 m of water. Use M 20 concrete and Fe 415 steel.

Solution 4. This is a slab panel of liquid retaining structure. We have to

determine the depth of the slab and area of steel by trial and error. Let us assume D = 250 mm, and d = 200 mm giving a cover of 50 mm in the top liquid surface. The cover at the bottom surface is assumed to be 30 mm.


Step 1. Evaluation of loads Loads of 3 m height of water = 30000 N/m2

Loads of plastering (assumed) = 500 N/m2

Weight of reinforced concrete slab (of depth = 250 mm) = 25000 (250) (10-3) = 6250 N/m2

__________ Total loads = 36750 N/m2

Step 2. Calculation of bending moments

Table 26 of IS 456 gives the values of coefficients ∝x = 0.032 for the negative moment and ∝y = 0.024 for the positive moment. Thus, we have, negative moment = (0.032) (36750) (4) (4) = 18.816 kNm and positive moment = (0.024) (36750) (4) (4) = 14.112 kNm.

Step 3. Preliminary areas of steel Assuming fst = σst = 150 N/mm2, kb = 93.33/(150+93.33) = 0.38,

Lever arm = jd = (0.87) (200) = 174 mm for the negative moment and lever arm = jd = 0.87(220) = 191.4 mm for the positive moment at mid-span, the areas of steel are as follows:

At the support = M/σst jd = 18816000 / (150) (174) = 720.92 mm2

At the mid-span = M/σst jd = 14112000 / (150) (191.4) = 491.54 mm2.

Provide 10 mm diameter bars @ 100 mm c/c to give 785 mm2 at the support and 10 mm diameter bars @ 150 mm c/c at the mid-span to give 524 mm2.

Step 4. Checking for stresses and moment of resistance of the section (A) At the support


Figure 14.37.3 shows the section at the support, which is assumed as uncracked. The depth of neutral axis x is determined by taking moment of the areas of concrete and steel about the bottom, which gives:

x = {(1000) (250) (125) + (12.33) (785) (200)} / {(1000) (250) + (12.33) (785)} = 127.79 mm; D – x = 250 – 127.79 = 122.21 mm. The moment of inertia about the neutral axis is:

Iyy =(1000) (122.21)3 / 3 + (1000) (127.79)3 / 3 + (12.33) (785) (122.21 – 50)2 = (1.35) (109) mm4

The moment of resistance of the section is determined using ftb =

σtb = 1.7 N/mm2, as given below: M =ftb (Iyy) / (D-x) = 1.7 (1.35) (109) / 122.21 = 18.7 kNm < 18.816

kNm. Hence, the design has to be revised. Accordingly, the checking of stresses at the mid-span for positive moment is not carried out.

Step 5. Revised Section


(A) The depth of the section is increased to 270 mm. Accordingly, d =

220 mm for support section and d = 240 mm for mid-span section. With j = 0.87, the lever arm at the support section = 0.87 (220) =191.4 mm and at the mid-span section jd = 0.87 (240) = 208.8 mm. We have to add the extra self-weight of slab.

Earlier total loads in Step 1 = 36750 N/m2

Additional load of 20 mm thickness = 500 (25000) (20) / 103 = 500 N/m2

___________________ Total revised loads = 37250 N/m2


Revised negative moment = (0.032) (37250) (16) = 19.072 kNm Revised positive moment = (0.024) (37250) (16) = 14.304 kNm Revised negative steel = (19072000) / (150) (191.4) = 664.298 mm2 and Revised positive steel = (14304000) / (150) (208.8) = 456.7 mm2. Provide 10 mm diameter bars @100 mm c/c (= 785 mm2) at the support and 10 mm diameter bars @ 150 mm c/c (= 524 mm2) at the mid-span, as shown in Figs. 14.37.4a and 14.37.5a, respectively. Step 6. Checking of stresses and moment of resistance (A) At the support (Fig. 14.37.4b)

The depth of neutral axis x = {1000 (270) (135) + (112.33) (785) (220)} / {(1000) (270) + (12.33) (785)} = 137.94 mm. (D-x) = 270 – 137.94 = 132.06 mm and (d-x) = 220 -137.94 = 82.06 mm. Iyy = {(1000) (137.94)3 / 3} + {(1000) (132.06)3 / 3} + {(12.33) (785) (82.06)2} = 1.71 (109) mm4

Moment of resistance = (1.7) (1.71) (103) / 132.06 = 22.01 kNm ⟩ 19.072 kNm. ftb = 19.072 (106) (132.06) / (1.71) (109) = 1.47 N/mm2 ⟨ 1.7 N/mm2. fcb = 19.072 (106) (137.94) / (1.71) (109) = 1.54 N/mm2 ⟨ 7 N/mm2. fst = (13.33)(19.072) (106)(82.06) / (1.71)(109) = 12.2 N/mm2 ⟨ 230 N/mm2

Hence, the section at the support is ok. (B) At the mid-span (Fig. 14.37.5b)

The depth of neutral axis x = {(1000) (270) (135) + (12.33) (524) (240)} / {(1000) (270) + (12.33) (524)} = 137.45 mm, D - x = 132.55 mm and d-x = 102.55 mm.


Iyy = {(1000) (137.45)3 / 3}+ {(1000) (132.55)3 / 3} + {(12.33) (524) (102.55)2} = 1.71 (109) mm4. Moment of resistance of the section (when ftb = 1.7 N/mm2) = 1.7 (1.71) (103)/(132.55) = 21.93 kNm ⟩14.304 kNm.

When ftb = 1.7 N/mm2, fcb = 1.7 (137.45) / (132.55) = 1.7642 N/mm2.

The moment of resistance (when fcb = 1.7642 N/mm2) = 1.7642 (1.71) (103)/(137.45) = 21.94 kNm ⟩ 14.304 kNm.

For the positive moment of 14.304 kNm, ftb = (14.304) (132.55) (106)

/ (1.71) (109) = 1.109 N/mm2 ⟨ 1.7 N/mm2 and fcb = (14.304) (137.45) (106) / (1.71) (109) = 1.145 N/mm2 ⟨ 7 N/mm2

Problem 5. Determine the moment of resistance of the slab of Fig. 14.37.6

subjected to moment alone using M 20 and Fe 415.

Solution 5. Since tension is on the remote face and the depth of the slab is

more than 225 mm, we use cracked section. The value of m = 93.33/7 = 13.33.

Taking moment of area of concrete and steel about the top face. 500 x2 = (13.33) (942) (250 - x) or, x2+ 25.11372 x – 6278.43 = 0 which gives x = 67.67 mm. Accordingly, d – x = 250 - 67.67 =

182.33 mm and D –x = 280 – 67.67 = 212.33 mm.


Moment of resistance from steel is determined using fst = σst = 190 N/mm2. Then M = σst Ast (d –x/3) = 190(942) (250 – 67.67/3) (10-6) = 40.71 kNm.

When fst = σst = 190 N/mm2, fcb = 190 (67.67) / (182.33) (13.33) =

5.29 N/mm2 ⟨ 7 N/mm2. The moment of resistance from compression concrete = 0.5(67.67)

(5.29) (1000) (250 - 67.67/3) = 40.707 kNm. Problem 6. Design the wall of a water tank using concrete of grade M 20

and steel of grade Fe 415, when subjected to Ft = 60 kN/m and M = 25 kNm/m.

Solution 6. Since both Ft and M are acting on the wall of a liquid retaining

structure, the design must satisfy Eq. 14.8 of Lesson 36, i.e., (ftd /σtd) + (ftb /σtb) ≤ 1.0 (14.8) where σtd = 1.2 N/mm2 and σtb = 1.7 N/mm2 for M 20 concrete, as

given in Table 14.1 of Lesson 36, when the tension is on the liquid face. The section is considered as uncracked.

Step 1. Preliminary values of d, D, Ast1 for moment and Ast2 axial

force

Assuming σst = 150 N/mm2, we get kb = 93.33/(150+93.33) = 0.38

and j = 1 - 0.38/3 = 0.87. Assuming pt = 0.4 per cent, we get from:


M = Ast1 σst jd = (pt jσst / 100) (bd2) or, d = (100 M/σst pt j b)1/2 = {(100) (25000 000)/(150) (0.4) (0.87)

(1000)}1/2 = 218.8 mm. Provide d = 250 mm with cover of 50 mm, D = 300 mm.

Ast1 for moment (0.4 per cent) = 0.4 (300) (1000) / 100 = 1200 mm2. Ast2 for Ft = 60000 / 150 = 400 mm2. Total Ast = Ast1 + Ast2 = 1200 + 400 = 1600 mm2. Provide 16 mm

diameter bars @ 120 mm c/c to give 1675 mm2, as shown in Fig. 14.37.7a.

Step 2. Sectional properties Figure 14.37.7a shows the section. x = {1000 (300) (150) + (12.33) (1675) (250)} / {1000 (300) +

(12.33) (1675)} = 156.44 mm. D – x = 300 – 156.44 = 143.56 mm, and d – x = 93.56 mm. Iyy = 1000 (143.56)3 / 3 + 1000 (156.44)3 / 3 + (12.33) (1675)

(93.56)2

= 2.443 (109) mm4. A = area of the section = 300 (1000) + (12.33) (1675) = 320652.75

mm2. Step 3. Values of stresses and checking of Eq. 14.8 ftb = (25000 000) (143.56) / (2.443) (109) = 1.469 N/mm2 ⟨ 1.7

N/mm2. ftd = 60000 / 320652.75 = 0.187 N/mm2 ⟨ 1.2 N/mm2

(ftb / σtb) + (ftd / σtd) = (1.469 / 1.7) + (0.187 / 1.2) = 1.02 ⟩ 1.0. Hence, the section has to be revised.

The revision can be done in two ways: (i) increasing Ast1 to 0.5 per

cent keeping the depth of the section = 300 mm, and / or (ii) increasing the depth of the section by 20 mm. We try with the first one in the next step.

Step 4. Revised section


With D as 300 mm, Ast1 = 0.5 (300) (1000) / 100 = 1500 mm2 and

Ast2 = 400 mm2 (as in earlier). So, Ast = 1900 mm2. Provide 20 mm diameter bars @ 160 mm c/c to give 1963 mm2 as shown in Fig. 14.37.8a.

x ={(1000) (300) (150) + (12.33) (1963) (250)}/{(1000) (300) +

(12.33) (1963)} = 157.47 mm. D – x = 300 – 157.47 = 142.53 mm and d –x = 250 – 157.47 =

92.53 mm. I = 1000 (157.47)3/3 + 1000 (142.53)3/3 + (12.33) (1963) (92.53)2 =

2.47(109) mm4

A = (300) (1000) + (12.33) (1963) = 324203.79 mm2. ftb = 25(106) (142.53) / 2.47(109) = 1.44 N/mm2

ftd = 6000 /324203.79 = 0.185 N/mm2

(ftb / σtb) + (ftd / σtd) = (1.44 / 1.7) + (0.185/1.2) = 1.0012 ⟩ 1.0. So, we further increase the depth of the concrete and repeat the

calculations of Step 4 in Step 5. Step 5. Further revision


Now, the revised section has: D = 320 mm, d = 270 mm and Ast

=1963 mm2 (20 mm diameter bars @ 160 mm c/c) as shown in Figs. 14.37.9a and b. The following are the results:

x = 167.77 mm D –x = 152.23 mm I = (3.003) (109) mm4

A = 344203.79 mm2

ftb = 1.267 N/mm2

ftd = 0.174 N/mm2

and (ftb /σtb) + (ftd / σtd) =0.745 + 0.145 = 0.89 ⟨ 1.0 Hence, the revised section of Fig. 14.37.9a is ok. Problem 7. Determine the area of tensile steel of the wall of a water tank of

250 mm depth and subjected to M = 35 kNm/m only. Use M 20 and Fe 415. Redesign, if needed.

Solution 7. Given data are D = 250 mm, d = 225 mm, M 20 and Fe 415. Since,

the tension face is at a distance of 225 mm away, we consider fst ≤190 N/mm2 and assume the section as cracked. We determine the preliminary area of steel first.


Step 1. Determination of preliminary area of steel

For M 20 concrete, m = 93.33/7 = 13.33, kb = 93.33 / (σst + 93.33)

= 93.33 / (190 + 93.33) = 0.329 and j = 1 – kb / 3 = 1 – 0.329/3 = 0.89.

Preliminary Ast = M/σst jd = 35(106) / (190) (0.89) (225) = 919.9

mm2. Provide 12 mm diameter bars @ 120 mm c/c to have 942 mm2, as shown in Fig. 14.37.10a.

Step 2. Properties of section Taking moment of the area about the top surface (Fig. 14.37.10a),

we have: 500 x2 = (13.33) (942) (225 – x) or, x2 + 25.11372 x – 5650.587 = 0. The solution is x = 76.21 mm. Iyy = 1000 (76.21)3/3 + (13.33) (942) (225 – 76.21)2 = 0.425 (109)

mm4. Step 3. Checking for the stresses


Compression stress in concrete in bending fcb = Mx/Iyy = 35 (106) (76.21) / (0.425) (109) = 6.28 N/mm2. Tensile stress in steel = fst = m(fcb) (d-x) / x = (13.33) (6.28)(148.79)/76.21 = 163.43 N/mm2 ⟨ 190 N/mm2. Hence, the section is ok.

Problem 8. Check the acceptability of the design of the base slab of a water tank as shown in Fig. 14.37.11, which is subjected to Ft = 60 kN/m and M = 6 kNm/m. Assume the section as uncracked. Use M 20 and Fe 415.

Solution 8. For M 20 concrete m = 93.33/7 = 13.33. Equations 14.2 and 14.4 of Lesson 36 are used in determining ftd and ftb, respectively. Since the section is symmetrically reinforced, the depth of the neutral axis = 85 mm.

Iyy = 1000 (170)3 / 12 + (12.33) (503) (2) (50)2 = 0.4404 (109) mm4

ftd = Ft /{(1000) (170) + (12.33) (2) (503)} = 0.329 N/mm2

ftb = 6(106) (85) / (0.4404) (109) = 1.158 N/mm2

(ftd / σtd) + (ftb / σtb) = (0.329 /1.2) + (1.158/1.7) = 0.224+0.681 = 0.955 ⟨ 1.0. Hence, the design is ok.


14.37.3 Practice Questions and Problems with Answers Q.1: Solve Problem 1 of sec.14.37.2 when Fe 250 is used. A.1: Given data are Ft = 470 kN, concrete grade is M 25 and steel grade is Fe

250. Modular ratio m = 10.98 (see solution 1 of sec. 14.37.2). The values of fst of Fe 250 ≤ 140 N/mm2 and ftd ≤ 3.2 N/mm2.

Step 1. Determination of Ast

From Eq. 14.1 of Lesson 36, we have Ast = Ft /σst = 470000/140 = 3357.14

mm2. Provide 6 bars of 25 mm diameter and 4 bars of 12 mm diameter to have Ast = 2945 + 452 = 3397 mm2.

Step 2. Determination of width and depth of the section From Ag = (Ft / ftd) – (m – 1) Ast, we have Ag = 112972.94 mm2. Provide

300 mm × 380 mm section. Step 3. Checking for tensile stress of concrete From Eq. 14.2 of Lesson 36, we have: ftd =Ft / (Ac + m Ast) = Ft / {Ag + (m –1) Ast} = 3.18 N/mm2 ⟨ 3.2 N/mm2.

Hence, the section of 300 mm × 380 mm with 6 bars of 25 mm diameter and 4 bars of 12 mm diameter is ok.


Q.2: Determine the moment of resistance of the section shown in Fig. 14.37.12.

Use M 20 and Fe 415. A.2: The section is considered uncracked as the liquid face is having tension.

The modular ration m = 93.33 / 7 = 13.33 and ftb ≤ 1.7 N/mm2. Step 1. Sectional properties The depth of the neutral axis, as shown in Fig. 14.37.12, is obtained by

taking moment of the areas about the top, which gives: x = {(300) (900) (450) + (1100) (290) (610 + 145) + (12.33) (2463) (900 –

85) + (1.5) (12.33) (942) (60)} / {(300) (900) + (1100)(290) + (12.33) (2463) + (1.5) (12.33) (942) = 609.53 mm,

D – x = 900 – 609.53 = 290.47 mm. The moment of inertia about the neutral axis yy is: Iyy = (300) (609.53)3 / 3 + (300) (0.47)3/3 + 1400 (290)3 / 12 + (1400) (290)

(145 + 0.47)2 + (12.33) (2463) (290.47 – 85)2 + (1.5) (12.33) (942) (609.53 – 60)2 = (40.626) (109) mm4.

Moment of resistance = (1.7) (40.626) (109) / 290.47 = 237.767 kNm/m. Q.3: Solve Problem 5 of sec. 14.37.2 (Fig. 14.37.6a) using Fe 250.


A.3: Given data are: D = 280 mm, d = 250 mm, Ast = 12 mm diameter bars @

120 mm c/c = 942 mm2, M 20 and Fe 250 (Fig. 14.37.6a). Modular ratio m = 93.33 / 7 = 13.33, x = 67.67 (as obtained in the solution

of Problem 5), σst = 125 N/mm2 and (d – x) = 250 – 67.67 = 182.33 mm. Therefore, M = σst Ast (d – x/3) = (125) (942) (250 – 67.67/3) = 26.78 kNm.

fcb = σst (67.67) / (13.13) (182.33) = 3.48 N/mm2. The moment of

resistance from compression concrete = 125 (942) (250 – 67.67/3) = 26.78 kNm ⟨ 40.71 kNm of the slab of Problem 5.

It is thus seen that though the slab of Problem 5 and this one have x =

67.67 mm but moment of resistance of this slab is much less when Fe 250 is used.

14.37.4 Reference


















minutes

TQ.1: The base slab of the water tank, as shown in Fig. 14.37.13a, is subjected

to Ft = 60 kN/m and M = 1.2 kNm/m. Determine the stresses in steel and concrete. Use M 20 and Fe 415.

[20 marks] A.TQ.1:Given D = 200 mm, As1 = 785 mm2, As2 = 785 mm2, M 20 and Fe 415.


The eccentricity of Ft = e = 1.2 (103) /60 = 20 mm, i.e., within the section. The whole section is in tension and the section is cracked. We have Eqs. 14.9 and 14.10 as

Ft = fst1 As1 + fst2 As2 (14.9) Ft (d1 – e) – fst2 As2 (d1 + d2) = 0

(14.10) From Eq. 14.10, we have: fst2 = Ft (d1 – e) / {As2 (d1 + d2)} = (60000) (50 -

20)/(785) (100) = 22.93 N/mm2 ⟨ 150 N/mm2. Using the value of fst2 in Eq. 14.9, we have fst1 = (Ft - fst2 As2) / As1 = {60000

– (22.93) (785)}/785 = 53.50 N/mm2 ⟨ 150 N/mm2. Modular ratio m for M20 = 93.33 / 7 = 13.33. ftd = Ft /{1000(200) + (13.33 –1) (785) (2)} = 0.27 N/mm2

ftb = fst1 / m = 53.50 / 13.33 = 4.0 N/mm2 ⟩ 1.7 N/mm2. It shows that the section is cracked. TQ.2: Determine the area of tensile steel of a singly-reinforced bunker wall of

depth 300 mm subjected to Ft = 60 kN/m and M = 42 kNm/m at the horizontal level. Use M 20 and Fe 415 grade of steel.

(30 marks) A.TQ.2: In this problem, the eccentricity of the tensile force Ft = M/Ft =42000 / 60

= 700 mm, i.e., e is outside the section. Given data are: D = 300 mm, d = 275 mm. For M 20 concrete, m = 93.33/7 = 13.33, fcb ≤ 7 N/mm2. Let us assume j = 0.87.

Step 1. Preliminary area of tension steel From Eq. 14.27 of Lesson 36, we have: Ast = (Ft /σst) {1 + (e + 0.5 D - d) /

jd} = (60000 /230) {1 + (700 + 150 - 275) / 0.87 (275)} = 887.83 mm2. Provide 12 mm diameter bars @ 120 mm c/c to give 942 mm2.

Step 2. Depth of the neutral axis Eq. 14.23 of Lesson 36 gives: m (Ast) (d – x) (e + 0.5 D – d) = 0.5 bx2 (e +

0.5 D – x/3), which gives: (13.33) (942) (275 – x) (700 +150-275) = 500 x2 (700 + 150 – x/3)


or x3 – 2550 x2 – 43321.167 x + 11913320.93 = 0.The solution of the equation is x = 61.0265 mm by trial and error.

Step 3. Determination of stresses of concrete and steel Equation 14.22 gives: fcb = Ft / {m Ast (d – x) / x – (0.5 bx)} = 60000 / {(13.33) (942) (275 – 61.0265) / 61.0265 – 500 (61.0265)} = 4.44 N/mm2 ⟨ 7 N/mm2. Equation 14.15 gives: fst = m fcb (d – x) / x = (13.33) (4.44) (275 – 61.0265) / 61.0265 = 207.52 N/mm2 ⟨ 230 N/mm2 . Hence, the solution is o.k. 14.37.6 Summary of this Lesson This lesson illustrates the applications of the equations explained in

Lesson 36 for the analysis and design of tension members employing working stress method as stipulated in IS Codes. These structures may be either liquid retaining structures or may not have any contact with the liquid. The permissible stresses in concrete and steel depend on several factors as explained in Lesson 36. The section may be considered cracked or uncracked. Such tension structures may be subjected to axial tension only, moment only or combinations of them. Illustrative examples cover most of the categories of problems. For better understanding, some of the problems are solved from the first principle instead of using the equations directly. Understanding the illustrative examples and solutions of the practice and test problems will help in applying the equations in analysing and designing tension structures as per the stipulations of IS Codes.


Module 15

Redistribution of Moments


Lesson 38

Redistribution of Moments – Theory and

Numerical Problems Version 2 CE IIT, Kharagpur



• explain the concept of redistribution of moments in the design of statically

indeterminate reinforced concrete structures,

• explain the behaviour of statically indeterminate reinforced concrete

structures with increasing loads till the collapse of the structures,

• state the advantages of redistributing the moments in statically

indeterminate reinforced concrete structures,

• identify structures when such redistributions are to be made based on the

analysis of structures,

• mention the reasons why full redistribution is not allowed in reinforced

concrete structures,

• specify the stipulations of IS Codes about the redistribution separately when

the structure is designed by working stress method and by limit state

method,

• apply the theories in solving numerical problems of statically indeterminate

beams of one or more spans as per the stipulations of IS Code.


Statically indeterminate structures made of reinforced concrete like fixed ended one span beams, continuous beams and frames are designed considering internal forces like bending moment, shear force and axial thrust obtained from structural analysis. Either one or several sections of these structures may have peak values of the internal forces, which are designated as critical sections. These sections are dimensioned and reinforced accordingly. Flexural members, however, do not collapse immediately as soon as the loads at a particular section cause bending moment exceeding the maximum resisting moment capacity of that section. Instead, that section starts rotating at almost constant moment. This is known as formation of plastic hinge at that section reaching its maximum resisting moment capacity. The section then transfers loads to other sections if the applied loads are further increased. This process continues till the structures


have plastic binges at sufficient sections to form a failure mechanism when it actually collapses. However, significant transfer of loads has occurred before the collapse of the structure. This transfer of loads after the formation of first plastic hinge at section having the highest bending moment till the collapse of the structure is known as redistribution of moments. By this process, therefore, the structure continues to accommodate higher loads before it collapses.

The elastic bending moment diagram prior to the formation of first plastic

hinge and the final bending moment diagram just before the collapse are far different. The ratio of the negative to positive elastic bending moments is no more valid. The development of plastic hinges depends on the available plastic moment capacity at critical sections. It is worth mentioning that the redistribution of moment is possible if the section forming the plastic hinge has the ability to rotate at constant moment, which depends on the amount of reinforcement actually provided at that section. The section must be under-reinforced and should have sufficient ductility.

This phenomenon is well known in steel structures. However, the

redistribution of moment has also been confirmed in reinforced concrete structure by experimental investigations. It is also a fact that reinforced concrete structures have comparatively lower capacity to rotate than steel structures. Yet, this phenomenon is drawing the attention of the designers. Presently, design codes of most of the countries allow the redistribution up to a maximum limit because of the following advantages:

1) It gives a more realistic picture of the actual load carrying capacity of

the indeterminate structure. 2) Structures designed considering the redistribution of moment (though

limited) would result in economy as the actual load capacity is higher than that we determine from any elastic analysis.

3) The designer enjoys the freedom of modifying the design bending

moments within limits. These adjustments are sometimes helpful in reducing the reinforcing bars, which are crowded, especially at locations of high bending moment.


15.38.2 Two Span Beam


Let us take up a two-span beam of uniform cross-section, as shown in Fig. 15.38.1a, with the following assumptions:

a) The ultimate moments of resistance of sections at B and D are –Mun and +Mup, respectively.

b) There will not be any premature shear failure so that the sections can

attain the respective ultimate moment capacity.

c) The moment-curvature relationship for the ductile sections is the

idealised bilinear as shown in Fig. 15.38.2. It may be noted that for all practical purposes only a limited rotation will take place.

d) All sections of the beam have the same constant flexural rigidity up to

the ultimate moment and moment remains constant at the ultimate moment with increasing curvature.

e) The self-weight of the beam is neglected. The beam is subjected to two point loads of magnitude P each and acting

at distances of l/2 from the two supports A and C, as shown in Fig. 15.38.1a. The elastic bending moment diagram is shown in Fig. 15.38.1b. On increasing the loads P, the ultimate moment of resistance Mun of cross-section at B will be reached at the support first before reaching at other sections, which is shown in Fig. 15.38.1d. The plastic hinge will be formed at the cross-section B (Fig. 15.38.1c). The two point loads P can still be increased as long as the plastic hinge at B will rotate sufficiently. If the cross-section at B is brittle, the load will


decrease fast (see Fig. 15.38.2) and the beam will fail suddenly. Hence, the beam will not carry any additional load. The cross-section at B has to be ductile so that it undergoes rotation at the constant moment of resistance, which will enable the beam to carry additional loads. This increase of load will continue until the maximum positive moments in the span (at D and E) reach Mup, as shown in Fig. 15.38.1f, when plastic hinges will form at D and E also. The three plastic hinges at B, D and E will form the collapse mechanism (Fig. 15.38.1e). The structure will fail at this stage carrying much higher loads. It is important to note that the requirement of equilibrium will be satisfied at all stages, i.e., Mp and Mn , during the elastic phase, will follow the equation:

Mp + 0.5 Mn = Pl/4 (15.1)

From the structural analysis of the beam of Fig. 15.38.1a, we know that:

Mn = 6 Pl / 32 and Mp = 5 Pl / 32. Substituting these values in Eq. 15.1, we find that the equation is satisfied where Pl/4 is the maximum positive moment of a simply supported beam having a load P at the centre of the beam. From the above values of Mn and Mp (= 6 Pl/32 and 5 Pl/32, respectively), we also note that: Mn / Mp = 1.2 (15.2)


As the loads are increased till the formation of the first plastic hinge at B, this ratio of Mn /Mp as 1.2 will be maintained, as shown in Fig. 15.38.3. This phase is known as the elastic phase when the bending moments increases with increasing loads maintaining the ratio of Mn / Mp as 1.2. With further increase of the loads, the plastic hinge at B will rotate at the constant moment Mun and positive moments at D and E will increase as shown in Fig. 15.38.3. This phase is known as moment redistribution phase as the loads are now transferred to sections, which have less moment. However, the cross-section at B must have the ability to sustain the required plastic rotation at this stage with increasing loads. As the value of Mp is now increasing when Mn is remaining constant, the ratio of Mn / Mp will not be 1.2 any more. Thus, in the redistribution phase, the additional moments at higher loads are to be redistributed to the support and mid-span in such a manner that we get a similar equation like Eq. 15.1, i.e. Mup + 0.5 Mun =Pl/4 (15.3) which means that, after the redistribution Mun = Mn –M (15.4) where M is some amount of moment by which negative moment at the support is reduced. From Eqs. 15.3 and 15.4, we have: Mup = Pl/4 – 0.5 Mun = Pl/4 - 0.5 (Mn-M) = (Pl/4 – 0.5 Mn) + 0.5 M = Mp + 0.5 M, which we get by substituting Mp = Pl/4 – 0.5 Mn from Eq. 15.1. Therefore, we have: Mup = Mp + 0.5 M (15.5) Thus, in the moment redistribution, if we reduce some amount M from the negative moment Mun at support as in Eq. 15.4, we have to add 0.5 M to the span moment Mp to get the Mup as in Eq. 15.5. The redistributed moments Mun and Mup satisfy the equilibrium condition of Eq. 15.3. Similar equilibrium condition is also satisfied at the elastic stage by the Mn and Mp as in Eq. 15.1. The amount of moment M which will be reduced from the negative support moment depends on the rotational capacity as per the actual reinforcement provided in the cross-section. Furthermore, the deformation of the support should be within acceptable limits under service loads. In the extreme case, if the negative moment at the support Mun is reduced to zero, i.e., M = Mn , we have from Eq. 15.5: Mup = Mp + 0.5 Mn = Pl/4 (15.6)


That is, the unreinforced central support will crack and the continuous beam will behave as two simply supported beams. Thus, the choice of the bending moment diagram after the redistribution should satisfy the equilibrium of internal forces and external loads. Moreover, it must ensure the following:

1. The plastic rotations required at the critical sections should not exceed the amount the sections can sustain.

2. The extent of cracking or the amount of deformation should not

make the performance unsatisfactory under service loads. The redistribution of moments is permitted if the analysis of forces and moments is done following linear elastic behaviour. Analysis by linear elastic behaviour is a logical procedure in the working stress method of design where the design concept is based on the assumptions of linear elastic behaviour of materials up to the level of recommended safe stresses. However, analysis by linear elastic theory and design by the limit state of collapse appear to be somewhat contradictory. At the stress levels of 0.87 fy for steel and 0.67 fck for concrete, the strains are not linearly related. Unfortunately, till now there is no such analysis, which takes into account the complex behaviour of reinforced concrete just before the collapse. Accordingly, the elastic theory of analysis of forces and moments cannot be avoided at present except for the slabs. Moreover, unlike steel structures, cross-section forming the first plastic hinge in reinforced concrete members undergoes limited rotations, which is insufficient for other sections to attain the desired Mup. Therefore, full redistribution has been restricted in the design of reinforced concrete members. These restrictions and stipulations of IS Codes are taken up in the next section. 15.38.3 Recommendations of IS 456 IS 456 recommends the redistribution of moments during the analysis provided the analysis is done by linear elastic theory. Such redistribution of moments is not permitted when either simplified analysis is done or coefficients of cl.22.5 of IS 456 are used to determine the moments. In the working stress method, cl. B-1.2 of IS 456 stipulates that the moments over the supports for any assumed arrangement of loading, including the dead load moments may each be increased or decreased by not more than 15 per cent, provided that these modified moments over the supports are used for the calculation of the corresponding moments in the spans.


In the limit state of collapse method, cl.37.1 of IS 456 recommends the redistribution of the calculated moment in continuous beams and frames satisfying the following conditions:

1. Equilibrium between the internal forces and the external loads should be maintained.

2. The ultimate moment of resistance provided at any cross-section of a

member after the redistribution should not be less than 70 per cent of the moment at that cross-section obtained from an elastic maximum moment diagram covering all appropriate combinations of loads.

3. The elastic moment at any cross-section in a member due to a particular

combination of loads shall not be reduced by more than 30 per cent of the numerically largest moment given anywhere by the elastic maximum moment diagram for the particular member, covering all appropriate combinations of loads.

4. Cross-sections having moment capacity after redistribution less than that

of the elastic maximum moment shall satisfy the relationship: (xu / d) + (δM / 100) ≤ 0.6 (15.7) where xu = depth of the neutral axis, d = effective depth, and δM = percentage reduction in moment.

5. Thirty per cent reduction in moment permitted in 3 above shall be restricted to ten per cent for structures in which the structural frames provide lateral stability.

15.38.4 Explanations of the Conditions Stipulated in IS Code The first condition of maintaining the equilibrium between internal forces and external loads is explained in sec.15.38.2 taking up a two-span continuous beam. For the other conditions, let us use the following notations for the sake of brevity: Mew = elastic bending moment under working loads, Meu = elastic bending moment under design loads i.e., the factored loads

considering partial safety factor of loads γf, Mdu = design factored moment after redistribution, and


δM = percentage reduction in Meu. Since γf = 1.5, we can write Meu = 1.5 Mew (15.8) The second condition of cl.37.1 of IS 456, as given in sec. 15.38.3 of this lesson, is expressed as Mdu ≥ 0.7 Meu (15.9) Beeby has reported that the redistribution as given in Eq. 15.8 is possible before the lower bound rotation capacity is reached (Beeby, A.W, “The analysis of beams in plane frames according to CP 110” Cement and Concrete Association, U.K. Development Report No. 1, October 1978). Thus, Eq. 15.9 ensures the first criterion of the redistribution of moments in reinforced concrete structures as given in sec. 15.38.2. Equation 15.9 also satisfies other requirements as explained below.


Using the value of Meu from Eq. 15.8 into Eq. 15.9, we have Mdu ≥ 0.7 (1.5 Mew) ≥ 1.05 Mew (15.10) This ensures that the design factored moments are greater than the elastic moments everywhere in the structures. Figure 15.38.4 presents three bending moment diagrams: (i) elastic moment under working loads (Mew), (ii) elastic moment under ultimate loads (Meu) and (iii) design moment under ultimate loads (Mdu) of the two-span beam of Fig. 15.38.1a. The moment diagram after the redistribution (Mdu) satisfies the condition of Eq. 15.1 such that b + 0.5 a = b′ + 0.5 a′ (15.11)


However, Mdu diagram is not satisfying the requirement of Eq. 15.10 as in the zone GH of length x (Fig.15.38.4), the negative moment after redistribution is less than the negative moment under working loads. To satisfy Eq. 15.10, the bending moment in the zone should be as per Mew. We now take up the fourth condition of cl.37.1 of IS 456 and given in sec. 15.38.3 of this lesson, which is expressed in Eq. 15.7 as (xu/d) + (δM / 100) ≤ 0.6 (15.7) This equation ensures that the cross-section of the member is under-reinforced, as explained below. Allowing the maximum redistribution of thirty per cent as per the third condition of sec. 15.38.3, we have δM/100 = 0.3. Substituting this in Eq. 15.7, we get xu/d ≤0.3. For (δM/100) less than thirty per cent, the corresponding xu/d can be determined from Eq. 15.7. The values of xu,max/d for the three grades of steel are given in Table 3.2 of Lesson 5. It is easy to verify that permitting the redistribution of thirty per cent, the cross-sections remain under-reinforced. This ensures low amount of steel reinforcement to have smaller value of xu which will give higher value of the rotation φ as the rotation φ = 0.0035/xu. It is evident, therefore, that for the analysis of determining the values of bending moments after the redistribution, bending moment envelopes are to be drawn satisfying the requirements discussed above. This is explained with the help of illustrative examples in the next section.


15.38.5 Illustrative Examples


Problem 1. Draw the design bending moment diagram of the beam of Fig. 15.38.5a, clamped at both ends and carrying ultimate uniformly distributed load of 24 kN/m with full redistribution of 30 per cent as per IS 456.

Solution 1. Step 1: Elastic bending moment diagram The beam (Fig.15.38.5a) is statically indeterminate. The elastic

bending moment diagram is shown in Fig. 15.38.5b. The negative and positive moments at the supports (A and B) and mid-span (C), respectively, are:

MA = MB = - w l / 12 = - 24 (8) (8) / 12 = - 128 kNm B

2

MC = + w l2/24 = + 24 (8) (8) / 24 = + 64 kNm The distance x where the bending moment is zero is obtained by

taking moments of forces of the free body diagram (Fig. 15.38.5c) about D after determining the vertical reaction at A.

Vertical reaction at A, VA = 24 (8) / 2 = 96 kN. At section D, Mx =

96x – 128 – 12x2. So, the value of x when Mx = 0 is obtained from 96x – 128 – 12x2 = 0 or, 3x2 – 24x + 32 = 0. The solution of the equation is x =1.69 m and 6.31 m.

Step 2. Redistributed bending moment diagram From Fig.15.38.5e, the maximum negative moment at A after the

full redistribution of 30 per cent = 0.7 (128) = 89.6 kNm. So, the maximum positive moment at the mid-span C = w l2 / 8 – 89.6 = 102.4 kNm. The vertical reaction at A is: VA = 96 kN.

The point of inflection, i.e., the point where bending moment is

zero, is obtained by taking moment of forces about E of the free body diagram shown in Fig. 15.38.5d. Thus, we have: 96x – 12x2 – 89.6 = 0, which gives x = 1.08 m and 6.92 m. The bending moment diagram after the redistribution is shown in Fig. 15.38.5e.

Step 3. Envelope of bending moment diagrams At x =1.08 m, the elastic bending moment is Mx = 96x – 128 – 12x2

= - 38.32 kNm, which becomes 0.7 (- 38.32) = - 26.824 kNm after the redistribution. The envelope of the bending moment diagram is shown in Fig. 15.38.5f.


Problem 2. Draw the design bending moment diagram of the beam of Fig. 15.38.6a, clamped at both ends and carrying two point loads of 30 kN each at distances of 3 m from the supports. Assume full redistribution of 30 per cent as per IS 456.

Solution 2. Step 1. Elastic bending moment diagram The beam (Fig.15.38.6a) is statically indeterminate. The elastic

bending moment diagram is shown in Fig. 15.38.6b. The negative and positive moments at the supports (A and B) and at mid-span (E), respectively, are:

MA = MB = - 2 P l/9 = -2 (30) (9)/ 9 = - 60 kNm ME = 90 – 60 = 30 kNm. Vertical reaction at A = VA = 30 kN Taking moment of all the forces about F at a distance of x (0 ≤ x ≤ 3

m) from A of the free body diagram shown in Fig. 15.38.6c, we have 30 x – 60 = 0, which gives x = 2 m, where the bending moment is zero.

Step 2. Redistributed bending moment diagram Redistributed negative moment at A = 0.7 (60) = 42 kNm Redistributed positive moment at E = 90 - 42 = 48 kNm The point of zero moment (point G) is obtained by taking moments

of forces about G of the free body diagram shown in Fig. 15.38.6d, when x ≤ 0 ≤ 3 m; gives 30x – 42 = 0. Therefore, x = 1.4 m.

The redistributed moment diagram is shown in Fig. 15.38.6e. Step 3. Envelope of bending moment diagram At G when x = 1.4 m, elastic moment = 30(1.4) – 60 = - 18 kNm.

The redistributed moment at G = 0.7 (-18) = -12.6 kNm. The envelope of the bending moment diagrams is shown in Fig. 15.38.6f.


Problem 3. Draw envelope of the design moments of the two-span continuous

beam shown in Fig. 15.38.7a, carrying characteristic live load of 35 kN/m in addition to its characteristic self-weight. The cross-section of the beam is 300 mm × 700 mm.

Solution 3. The two-span continuous beam of Fig. 15.38.7a is statically

indeterminate and the method of moment distribution is employed for the analysis. Clause 22.4.1 of IS 456 stipulates the arrangements of live loads, which are:

1) Design dead load on all spans with full design imposed loads on

two adjacent spans to get the maximum negative moment over the support, and

2) Design dead load on all spans with full design imposed loads on

alternate spans to get the maximum span moment (positive moment).

In this case of two-span continuous beam, therefore, we have the following three load cases:

(i) For the maximum negative moment at the support B, both the spans are loaded with self-weight and live loads,


(ii) For the maximum positive moment in span AB, only the span AB is loaded with live load and self-weight of the beam in both the spans, and

(iii) For the maximum positive moment in span BC, only

the span BC is loaded with live loads and self-weight of the beam in both the spans.

However, we are considering only cases (ii) and (iii) as they are the

mirror image of each other. Step 1. Evaluation of design loads Characteristic dead load of the beam = (0.3) (0.7) (25) = 5.25 kN/m.

Therefore, the design dead load with γf = 1.5 is (1.5) (5.25) = 7.875 kN/m. The design live load = 1.5 (35) = 52.5 kN/m.

Step 2. Load case (i): Elastic bending moment diagram Figure 15.38.7b shows the load case (i) of the beam. The

calculations of the moment distribution are presented in Table 15.1. Figure 15.38.7c presents the elastic bending moment diagram and Fig. 15.38.7d presents the free body diagram and the calculations of determining the reactions. The final values of the vertical reactions are: VA = + 181.125 kN, VB1 = + 301.875 kN, VB2 = + 301.875 kN and VC = +181.125 kN (+ means upward). Shear force is zero at D (x = 3 m), where the moment is 271.6875 kNm, obtained from the following equations:

(a) 181.125 – 60.375 x = 0 gives x = 3 m, and (b) Mx (at x = 3.0 m) = 181.125 (3) – 60.375 (3) (3) /2 = 271.6875

kNm. At point E (mid-span), Mx = 181.125 (4) – 60.373 (4) (4)/2 = 241.5 kNm. At point G, where x = 6 m, the bending moment is zero as obtained from equation 181.125x – 30.1875x2 = 0. For the span BC, the bending moment diagram is the mirror image of the bending moment diagram of span AB due to symmetrical beams and loadings.

Table 15.1 Moment Distribution for Load Case (i)

Joint A B C Member AB BA BC CB


D.F 1.0 0.5 0.5 1.0 F.E.M + 322.0 - 322.0 +

322.0 - 322.0

Balance moment - 322.0 - - + 322.0 Carry over moment - - 161.0 +

161.0 -

Total balanced moment 0 - 483.0 + 483.0

0

Note: D.F. = Distribution Factors; F.E.M = Fixed End Moments Step 3. Load Case (i): Redistributed bending moment diagram Redistributed moment at support B = 0.7 (483) = 338.1 kNm.

Vertical reaction at A = 199.2375 kN (Fig. 15.38.7f). Shear is zero at F where x = 3.3 m as obtained from: 199.2375 –

60.375 x = 0. Maximum bending moment at F (x = 3.3 m) is: 199.2375 (3.3) -

60.375 (3.3) (3.3) /2 = 328.742 kNm. At the mid-span E (x = 4 m), the moment is 60.373 (8) (8) / 8 –

338.1 /2 = 313.95 kNm. Moment is zero at H where x = 6.6 m is obtained from: 199.2375 x

– 60.373 x2/2 = 0. Calculations for the span BC are not performed due to symmetrical

loadings. The redistributed bending moment diagram is shown in Fig. 15.38.7e.


Step 4. Load Case (ii): Elastic bending moment diagram This load case is for the maximum positive moment in span AB.

The design loads consist of factored dead load and factored live load = 1.5 (5.25 + 35) = 60.375 kN/m for the span AB. For the span BC, we consider only the dead load with a load factor of one only. So, the load in span BC is 5.25 kN/m. The loads are shown in Fig. 15.38.8a. Bending moments are determined by moment distribution method and presented in Table 15.2. Bending moment diagram is shown in Fig. 15.38.8b and the vertical reactions are shown in Fig. 15.38.8c.

Table 15.2 Moment Distribution for Load Case (ii)

Joint A B C Member AB BA BC CB D.F 1.0 0.5 0.5 1.0 F.E.M + 322.0 - 322.0 + 28.0 - 28.0 Balanced Moment - 322.0 + 147.0 + 147.0 + 28.0 Carry over Moment - - 161.0 + 14.0 - Balanced Moment - + 73.5 + 73.5 - Total Moment 0.0 - 262.5 + 262.5 0.0

Note: D.F. = Distribution Factors, F.E.M = Fixed End Moments Final values of the reactions are: VA = + 208.6875 kN, VB1 = +

274.3125 kN, VB2 = + 53.8125 kN and VC = - 11.8125 kN (+ means upward).

In the span AB, shear force is zero at I where x = 3.46 m and the

maximum moment is 360.666 kNm, obtained from the following equations:

(a) 208.6875 – 60.375 x = 0, gives x = 3.46 m and (b) Mx (at x = 3.46 m) = 208.6875 (3.46) – 60.375 (3.46) (3.46)/2 =

360.666 kNm. At point E (mid-span), Mx = 208.6875 (4) – 60.375(4) (4)/(2) = 351.75 kNm. The bending moment is zero at J (x = 6.91 m), obtained from the equation 208.6875 x – 60.375 (x2/2) = 0 where 0 ≤ x ≤ 8 m in the span AB. In the span BC, the bending moment is zero only at C (x = 8 m from B), as obtained from the equation: - 262.5 – 5.25 (x2/2) + 53.8125 x = 0, where 0 ≤ x ≤ 8 m in the span BC. The other value


of x where the negative moment is zero is x = 12.5 m. This point is outside the span 8m of BC. Thus, the bending moment is negative in the whole span BC.

Step 5. Load case (ii): Redistributed bending moment diagram While redistributing the moments for the load case (ii), we have to

take care of the static equilibrium of span AB regarding the redistributed moments. Figure 15.38.7d shows that the redistributed positive moment is the maximum at F (x = 3.3 m) and the value is 328.742 kNm. The maximum redistributed negative moment is 338.1 kNm at B. So, let us reduce the elastic bending moment at F (x = 3.3 m) to 328.742 kNm for the load case (ii) also. It should be checked if this reduction is within the maximum limit of 30 per cent or not.

Elastic bending moment at F (x = 3.3 m) for the load case (ii) = VA x

– wx2/2 = 208.6875 (3.3) – 60.375 (3.3) (3.3)/2 = 359.93 kNm (Fig. 15.38.8b). Hence, the reduction of 359.93 kNm to 328.742 kNm is (100) (359.93 – 328.742) / 359.93 = 8.665 per cent, which is within the limit of 30 per cent.

To have the redistributed moment at F = 328.742 kNm, the

magnitude of VA is determined from the equation: VA (3.3) – w (3.3) (3.3)/2 = 328.742 kNm, which gives VA = 199.237

kN. At F where x = 3.3 m, the shear force = 199.237 – 60.375 (3.3) = 0.

Therefore, the moment at F is the maximum positive redistributed moment of magnitude = 328.742 kNm.

At B where x = 8 m, the redistributed moment for the load case (ii)

is obtained from: (199.237) (8) – 60.375 (8) (8) / 2 = - 338.1 kNm, which is the same moment at B in the load case (i).

At E, (mid-span) where x = 4 m, the moment is: (199.237) 4 –

60.375 (4) (4)/2 = 313.95 kNm. The value of x where the redistributed bending moment = 0, in the span AB, is obtained from: 199.37 x – 60.375 (x2/2) = 0, which gives x = 6.6 m.

For the span BC, the redistributed negative moment is – 338.1

kNm, as obtained in the span AB for this load case. The vertical reaction VB2 in the span BC is (5.25) (4) + 338.1/8 = 63.2625 kN. The location of the point where the moment is zero in the span BC is obtained from the equation: – MB - wx2/2 + VB2 x = 0 or – 338.1 –


5.25 (x2/2) + 63.2625 x = 0, which gives x = 8 m. The other value of x where moment is zero is 16.1 m (outside the span BC). Therefore, the redistributed moment is negative in the entire span BC. The redistributed bending moment diagram is shown in Fig. 15.38.8d and calculations of reaction are shown in Fig. 15.38.8e.

As mentioned earlier, the load case (iii) is not separately taken up

since it is the mirror image of load case (ii).


Step 6. Superimposition of the elastic bending moment diagrams

The envelope of the positive and negative bending moments of

elastic analyses is to be prepared considering all three load cases from Figs. 15.38.7c and 15.38.8b. The bending moment diagram for the third load case is the mirror image of Fig. 15.38.8b. The positive moment is to be taken from the diagram of span AB of Fig. 15.38.8b for both the spans as that is more than the values of Fig. 15.38.7c.

For the negative moment in span BC, the diagram for the load case

(i) will intersect that of the load case (ii). Similarly, for the negative moment in span AB, the diagram of load case (i) will intersect that of load case (iii). Due to the mirror image, the point of intersection is determined for span BC only.

For the load case (i) in span BC, the bending moment at a distance

of x from the support B is given by: Mx = - 483 + 301.875 x – 60.375 (x2/2) Similarly, for the load case (ii) in span BC, the bending moment at a

distance of x from the support B is given by: Mx = - 262.5 + 53.8125 x – 5.25 (x2/2) We can find the value of x, where these two negative moments are

the same by equating the two expressions: - 483 + 301.875x – 60.375 (x2/2) = - 262.5 + 53.8125x – 5.25

(x2/2) or 27.5625 x2 – 248.0625 x + 220.5 = 0 The values of x are 1.0 m and 8.0 m. So, at a distance of 1.0 m

from the support B, the two negative moments intersect. The value of the moment is obtained from either of the two equations, which comes out to be 211.3125 kNm. Accordingly, Fig 15.38.9a presents the envelope of the elastic bending moment diagrams making use of symmetry of the load in the two spans.


Step 7. Superposition of redistributed bending moment diagrams

The procedure is exactly the same as in Step 6. The positive

bending moment diagram is taken from that of span AB of load case (ii) and the negative moment envelope is obtained combining the negative bending moments of span BC for the two load cases. In this redistributed negative moments, there is no intersection of the diagram as in Step 6. Accordingly, Fig. 15.38.9b presents the envelope of the redistributed bending moments using symmetry of the loads in the two spans.

15.38.6 Practice Questions and Problems with Answers Q.1: Explain the concept of redistribution of moments in statically indeterminate

reinforced concrete structures. A.1: Paragraphs 1 and 2 of sec. 15.38.1 Q.2: Mention three advantages of considering the redistribution of moments for

the design of statically indeterminate reinforced concrete structures. A.2: Paragraph 3 of sec. 15.38.1 Q.3: State the assumptions of considering the redistribution of moments in the

design of statically indeterminate reinforced concrete structures. A.3: Paragraph 1 of sec. 15.38.2 Q.4: What are the recommendations of IS 456 regarding the redistribution of

moment in the design of statically indeterminate structures employing working stress and limit state methods?

A.4: Section 15.38.3 is the full answer.


Q.5: Solve Problem 1 of sec. 15.38.5 (Fig. 15.38.5a) when the redistribution is

limited to 20 per cent. A.5: Step 1 is the same as that of Problem 1 of sec. 15.38.5. Please refer to

Figs. 15.38.5b and c for the elastic bending moment diagram and free body diagram, respectively.

Step 2. Redistributed bending moment diagram and envelope of

moment diagrams.


Maximum negative moment at A after the redistribution of 20 per cent = 0.8 (-128) = - 102.4 kNm. So, the maximum positive moment at the mid-span C = 192-102.4 = 89.6 kNm (Please refer to Fig. 15.38.10b). VA = 96 kN. The point of inflection is obtained from: 96 x – 12x2 – 102.4 = 0, which gives x = 1.27 m (Fig. 15.38.10a). At x = 1.27 m, elastic moment = 96x – 12x2 – 128 = - 25.435 kNm. The redistributed moment at E (x = 1.27 m) = 0.8 (- 25.435) = - 20.35 kNm.

The redistributed bending moment diagram is shown in Fig. 15.38.10b and the envelope of the moment diagram is shown in Fig. 15.38.10c. 15.38.7 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 15.38.8 Test 38 with Solutions

Maximum Marks = 50 Maximum Time = 30

minutes TQ.1: Mention three advantages of considering the redistribution of moments for

the design of statically indeterminate reinforced concrete structures. (10 marks)

A.TQ.1: Paragraph 3 of sec. 15.38.1 TQ.2: What are the recommendations of IS 456 regarding the redistribution of

moment in the design of statically indeterminate structures employing working stress and limit state methods?

(10 marks) A.TQ.2: Section 15.38.3 is the full answer

TQ.3: Solve Problem 2 of sec. 15.38.5 (Fig. 15.38.6a) considering the

redistribution up to (a) 20 per cent and (b) 10 per cent, separately. (30 marks)


A.TQ.3: Step 1 is the same as that of Problem 2 of sec. 15.38.5. Please refer to Figs. 15.38.6b and c for the elastic bending moment diagram and free body diagram, respectively.

Step 2. Redistributed bending moment diagram (20 per cent

redistribution) and envelope of the moment diagrams Redistributed negative moment at A = 0.8 (- 60) = - 48 kNm. Redistributed positive moment at E = 90 – 48 = 42 kNm The point of zero moment (point G) is obtained by taking moment of forces

of the free body diagram about G, shown in Fig. 15.38.11a, when 0 ≤ x ≤3 m, gives 30x – 48 = 0 or x = 1.6 m. The redistributed moment diagram is shown in Fig. 15.38.11b.

At G where x = 1.6 m, the elastic moment = 30 (1.6) – 60 = - 12kNm.

Therefore, the redistributed moment at G = 0.8 (-12) = - 9.6 kNm. The envelope of the moment diagrams is shown in Fig. 15.38.11c.

Step 3. Redistributed bending moment diagram (10 per cent

redistribution) and envelope of the moment diagrams. Redistributed negative moment at A = 0.9 (- 60) = - 54 kNm. Redistributed positive moment at E = 90 - 54 = 36 kNm.


The point of zero moment (point G) is obtained by taking moment of forces

about G of free body diagram shown in Fig. 15.38.12a, when 0 ≤ x ≤ 3 m, gives 30x – 54 = 0, or x = 1.8 m. The redistributed moment diagram is shown in Fig. 15.38.12b.

At G, where x = 1.8 m, the elastic moment = 30 (1.8) – 60 = - 6 kNm.

Therefore, the redistributed moment at G = 0.9 (- 6) = - 5.4 kNm. The envelope of the moment diagrams is shown in Fig. 15.38.12c.


This lesson explains the concept of redistribution of moments in statically indeterminate reinforced concrete structures designed either by the working stress or by the limit state methods. The behaviour of such structures with increasing loads after reaching the respective design loads is important to understand the inclusion of redistribution of moments. The advantages of such inclusion are mentioned in this lesson. The formations of first plastic hinge and subsequently other plastic hinges leading to the formation of collapse mechanism will not only increase the load carrying capacity of the structure but also help the designer to have flexibility in reducing or increasing moments to a limited extent. By this, the detailing of reinforcement helps to avoid congestion of the bars at sections having high moments. Due to the limited rotation capacities of reinforced concrete cross-sections, the reductions / additions of moments are somewhat restricted, which are also explained as stipulated in IS 456. Several numerical problems are solved for the purpose of illustration considering full or partial redistribution in clamped or two-span continuous beams taking up different combinations of dead and live loads. Understanding the illustrative examples and solving the practice and test problems will help in utilising the redistribution to get the advantages in designing statically indeterminate reinforced concrete structures.


Module 16

Earthquake Resistant Design of Structures


Lesson 39

Seismic Effects, Material Behaviour and

General Principles of Earthquake Resistant Design of Structures




• state the direct and indirect seismic effects,

• explain the behaviour of plain concrete, steel and reinforced concrete with

high intensity repeated axial cyclic loads,

• define common terminologies needed for earthquake engineering,

• name different standards of BIS regarding the earthquake resistant design

of structures,

• state the assumptions of earthquake resistant design of structures,

• determine the horizontal seismic coefficient, design seismic base shear and

distribution of design force,

• state the steps of static elastic design of such structures,

• mention the objectives of earthquake resistant design of structures,

• explain the need for ductility and ductile detailing of reinforcement in the

earthquake resistant design of structures.


Earthquakes, Tsunamis, Seiches, Landslides, Floods and Fires are natural calamities causing severe damage and sufferings to persons by collapsing the structures, cutting off transport systems, killing or trapping persons, animals etc. Such natural disasters are challenges to the progress of development. However, civil engineers as designers have a major role to play in minimising the damages by proper designing the structures or taking other useful decisions. Because of the vastness of the topic, “Disaster management and mitigation”, this module includes understanding the earthquakes, behaviour of the materials of construction and structures and the extent to which structural engineers make use of the knowledge in taking proper decisions in designing the structures made of reinforced concrete.


Earthquakes have many other effects besides vibrating the structures in response to ground shaking at its foundation. These other effects may even exceed that due to vibration. Unfortunately, the procedure of their estimation and the required steps for the design are considered outside the scope of structural engineering. Different seismic resistant design codes have provisions to take into account the vibration of structures. But, these codes do not have any provision to take care of other effects. However, structural engineers should be aware of the intensity of the hazards with a view to taking precautionary measures either in the design of structures, advising clients in selecting proper sites in such zones or making them aware of the importance of proper maintenance of the structures and other considerations the clients should follow up while using the designed structures. The different direct and indirect effects of earthquakes are mentioned in the following section. 16.39.2 Direct and Indirect Seismic Effects A. Direct effects. The following are the direct seismic effects: 1. Damage due to surface faulting

The damage due to surface faulting varies widely. It may totally demolish houses, rupture the foundations, tilt the foundation slabs and walls or may cause minor damage to the houses.

2. Damage due to liquefaction Liquefaction of soil may cause instability due to internal seismic waves

and thereby may significantly damage in form of settlement, tilting and rupture of the structures. The extent of damage depends on properties of soils of different layers, depth of the water table and the intensity, magnitude and duration of the earthquake. Accordingly, there may be either large settlement or differential settlement of the ground surface. Please refer to sec. 16.39.4(h) of this lesson for the definition of liquefaction as per IS Code.

It is always desirable to avoid construction in such areas than to design

the structures following codal provisions, which may be insufficient, though ensure effective design against vibration of structures due to shaking at the foundation level.

3. Damage due to ground shaking As the state of the art of this subject is still developing, integrated field

inspection of structures damaged by earthquakes and their analyses are useful in further adding/improving the expert knowledge in the seismic


resistant design and construction. Earlier inspections and analyses established the types of foundation, configurations of structures, materials of construction and design and detailing of construction. Such data are being continuously updated.

4. Damage due to sliding of superstructure on its foundation It is essential that the whole structure and the foundation should work as a

unit especially for the seismic resistant design and construction of structures. For this the superstructures should be anchored properly to the foundation.

5. Damage due to structural vibration The extent of the damage due to structural vibration depends on the

materials of construction. Wood, reinforced concrete and steel are widely used in civil engineering structures. It is well-known that inertia forces are developed as vibration response of a structure due to earthquake ground shaking. The intensity of such inertia forces is directly proportional to the product of mass and acceleration. Hence, reduction of mass is very effective to minimise the inertia forces. In this respect, timber has the maximum advantage as a potential construction material due to its low mass. Concrete, though a heavy material when reinforced with steel bars, has good strengths both in compression and in tension. Accordingly, reinforced concrete can be used effectively by providing proper amount of reinforcement and correct detailing of them as they play significant roles in the seismic resistant design of reinforced concrete structures. Steel has the additional advantages of ductility, strength and toughness per unit weight than concrete. The major problem of steel is the buckling. Behaviour of plain and reinforced concrete and steel with high intensity repeated axial cyclic loads is taken up in the next section.

B. Indirect effects Tsunamis, seiches, landslides, floods and fires are the indirect effects of

earthquakes. These may occur either alone or in combinations to add to the damages during an earthquake.

16.39.3 Behaviour of Concrete and Steel with High Intensity

Repeated Axial Cyclic Loads. Earthquakes cause ground motions in a random fashion in all directions having significant horizontal and vertical ground accelerations as function of time. Structures subjected to ground motions respond in a vibratory fashion. The maximum response acceleration during the elastic stage depends on the natural frequency of vibration of the structure and the magnitude of the damping. The


maximum inertia loads acting on a structure during an earthquake is determined by multiplying the mass by the acceleration. In order to understand the behaviour of reinforced concrete structure under such repeated cyclic loads, we first discuss the behaviour of plain concrete and reinforcing steel below. (a) Plain concrete Plain concrete has tensile strength less than twenty per cent of its compressive strength. The tensile strength of plain concrete is normally not obtained from the direct tension test because of difficulties in holding the specimens and the uncertainties of developing secondary stresses due to holding devices. Hence, this is measured indirectly either by split cylinder test or by bending test conducted on plain concrete prisms. Prism tests give the tensile strength of concrete in flexure, known as the modulus of rupture. Plain concrete specimens are not tested under repeated axial cyclic tensile loads.

However, earlier experiments on plain concrete cylinders with high intensity repeated axial compressive loads at slow rate of straining give the stress-strain curve as shown in Fig. 16.39.1. The stress-strain curve shows pronounced hysteresis effect and the envelope curve is almost identical to that of a single continuous load. Furthermore, such a curve obtained from the static test is also the same as that of the first cycle. Figure 16.39.1 shows that the slope of the stress-strain curve and the maximum attainable stress decrease with the number of cycles indicating the reduction of strength and stiffness of concrete due to the formation of cracks. With the increase of rate of loading, the compressive strength of concrete increases while the strain at the maximum stress decreases.


(b) Reinforcing steel

Figure 16.39.2 represents a typical stress-strain curve based on experiments with steel rods with repeated axial loads in tension and compression. Initially, the curve is similar to that obtained in static test in tension. On reversing the load after reaching the yield strain in tension, the unloading curve is seen to be curvilinear forming loops instead of the straight line as shown in Fig. 16.39.2, showing Bauschinger effect. The loop formed by one complete cycle is known as hysteresis loop, the area of which is the energy absorbed by the specimen in that cycle. Practically the same path is repeated in subsequent cycles. Thus, we find that the stress-strain curve for steel is cycle independent. The loop, therefore, continues for number of cycles till the specimen buckles or fails due to fatigue. It is also worth mentioning that the same hysteresis loops are obtained for specimens loaded first in compression, unloaded and reloaded in tension. However, the yield strength of steel is dependent on the rate of loading.


(c) Reinforced Concrete

Similar tests of loading and unloading in tension and compression on doubly-reinforced cantilever beam loaded with a point load P at the free end reveal the following:

(i) Formation of large cracks when the load P acting downward

causes post-elastic range of stress in tension steel (Fig. 16.39.3a).


(ii) Due to residual plastic strains in the steel, these cracks do not completely close on unloading but they remain open (Fig. 16.39.3b).

(iii) On loading the beam in a reverse direction (upward), as shown

in Fig. 16.39.3c, the resistance to rotation is decreased with respect to that during the first loading. This reduction is due to the presence of open cracks in the compression zone. As a result, the whole of compression is carried by compression steel. Thus, the flexural rigidity of the section is only that of steel.

(iv) At higher load (upward) when Bauschinger effect starts and

behaves inelastically, the flexural rigidity further reduces (Fig. 16.39.3d). The cracks in the compression zone may close depending on the magnitude of the load and the relative amounts of tension and compression steel. As a result, the stiffness of the member increases since concrete now can carry some compression load due to the closing of the cracks. However, if the cracks do not close and the member is unloaded, the cracks may be throughout the whole depth at critical sections. The width of these full-depth cracks depends on the amount of yielding and the effectiveness of the bond.

(v) If the member is again loaded downward, the member behaves

as a steel beam initially as the concrete is not in contact with the steel at the face of the crack.


Due to such reversal of loading, opening and closing of cracks in

the alternate tension and compression zones will gradually reduce the compressive strength of concrete either due to less or no contact because of slight relative lateral movement or presence of debris in the crack. This along with Bauchinger effect of steel gives the moment-curvature relation for the doubly-reinforced section as shown in Fig. 16.39.4. This diagram is far different from the idealised moment-curvature relation as shown in Fig. 16.39.5. The rounding and pinching of the loops of Fig. 16.39.4 clearly show smaller area than that of Fig. 16.39.5. Therefore, less energy will be dissipated per cycle in the actual case than the assumed idealised one. Accordingly, the response of frames to severe earthquake motions will be influenced by such effects.

Furthermore, the presence of high shear causing large shear

displacement will split the concrete longitudinally along the flexural steel bars leading to further loss of bond and stiffness.

Thus, the influential factors of load-displacement relationship of

reinforced concrete members during severe earthquake subjected to reversed inelastic deformations are summarised as:


1. Inelastic behaviour of steel reinforcement, 2. The extent of cracking of concrete,

3. Effectiveness of bond and anchorage, and

4. Presence of high shear.

Accordingly, realistic dynamic analyses shall be based on more accurate moment-curvature loops. The moment-curvature relationships, shown in Figs. 16.39.4 and 16.39.5, reveal that the deformation changes from cycle to cycle after the yielding starts.

16.39.4 Terminology for Earthquake Engineering (a) Design Basis Earthquake (DBE) It is an earthquake which can reasonably be expected to occur at least

once during the design life of the structure, as defined in cl.3.6 of IS 1893 (Part 1): 2002.

(b) Epicentre Epicentre is the geographical point on the surface of earth vertically above

the focus of the earthquake (cl.3.10 of IS 1893 (Part 1): 2002). (c) Focus Focus is the source of the elastic waves of the originating earthquake

inside the earth which cause shaking of ground (cl. 3.13 of IS 1893 (Part 1): 2002).

(d) Intensity of earthquake The intensity of an earthquake indicates the strength of shaking during the

earthquake and is expressed by a number according to the modified Mercalli Scale or M.S.K Scale of seismic intensities (cl. 3.15 of IS 1893 (Part 1): 2002).

(e) Magnitude of earthquake (Richter’s Magnitude) The magnitude of an earthquake is expressed by a number, which is a

measure of the energy released in an earthquake. The magnitude of an earthquake is defined as logarithm to the base 10 of the maximum trace amplitude, expressed in micron, which the standard short-period torsion


seismometer (with a period of 0.8 second, magnification of 2800 and damping nearly critical) would register due to the earthquake at an epicentral distance of 100 km (cl. 3.18 of IS 1893 (Part 1): 2002).

(f) Critical damping Critical damping is the damping beyond which the free vibration motion

will not be oscillatory (cl. 3.3 of IS 1893 (Part 1): 2002). (g) Maximum Considered Earthquake (MCE) Maximum Considered Earthquake is the most severe earthquake whose

effects are considered by IS 1893 (Part 1): 2002, as given in cl. 3.19 of this standard.

(h) Liquefaction Liquefaction is a state in saturated cohesionless soil wherein the effective

shear strength is reduced to negligible value for all engineering purposes due to pore pressure caused by vibrations during an earthquake when they approach the total confining pressure. In this condition the soil tends to behave like a fluid mass (cl. 3.16 of IS 1893 (Part 1): 2002).

16.39.5 Bureau of Indian Standards for Earthquake Design In our country, several major earthquakes have occurred in the Himalayan-Nagalushai region, Indo-Gangetic Plain, Western India, Kutch and Kathiawar regions. Taking into account seismic data from studies of these Indian earthquakes, Bureau of Indian Standard first published IS 1893 “Recommendations for earthquake resistant design of structures” in 1962 and revised in 1966. Considering the local seismology, accepted level of seismic risk, building topologies and materials and methods used in construction, presently the Bureau of Indian Standards has the following seismic codes:

1. IS 1893 (Part 1), 2002; Indian Standard Criteria for Earthquake Resistant Design of Structures (5th Revision),

2. IS 1893 has other four parts: (a) Part 2 for liquid retaining tanks–

elevated and ground supported, (b) Part 3 for bridges and retaining walls, (c) Part 4 for industrial structures including stack like structures and (d) Part 5 for dams and embankments. However, they are yet to be finalised. Hence, provisions of Part 1 will be read along with relevant clauses of IS 1893: 1984 for structures other than buildings.


3. IS 4326: 1993, Indian Standard Code of Practice for Earthquake Resistant Design and Construction of Buildings, (2nd Revision),

4. IS 13827: 1993, Indian Standard Guidelines for Improving Earthquake

Resistance of Earthen Buildings,

5. IS 13828: 1993, Indian Standard Guidelines for Improving Earthquake Resistance of Low Strength Masonary Buildings,

6. IS 13920: 1993, Indian Standard Code of Practice for Ductile Detailing

of Reinforced Concrete Structures Subjected to Seismic Forces, and

7. IS 13935: 1993, Indian Standard Guidelines for Repair and Seismic Strengthening of Buildings.

The regulations of these standards will not result in structures having no damage during earthquake of all magnitudes. However, the regulations shall ensure that, as far as possible, structures will be able to respond without structural damage to shocks of moderate intensities and without total collapse to shocks of heavy intensities.

16.39.6 General Principles of Earthquake Resistant

Design of Structures

a) Ground motion

The characteristic parameters like intensity, duration etc. of seismic ground vibrations depend upon the magnitude of the earthquake, its depth of focus, distance from the epicentre, properties of soil or medium through which the seismic waves travel and the soil strata where the structure stands. The random earthquake motions can be resolved in any three mutually perpendicular directions. The horizontal direction is normally the prominent direction. Vertical acceleration is considered in large-span structures.

The response of a structure to ground vibrations depends on the nature of

foundation soil, form, material, size and mode of construction of structures and the duration and characteristics of ground motion.

(b) Assumptions

The following are the assumptions in the earthquake resistant design of structures:


1. Impulsive ground motions of earthquake are complex, irregular in

character, changing in period and time and of short duration. They, therefore, may not cause resonance as visualised under steady-state sinusoidal excitations, except in tall structures founded on deep soft soils.

2. Wind, maximum flood or maximum sea waves will not occur

simultaneously with the earthquake.

3. For static analysis, elastic modulus of materials shall be taken unless otherwise mentioned.

16.39.7 Design Lateral Forces (a) Design horizontal seismic coefficient The design horizontal seismic coefficient Ah (cl. 6.4.2 of IS 1893 (Part 1): 2002) for structure is determined from Ah = (Z I Sa / 2 R g) (16.1) where Z = the zone factor as given in Table 2 of IS 1893 (Part 1): 2002, based on

classifying the country in four seismic zones, I = Importance factor, depending upon the functional use of the structure

as given in Table 6 of IS 1893 (Part 1): 2002, R = Response reduction factor, depending on the perceived seismic

damage, performance of the structure, characterised by ductile or brittle deformations, as given in Table 7 of IS 1893 (Part 1): 2002. However, the value of I/R shall not be greater than 1.0, and

Sa/g = Average response acceleration coefficient for rock or soil sites as

given in Fig. 2 and Table 3 of IS 1893 (Part 1): 2002. It is further stipulated in cl.6.4.2 of IS 1893 (Part 1): 2002, that for any structure with undamped natural period of vibration of the structure (in seconds) T ≤ 0.1 second, the value of Ah will not be taken less than Z/2 whatever be the value of I/R.


(b) Design seismic base shear The total design lateral force or design seismic base shear VB along any

principal direction (cl.7.5.3 of IS 1893 (Part 1): 2002) shall be determined from the following equation:

VB = Ah W (16.2) where Ah is as given in Eq. 16.1, and W is the seismic weight of the building

as given in cl. 7.4.2 of IS 1893 (Part 1): 2002. (c) Distribution of design force The design base shear VB of Eq. 16.2 shall be distributed along the height of

the building as per the following equation (cl.7.7 of IS 1893 (Part 1): 2002): B

2j

(16.3) 2

1

n

i B i i ij

Q V (W h ) / W h=

= ∑where Qi = Design lateral force at floor i, Wi = Seismic weight of floor i, hi = Height of floor i measured from base, and n = Number of storeys in the building at which the masses are located 16.39.8 Static Elastic Design The various steps, based on the elastic design are:

1. Dead load and imposed loads are considered as per cl.6.3 of IS 1893 (Part 1): 2002.

2. The horizontal loads are determined as explained in sec. 16.39.7.

3. Bending moments and shear forces are determined on columns and floor

beams in a manner similar to that of frames subjected to wind loads.

4. All parts must be efficiently bonded together so that the structure acts as a unit.


5. Panel walls, finishes and ornaments should be permanently attached to the frame to ensure that they do not collapse independently in the event of a shock.

6. Separate footings of columns shall be connected by ties to resist the

compressive thrust or tensile pull of magnitude 1/10th of the load on the footing.

Structures designed on this simple elastic principle may even survive when subjected to severe earthquake due to the following:

a) Yielding at critical sections increases the period of vibration and helps to

absorb greater amounts of input energy. b) Assistance of non-structural partitions and the dissipated energy are

helpful as they crack.

c) Yielding of foundations helps to reduce the predicted response. 16.39.9 Dynamic Analysis Though static elastic analysis is considered sufficient for smaller building, dynamic analyses shall be performed to determine the seismic force and its distribution to different levels for regular and irregular buildings, as defined in cl. 7.1 of IS 1893 (Part 1): 2002, following the recommendations of cl. 7.8 of the same IS Code.

(a) Regular buildings – Those greater than 40 m in height in Zones IV and V and those greater than 90 m in height in Zones II and III,

(b) Irregular buildings – All framed buildings higher than 12 m in Zones IV and

V and those higher than 40 m in Zones II and III.

Though not mandatory, the code also recommends dynamic analysis for buildings lesser than 40 m for irregular building in Zones II and III (see Note below cl.7.8.1 of IS 1893 (Part 1): 2002).

The dynamic analysis may be carried out either by Time History Method or by Response Spectrum Method. More about the dynamic analysis is beyond the scope of this module.


16.39.10 Objectives of Earthquake Resistant Design of Structures It is uneconomical to design structures to withstand major earthquakes elastically. Therefore, the trend of the design is that the structure should have sufficient strength and ductility to withstand large tremors elastically. For this the interconnections of the members must be designed particularly to ensure sufficient ductility. Accordingly, the design approach adopted in IS 1893 (Part 1): 2002 is stated in cl 6.1.3 of the standard which is as follows: The design approach is to ensure the following:

(a) that structures possess at least a minimum strength to withstand minor earthquakes (⟨DBE), which occur frequently, without damage;

(b) that structures resist moderate earthquakes (DBE) without significant

structural damage though some non-structural damage may occur; and

(c) that structures withstand major earthquakes (MCE) without collapse.

16.39.11 Ductility and Ductile Detailing of Reinforcement Actual forces caused by earthquakes on structures are much greater than the design forces determined following IS 1893 (Part 1): 2002. However, ductility arising from inelastic behaviour of reinforced concrete and steel, as explained in sec.16.39.3 and detailing of reinforcement and over-strength due to reserve strength in structures beyond the design strength are to be relied upon to take care of this difference between the actual and design lateral loads. Thus, ductility and ductile detailing of reinforced concrete structures are very important when the structures are subjected to seismic forces. Accordingly, IS 13920: 1993 stipulates the recommendations of ductile detailing of reinforced concrete structures subjected to seismic forces. These two aspects, therefore, are taken up in the next lesson. 16.39.12 Practice Questions and Problems with Answers Q.1: State the direct and indirect seismic effects.


A.1: Section 16.39.2 is the complete answer Q.2: Explain the behaviour of plain concrete with high intensify repeated axial

cyclic loads. A.2: Part (a) of sec. 16.39.3 Q.3: Explain the behaviour of steel with high intensity repeated axial cyclic loads. A.3: Part (b) of sec. 16.39.3 Q.4: Explain the behaviour of reinforced concrete with high intensity repeated

axial cyclic loads. A.4: Part (c) of sec. 16.39.3 Q.5: Define the following terminologies as per IS 1893 (Part 1): 2002: (a) Design

Basis Earthquake (DBE), (b) Epicentre, (c) Focus, (d) Intensify of earthquake, (e) Magnitude of earthquake, (f) critical damping, (g) Maximum Considered Earthquake and (h) Liquefaction.

A.5: Section 16.39.4 is the complete answer. Q.6: Name the different BIS Standards for earthquake design. A.6: Section 16.39.5 is the complete answer. Q.7: Write the expression for determining (a) Horizontal seismic coefficient, (b)

Design seismic base shear and (c) Distribution of design force. A.7: Section 16.39.7 is the complete answer. Q.8: Write the steps of performing static elastic design of earthquake resistant

structures. A.8: Section 16.39.8 is the complete answer.


Q.9: When should we have to perform dynamic analysis? A.9: Section 16.39.9 is the complete answer. Q.10: State the objectives of earthquake resistant design of structures. A.10: Section 16.39.10 is the complete answer. Q.11: Explain the need of ductility and ductile detailing of reinforced concrete

structures subjected to seismic forces. A.11: Section 16.39.11 is the complete answer.

16.39.13 References

















16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi. 17. Structural Engineering Slide Library, W.G.Godden, Editor; Set J:

Earthquake Engineering V.V.Bertero, National Information Service for Earthquake Engineering, University of California, Berkeley, USA. ( http://nisee.berkeley.edu/bertero/html/damage_due_to_ground_failure.html )

16.39.14 Test 39 with Solutions Maximum Marks = 50 Maximum Time = 30

minutes Answer all questions. TQ.1: State the direct and indirect seismic effects.

(10 Marks) A.TQ.1: Section 16.39.2 is the complete answer TQ.2: Explain the behaviour of reinforced concrete with high intensity repeated

axial cyclic loads. (10 Marks)

A.TQ.2: Part (c) of sec. 16.39.3 TQ.3: Name the different BIS Standards for earthquake design.

(15 Marks) A.TQ.3: Section 16.39.5 is the complete answer. TQ.4: Write the steps of performing static elastic design of earthquake resistant

structures. (10 Marks)

A.TQ.4: Section 16.39.8 is the complete answer. TQ.5: State the objectives of earthquake resistant design of structures.

(5 Marks)


http://nisee.berkeley.edu/bertero/html/damage_due_to_ground_failure.html

http://nisee.berkeley.edu/bertero/html/damage_due_to_ground_failure.html

A.TQ.5: Section 16.39.10 is the complete answer. 16.39.15 Summary of this Lesson This lesson explains the direct and indirect effects of earthquakes. Behaviour of plain concrete, steel bars and reinforced concrete are discussed when subjected to high intensity repeated axial loads. Some common terminologies are defined as given in IS 1893 (Part 1): 2002. Different BIS standards pertaining to earthquake design are given. General principles and assumptions are given in this lesson. Determinations of horizontal seismic coefficient, design seismic base shear and distribution of design force are explained. Steps of the static elastic design are given mentioning the particular situations when dynamics analysis should be done. The objectives of earthquake resistant design of structures are explained. The need for ductility and ductile detailing of reinforcement is explained in this lesson.


Module 16

Earthquake Resistant Design of Structures


Lesson 40

Ductile Design and Detailing of Earthquake

Resistant Structures Version 2 CE IIT, Kharagpur


• define and explain ductility factor with respect to displacement, curvature or

rotation,

• state the advantages of ductility in the design of reinforced concrete

members,

• derive expressions of ductility factor of singly and doubly-reinforced

rectangular beams,

• mention the factors influencing ductility,

• give general specification of materials for the ductile design of reinforced

concrete members,

• state the general guidelines in the design and detailing of structures having

sufficient strength and ductility,

• identify the situation when special confining reinforcement is needed,

• to determine the ductility factor of singly and doubly-reinforced rectangular

beams applying the expressions,

• to apply the knowledge in designing and detailing beams, columns and

beam-column joints as per IS 13920:1993.


As mentioned in sec. 16.39.10 of Lesson 39, it is uneconomical to design structures to withstand major earthquakes. However, the design should be done so that the structures have sufficient strength and ductility. This lesson explains the requirements and advantages of ductility in the design of reinforced concrete members which can be expressed with respect to displacement, curvature or rotation of the member. The expressions of ductility of singly and doubly-reinforced beams with respect to curvature are derived. The influencing parameters of the ductility are explained. Several aspects of design for ductility are explained mentioning detailing for ductility, as stipulated in IS 13920:1993, for


flexural members and columns. Illustrative examples are solved to determine the ductility with respect to curvature of singly and doubly-reinforced beams. Moreover, numerical problems are solved to illustrate the design of beams, columns and beam-column joints. 16.40.2 Displacement Ductility


It is essential that an earthquake resistant structure should be capable of

deforming in a ductile manner when subjected to lateral loads in several cycles in


the inelastic range. Let us take a single degree of freedom oscillator, as shown in Fig. 16.40.1. In the elastic response, the oscillator has the maximum response at a. The area oab represents the potential energy stored when maximum deflection occurs. The energy is converted into kinetic energy when the mass returns to zero position. Figure 16.40.2 shows the oscillator forming a plastic hinge at a much lower response c when the deflection response continues along cd, d being the maximum response. The potential energy at the maximum response is now represented by the area ocde. When the mass returns to zero position, the part of the potential energy converted to kinetic energy is represented by fde, while the other energy under the area ocdf is dissipated by the plastic hinge by being transferred into heat and other forms of energy, which are irrecoverable. It is thus evident that, elastically, the full potential energy is returned to kinetic energy, while elastoplastically a part of the energy is converted into kinetic energy. Hence, the potential energy stored in the elastoplastic structure may not be equal to that in elastic structure and the maximum deflection of the elastoplastic structure may not be equal to that of elastic structure. Figures 16.40.3 and 4 present equal maximum deflection and equal maximum potential energy responses of two structures.

The displacement ductility factor μ, a measure of ductility of a structure, is

defined as the ratio of Δu, and Δy, where Δu, and Δy are the respective lateral deflections at the end of post elastic range and when the yield is first reached. Thus, we have

μ (with respect to displacement) = Δu / Δy

(16.4) The values of displacement ductility factor should range from 3 to 5. 16.40.3 Curvature Ductility The curvature ductility factor μ is defined as the ratio of φu and φy, where φu and φy are the respective curvatures at the end of postelastic range and at the first yield point of tension steel, as stipulated in cl. 3.3 of IS 13920:1993. Thus, we have

μ (with respect to curvature) = φu / φy, (16.5) It should be noted that the curvature ductility factor is significantly different

from the displacement ductility factor. At the start of yielding in a frame, the deformations concentrate at the positions of plastic hinge. Therefore, when a frame is deflected laterally in the postelastic range, the φu / φy, ratio in a plastic hinge may be greater than Δu, / Δy ratio.


16.40.4 Rotational Ductility

In a similar manner, the rotational ductility factor μ is defined as the ratio of θu and θy, where θu and θy are the respective rotations of at the end of postelastic range and at the first yield point of tension steel. Thus, we have

μ (with respect to rotation) = θu / θy

(16.6) Thus, there are three methods of defining the ductility. In general, it can be stated that the ductility is the ratio of absolute maximum deformation at the end of postelastic range to the yield deformation. Accordingly, the ductility can be defined with respect to strain, rotation, curvature or deflection. Rotation and curvature based ductility factors take into account shape and size of the member. Though there is no special advantage of one or the other definition, we will consider curvature ductility, as stipulated in cl. 3.3 of IS 13920:1993 and explained in sec. 16.40.3 (Eq. 16.5). 16.40.5 Advantages of Ductility

The following are the advantages of a reinforced concrete structure having sufficient ductility:

(i) A ductile reinforced concrete structure may take care of

overloading, load reversals, impact and secondary stresses due to differential settlement of foundation.

(ii) A ductile reinforced concrete structure gives the occupant

sufficient time to vacate the structure by showing large deformation before its final collapse. Accordingly, the loss of life is minimised with the provision of sufficient ductility.

(iii) Properly designed ductile joints are capable of resisting forces

and deformations at the yielding of steel reinforcement. Therefore, these sections can reach their respective moment capacities, which is one of the assumptions in the design of reinforced concrete structures by limit state method.

16.40.6 Expressions of Ductility of Reinforced Concrete

Rectangular Beams (A) Singly-reinforced rectangular section


igures 16.40.5a, b and c show the cross-section of a singly-reinforced

rectangular beam, strain profile and curvature at yield and ultimate stages. From Fig. 16.40.5b, we have: φy = ∈y / (d – kd) (16.7) where ∈y = fy/Es (16.8)

k = - (mp/100) + {(mp/100)2 + 2 (mp/100)}1/2 (16.9)

m = 280 / 3σcbc (16.10) and σcbc = permissible stress of concrete in bending compression. From Fig. 16.40.5c, we have: φu = ∈uc / xu (16.11) where ∈uc = 0.0035, as given in cl. 38.1b of IS 456:2000. From Eq. 3.17 of Lesson 5, we have for singly-reinforced sections, xu / d = 0.87 fy p / (36) (fck) (16.12)


where p = 100 Ast / bd (16.13) It is also known that xu / d ≤ xu, max / d for under-reinforced sections. Substituting Eqs. 16.7 and 16.11 in Eq. 16.5, we have: Curvature ductility μ = (∈uc / ∈y) {(1 – k) / (xu/d)} (16.14) Substituting ∈y, k and (xu/d) from Eqs.16.8, 9 and 12, respectively, and using ∈uc = 0.0035 and Ey = 200000 N/mm2, we have:

μ = (25200/0.87 fy) (fck / p fy) ⎭⎬⎫

⎩⎨⎧ +−+ )/mp()/mp()/mp( 10021001001 2

(16.15) From Fig. 16.40.5c, we also have xu /(d-xu) = ∈uc / ∈ustwhich gives xu / d =∈uc / (∈uc + ∈ust) (16.16) where ∈ust = maximum strain in tension steel = μs ∈y, where μs = Ductility in steel with respect to strain. Thus, the curvature ductility μ = φu/φy (Eq. 16.5) can be determined in any of the following ways:

(i) Employing Eq. 16.14 using ∈uc = 0.0035 and determining ∈y, k and (xu/d) from Eqs. 16.8, 16.9 and 16.12, respectively from the given fck, fy and p.

(ii) Employing Eq. 16.15 from the given fck , fy and p.

The value of k can be determined by finding the depth of the neutral axis taking moment of the compression concrete and the tensile steel about the neutral axis directly in place of employing Eq. 16.9. Numerical problems are solved in sec. 16.40.12 to illustrate the determination of ductility factor with respect to curvature. The value of xu/d can also be determined from Eq.16.16 if the value of μs, ductility in steel with respect to strain is given.


(B) Doubly-reinforced rectangular sections. Doubly-reinforced rectangular sections have the area of compression steel = Asc in addition to the area of tension steel = Ast. Accordingly, the value of k is obtained after determining the depth of the neutral axis by taking moment of the compression concrete, compression steel and tension steel about the neutral axis. The expression of xu/d is derived below. Equating the total compressive force C due to concrete (= 0.36 fck b xu) and compression steel (= fscAsc) with the tensile force T (= 0.87 fy Ast) for a doubly-reinforced rectangular section, we have: 0.36 fck b xu + fsc Asc = 0.87 fy Ast (16.17) which gives

bdf.Af

bdf.Af.

dx

ck

scsc

ck

styu360360

870−=

(16.18) Using Ast = p bd/100 and Asc = pc bd/100 in Eq. 16.18, we have: (xu/d) = (fy /36fck) {0.87p – fsc pc / fy} (16.19) When fsc = 0.87 fy (16.20) we have from Eq. 16.19: (xu/d) = (0.87 fy/36 fck) (p – pc) ≤ (xu, max/d) (16.21) Accordingly, the ductility factor with respect to curvature for the doubly-reinforced rectangular sections can be determined as explained through illustrative examples in sec. 16.40.12. 16.40.7 Factors Influencing Ductility The following factors influence the ductility of reinforced concrete sections.


(i) Figure 16.40.6 shows the plots of ductility μ with respect to

curvature versus p or (p – pc) for singly / doubly-reinforced rectangular sections for Fe 415 steel and three grades of concrete. It is evident from the plots that the ductility decreases with the increase of p or (p – pc). Accordingly, cl. 6.2.2 of IS 13920:1993 stipulates the maximum percentage of steel on any face at any section as 2.5. Providing high percentage of tension steel will cause brittle failure by crushing of concrete. Thus, the sections should be designed as under-reinforced.

(ii) Equation 16.15 reveals that the ductility decreases with increasing

grade of steel for a particular grade of concrete and specific percentage of steel. Hence, Fe 250 (mild steel) is the most preferred steel from the point of ductility of reinforced concrete sections. Clause 5.3 of IS 13920:1993 recommends steel reinforcement of grade Fe 415 or less. However, Fe 500 and Fe 550, produced by thermo-mechanical treatment process are also recommended by the code as they have elongation more than 14.5 per cent provided they conform to other requirement of IS 1786:1985.


(iii) Equation 16.15 and Fig. 16.40.6 show that the ductility increases with increasing grade of concrete for a particular grade of steel and specific percentage of steel. Accordingly, cl.5.2 of IS 13920:1993 prescribes the minimum grade of concrete as M 20 for all buildings which are more than three storeys in height.

(iv) Lower values of k and xu will have higher values of ductility as

evident from Eq. 16.14. Accordingly, T beams have more ductility than that of rectangular beams because of the reduced depth of neutral axis due to the presence of enlarged compression flange.

(v) The presence of lateral reinforcement prevents premature shear

failure to enable the sections undergoing sufficient deformation. Moreover, lateral reinforcement in the compression zone arrests the buckling of compression reinforcement. Thus, lateral reinforcement, though cannot increase the ductility directly, helps to reach the attainable ductility as per the design.

16.40.8 Design for Ductility

The objectives of the ductile design of reinforced concrete members are to ensure both strength and ductility for the designed structures or members. Strength of members can be assured by proper design of the sections following limit state method as explained earlier. However, for ensuring ductility, specific recommendations are to be followed as given in IS 13920:1993 regarding the materials, dimensions, minimum and maximum percentages of reinforcement. Further, detailing of reinforcement plays an important role. Accordingly, some of the major steps to be followed in the design are given below which will ensure sufficient ductility in the design.

(i) General specification of materials

(a) The minimum grade of concrete shall be M 20 for all buildings, which are more than three storeys in height (cl. 5.2 of IS 13920:1993).

(b) Steel reinforcing bars of grade Fe 415 or less shall be used.

However, steel bars of grades Fe 500 and Fe 550 may be used if they are produced by thermo-mechanical treatment process having elongation more than 14.5 per cent (cl. 5.3 of IS 13920:1993).


(ii) General guidelines in the design and detailing

(a) Simple and regular layout should be made avoiding any offsets of beams to columns or offsets of columns from floor to floor. Changes of stiffness of columns should be gradual from floor to floor.

(b) In a reinforced concrete frame, beams and columns should be designed such that the inelasticity is confined to beams only, while the columns should remain elastic. To satisfy this requirement, the sum of the moment capacities of the columns at a beam-column joint for the design axial loads should be greater than the sum of the moment capacities of the beams along each principal plane. Therefore,

∑ Mcolumn ⟩ 1.2 ∑ Mbeams (16.22)


The moment capacities of beams and columns are such that the column moments oppose the beam moments as shown in Fig. 16.40.7.

(c) The beams and columns should be designed taking into account

the reversal of stresses due to the nature of earthquake forces. (d) Beam-column connections should be made monolithic.

(e) The following are the requirements of flexural members:

• The factored axial stress on the member due to earthquake

loading shall not exceed 0.1 fck (cl. 6.1.1 of IS 13920:1993). • The width to depth ratio of the member shall preferably be

more than 0.3 (cl. 6.1.2 of IS 13920:1993).


• The width of the member should not be less than 200 mm (cl. 6.1.3 of IS 13920:1993).

• The total depth D of the member shall preferably be not

more than 1/4 of the clear span (cl. 6.1.4 of IS 13920:1993).

• The minimum percentage of tension steel on any face at any section is yck f/f24 , where fck and fy are in N/mm2 (cl. 6.2.1b of IS 13920:1993).

• The maximum percentage of steel on any face at any

section is 2.5 (cl. 6.2.2 of IS 13920:1993).

• The positive steel at a joint face must be at least equal to half the negative steel at that face (cl. 6.2.3 of IS 13920:1993).

• The redistribution of moments shall be used only for vertical

load moments and not for lateral load moments (cl. 6.2.4 of IS 13920:1993).


• The anchorage length of top and bottom bars of the beam in an external joint shall be measured beyond the inner face of the column and should be equal to development length in tension plus 10 times the bar diameter minus the allowance for 90 degree bends (Fig. 16.40.8). In an internal joint, both face bars of the beam shall be taken continuously through the column (cl. 6.2.5 of IS 13920:1993).

• Longitudinal bars shall be spliced if hoops are provided over

the entire splice length at a spacing not exceeding 150 mm, as shown in Fig. 16.40.9. The lap length shall be at least equal to the development length in tension. However, lap splices should not be provided (a) within a joint, (b) within a distance of 2d from the face of the joint, and (c) within a quarter length of the member where flexural yielding may generally occur under the effect of earthquake forces. Moreover, at one section not more than 50 per cent of the bars should be spliced (cl. 6.2.6 of IS 13920:1993).

16.40.9 Design for Shear in Flexural Members


Web reinforcement shall consist of either (a) vertical hoops or (b) U-stirrup with a 135 degree hook and a 10 diameter extension (but not ⟨ 75 mm) at each end and a crosstie, as shown in Figs. 16.40.10a and b, respectively. The crosstie is a bar having a 135 degree hook with a 10 diameter extension (but not ⟨ 75 mm) at each end (cl. 6.3.1 of IS 13920:1993). Normally, vertical hoops shall be used except in compelling circumstances when U-stirrups and crosstie shall be used. The minimum diameter of the bar forming a hoop shall be 6 mm, except for beams with clear span exceeding 5 m, where the minimum diameter of the bar forming a hoop shall be 8 mm (cl. 6.3.2 of IS 13920:1993).


As per cl. 6.3.3 of IS 13920:1993, the shear force to be resisted by the vertical hoops shall be the maximum of:

(i) calculated factored shear force as per analysis, and (ii) shear force due to formation of plastic hinges at both ends of the

beam plus the factored gravity load on the span. The expressions are given below (Fig. 16.40.11 a and b):

(a) for sway to right:

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ +−= +

AB

Bhlim,u

Aslim,uLD

aa,u L

MM.VV 41

(16.23) and


⎥⎦

⎤⎢⎣

⎡ ++= +

AB

Bhlim,u

Aslim,uLD

bb,u LMM

.VV 41 , and

(b) for sway to left

⎥⎦

⎤⎢⎣

⎡ ++= +

AB

Bslim,u

Ahlim,uLD

aa,u LMM

.VV 41

(16.24) and

⎥⎦

⎤⎢⎣

⎡ +−= +

AB

Bslim,u

Ahlim,uLD

bb,u LMM

.VV 41

where

, and , = sagging and hogging moments of

resistance of the beam section at ends A and B, respectively, L

Aslim,uM Ah

lim,uM Bslim,uM Bh

lim,uM

AB = clear span of the beam, and = shear forces at ends A and B, respectively, due to vertical loads with a partial safety factor of 1.2 on loads.

LDaV + LD

bV +

Out of the two values of Vu,a , the design shear at the end A shall be the larger. Similarly, out of the two values of Vu,b , the design shear at the end B shall be the larger. The expressions of Eqs. 16.23 and 16.24 are based on the assumption that the ratio of the actual ultimate tensile stress to the actual tensile yield strength of steel is not less than 1.25. Accordingly, 1.25 times the yield strength of steel divided by 0.87 is 1.43, which has been taken as 1.4 in Eqs. 16.23 and 16.24.


The contributions of bent up bars and inclined hoops, if any, shall not be considered to shear resistance of the section, as stipulated in cl. 6.3.4 of IS 13920:1993. Figure 16.40.12 shows the spacing of hoops and the following recommendations of cl. 6.3.5 of IS 13920:1993 are to be followed:

• Over a length of 2d at either end of the beam, the spacing of hoops shall not exceed (a) d/4, and (b) 8 times the diameter of the smallest longitudinal bar. However, the spacing shall not be less than 100 mm.

• The first hoop shall be at a distance not exceeding 50 mm from the joint

face.

• Vertical hoops at the same spacing as mentioned above shall also be provided over a length of 2d on either side of a section where flexural yielding may occur under the effect of earthquake forces.

• At other places, the beam shall have vertical hoops at a spacing not

exceeding d/2.


16.40.10 Column and Frame Members Subjected to Bending and Axial Load The following requirements are applicable to frame members having factored axial stress more than 0.1 fck under the effect of earthquake forces. The dimension of the member should be at least 200 mm. However, in frames having beams with centre-to-centre span exceeding 5 m or columns of unsupported length exceeding 4 m, the shortest dimension of the column shall be at least 300 mm (cl. 7.1.2 of IS 13920:1993). (A) Longitudinal reinforcement Longitudinal reinforcing bars shall be spliced only in the central half of the member length. Hoops shall be provided over the entire splice length at spacing not exceeding 150 mm centre to centre. A maximum of 50 per cent of the bars shall be spliced at one section (cl. 7.2.1 of IS 13920:1993).

The detailing of any area of column extending more than 100 mm beyond the confined core due to architectural requirement (cl. 7.2.2 of IS 13920:1993) shall be as follows:


(a) If the contribution of this area to strength is considered, then it should be the minimum longitudinal and transverse reinforcement as per IS 13920:1993.

(b) If this area is treated as non-structural, minimum longitudinal

and transverse reinforcement shall be provided as per IS 456:2000, as shown in Fig. 16.40.13.

(B) Transverse reinforcement

Transverse reinforcement shall consist of spiral or circular hoops for circular columns and rectangular hoops for rectangular columns. Rectangular hoops shall be a closed stirrup having a 135 degree hook with a 10 diameter extension (but not ⟨ 75 mm) at each end, as shown in Fig. 16.40.14 (cl. 7.3.1 of IS 13920:1993).


The parallel legs of rectangular hoop shall be spaced not exceeding 300 mm c/c. For spacing more than 300 mm, a crosstie shall be provided as shown in


Fig. 16.40.15. Alternatively, a pair of overlapping hoops may be provided within the column, as shown in Fig. 16.40.16. Hooks shall engage peripheral longitudinal bars (cl. 7.3.2 of IS 13920: 1993). The maximum spacing of hoops shall be half the least lateral dimension of the column, except where spiral confining reinforcement is provided as per cl. 7.4 of IS 13920:1993 (cl. 7.3.3 of IS 13920:1993).


The design shear force for columns, as recommended in cl. 7.3.4 of IS 13920:1993, shall be the maximum of:

(i) calculated factored shear force as per analysis, and (ii) a factored shear force given by

⎥⎦

⎤⎢⎣

⎡ +=

st

bRlim,u

bLlim,u

u hMM

.V 41

(16.25) where

and = moments of resistance, of opposite sign, of beams

framing into the column from opposite faces, as shown in Fig. 16.40.17, and

bLlim,uM bR

lim,uM

hst = storey height.


16.40.11 Special Confining Reinforcement


The following requirement shall be met with, unless a larger amount of transverse reinforcement is required from shear strength considerations. Special confining reinforcement shall be provided over a length lo from each joint face towards mid-span, and on either side of any section, where flexural yielding may occur under the effect of earthquake forces, as shown in Fig. 16.40.18. The length lo shall not be less than (a) larger lateral dimension of the member at the section where yielding occurs, (b) 1/6 of clear span of the member, and (c) 450 mm (cl. 7.4.1 of IS 13920:1993).

Special confining reinforcement shall extend at least 300 mm into the footing or mat when a column terminates into a footing or mat, as shown in Fig. 16.40.19 (cl. 7.4.2 of IS 13920:1993). Special confining reinforcement shall be provided over the full height of the column for the following situations (cls. 7.4.3 to 5 of IS 13920:1993):

(i) when the calculated point of contra-flexure, under the effect of gravity and earthquake loads, is not within the middle half of the member clear height,

(ii) columns supporting reactions from discontinued stiff members,

such as walls, and

(iii) columns which have significant variation in stiffness along their heights.


The maximum spacing of hoops used as special confining reinforcement shall be 1/4 of minimum member dimension. However, the spacing of hoops should not be less than 75 mm nor more than 100 mm (cl. 7.4.6 of IS 13920: 1993). The area of cross-section Ash of the bar forming circular hoops or spiral, to be used as special confining reinforcement, shall not be less than (cl.7.4.7 of IS 13920:1993).

⎥⎦

⎤⎢⎣

⎡−= 1090

k

g

y

ckksh A

A

ff

D S .A

(16.26) where Ash = area of the bar cross-section, S = pitch of spiral or spacing of hoops, Dk = diameter of core measured to the outside of the spiral or hoop, fck = characteristic compressive strength of concrete cube, fy = yield stress of steel (of circular hoop or spiral), Ag = gross area of column cross-section, and Ak = area of concrete core = (π/4) D2

k

The area of cross-section Ash of the bar forming rectangular hoop, to be used as special confining reinforcement, shall not be less than (cl.7.4.8 of IS 13920:1993):

⎥⎦

⎤⎢⎣

⎡−= 01180 .

A

A

ff

h S .Ak

g

y

cksh

(16.27) where

h = longer dimension of the rectangular confining hoop measured to its outer face. It shall not exceed 300 mm (see Figs. 16.40.14, 15 and 16), and

Ak = area of confined concrete core in the rectangular hoop

measured to its outside dimensions.



Problem 1: Determine the ductility with respect to curvature of the beam shown

in Fig. 16.40.20 using M 20 and Fe 250. Solution 1: This is a singly-reinforced rectangular beam of b = 300 mm, d =

540 mm, D = 600 mm, Ast = 942 mm2 (3-20T), fck = 20 N/mm2 and fy = 250 N/mm2.

Step 1: Checking for minimum and maximum percentages of

reinforcement (sec. 16.40.8 (ii) e) Minimum percentage of Ast = 100 (0.24) ckf / fy = 0.429

Maximum percentage of Ast = 2.5 For this problem p = 94200/(300) (540) = 0.581. Hence, Ast is within

minimum and maximum percentages i.e., 0.429 < 0.581 < 2.5). Step 2: Determination of ∈y, k and xu/d (Eqs. 16.8, 16.9 and 16.12) From Eq. 16.8, ∈y = fy / Es = 250/200000 (1) From Eq. 16.9, k = - (mp/100) + {(mp/100)2 + 2 (mp/100)}1/2 where

m = 280/3σcbc = 280/3(7) = 13.33 and p = 0.581. Therefore, k = 0.3236.

Taking moment of compression concrete and tension steel about

the neutral axis, we have: 300 (x2/2) = 942 (13.33) (d - x),


or x2 + 83.7124x – 45204.696 = 0, which gives x = 174.742 mm. Therefore, k = x/d = 0.3236.

From Eq. 16.12, xu/d = 0.87 fy p/(36 - fck) ≤ (xu, max/d) , which gives: xu/d = 0.87(250) (0.581)/(36) (20) = 0.1755 < 0.53, (as xu, max/d = 0.53). Step 3: Determination of ductility with respect to curvature (Eq. 16.14)

From Eq. 16.14, μ = (∈uc /∈y) {(1 – k) / (xu/d)}. Using ∈uc = 0.0035 and substituting the values of ∈y, k and (xu/d) from step 2,

μ = {0.0035 (200000)/250} {(1 –0.3236) / 0.1755} = 10.79

Hence, the ductility of this beam μ = 10.79.

Problem 2: Compare the ductility with respect to curvature of the cross-section

of the beam of Fig. 16.40.21 using (a) M 20 and Fe 250, and (b) M 20 and Fe 415.

Solution 2: This is a doubly-reinforced rectangular beam of b = 300 mm, d =

540 mm, D = 600 mm, d′ = 50 mm, Ast = 1884 mm2 (6-20T), Asc = 942 mm2 (3-20T), fck = 20 N/mm2 and fy = 250 N/mm2 for (a); and 415 N/mm2 for (b). The amounts of steel are within minimum and maximum percentages. Here, p = 1.162 and pc = 0.581. The modular ratio m for M 20 for both (a) and (b) parts of the problem = 13.33 as in Problem 1.


Step 1: Determination of k for both parts a and b Taking moment of compression concrete, compression steel and

tension steel about the neutral axis, we have: 300 (x2/2) + 942 (1.5 m – 1) (x – 50) = 1884 (13.33) (d –x)

or x2 + 286.7134x – 96373.822 = 0, which gives: x = 198.5862272 mm. Hence, k = x/d = 198.5862272/540 = 0.3677

Step 2: Determination of xu/d (Eq. 16.21) Equation 16.21 is: xu/d = (0.87 fy/36 fck) (p – pc) ≤ (xu, max/d) For (a), when fy = 250 N/mm2: xu/d = {0.87 (250)/36(20)} (0.581) = 0.1755 < 0.53, (as xu,max/d = 0.53) For (b), when fy = 415 N/mm2: xu/d = {0.87 (415)/36(20)} (0.581) = 0.2913 < 0.48, (as xu,max/d = 0.48) Step 3: Determination of ductility with respect to curvature (Eq. 16.14) Equation 16.14 is: μ = (∈uc / (∈y) {(1 – k) / (xu / d)}. Using ∈uc =

0.0035, ∈y = fy/Es, and substituting the values of k and xu/d from steps 1 and 2, we have,

For (a): μ = {(0.0035) (200000)/250} {(1 – 0.3677) / 0.1755} = 10.088 For (b): μ = {(0.0035) (200000)/415} {(1 – 0.3677) / 0.2913} = 3.66

Hence, the ductility of this doubly-reinforced beam is 10.088 when Fe 250 is used and is 3.66 when Fe 415 is used. The loss of ductility when opting for Fe 415 in place of Fe 250 is noted.

Problem 3: Design an inner beam of span 6 m of one reinforced concrete

frame for ductility using M 25 and Fe 415. The beam has negative bending moment of 300 kNm and shear force of 250 kN at the face of beam-column joint due to gravity and earthquake loads.

Solution 3: Step 1: Selection of dimensions and determining areas of steel


Let us assume the dimensions b = 300 mm and D = 600 mm. With

effective cover of 55 mm, d = 545 mm and d′ /D = 0.1 and using partial safety factor for load as 1.2, we have: Mu/bd2 = 1.2(300)106/300 (545) (545) = 4.04 N/mm2

The design for ductility has been given in sec. 16.40.8 (ii) following

the general guidelines in the design and detailing. Following the general guidelines, Design Tables 20.2 and 20.3 have been presented in the book, “Reinforced Concrete Limit State Design” (6th Edition) by A.K. Jain. These tables give the percentages of tension and compression steel for M 20 and M 25 grades up to p/pc

≤ 2.0 for four different values of d′/D as 0.05, 0.10, 0.15 and 0.20, and Mu/bd2. Accordingly, we refer to Table 20.3 for this problem, when Mu/bd2 = 4.04 N/mm2 and d′/D = 0.1 to get p = 1.2554 and pc = 0.6276.

The minimum percentage of steel = 100 (0.24) 25 / 415 = 0.289

and maximum percentage of steel = 2.5. Thus, the determined percentages are within the respective limits. So, Ast = 1.2554 (300) (545) / 100 = 2052.58 mm2. Provide 2-28T + 2-25 T (1231 + 981 = 2212 mm2) giving p provided = 1.353 per cent. The compression steel Asc = 0.6276 (300) (545)/100 = 1026.13 mm2. Provide 2-28T (1232 mm2), giving pc provided = 0.753 per cent.

Step 2: Design for shear With partial safety factor for loads as 1.2, the factored shear force

Vu = 1.2 (250) = 300 kN. τv = Vu/bd = 300/(0.3) (545) = 1.83 N/mm2, τc = (from Table 19 of IS 456 with p = 1.353%) = 0.716 N/mm2, and τmax (from Table 20 of IS 456) = 3.1 N/mm2

Since τc ⟨τv ⟨τmax, we have to provide vertical hoops (stirrups). Providing 10 mm–2 legged hoops (Asv = 157 mm2), we have the spacing of hoops as

Sv = 0.87 fy Asv d/(Vu - τc bd) = 0.87 fy Asv / (τv - τc)b = 0.87(415) (157) / (1.83 – 0.716) 300 = 169.61 mm.


The arrangement, minimum and maximum spacing of hoops are given in sec. 16.40.9. The maximum spacing is (a) d/4 = 136.25 mm, (b) 8 (25) = 200 mm and the spacing shall not be less than 100 mm, for a length of 2 d at either end of the beam. So, provide 10 mm – 2 legged hoops @130 mm c/c up to 1200 mm from the face of the column and then @ 260 mm c/c up to the centre-line, to be placed symmetrically. It may be noted that the maximum spacing at other places is d/2 = 272 mm. The first hoop shall be provided at a distance of 30 mm (⟨ 50 mm as per the code) from the face of the column. There will be ten hoops in 1200 mm. Figure 14.40.22 shows the cross-section of the beam.

Problem 4: Check (i) if the inner beam-column joint of a reinforced concrete

frame satisfies weak girder–strong column proportion and (ii) shears


in beam and column, using M 25, Fe 415 and other details as given below:

Clear span of the left beam = 5 m, clear span of the right beam = 4

m, slab thickness = 120 mm, finish on slab = 50 mm, live loads on floor = 2.0 kN/m2, axial load on column = 800 kN, beam dimensions = 300 mm × 600 mm, steel bars at top on either side = 4-25T + 2-20T (1.6 per cent), steel bars at bottom on either side = 2-25T + 1-20T (0.8 per cent), column dimensions = 300 mm × 600 mm, column reinforcement bars = 8-28T + 4 -25T (3.827 per cent), and height of storey = 3.6 m (Fig. 16.40.23).

Solution 4: (i) Beam-column Joint As given in sec. 16.40.8 (ii), the requirement of the design is to

satisfy Eq. 16.22, i.e., ∑ Mcolumn ⟩ 1.2 ∑ Mbeam

along each principal plane. We consider the joint in bending about the weaker axis and use tables of SP-16 for the computation. Assuming d′ / d = 0.1 and with p = 1.6%, Table 51 of SP-16 gives Mu/bd2 = 4.77 for doubly-reinforced section. This gives the hogging moment capacity as Mu = 4.77 (300) (545) (545) = 425.04 kNm. The sagging moment capacity of singly-reinforced section is obtained from Table 3 of SP-16, where Mu / bd2 = 2.503 for p = 0.8%. Thus, Mu = 2.503 (300) (545) (545) = 223.04 kNm. Therefore, ∑ Mbeam = 425.04 + 223.04 = 648.08 kNm

For columns Pu = 1.2 (800) = 960 kN gives Pu/fck bD = 1.2 (800) /

(25) (3) (60) = 0.213 and p / fck = 3.827/25 = 0.153. Chart 32 of SP-16 gives Mu/fck bD2 = 0.256. The column moment Mu = 0.256 (25) (300) (600) (600) = 691.2 kNm. Therefore,

∑ Mcolumn = 2 (691.2) = 1382.4 kNm. Here, ∑ Mcolumn ⟩ (1.2) ∑ Mbeam (= 1.2 (648.08) = 777.7 kNm) Hence, the requirement of Eq. 16.22 is satisfied. (ii) Shear in beam and column Step 1: Shear capacity of left beam having spacing of 4m c/c


Live loads = 4 m (2 kN/m2) = 8 kN/m Dead load of slab = 4 (0.12 + 0.05) (25) = 17.0 kN/m Dead load of web of beam = 0.3 (0.48) (25) = 3.6 kN/m Total dead load = 20.6 kN/m. Factored shear force due to gravity loads = 1.2 (20.6 + 8) (5)/ 2 = 85.8 kN. Factored shear force due to plastic hinge = 1.4 (648.08) / 5 = 181.46 kN Thus, Vu = 85.8 + 181.46 = 267.26 kN τv = 267.26 / 300 (0.545) = 1.63 N/mm2

τc (from Table 19 of IS 456 for p = 1.6 %) = 0.756 N/mm2. τcmax (from Table 20 of IS 456) = 3.1 N/mm2.

Since τc ⟨τv ⟨τcmax, we provide 8 mm-2 legged hoops (Asv = 100 mm2) of spacing sv where sv is as follows:

sv = 0.87 fy Asv / (τv - τc)b = 0.87 (415) (100) / (1.63 – 0.756) (300) = 137.7 mm c/c.

For a distance of 2 d (= 1090 mm) from the face of the column, the maximum spacing is: (a) 0.25 (545) = 136.25 mm and (b) 8 (20) = 160 mm. The spacing should not be less than 100 mm. Beyond 1090 mm, the spacing = d/2 = 272.5 mm.

Let us change the diameter of hoop as 10 mm beyond 1090 mm, for which the spacing is 0.87 (415) (157)/(1.63 – 0.756) (300) = 216.36 mm c/c.

So, let us provide 8 mm-2 legged hoops @ 130 mm c/c up to 1.2 m

from the face of the column at either end. The first hoop shall be at a distance of 30 mm from the face of the joint. We need ten hoops. Beyond 1.2 m, provide 10 mm-2 legged hoops @ 200 mm c/c, symmetrically.

Step 2: Checking of column for storey height = 3600 mm Factored shear in column Vu = 1.4 (648.08)/3.6 = 252.03 kN. This

gives τv = 252.03/3 (60) = 1.4 N/mm2. τc (from Table 19 of IS 456: 2000 with 1.913% Ast at each face) =

0.806 N/mm2. This shall be multiplied by a factor = 1+3 (1.2) (800)/(30) (6) (25) = 1.64. However, the multiplying factor is limited


to 1.5 (see cl. 40.2.2 of IS 456: 2000). So, τc = 1.5 (0.806) = 1.209 N/mm2.

τcmax (from Table 20 of IS 456: 2000) = 3.1 N/mm2. Since τc ⟨τv ⟨τcmax, we provide 8 mm – 2 legged hoop of mild steel (Fe 250), for which the spacing is

sv = 0.87fyAsv/(τv - τc)b = 0.87(250)(100)/(1.4 – 1.209) (300) = 379.58 mm.

The maximum spacing is 0.5 (300) = 150 mm (cl. 7.3.3 of IS 13920:1993). Hence, provide 8 mm-2 legged hoops of Fe 250 @150 mm c/c except in the confining steel zone.

Step 3: Confining steel Here, axial stress = 1.2 (800) / 30(6) = 5.33 N/mm2 ⟩ 0.1 fck (=2.5

N/mm2). So, we provide confining steel of rectangular closed loops of spacing lesser of (a) 0.25 (300) and (b) 100 mm (cl.7.4.6 of IS 13920: 1993). So, the spacing is 75 mm c/c. From Fig. 16.40.23, we have h = 200 mm.

From Eq. 16.27, we have: Ash = 0.18 S h {(Ag / Ak) –1} (fck / fy) = 0.18 (75) (200) [(300) (600) / (184) (504)}-1](25/415) = 153.05 mm2

Provide 2-10 mm rectangular hoops (157 mm2). These confining hoops shall be provided for a distance greater of: (i) 600 mm, (ii) 3600/6 = 600 mm and (c) 450 mm. So, we provide 8 numbers of 2 legged 10 mm diameter hoops @ 75 mm c/c up to 630 mm from the joint. The first one is at a distance of 30 mm from the joint (see sec.16.40.11). 16.40.13 Practice Questions and Problems with Answers Q.1: Define and explain ductility with respect to (a) displacement, (b) curvature

and (c) rotation of a reinforced concrete structure. A.1: Secs. 16.40.2, 3 and 4 are the respective answers of part (a), (b) and (c). Q.2: State the advantages of ductility in reinforced concrete structures. A.2: Sec. 16.40.5 is the complete answer. Q.3: Derive the expressions of ductility of (a) singly-reinforced and (b) doubly-

reinforced concrete beams.


A.3: Part (A) and Part (B) of sec. 16.40.6 are the answers of (a) and (b), respectively. Q.4: State the factors influencing ductility of reinforced concrete sections. A.4: Sec. 16.40.7 is the complete answer.

Q.5: Compare the ductility with respect to curvature of the cross-section of the

beam of Fig. 16.40.23 using (a) M 25 and Fe 250, and (b) M 25 and Fe 415.

A.5: This is a doubly-reinforced rectangular beam of b = 300 mm, d = 540 mm,

D = 600 mm, d′ = 50 mm, Ast = 3694 mm2 (6-28T), Asc = 1847 mm2 (3-28T), fck = 25 N/mm2 and fy = 250 N/mm2 for (a) and 415 N/mm2 for (b). The percentages of tension steel p = 2.28% and pc = 1.14%. The reinforcing steel is within the range of minimum and maximum percentages. The modular ratio m = 280/3σcbc = 280/3(8.5) = 10.98.

Step 1: Determination of k for both parts a and b

Taking moment of compression concrete, compression steel and tension steel about the neutral axis, we have:

300 (x2/2) + 1847(1.5 m – 1) (x-50) = 3694 (m) (d-x) or x2 + 460.8880667x – 155540.7953 = 0 or x = 226.333 mm giving k = x/d = 0.419

Step 2: Determination of xu/d (Eq. 16.21) Equation 16.21 is: xu/d = (0.87 fy/36 fck) (p – pc) ≤ (xu, max /d)


For (a), when fy = 250 N/mm2: xu/d = {0.87 (250) / 36(25)} {1.14} = 0.2755 ⟨0.53 (as xu,max / d = 0.53)

For (b), when fy = 415 N/mm2: xu/d = {0.87 (415) / 36(25)} {1.14} = 0.457 ⟨0.48

(as xu,max / d = 0.48)

Step 3: Determination of ductility with respect to curvature (Eq. 16.14) Equation 16.14 is: μ = (∈uc / ∈y) {(1- k)/(xu / d)} Using ∈uc = 0.0035, ∈y = fy / Es and substituting the values of k and xu /d

from steps 1 and 2, we have For (a): μ = {(0.0035) (200000)/250} {(1 – 0.419) / (0.2755)} = 5.905 For (b): μ = {(0.0035) (200000)/415} {(1 – 0.419) / (0.457)} = 2.144

Hence, the ductility of this doubly-reinforced beam is 5.905 when Fe 250 is used and is 2.144 when Fe 415 is used.

Q.6: Design the column of a mulitstoreyed building for ductility with M 25 and

Fe 415 subjected to an axial force of 2000 kN and bending moment of 416.67 kNm.

A.6:


Step 1: Dimensions and area of steel of column Let us take the column dimensions as 500 mm × 500 mm. Considering

partial safety factor for loads = 1.2, we have: Mu = (1.2) (416.67) = 500 kNm Pu = (1.2) (2000) = 2400 kN Pu/fck bD = 2400 / (25) (5) (50) = 0.384 Mu/fckbD2 = 500 / (25) (125) = 0.16 d′/D = 75/500 = 0.15

Chart 45 of SP-16 gives p/fck = 0.16. With p = 0.16 (25)= 4%, we have Ast = 4(500)(500)/100 = 10000 mm2. Provide 4 - 32T + 8 – 28T + 4 – 25T (= 3217 + 4926 + 1963 = 10106 mm2) as shown in Fig. 16.40.25.

Step 2: Lateral ties

Diameter of lateral tie is the greater of 32/4 = 8 mm or 6 mm. Let us take 8 mm diameter bars.


Maximum pitch is 0.5 (500) = 250 mm (cl. 7.3.3 of IS 13920:1993). So, provide 8 mm ties @ 250 mm c/c.

Step 3: Confining reinforcement

Here, Pu/A = 2400 (103)/(500) (500) = 9.6 N/mm2 ⟩ 0.1 fck (=2.5 N/mm2). Hence, confining reinforcement shall be provided. From Eq. 16.27, we have

0 18 1gcksh

y k

AfA . S h

f A⎡ ⎤

= −⎢ ⎥⎣ ⎦

Ash = 113 mm2 for 12 mm diameter hoop. h = 500 – 2 (40 + 12) = 396 mm ⟩300 mm. So, h is revised as 396/2 = 198 mm ⟨ 300 mm. Ag = 500 (500) = 250000 mm2

Ak = {500 – 2 (40 + 12 + 8)} (380) = 144400 mm2 and {(Ag/Ak) – 1} = 0.7313. Hence, S = 113 (415)/{0.18(198) (0.7313) (25)} = 71.9 mm Provide confining hoops of 12 mm diameter bar @ 70 mm c/c, as shown

in Fig. 16.40.25. The distance of the confining reinforcement is the largest of: (i) clear span

of column / 6 = 3600/6 = 600 mm, (ii) larger lateral dimension = 500 mm and (iii) 450 mm. So, provide confining hoops for a distance of 700 mm from the face of the joint (eleven numbers of hoop).

16.40.14 References


















minutes Answer all questions. TQ.1: State the advantages of ductility in reinforced concrete structures.

(10 Marks) A.TQ.1: Sec. 16.40.5 is the complete answer. TQ.2: Derive the expressions of ductility of (a) singly-reinforced and (b) doubly-

reinforced concrete beams. (15 Marks)

A.TQ.2: Part (A) and Part (B) of sec. 16.40.6 are the answers of (a) and (b), respectively.


TQ.3: Compare the ductility with respect to curvature of the cross section of the

beam of Fig. 16.40.25 using (a) M 30 and Fe 250 and (b) M 30 and Fe 415.

A.TQ.3: This is a doubly-reinforced rectangular beam of b = 300 mm, d = 540

mm, D = 600 mm, d′ = 50 mm, Ast = 1206 mm2 (6 – 16T), Asc = 603 mm2 ( 3 – 16T ), fck = 30 N/mm2 and fy = 250 N/mm2 for (a) and 415 N/mm2 for (b). The modular ratio m = 280/3(10) = 9.33, minimum percentage of steel = 250/3024 = 0.526 for (a) and = 415/3024 = 0.317 for (b). Here, p = 120600/300(540) = 0.7444 per cent. Thus, percentage of Ast is within the range of minimum and maximum limits (2.5%).

Step 1: Determination of k for both parts (a) and (b) Taking moment of compression steel and tension steel about the neutral

axis, we have: 300 (x2/2) + 603(1.5 m-1) (x - 50) = 1206 (m) (d - x) or x2 + 127.2531x – 43119.123 = 0 which gives x = 153.554 mm, which gives: k = x/d = 0.284.

Step 2: Determination of xu/d (Eq. 16.21) Equation 16.21 is: xu/d = (0.87 fy /36 fck) (p – pc) ≤ (xu, max/d) For (a), when fy = 250 N/mm2: xu/d = {0.87 (250) / 36(30)} (0.3722) = 0.0749 ⟨ 0.53 (as xu,max / d = 0.53)


For (b), when fy = 415 N/mm2: xu/d = {0.87 (415) / 36(30)} (0.3722) = 0.124 ⟨

0.48 (as xu,max / d = 0.48) Step 3: Determination of ductility with respect to curvature (Eq. 16.14) Equation 16.14 is: μ = (∈uc / ∈y) {(1 – k) / (xu/d)}

Using ∈uc = 0.0035, ∈y = fy /Es and substituting the values of k and xu / d from Steps 1 and 2, we have:

For (a): μ = {(0.0035) (200000)/250} {(1 – 0.284) / (0.0749)} = 26.766 For (b): μ = {(0.0035) (200000)/415} {(1 – 0.284) / (0.124)} = 9.74

Hence, the ductility of this doubly-reinforced beam (Fig. 16.40.26) is 26.766 when Fe 250 is used and is 9.74 when Fe 415 is used. 16.40.16 Summary of this Lesson

This lesson defines and explains the ductility factor with respect to displacement, curvature and rotation. Stating and explaining the advantages of ductility in the design of reinforced concrete members, expressions of ductility factor are derived for singly and doubly-reinforced concrete rectangular beams. Factors influencing the ductility are explained. General specifications of materials are given and the general guidelines in the design and detailing of structures are stated as stipulated in IS 13920:1993. The situations requiring special confining reinforcement are explained. Several illustrative examples are solved determining the ductility of singly and doubly-reinforced rectangular beams. Furthermore, numerical problems are also solved to explain the design of beams, columns and beam-column joints as per IS 13920:1993.


objectives and methods of analysis and design

Documents