DESIGN OF REINFORCEDCONCRETE STRUCTURES

Volume 1

Mashhour Ahmed GhoneimProfessor of Concrete Structures

Cairo University

Mahmoud Tharwat EI-MihilmyAssociate Professor of Concrete Structures

Cairo University

Second Edition2008

PREFACETeaching reinforced concrete design, carrying out research relevant to the behavior ofreinforced concrete members, as well as designing concrete structures motivated thepreparation of this book. The basic objective of this book is to furnish the reader withthe basic upderstanding of the mechanics and design of reinforced concrete. Thecontents of the book conform to the latest edition of the Egyptian Code for the Designand Construction of Concrete Structures ECP-203. The authors strongly recommendthat the Code be utilized as a companion publication to this book.

The book is aimed at two different groups. First, by treating the material in a logicaland unified form, it is hoped that it can serve as a useful text for undergraduate andgraduate student courses on reinforced concrete. Secondly, as a result of the continuingactivity in the design and construction of reinforced concrete structures, it will be ofvalue to practicing structural engineers.

Numerous illustrative examples are given, the solution of which has been supplied soas to supplement the theoretical background and to familiarize the reader with thesteps involved in actual design problem solving.

In writing the book, the authors are conscious of a debt to many sources, to friends,colleagues, and co-workers in the field. Finally, this is as good a place as any for theauthors to express their indebtedness to their honorable professors of Egypt, Canadaand the U.S.A. Their contributions in introducing the authors to the field will always.be remembered with the deepest gratitude.

This volume covers the following topics

Reinforced Concrete Fundamentals. Design of Singly Reinforced Sections

Design of Doutily Reinforced Sections Design of T-Beams Design for Shear Bond and Development length .Design of Simple and Continuous Beams Truss Models for the Behavior of RIC Beams Design for Torsion

It also includes appendices containing design aids.

TABLE OF CONTENTS

1. REINFORCED CONCRETE FUNDAMENTALS

1.1 Introduction 11.2 Reinforced Concrete Members , 31.3 Reinforced Concrete 5104 Reinforced Concrete Behavior 51.5. Mechanical Properties ofConcrete 71.5.1 Compressive Strength 71.5.2 Tensile strength 111.5.3 Modulus of Elasticity ; 131.504 Strength ofConcrete Under Biaxial Loading 141.5.5 Shrinkage 161.5.6 Creep 171.6 Reinforcing Steel 181.7 Limit States Design Method 201.8 Strength Reduction Factors 201.9 Classification of Loads 231.10 Load Combinations ..: 25

Example 1.1 28

2. DESIGN OF SINGLY REINFORCED SECTIONS

2.12.22.32.3.12.3.22.3.2.12.3.2.22.3.32042.52.62.72.8

Introduction 31Reinforced Concrete Beam Behavior : 32Flexure Theory of Reinforced Concrete ~4

Basic Assumptions of the Flexure Theory 34Stress-StrainRelationships ....................................................................... 35Concrete in Compression ; 35Reinforcing Steel 36The Equivalent Rectangular Stress Block 37

Analysis ofSingly Reinforced Sections 39Maximum Area of Steel of a Singly Reinforced Section .42Balanced, Under, and Over Reinforced Sections .47Minimum Area ofSteel. 48Factors Affecting Ultimate Strength ;..49.

Example 2.1 52Example 2.2 54Example 2.3 56Example 204 ~ 58

ii

Example 2.5 60Example 2.6 62Example 2.7 65Example 2.8 67

2.9 Design of Singly Reinforced Sections by First Principles 69Example 2.9 70Example 2.10 73

2.10 Design of Singly Reinforced Sections Using Curves ..; 752.10.1 Design Charts (R-J..l) , 752.10.2 Design Chart (R- oi) 77

Example 2.11 80Example 2.12 , 81Example 2.13 83Example 2.14 84Example 2.15 : 85

5 BOND, DEVELOPMENT LENGTH AND SPLICING OFREINFORCEMENT

5.1 Introduction 1865.2 Average Bond Stresses in a Beam 1875.3 True Bond Stresses in a Beam : 1895.4 Development Length ; 1905.4.1 Theoretical Considerations 1905.4.2 Development.Length According to ECP 203 1925.5 Bar Cutoffs in Flexural Members 1965.5.1 The Moment of Resistance ofa RIC Beam 1965.5.2 Curtailment of Bars in Beams 1985.5.3 Egyptian Code's Requirements for Curtailment ..; 2005.6 Beams with Bent-up Bars : 2035.7 Anchorage of Web Reinforcement 2035:8 Splicing of Reinforcement 2045.8.1 Lap splices 2045.8.2 Welded and Mechanical Connections 206

3.3.13.1.13.1.23.1.3

3.1.4

3.1.5

3.23.2.13.2.23.2.33.2.43.2.53.2.6

DOUBLY REINFORCED BEAMS AND T-BEAMSDoubly Reinforced Sections 86Introduction 86

Analysis of Doubly Reinforced Sections 88Maximum Area of Steel for Doubly Reinforced Sections 92Example 3.1(compression steel yields) 95Example 3.2 97Example 3.3 (compression steel does not yield) 98

Design of Doubly Reinforced Sections Using First Principles 100Example 3.4 101Example 3.5 103Example 3.6 105Example 3.7 107

Design ofDoubly Reinforced Sections Using Curves 108Example 3.8 111Example 3.9 ; 112Example 3.10 113

T-Bearns 115Application ofT-Beams 115Effective Flange Width , 115Analysis ofT-Beams : 119Minimum Area of Steel for T-sections 122Maximum Area of Steel for T-sections 123Design ofT-sections Using First Principles 126Example 3.11 127Example 3.12 129Example 3.13 T Sections (ats) 133Example 3.15 135Example 3.16 : 137

3.2.73.2.7.13.2.7.2

3.3

4.

4.14.24.34.44.4.14.4.24.4.34.54.5.14.5.24.5.34.5.44.5.5

Design of T-sections Using Curves 139Development of the Curves ; 139Using the Design Aids (charts CI-J and RT-J) 142Example 3.17 (ats) 145

Design ofL-Sections : 147Example 3.19 149Example 3.20 151

SHEAR IN RIC BEAMS

Introduction 153Shear stresses in Elastic Beams ; 154Shear Stresses in Cracked RIC Beams 158Behavior of Slender Beams Failing in Shear 159

Inclined Cracking 159Internal Forces in Beams without stirrups 160Behavior of Slender Beams with Stirrups : 162

Egyptian Code's Procedure for Shear Design 164Critical Sections for Shear 164Upper limit of Design Shear Stress 166Shear Strength Provided by Concrete 166Shear Strength Provided by Shear Reinforcement.. 167Code Requirements for Shear Reinforcement.. 170Example 4.1 172Example 4.2 176Example 4.3 180Example 4.4 183

iii iv

REINFORCED CONCRETE BEAMSIntroduction 207Statical Systems of RiC Beams 208The effective span 209Loads Acting on Beams 210

Own weight of beams 210Slab loads 211Wall loads 216

Slenderness limits for beams 219Linear Elastic Analysis of Continuous Beams 220 .Reinforcement Detailing in'RiC Beams 221

Bar Spacing 222Egyptian Code Recommendations 223Example 6.1 224Example 6.2 231Example 6.3 235Example 6.4 241Example 6.5 245Example 6.6 254Example 6.7 266Example 6.8 273Example 6.9 283

DESIGN FOR TORSION

APPENDIX A: DESIGN AIDS 391REFERENCES : 409

66.16.16.26.36.3.16.3.26.3.36.46.56.66.6.26.6.3

7

7.17.27.2.17.2.27.2.2.17.2.2.27.37.47.4.17.4.27.4.37.4.47.4.57.4.6

7.67.6.17.6.2

TRUSS MODEL FOR BEAMS FAILING IN SHEAR

Introduction 290Background 291

Slender Beams Versus Deep Beams 291Analysis of Forces in RiC Slender Beams 293

Sectional Analysis 294Mechanical- Mathematical Models 295

Truss Model for Slender Beams.. 296Traditional 45-Degree Truss Model 297

Formation ofthe45-Degree Truss 297Evaluation of the Forces in the Stirrups 298The Compression Force in the Diagonals 299The Axial (Longitudinal) Force Due to Shear 302Comments on the 45-Degree Truss-Model ; 303Comparison of the Truss Model and ECP 203 303Example 7.1 304Example 7.2 ; ~ 310

The Variable-Angle Truss ModeL. ................................................................... 316General 316Analysis of the Variable Angle Truss ModeL 3l7Example 7.3 ~ 320

8

8.18.28.2.18.2.28.2.38.38.48.58.5.18.5.28.5.38.5.48.68.6.18.6.28.6.38.6.48.6.58.6.5.18.6.5.28.6.68.6.7

8.78.88.8.18.8.28.8.38.8.48.8.58.98.10

Introduction 326Equilibrium Torsion and Compatibility Torsion 327

General 327Equilibrium Torsion 327Compatibility Torsion 329

Principal Stresses due to Torsion 330Thin-Walled Tube in Torsion 331Space-Truss Model for Torsion 333

Components of the Space Truss 333Diagonal Compressive Stresses ................................................................ 335Forces in Stirrups 337Longitudinal Force 337

The Design for Torsion in the Egyptian Code : 339General 339Calculation of the Shear Stress due to Torsion 339Consideration of Torsion 341Adequacy ofthe Concrete Cross-Section " 341Design of Torsional Reinforcement.. 341Closed Stirrups 341

Longitudinal Reinforcement. 342Code Requirements for Reinforcement Arrangement.. 342Summary of Torsion Design According to ECP 203 .. 345

. Example 8.1 347Example 8.2 351Example 8.3 ; 355

Combined Shear and Torsion 359The Design for Shear and Torsion inECP 203 359

Consideration of Torsion : , 359Adequacy of the Concrete Cross-Section 360Design of Transverse Reinforcement : 361Design ofLongitudinal Reinforcement. 361Summary ofthe Design for Shear and Torsion 362

Compatibility Torsion .: 365Torsional Rigidity 365

Example 8.5 372Example 8.6 376Example 8.7.............................................................................................. 382

vvi

1REINFORCED CONCRETE FUNDAMENTALS

Photo 1.1 Nile City Towers, Cairo-Egypt.

1.1 IntroductionReinforced concrete is one of the most important available materials forconstruction in Egypt and all over the world. It is used in almost all structuresincluding; buildings, bridges, retaining walls, tunnels, tanks, shells. and evenships.

Concrete is a mixture of sand and gravel held together with a paste of cementand water. Sometimes one or more admixture is added to change certaincharacteristic of the concrete such as its workability, durability, and time ofhardening. Concrete has a high compressive strength and a very low tensilestrength.

Reinforced concrete is a combination of concrete and steel wherein the steelreinforcement provides the tensile strength lacking in the concrete. Steelreinforcement is also capable of resisting. compression forces and is used incolumns as well as in other situations to be described later.

The tremendous success of reinforced concrete can be understood if itsnumerous advantages are considered. These include the following:

It is a low maintenance material. It has great resistance to the action of fire provided that there is adequate

cover over the reinforcing steel.

A special nature of concrete is its ability to be cast in to a variety ofshapes from simple slabs, beams, and columns to great arches and shells.

A lower grade of skilled labor is required for erection as compared toother materials such as structural steel.

In. most areas, concrete takes advantage of inexpensive local materials(sand, gravel, and water) and requires a relatively small amount of cementand reinforcing steel.

To use concrete successfully, the designer must be completely familiar with itsweak points and its strong ones. Among its disadvantages are the following:

Concrete has a very low tensile strength, requiring the use of tensilereinforcing.

Forms are required to hold the concrete In place until it hardenssufficiently. Formwork could be expensive.

The properties of concrete could vary widely due to variations in itsproportioning and mixing. Furthermore, the placing and curing of concreteis not as carefully controlled, as is the production of other materials suchas structural steel.

In general, reinforced concrete members are relatively large, as comparedto structural members, an important consideration for tall buildings andlong span bridges.

2

1.2 Reinforced Concrete Members

Reinforced concrete structures consist of a series of members. The first and thesecond floors of the building shown in Fig. 1.1 have a slab-and-beam system, inwhich the slab spans between beams, which in tum apply loads to the columns.Again, the columns' loads are applied to footings, which distribute the load overa sufficient area of soil.

The structure shown in Fig 1.2 is a typical framed structure. The slab carries itsown weight, flooring and live loads. The load is then transferred to secondarybeams. The reactions of the secondary beams are transferred to the girders,which in tum are supported by the columns. Finally, the columns' loads areapplied to the footings, which distribute the load to the soil.

Photo 1.2 Reinforcement placement during construction

3

The disadvantage of steel is corrosion, but the concrete surrounding thereinforcement provides an excellent protection. Moreover, the strength of theexposed steel subjected to fire is close to .zero, but again the enclosure of thereinforcement in the concrete produces very satisfactory fire protection. Finally,concrete and steel work very well together in temperature changes because theircoefficients of thermal expansion are almost the same. The coefficient ofthermal expansion for steel is 6.5xI0-6, while that for the concrete is about5.5xlO-6.

The addition of steel reinforcement that bonds strongly to concrete produces' arelatively dnctile material capable of transmitting tension and suitable for anystructural elements, e.g., slabs, beam, columns. Reinforcement should be placedin the locations of anticipated tensile stresses and cracking areas as shown in Fig1.3. For example, the main reinforcement in a simple beam is placed at thebottom fibers where the tensile stresses develop (Fig. 1.3A). However, for acantilever, the main reinforcement is atthe top of the beam at the location of themaximum negative moment (Fig.l.3B). Finally for a continuous beam; a part ofthe main reinforcement should be placed near the bottom fibers where thepositive moments exist and the other part is placed at the top fibers where thenegative moments exist (Fig. 1.3C).

1.4 Reinforced Concrete Behavior

1.3 Reinforced Concrete

Concrete and steel work together beautifully 'in reinforced concrete structures.The advantages of each material seem to compensate for the disadvantages ofthe other. The great shortcoming oflow concrete tensile strength is compensatedfor by the high tensile strength of the steel. The tensile strength of the steel isapproximately equal to 100-140 times the tensile strength of the usual concretemix. Also, the two materials bond together very well with no slippage, and thusact together as one unit in resisting the applied loads.

It is a well-known fact that plain concrete is strong in compression and veryweak in tension. The tensile strength of concrete is about one-tenth itscompressive strength. As a result, a plain concrete beam fails suddenly as soonas the tension cracks start to develop. Therefore, reinforcing steel is added in thetension zone to carry all the developed tensile stresses; this is called a reinforcedconcrete beam. .

R.C.Beam

Column

Footing

Secondary beam

Loads

Fig. 1.1 Slab and beam system in a building

Fig. 1.2 Typical reinforced concrete structural framing system

4 5

Cracks

Cracks

A- Simple beam

B-Cantilever beam

Reinforcement

Reinforcement

Cracks

1.5. Mechanical Properties of Concrete1.5.1 Compressive StrengthMany factors affect the concrete compressive strength such as the water cementratio, the type of cement, aggregate properties, age of concrete, and time ofcuring. The most important factor of all is the water cement ratio. The lowerwater content with good workability leads to higher concrete compressivestrength. Increasing the water cement ratio from 0.45 to 0.65 can decrease thecompressive strength by 30-40 percent. Currently, high-range water-reducingadmixtures (super plasticizers) are available and they allow engineers toproduce fluid concrete mixes with a sharply reduced amount of water.In Egypt, the compressive strength of concrete is usually determined by loadinga158 mm cube up to failure in uniaxial compression after 28 days of casting andis referred to as !cu. Additional details covering the preparation and testing ofcubes are covered by the Egyptian Code for Design and Construction ofConcrete Structures (ECP-203) including correction factors that can be used ifthe tested specimen is not the same dimension or shape as the standard cube.This is the strength specified on the construction drawings and used in thedesign calculations.It should be mentioned that in other countries such as the United States andCanada, the compressive strength is measured by compression tests on 150 mmx 300 mm cylinders tested after 28 days of moist curing. In the case of usingspecimens other than the standard cube, the ECP 203 gives the correction factorsshown in Table 1.1 to obtain the equivalent compressive strength of the standardcube.Table 1.1 Correction factors to obtain the equivalent.fcu=.fc x factor

C-Continuous beam

Fig. 1.3 Reinforcement placement for different types of beams

6

I.

Shape Size (rrun) Correction factorCube 100 xlOO x 100 0.97Cube (158 x 158 x 158) or (150 x 150 x 150) 1.00Cube 200 x 200 x 200 1.05Cube 300 x 300 x 300 1.12Cylinder 100 x200 1.20Cylinder 150 x 300 1.25Cylinder 250 x 500 1.30Prism (150 x 150 x 300) or (158 x 158x 316) 1.25Prism (150 x 150 x 450) or (158 x 158 x 474) 1.3Prism 150 x 150 x 600 1.32

7

The ECP 203 states in clause (2.5.2) that a concrete strength of 18 N/mm 2should be used to qualify for reinforced concrete category,15 N/mm2 for plainconcrete, and 30 Nzmrrr' for prestressed concrete. Table 1.2 illustrates the gradesof reinforced concrete RlCand prestressed concrete PIS as permitted by thecode.Table 1.2 Grades of reinforced and prestressed concrete (Nzmm! )RiC 18 120 125 30 35 40 45

PIS 30 35 40 45 501

551

60

Field conditions are not the same as those in the laboratory, and the specified28-days strength might not practically be achieved in the field unless almostperfect mixture, vibration, and perfect curing conditions are present. As a result,section 2-5-3 of the ECP 203 requires that the target concrete compressivestrength, h, must exceed the characteristic strength feu by a safety margin (M).The safety margin for a concrete mix design depends on the quality control ofthe concrete plant and can range from 4 N/mm2 to 15 Nzmm", Table 1.3 (2-15 ofthe Code) lists the values of the safetymargin M according to the number of theperformed tests and the characteristic strength feu. Therefore the targetedconcrete compressive strengthf", is given by

f,,,=fcu+M (1.1)

Table 1.3 Value ofthe safety margin M (Nzmm")Statisticaldata Safety margin M

feu < 20 N/mm2 20-40N/mm2 40-60N/mm2

40 test data1.64 SD ~ 4 N/mm2 1.64 SD ~ 6 N/mm2 1.64 SD > 7.5or more -

N/mm2

less than 40 ~ 12N/mm2 ~ 15 N/mm2test data Not less than 0.6 feu

One test data is an average of 3 cube testsSD: Standard deviation

8

Since concrete is used mostly in compression, its compressive stress-strain curveis of a prime interest. Figure1.4 shows a typical set of such curves obtained fromuniaxial compression test of cylinders. All curves have somewhat similarcharacteristics. They consist of an initial relatively straight elastic portion inwhich stresses and strains are closely proportional, then begin to curve to reach amaximum value at a strain of 0.002 to 0.003. There is a descending branch afterthe peak stress is reached. It can be noticed that the weaker grades of concreteare less brittle than the stronger ones. Thus, they will take larger stains anddeformations before breaking.

70

60

.--.. 50N EE--. 406..,..,

0.72j,,"

0.10j""

__________________;.;-_...,. --:;,,;;,-!.:-l:=~~~-- ~T

0.0038

1.5.2 Tensile strengthExperimental tests indicate that the tensile strength of concrete is highly variableand ranges from about 8-12% of its compressive strength. The actual valuedepends on the type of test and crack propagation pattern at failure.

Tensile strength is usually determined by the bending test (Fig. 1.6) or by thesplit cylinder test (Fig 1.7). The ECP 203 states that the value of concretetensile strength can be taken from experimental tests as follows:

60% from the concrete tensile strength determined from bending test.85% from the concrete tensile strength determined from split cylinder test.

In the bending test (modulus ofrapture test), a plain concrete beam is loaded inflexure up to failure as shown in Fig. 1.6. The flexure tensile strength or themodulus ofruptureJ,. is computed from the following equation

Fig. 1.5 Modified Hognestad curve for concrete stress-strain relation6Mt, =-2 (1.2)bxt

Photo 1.3 Milwaukee Art Museum, USA.

10

I~

P/2 P/2

~I

Fig. 1.6 Bending tensile test

11

iDI~.~

The split cylinder test is performed on a 150x300 mm cylinder placed on its sideand loaded in compression along its length as shown in Fig. 1.7.A The stressesalong the diameter are nearly uniform tension perpendicular to the plan ofloading as shown in Fig. 1.7.b The splitting tensile strength let is calculated fromthe following expression

2Pfet =-- (1.3)Jr d L .

The parameters in Eq. 1.3 are defined in Fig. 1.7.

1.5.3 Modulus of ElasticityIt is clear from the stress-strain curve of the concrete shown in Fig.1.3 that therelation between the stress and the strain is not linear. Thus, the modulus ofelasticity changes from point to point. Furthermore, its value varies withdifferent concrete strengths, concrete age, type of loading, and thecharacteristics of cement and aggregate. The initial tangent is sometimes used toestimate the concrete modulus of elasticity, in which the slope of the stress-strain curve of concrete at the origin is evaluated as shown in Fig. 1.8. The ECP-203 gives the following formula for estimating the concrete modulus ofelasticity

fell' = 0.6 .JJ:: (1.4)

Secantmodulus

where leu is the concrete compressive strength in N/mm2The magnitude of the modulus of elasticity is required when calculatingdeflection, evaluating bracing condition, and cracking of a structure.

Ee = 4400.JJ:: (1.5)

C: Stresses on an element

-~hlh~j;,fi

B: Force system

Fig 1.7 Split cylinder test

A: Test setup

The tensile strength computed using the modulus of rupture is always higherthan the split cylinder tension tests. The tensile strength of the concrete can bedetermined using its compressive. strength. The tensile strength does notcorrelate well with the concrete compressive strength but rather with its squareroot. The ECP-203 gives an expression for estimating the concrete tensile

.strength letr as a function of its compressive strength as follows:

Strain

Fig. 1.8 Initial tangent modulus of concrete

12 13

1.5.4 Strength of Concrete Under Biaxial LoadingPortions of many concrete members may be subjected to stresses in twoperpendicular directions (biaxial state). The strength ofthe concrete is affectedgreatly by the applied stress in the perpendicular direction as shown in Fig. 1.9.In Fig. 1.9, all the stresses are normalized in terms of the uniaxial compressivestrength feu. The curve has three regions; biaxial compression-compression,biaxial tension-tension, biaxial tension-compression.In the compression-compression zone, it can be seen that the compressivestrength of the concrete can be increased by 20-25% when applying compressivestress in the perpendicular direction.In the tension-tension zone, it is clear that the tensile strength of the concrete isnot affected by the presence of tension stresses in the normal direction. Forexample, a lateral tension of about half the value of the uniaxial tensile strengthwill reduce the compressive strength to 50% of the uniaxial compressivestrength.

The biaxial state may occur in beams as shown in Fig. 1.10 where the principletensile and compressive stresses lead to biaxial tension compression state ofstress. The split cylinder test illustrated in Fig. 1.7C is a typical example ofbiaxial state of stress, where the compressive stresses develop in the verticaldirection and tensile stresses develop in the horizontal direction. This is the mainreason that splitting tensile strength is less than flexural tensile strength.

}?ig. 1.10 Biaxial state of stress in beams

+-I

1.41.210.80,60.40.2o

.----- r----

feu V f1 I~0-0

0'

1\J ..-

0

I II f, ..-

I - II Ir- ....-j ..-.,.

.0'

..-0

. compr ssion.'.. V..-

-0

/..-t; .---

.

.

..- feuension It~tr

o

-0.2-0.2

0.2

0.6

1.4

1

1.2

0.4

0.8

Fig. 1.9 Strength of concrete in biaxial stress . Photo 1.4 Typical reinforced concrete structure

14 15

1.5.5 ShrinkageAs the concrete dries it shrinks in volume due to the excess water used inconcrete mixing. The shortening of the concrete per unit length due to moistureloss is called shrinkage strain. The magnitude of the shrinkage strain is afunction of the initial water content, the composition of the concrete and therelative humidity of the surroundings. Shrinkage is also a function of member'ssize and shape. Drying shrinkage occurs as the moisture diffuses out of theconcrete. As a result, the exterior shrinks more rapidly than the interior. Thisleads to tensile stresses in the outer skin of the concrete member andcompressive stresses in its interior. The rate of the shrinkage increases as theexposed area to the volume increases.

The ECP-203 gives the following formula to estimate the virtual memberthickness

B = 2;< (1.6)c

where B is the virtual member thickness, Ac area of the cross section, P, is thesection perimeter subjected to shrinkage.

Values of final shrinkage for ordinary concrete are generally of the order of0.00016 to 0.00030 and can be taken from table 104.

Table 1.4 Values of shrinkage strain for concrete (x 10.3)

weather Dry weather Humid weather

condition Relativehumidity==55% Relativehumidity== 75%

Time by Virtual thickness B Virtual thickness B

B ~ 600 600 < B > 200 B 5. 200 B ~ 600 600 < B > 200 e s 200days

3-7 0.31 0.38 0043 0.21 0.23 0.26

7-60 0.30 0.31 0.32 0.21 0.22 0.23

>60 0.28 0.25 0.19 0.20 0.19 0.16

almost flat curve

Fig. 1.11 Variation of shrinkage with time for a typical concrete mix

Although shrinkage continues for many years as shown in Fig. 1.11,approximately 90% of the ultimate shrinkage occurs during the first year. 1.5.6 Creep

When a reinforced concrete member is loaded, an initial deformation occurs asshown in Fig. 1.12. Experimental studies show that this initial deformationincreases with time under constant loading.The total deformation is usually divided into two parts: (1)initial deformation (2)a time dependent deformation named creep.After the occurrence of the immediate deformation (point A" to point A), thecreep deformation starts rapidly (point A to pint B) and then continues at a muchlower rate till almost it becomes a flat curve at infinity. More than 75% of thecreep deformation occurs during the first year and 95% in the first five years. Ifthe load is removed at point B, immediate recovery occurs (point C), followedby a time dependent recovery till point D (creep recovery). The member willnever recover all the developed deformation and there will be a non-recoverabledeformation called permanent deformation.The creep deformations are within a range of one to three times theinstantaneous elastic deformations. Creep causes an increase in the deflectionwith time that may lead to undesirable deformation of the member. Thus, thedeflection must be investigated to ensure that the deformations are within theallowable limits ofthe code.

...

!

Timet=oo

16 17

1.6 Reinforcing SteelThe most common types of reinforcing steel are bars and welded wire fabrics.Deformed bars are the most widely used type and manufactured in diametersfrom 10 mm to 40 mm, They are produced according to the Egyptian standards262/1999. Bars are supplied in lengths up to 12m, however, longer bars may bespecially ordered. Reinforcing bars are available in four grades with a yieldstrength of 240, 280, 360, and 400N/mm2. The cost of steel having a yield stressof 400 N/mm2 is slightly higher than that of steel with a yield point of 240N/mm2. However, the gain in strength and accordingly the reduction in therequired steel area is obvious. It should be mentioned that grade 400 N/mm2 isthe highest steel gradeallowed by the Code for reinforced concrete structures.The ultimate tensile strength, the yield strength and the modulus of elasticity aredetermined from the stress-strain curve of a specimen bar loaded in uniaxialtension up to failure. The modulus of elasticity of steel (the slofe Ofthe stress-strain curve in the elastic region) is 200 GPa (200,000 N/mm ). The specifiedstrength used in design is based on the yield stress for mild steel, whereas for

yield plateau .

:elastic rangeI

start of strain hardening

r---i-,--t~ mild steel

0.002

II

, , A: .: plastic range ~v--- Strain

I

1;.

0.2% proofstress

high yield steel the strength is based on a specified proof stress of 0.2% asshown in Fig. 1.13. . .The major disadvantage of using steel in beams and columns IS COrr?s.lOn. Thevolume of the corroded steel bar is much greater than that of the ongmal one.The results are large outward pressure, which causes severe cracking andspallingof the concrete cover. The ECP-203 requires the increase o~ concretecover in corrosive environments. Epoxy coated bars are a perfect solution for theproblem of corrosion of the reinforcement. They are expensive and need to behandled very carefully to protect the coating layer from damage. However, theyare not as efficient as uncoated bars in developing full bond with surroundingconcrete.

Fig. 1.13 St~ess-Strain curve for mild and high grade steel

high grade steel

Stress

permanentdeformation

D...... -

---

elastic recovery

creep recovery

deformation underconstant loading

\

"......

....

I1I'

:C

Time

.load removal B

~'\~:;=::==:-

Elastic deformation

Fig. 1.12 Elastic and creep deformation of concrete

A

time ofloadi g

18 19

1.7 Limit States Design MethodMembers are designed with a capacity that is much greater than required tosupport the anticipated set of loads. This extra capacity not only provides afactor of safety against failure by an accidental overload or defectiveconstruction but also limits the level of stress under service loads to controldeflection and cracking. The Egyptian code permits the use of two designmethods, namely, the allowable working stress design method and the ultimatelimit states design method. In the present time, the former is the most commonlyused in the design of reinforced concrete structures.When a structure or a structural member becomes deficient for its planned use, itis said to have reached a limit state. The limit states of concrete structures can bedivided into the following three groups:

A. Ultimate Limit states

The strength reduction factors vary according to the applied compression force.As the compression force increases, the strength reduction factor in tumincreases. One of the reasons for that, is the nature of the brittle failure thataccompanies the compression forces. The strength reduction factor for concreteyc ranges from 1.73 for sections subjected to almost pure compression and 1.5for sections subjected to pure bending. The strength reduction factor for steelreinforcement "Is ranges from 1.32 for sections subjected to compression and1.15 for section subjected to pure bending.

For sections subjected to combined compression forces and bending (eccentriccompression sections) with at least 0.05t eccentricity, the ECP-203 gives thefollowing values for the strength reduction factors

Yc

=1.5XH-(e~t)}21.5 (1.7)These limit states are concerned with the failure of a structural member or thewhole structure. Such a failure should have a very low probability of occurrencesince it may lead to loss of human lives.

r, = 1.15X{~- (e~t)} 2 1.15 (1.8)

where e is the eccentricityand t is the member thickness and !:. 2 0.05. t

For other cases the strength reduction factors can be taken as

r, =1.5 }pure bending, shear and torsioneccentric and concentric tensile forces

rs

=1.15 bond and bearing

. Fig. 1.11 Concrete and steel strength reduction factors

eft

steel strength reduction factor Ys

concrete strength reduction factor Yc

/.15

1.50

0.500.05

. ,-..

~.8 1.73o

for serviceability limit states the reduction factors can be taken as

r, =1.0}for calculation of cracking, deflection and deformation

r, =1.0

Photo 1.5 Queensland, Australia, 322 meters 78 stories (2005).

22

1.9 Classification of LoadsThere are several types of loads that may act on a structure and can becategorized as:

Dead Loads: These are constant in magnitude and fixed in location for thelifetime of the structure. A major part of the dead loads results from the ownweight of the structure itself. The dead loads also include sand required forleveling of the flooring, flooring material and brick walls.Live loads depend mainly on the use of the structure. For buildings, live loadsare the results of occupants and furniture. In bridges, vehicle loads represent themajor live load. Their magnitude and location are variable. Live loads must beplaced in such a way to produce the maximum straining actions on thestructures. But rather by placing the live loads on the critical locations that causemaximum stresses for that member.Table 1.5 gives examples of the values of live load on some structures asmentioned in the Egyptian Code for Calculation of loads on Structures.

Table 1.5 Live loads value according to building type (kN/m2) .Structure Type Location/usage Live load

----

Rooms 2Residential buildings

Balconies, stairs, kitchen 3Offices 2.5

Office buildings Archives 5-10Balconies and stairs 4Patient rooms 2.5

Hospitals Surgery/lab 4 or moreBalconies and stairs 4Classrooms 3Labs 4 or moreSports centers 5

Schools and facultiesBook shelf area 10Lecture rooms 4Balconies and stairs 4Gust rooms 2

HotelsPublic area/restaurants/stairs 4

23

._-

Seated area 4Public area unseated 5

Cinemas and theaters5Balconies

Stairs and corridors _ 6--

Mosque / church / Seated area 4Halls Unseated area 5

Inaccessiblehorizontal flexible roof 0.6Roofs Inaccessiblehorizontal rigid roof 1.0

Accessible horizontalroof 2Parking area (small cars) 3

garages Buses 4Garage corridor 5

For residential buildings with more than five stories, the live loads may bereduced according to the Table 1.6

Table 1.6 Reduction of live load in multistory residential buildings

Location of the floor Live load valueRoof PFrom 1 to 4 under the roof PFifth floor under the roof 0.9PSixth floor under the roof 0.8 P--Seventhfloor under the roof 0.7PEighth floor underthe roof 0.6PNinth floor and more under the roof 0.5 P

Lateral loads These are the loads resulting from wind pressure, earthquakeloads, soil pressure, and fluid pressure. In recent years, significant progress hasbeen made to accurately estimate the horizontal forces due to wind orearthquake.TheECP 203 states a series of load factors and load combination cases to beused in designing reinforced concrete sections.

24

1.10 Load Combinations

- For members that are subjected to live loads and where the lateral loadscan be neglected, the ultimate factored loads U are computed from

U=I.4D+1.6L (1.9)where D are the working dead loads, and L are the working live loadsAlternatively if the live loads are the less than 75% of the dead load, thefollowing equation can be used

U=1.5(D+L) (1.10)If the member is subjected to earth or fluid pressure (E), the ultimate load isgiven by

U = 1.4 D+1.6 L+1.6 E (1.11)In the case oflateral pressure in closed spaces such as tanks and small pools, theultimate load is taken from

U = 1.4 D + 1.6 L + 1.4E

-Ifthe structure is subjected to wind loads W or earthquake loads S, theultimate load U is taken as the largest from the following two equations

U = 0.8 (1.4 D+ 1.6 L+ 1.6 W) (1.12)U =1.l2D+aL+S .. , (1.13)

Where a is a coefficient that takes into account the effect oflive load that mightexists on the building during an earthquake and is taken as follows

~ a=1/4 in residentional buildings.~ a=1I2 in public buildings and structures such as malls, schools, hospitals,

_garages and theaters.~ a=1 in silos, water tanks, and structures loaded with sustained live loads

such as public libraries, main storage areas and garages for public cars.

- In load cases in which reduction of live loads shall lead to increasing thevalue ofmaximum forces in some sections, the live load factor shall betaken to 0.9.

25

For cases in which the effects of the dead loads stabilize the structure, theultimate loads should be taken from the following set of equations

U = 0.9 D (1.14)

U =0.9 D+1.6 E (1.15)

U =0.9 D+1.4 E (for tanks andpools) (1.16)

U =0.9 D+1.3 W (1.17)

U=0.9D+1.3S (1.18)

Photo 1.6: Opera Sydney in Australia

26

Table 1.7 Load factors according to ECP 203

Condition Factored Load U

U = 1.4 D + 1.6 L

U = 1.5 (D+L) L '5,0.75 DBasic

U=0.9 D

U = 0.9 D + 1.6 L

U = 0.8 (1.4 D + 1.6 L 1.6 W)Iwind

U = 0.9 Dl.3 W

U = 1.12 D + a L + SEarthquake

U =0.9D S

U = 1.4 D + 1.6 L + 1.6 EEarth pressure

U = 0.9 D + 1.6 E

U = 1.4 D + 1.6 L + 1.4 EClosed tanks

U = 0.9 D + 1.4 E

Sett1ement,creep, or U = 0.8 (1.4 D + 1.6 L + 1.6 T)

temperature U = 1.4 D + 1.6 T

U = 1.4 D + 1.6 L + 1.6 K~ynamic loading

U = 0.9 D + 1.6 K

where D, L,W, S, E, T, K are the dead, live, wind, soil, earthquake, temperatureand dynamic loads respectively. .

27

dead or

-35.20-35.20

Bending moment (dead loads)Normal force (dead loads)live loads

95 kN

Example 1.1Using the load combinations of the ECP 203, determine the ultimate axial forceand bending moment combinations for the column CD at point C. The frame issubjected to the following working loadsD=15 kN/m'(uniform)L=30 kN/m'(uniform)Wind load of95 kN (may act in either direction)

DA

Normal force (live loads) Bending moment (live loads)

Solution:since the structure is indeterminate, a computer program was used to calculatethe axial and bending on the frame. The following figures summarize the results.

-52.4

-91.25 87.9

Normal force (wind loads) Bending moment (wind loads)

28 29

U =-165.6022.60

Load combinations for member CD

=-188.2kN and -143.0kN

2.1 IntroductionUntil the late 1980s, nearly all reinforced concrete buildings in Egypt weredesigned according to the working-stress design method. However, since 1989the ultimate limit states design method has gained popularity and has beenadopted by the Egyptian Code for Design arid Construction of ConcreteStructures. In this chapter, the basic design concepts of the ultimate limit statesdesign methods are discussed.

Photo 2.1: Alamillo Cartuja suspended bridge, Spain

2,

DESIGN OF SINGLY REINFORCED SECTIONS

-63.4 -99.8

-143.1 -62.5-188.2 IC -196.6")

e-207.0~ -161.9

Axial force Bendingcombination combination I

C -17.6:> Ie +36.4:>

Axial bending case Equationload moment No.

D -45.0 -35.2 1 U=1.4D+1.6L

L -90.0 -70.4 2 U =0.8(1.4 D + 1.6 L +1.6W)

W -17.64 -52.4 3 U =0.8 (1.4 D + 1.6 L - 1.6W)

4 U =0.9 D+1.3 W

5 U =0.9 D-1.3 W

U =0.8[1.4D +1.6L 1.6W]

U =0.8 [1.4'(-45) + 1.6 (-90) 1.6 (-17.64)]

An example of the calculation for the axial force for the case of (D+L+W) isgiven by

To compute the ultimate loads and according to the ECP-2003, fivecombinations were used as shown in the following table.

From the table, the maximum and minimum ultimate axial force on the columnis -207.0 and -17.6 respectively. The maximum and minimum ultimate bendingmoment at C is -196.6 and +36.4.

It is very important to notice that the design should be carried out based onstraining actions resulting from the same load combination not the maximumfrom each case. Thus, it is wrong to design the column for an axial compression

. force of -207.0 and bending moment of -196.6. Instead, the Section must bedesigned to withstand (an axial compression force of -207 and a bendingmoment of -161.9) and (-axial -188.2, bending -196.6). In addition, it should bedesigned for an axial force of-17.6 and a bending moment of+36.4.

30 31

uniform load

service load

2.2 Reinforced Concrete Beam Behavior

At moderate loads (ifthe concrete stresses do not exceed approximately one-third the concrete compressive strength), stresses and strains continue to be veryclose to linear. This is called the working loads stage, which was the basis of theworking-stress design method. When the load is furthered increased, morecracks are developed and the neutral axis is shifted towards the compressionzone. Consequently, the compression and tension forces will increase and thestresses over the compression zone will become nonlinear. However, the straindistribution over the cross section is linear. This is called the ultimate stage. Thedistribution of the stresses in the compression zone is of the same shape of theconcrete stress-straincurve. The steel stxese ], in this stage reaches yieldingstressfy. For normally reinforced beams, the yielding load is about 90%-95% ofthe ultimate load;'

When the load is further increased, the developed tensile stresses in the concreteexceed its tensile strength and tension cracks start to develop. Most of thesecracks are so small that they are not noticeable with the naked eye. At thelocation of the cracks, the concrete does not transmit any tension forces and steelbars are placed in the tension zone to carryall the developed tensile forcesbelow the neutral axis. The neutral axis is an imaginary line that separates thetension zone from the compression zone. Therefore, by definition the stress atthe neutral axis is equal to zero as shown in' Fig. 2.1. Thus, the part of theconcrete' below the neutral axis is completely neglected in the strengthcalculations and the reinforcing steel is solely responsible for resisting the entiretension force.

crackedsection

. cracks'

.....~~~~~~~~~~~~~~y.-~~:~~:;~IK'~~A,t=:::::::===:=:l:::::::::1:::::::l::::::::===:::::::::!,l ~ LL L.s.

a: before cracking

b: cracking stage, before yield, working load

Consider that a reinforced concrete beam as the one shown in Fig. 2.1. issubjected to an increasing load that will cause the beam to fail. Several stages ofbehavior can be clearly identified.At low loads, below the cracking load, the whole of the concrete section iseffective 'in resisting compression and tension stresses. In addition. since thesteel reinforcement deforms.thesame amount as the concrete, it will contributein carrying the tension stresses. At this stage, the distributions of strains and

.stresses are linear over the cross section.

j j j ! ! 'j r ~... ..... .. ...;-'------------------------------i------------']~----~V- ~~A-= =-- ~" - .A LJuncrackedsection strains stresses

ultimate load

. ~ . .

. Fig. 2.1 Reinforced concrete beam behavior at different stages of loading

c: ultimate and failure stage

0.003

~.'... ~~~_-_~_~~.~~--::~~__~~~~~-~~~NAU~L.J~L

." ... cs>ev j,=J;I/./5cracked sectionat ultimate

At the ultimate stage, two types of failure can be noticed. If the beam isreinforced with a small amount of steel, ductile failure will occur. In this type offailure, the steel yields and the concrete crushes after experiencing largedeflections and lots of cracks. On the other hand, if the beam is reinforced with alarge amount of steel, brittle failure will occur. The failure in this case is suddenand occurs due to the crushing of concrete in the compression zone withoutyielding of the steel and under relatively small deflections and cracks. This isnot a preferred mode of failure because it does not give enough warning beforefinal collapse.

32 33

................. (2.1.A)for s, < 0.002

for a.002::; e, ::;0.003

1'-------,----'-----1----,.-----1Concrete strain e,

_~~e_h!yc. .__,;, -....------1~ -enen

2.3.3 The Equivalent Rectangular Stress BlockTo compute the compression force resisted by concrete, the Egyptian Codereplaces the curved stress block shown in Fig 2.4C by an equivalent stress blockof an average intensity of 0.67 feuiYe and a depth a= peas shown in Fig. 2.40.The magnitude and location of the force calculated using the equivalent stressblock should be equal to that of the curved one.

neutral axis

f:' =0.67 fc" Ie' =0.67 j,."J,. y

0003 " C Y',,-; :_0).---:~iB~ I C'

S :C'l I

I

As!lsec~0 n....". ._ - _ - _ .

B: strain C: parabolic stress D: equivalent rectangulardistribution stress block

-----

c

Fig 2.3 Idealized stress-strain curve for steel

h/~'~ tension~.,;~ E.=200,OOO Nzmrn'ene!ys ,. ey/Ys Strain, Cs,

~compression-Iv/Y,--

2.3.2.2 Reinforcing SteelThe behavior of the steel reinforcement is idealized by the Egyptian code(section 4.2.1.1 )as an elastoplastic material as shown in Fig 2.3. The reinforcingsteel stress can be calculated using Eq. 2.1.B.

I, = e, x E, when c. < cy / r, (2.1.B)I, = i, / r, when e, ~ e! r,

Fig. 2.4 Equivalent rectangular stress block calculation.

To calculate the depth "a" of the stress block, one equates the compression forceobtained using the stress-strain curve of the Egyptian Code, shown in Fig. 2.4C,to that using the equivalent stress block (Fig. 2.40).The total compression force (C=C1+C2) obtained using the stress-strain curve ofthe Egyptian Code can be calculated as follows:

C. =bx(~XIe') ~ (2.2)

2 2e , (4 f')' (23)C =bx-x-xf =bx -eX .2 33 c 9 c

e ,4 c ,7 f' (24)C=C +C =-xbxf. +-xbxf. =-exbx .. 2 3 '9 '9 '

The compression force obtained using the stress block C' equals

C' =bxaxje' = fJ eXbxf.~ (2.5)

Photo 2.2 High grade steel Reinforcement

3637

By definition, C must be equal to C', thus solving Eq. 2.4 and Eq. 2.5 for /3 gives7P=-=0.7779

The code approximates the previous value to /3=0.8, thus the rectangular stressblock depth (a=0.8 c).To find the location of the total compression force C', take the moment of theforces at point "0" and note that the e.G of the force F2 is at 3/8 of the distance(2/3c)

2.4 Analysis of Singly Reinforced SectionsConcrete beams subjected to pure bending must resist both tensile and compressivestresses. However, concrete has very low tensile stresses, and therefore tension steel isplaced in these locations (below neutral axis) as shown Fig. 2.5. The most economicsolution is to place the steel bars as far as possible from the neutral axis except for theconcrete cover, which is normally assumed 50 mm from the external surface.

cracked section A-A

crackedzonesteel bars

compressed zone

concrete cover

Fig. 2.5 Reinforcement placement in reinforced concrete beam

The compressive stresses in concrete are replaced by a uniform stress block assuggested by the Egyptian Code (section 4.2.1.1.9) with distance "a" from theconcrete surface as shown in Fig. 2.6.

i c xb x f, x k, c= ~ XbXjc'(~) +*c xb x f, L72 c) (2.7)k1=0.404The code simplifies the value of k, with /3/2=0.4 (i.e. the resultant is at themiddle of the stress block)

Photo 2.3 Metropolitan Government Building in Tokyo38 39

The analysis of the cross section is carried out by satisfying two requirements:

EquilibriumI. L Forces(internal) = L Forces(externaI)

For sections subjected to pure bending, the external forces equal tozero. This leads toL Forces(internal) = 0 ::::> T - C = 0 ::::> T = C;

2. LMu (internal) = LMu (external) (taken about 'anypoint in the section) Compatibility of Strains ,I. The strain at any point is proportional to its distance from the

neutral axis.

Therefore, if the design problem has more than two unknowns, assumptionshave to be made to reduce them to exactly two. The stress in the tension steel isassumed to be equal to the yield strengthf;,. This assumption should be verifiedafter determining the neutral axis position. The equilibrium of the internal forcesis used to determine the stress block distance "a" as follows: .

C=T (2.8)

0.67 feu b a = As f y : : (2.9.A)1.5 1.15

If the tension steel does not yield Eq. 2.9.A becomes

Having determined the stress block distance a, the assumption of the tensionsteel yielding can be verified using compatibility of strains as follows (c=a/0.8and Es=200,000 N/mm

2 )

t. = E,

2.5 Maximum Area of Steel of a Singly ReinforcedSection

Cmu., = 460d 690+ f)' ............................................ (2.15)

The balanced failure occurs when the concrete strain reaches a value of 0.003 atthe same time that the steel reaches the yield strain divided by the reductionfactor (E!Ys) as shown in Fig. 2.8.

C

forces

T=A smax fl 1.15

0.003 0.67 fc,,/1.5IH..---7_~Lr----,- _.~

Ero

I"d

IEs>E/1.lS Istrains

2cmax = 3" cil

a max = 368d 690+ J; (2.16)

b

Asmax

,.

I----...T

'\b

As _..___ ~.==:i:==::

1'8s=S,IYs

0.003 0.67 fcull.5I' 'I. H

............................._ _~~_~~[~C\

____._J'I' 0.003 'I

d

Fig.2.S Neutral axis position at the balanced condition Fig. 2.9 Neutral axis position for calculating the maximum valuesallowed by the code

From similar triangles shown in Fig. 2.8, one can conclude that

C/> = 0.003 ,..;.. (2.12)d 0.003+ G)'

y,. where c, is the neutral axis at the balanced failure. The steel Young's modulus

Es equals

E, =f)' =f)'lys (2.13)G)' G)'lys

Substituting with steel Young's modulus E, =200,000 N/mm2 and Ys=1.15 gives

The ratio of the reinforcement in the concrete section (u) is an indication toshow if the section is lightly reinforced or heavily reinforced and can beexpressed as:

j.1= :~ '" (2.17)After finding the maximum neutral axis position Cmax, it is beneficial to computethe maximum area of steel As.max recommended by the code. To find themaximum area of steel, apply the equilibrium equation (C=T) with neutral axisat Cmaxas shown in Fig. 2.9.

~ 69~~f, (2.14) 0.67 fell b amax = Asmax f; (2.18)1.5 1.15 .Diving both sidesby (b x d) gives

If C < Cb, then the strain in the tension steel is greater than Esl'Ys and that thetension steel yields. To ensure ductile failure the ECP 203 requires that the valueof Cmax be limited to 2/3 Cb' Substitution in Eq. 2.14 and referring to Fig. 2.9

'. gives the following equation

0.6~.{cu a~ax = ,u~~;;' (2.19)

42 43

substituting with Eq. 2.16 into Eq. 2.19 gives Defining ((I = J1 i.Lfeu

((1m., = J1max

f.fy (2.22)cu

189690 f, + 1,; feu (2.20)

~----- ----,Steel cb/d cmax/d amax/d Rmax Rl max f.!max* rom~~

...~~05240/350 0.74 0.50 0040 0.214 0.143 8.56x10-4 feu280/450 0.71 0048 0.38 0.208 0.139 7.00x10-4 feu 0.196

'---_._-- ---_.

360/520 0.66 0044 0.35 0.194 0.129 5.00x10-4 feu 0.180

400/600 0.63 0042 0.34 0.187 0.125 4.~1x10-4 feu 0.172

450/520** 0.61 . 0040 0.32 0.180 0.120 3.65x104 feu 0.164

Table 2.1 Values of cmax/d, f.!mm ffimax

* feu in N/mm2** for welded' mesh

(2.23)690+ I,(j)max

Substituting with the value off.!max determined from Eq. 220 gives

189

3.5

3.0

2.5,--..

~0'-'

~ 2.0E:i

1.5

1.0

0.5240 280 320 360 400

Fig. 2.10 Effect offcu and fy on f.!max..h N/mnl

The ECP 203 limits the reinforcement ratio f.! to f.!max given by Eq. 2.20 toensure ductile failure. Moreover, it is a good practice, from the economic pointof view, to limit the area of steel reinforcement in beams to only 0.5-0.7 f.!max. Itcan be noticed that steel with smaller fy will have smaller yield strain Ey leadingto larger neutral axis .distance Cmax as shown in Table 2. L Thus, the smaller thesteel yield strength, the larger. the maximum permissible-steel ratio f.!max asshown in Fig. 2.10. .

It should be clear that if for a given section the neutral axis distance "c" is lessthan neutral axis maximum value ema:" then the steel is yielded, the actualar~a ofsteel As, and the applied moment M; is less than code maximum limits as

.. indicated in Eq. 2.21.

.

If' < Cmax then.I d d

Jl < Jlma,As

Maximum Moment CapacityTo determine the maximum moment for a singly reinforced section, one cancompute the moments of the tension force about the compression force (refer toFig. 2.9) at C=Cmax

M".max = \~~5J; (d - a;ax ) (2.24)D fini R' 1.5 x M".maxe InIng max as Rmax = . 2L; b d

: =1.304 JlI~"J;. (1- 0.4 c;x ) (2.26a)

2.6 Balanced, Under, and Over Reinforced Sections

In general, an under-reinforced section is the one in which reinforcing steelyields before the crushing of concrete. An over-reinforced section is the one inwhich failure occurs due to the crushing of concrete in the compression zonebefore the yielding of the steel. On the other hand, a balanced section is the onein which yielding of steel and crushing of concrete occur simultaneously.

According to the analysis carried out in section 2.5, one can conclude that if thesection is reinforced with Jl. less than Jl.b (=1.5 Jl.max) it is called "underreinforced". On the contrary, if the section is reinforced with Jl. greater than Jl.b'it is called "over reinforced". The under-reinforced sections are preferredbecause they fail in a ductile manner, in which the member will experience largedeflections, large strains, and wide cracks. This gives enough warning so thatrepair can be performed on that member. On the other hand, over reinforcedsections will fail suddenly without enough warnings. Figure 2.11 gives thestrain distributions and the related values of the three sections

0.003s.; = 1.304wmax (1- 0.4Cd,,-,) (2.26b)Substituting with the value of Jl.max calculated from Eq. 2.20 gives

: = 69~~J;, (1_0.4C~ax ) (2.27)

Rlmax = M"max 2 = Rmax ; (2.28)I; b d 1.5

I'

EsCmaxand (C>Cb)J1.>Jl.max and (Jl.>Jl.b)fs

2.7 Minimum Area of Steel

In some cases, and mainly due to architectural considerations, the member couldbe cho~en with con.crete dimensions bigger than those required by strengthcalculations, Accordmgly, the required area of steel could be very small. Thismay lead to situations where the strength of the section using cracked sectionanal?,sis is less than the, strength Of the uncracked section computed using thetensile strength ofconcrete. '

The failure of such sections is brittle and wide cracks tend to develop. Thus, tocontrol cracks,' to ensure ductility, and to avoid sudden failure in tension, theEgyptian code (4.2.1.2.g) requires that the actual area of steel As in any sectionshould be greater than Asmin given by: '

, . 10.225 .JJ: b d ~.!:!b dA'lnin = smaller of f" f). . :................. (2.29)

1.3A, '

{

0.25 b )-- d(mild steel)

but notless than 100 ' ,0.15 '- b d(highgrade)100

2.8 Factors Affecting Ultimate StrengthThere are several factors that affect the ultimate strength of a beam subjected tobending. These factors can be summarized as '

Yield strength of reinforcing steel.jj, Concrete compressive strength.j., Beam depth, d Beam width, b Reinforcement ratio, !-t.

The effect of steel yield strength on ultimate strength is shown in Fig. 2.12A. Itis clear that steel yield strength has a big impact on its ultimate capacity.Increasing the steel yield strength from 240 N/mm2 to 400 N/mm2 increases theultimate capacity by 55%. On the other hand, concrete compressive strength hasa little effect on the ultimate strength as shown in Fig. 2.12B. Changing concretecompressive strength from 20 N/mm2 to 40 N/mm 2 increases the ultimatestrength by only 5%.

Comparing Fig. 2.12C and Fig. 2.12D shows that increasing beam depth affectsthe ultimate capacity more than increasing beam width. Increasing beam depthfrom 500 mm to 1000 mm increases the capacity of the beam by almost threetimes. Finally, increasing steel reinforcement ratio has a significant effect on theultimate capacity as illustrated in Fig. 2.12E.

Iffcll is greater or equal to 25 N/mm 2 the term (0.225.JJ:: If" b d) is bigger than(0.25% b d) and (0.15% b d). Thus, there is no need to check the third conditionin Eq. 2.29, if 0.225 JJ:: If,. < 1.3A, . The minimum area of steel in this case canbe simplified to: .

48

0.225, if__-"vJ...",,-,uub d

fvI.3A,

....................... (2.30)

Photo 2.4 Interior reading halls in the Library of Alexandria

49

0.5

o240

---_.

L----~-

0.5 -1----1---+---+--------1

ANALYSIS PROBLEMGiven : b, t ,As feu, fyRequired : MuUnknowns: a, Mu'

Analysis SummaryIn this type of problem all the cross section information is known includingbeam cross section dimensions, steel yield strength and concrete strength. It isrequired to calculate the moment capacity Mu

35 feu 403025O-l-----I-----!----+----I

20

3.5 ,------,,..---.--.,.-.---.,----,

3 +---+---+---+---j2.5 -1----1---+---+--------1

} 2+---+---+---+----ji 1.5 t=:::==l=:::::=t:==::::j:==~

360 fy 400320280

3

3.5

2.5'\,:E 2'i

1.5

A-Effect of fy B-Effect of feu

Procedureo Step 1: Apply the equilibrium equation T=C to find the depth of

the stress block, "a" and the neutral axis depth "c"assuming that tension steel has yieldedfs=f/I.15.

o Step 2: Check that tension steel has yielded (fs ~ f)U5) byensuring the.c-cc, or by using Eq.2.1 O.

o Step 3: Compute the bending moment capacity M, by taking themoment about the concrete compression force.

800 1000d(mm)

600

-:V

-:-:V

/100

400

150

350

300

E250

i'i 2oo

350 450b(mm)

250

I -

~

350

150

100150

300

E 250!'i 200

E-Effect of reinforcement ratio l.l

-

----------

j

-----

I

--iI

C-Effect ofbeam width (b)

1.201.00

D-Effect ofbeam depth (d)

0.800.60

3.53.02.5

"'0.c 2.0

~ 1.51.00.50.0

0.40

Fig. 2.12 Parametric study on the ultimate moment capacity Photo 2.5 Cantilever box section in a reinforced concrete bridge

5051

Example 2.1Determine whether the section shown in figure is under-or-over-reinforced section and check code maximum permissible area ofsteel for the following cases .

1. As= 500 mm2. 2

2. As=1000 mm3. As =1500 mnr'4. As =2000 mrrr'fcu=25 Nzmrrr' , fy=360 N/mm2

Case 1: As= 500 mnr' under reinforced (As

SolutionStep1: Apply equilibrium equation T=C

Assume tension steel yields

Step 2: Check f s

From Table 2.1 for/y=400N/mm2 ~ cb/d=0.63Since c/d(0.311) < cJd (0.63) then fs=f/1.15Since c/d (0.311 )Y/1.15 T=As f/1.I5= 1200x400/1.l5

-

Stress and strain distribution in the

54 55

M =600 x240 (450- 62.3) =52.44 x106 =52.44 kN mu 1.15 2 .

Since M u(52.44 kN.m) is less than the applied moment (80 kN.m), thecross-section can not withstandthe applied moment (unsafe).

Step 2: Check fsFrom Table 2 -> cb/d=0.74 and cmax/d=0.5c/d = 0.173 < cJd (0.74) then fs=fJ1.l5Since c/d (0.173)

Example 2.4Calculate the maximum moment that the beam shown in figure can sustain.Check whether the cross-section meets the code requirements regarding themaximum area of steel.The material properties arefcu=25 N/mm2 and fy=400 N/mm2

IE200 )1

0'0('

IE >1

Example 2.5A 3 mm steel plate with a yield strength of 400 N/mm2 is glued to a concretebeam reinforced with steel bars (4l/J16, /y=360 Nzmnr' ) as shown in figure.Determine the bending moment that the reinforced concrete sectioncan resist.The concrete compressive strength of the beam is 20 N/mm2

200

e~Ie eo 0V) 0t- t-

iL .lh_

150mm

Step 2: Check cmax/dFrom the code cmax/d for (1;,=360 N/mm2 )=0.44Cmax = 0.44 x 700 = 308 mm .

c-(steelplateyieldsl =400/1.15). 285.6. 1.15 sp

Step 3 : Calculate the ultimate moment MuTake the m.oment about the concrete force C

SolutionStep1: Apply equilibrium equation T=C

Area of the plate Ap= 3 x 150 = 450 mm'Area of the steel bars = 416 = 804 mnr'C=T1+ T~Assume that both the plate and steel bars yield

0.67/e.u b a As 1;, AI' In>---"=-.- = --+--

1.5 . 1.15 1.15

0.67 x 20 x200 x a 804 x 360 450 x 400---+---

1.5 1.15 1.15

a= 228.48 mm , c= 3:... = 228.480 = 285.6 mm0.8. 0.8

M=T (d -::....)+T(d. _::...)u t 2 2 I' 2

M = As I), (d _::...)+ AI' In> (d _::...)u 1.15 2' 1.15 I' 2

M = 804 X360(700 _ 228.48) + 450 x 400 (. 1 _ 228.48 )._u 1 15 75 .5 - 247.17 kN.m

. 2 1.15 2

!Final results M u=247.7 kN.l1lI

TI~As f/1.15T2=Ap fypll.15

As=804..... ,"e __ ._

Ap=450

~ . 200 >1 0.003

o . __ ._._._._._. __ ._._~.I._.._._._~It=a--._, Co 0V) .t-t-

60 61

Example 2.6A reinforced concrete beam has a cross section of concrete dimensionsb=200mm and d=450 mm. Calculate the moment capacity and the area of steelusing the idealized curves for concrete & steel, without applying safety factors(Ys=Yc=l) for the strain distribution shown in cases A&B. The idealized stress-strain curve for the concrete and steel isgiven below.

Force in the steelsince 1>.(0.005) > 0.001 then from steel curvefs=320 N/mm2

Force in the concreteThe force in the concrete equals the stressed area multiplied by the width b.The concrete area can be divided into two parts as shown in the figure below

0.002x = --185.29 = 105.88mm

0.0035

T

C2

Isometric forStress distribution

1:,=0.005

A,

200mm

C1

= 23 xl 05.88 200 = 243524 N2 .

xl =185.29-105.88 =79.41 mm

Concrete~

0.00350.002

Case B Strain

200I I 0_001

o{ ;r'"'

Case B

c 0.001 =0.5556d 0.001 + 0.0008c=0.5556 x 450 = 250 mme Force in the steelsince Es(0.0008) < 0.001 then find steel stress from graphf, =0.0008 320 =256 N / mm?

s 0.001eForce in the concrete

The concrete force is equal to the compressed area of concretemultiplied by the width b.The stress in the concrete is a triangular shapefrom the concrete curve with strain=O.OOI ~ fc=I1.5 N/mm2

Example 2.7Find the ultimate moment capacity for the cross-section shown in the figurebelow.fcu=30 Nzmnr' , and fy=360 N/mm2

1< 400

~ISolution

In this problem we have two unknowns a and M,..

Step 1: Compute a.It should be noted that the code permits the use of the stress block fortrapezoidal sectionsThe total compression force C equals to the concrete stress (0.67 fcull.5)multiplied by the compressed area Ac Assume that tension steel hasyielded(fs=fyl1.15)0.67 feu Ac As f y

1.5 1.15

0.67 x30 xAc = 1600x3601.5 1.15

0.0035

Concrete

Nooo

23 N/mm2 ------ i

f,= 11.5 N/mm'

ooo

c = 11.5 x 250 200 = 287500 NI 2

. Y t= 450-250/3=366.667 mm

Mu= C 1 X Y1=(287500x 366.667)/106 =105.41 kN.m

A, fy/1.15

I- "Is=400-2 a tan 10

4001--------1--..~n 10 0.003

......1.....j~...."oo~

Ac=37378 mm2

Forces

f,=11.50.001 t--l 1Ct .c/3

C') T--UI

""dII

;>

400+sAe = a--. = a [400-a x tan(10))2

37378 = 400 a- 0.176 aZ

Solving for a

a=97.65 mma 97.65

c =-=--=122.06 mm0.8 0.8

aid = 0.195 > ald)min(O.I) ....ok

Step 2: Check fs

Example 2.8Find the ultimate moment capacity that this cross-section can resist. Thematerial properties for the beam arefcu=20 Nzrnm'', and fy==400 N/mmz

~~180

200

400

As=1250 mm2

SolutionStep 1: Compute a.

Assume that tension steel has yielded. Since we have two unknowns M,

and (a), solving the equilibrium equations gives

Since c/d (0.244) < cb/d (0.66), thus steel yields f5=f/1.15Since c/d (0.244)

Step 2: Check steel yield stress, fsc= alO.8 =256.675 mmd=180+200+400-50(cover)=730 mmSince c/d(O.35)

Pro~edure

Assume f.!E y

C0.003

"I250

2.5 d x 400US.

I.

Calculation of As, d

0.67 x 30 x 250 x a1.5

a=0.2596 d

a/d= 0.2596 > a/d)min(O.I) ....ok

Step 3: Calculate dM = As 1;. (d _!!:-)

u 1.15 2Photo 2.7 Reinforcement placement in a slab-beam roof

7071

--d=597mm270x 106= 2.5 d400 (d 0.2596 d)1.15 2

From Eq. (1) As = 2.5 d =1493 mm2

Rounding d to the nearest 50mm, d=600 mm and As=4 As o.kb=200

-I0.67 feu

1.50.003 I---l

1 J cu '1._._._._._.- .. _.- _._ ..-.- _.-._.-.-.-._.._._._.As

, ...---,-,-'"-,-Es>Ey

A =O.l1~Mub =0;11 330 x106x200=1689mm 2 ~ 2W

337 kN.m0.187 x 30x 250x 6002

1.5xl06. . 2

M = Rmax feu b du.max L5

250I~ -I0.67 x30

0.003 ~Step 2: Calculate a

O' 0

Step 3: Calculate d

M = As 1;, (d _!!-)u 1.15 2

330X106=1689X280(d_184.l2) ---? d=895mm1.15 2

aid = 0.205 > ald)min(O.I) ....ok

Use d=900 mm t=950 mm and As= 1689 mrrr'

2.10 Design of Singly Reinforced Sections UsingCurvesDesign aids are very useful tools in designing reinforced concrete sections. Toprepare the design aids, equilibrium equations and compatibility of strains areutilized. There are several charts that can be used in the design process. Weshall present several design charts followed by design examples to explain howto use such design aids.

2.10.1 Design Charts (R-~)Applying the equilibrium equation for the forces shown in Fig. 2.13

da 1.5xJ-lxJ;, J-lxf),-= =1.9468-- (2.32)

0.67 x 1.15 x leu feu

0.67 feu b a As f)'L5 1.15

Dividing by (b x d) and noting that I-l=A/(b .d)

= 719 mm?

IIi

I I'

SummaryThe following table illustrates the use of the charts depending on the exampleinformation:

d is given, As required1. Calculate Rl or n, or Ku2. Use charts (R-ro or R-I!

or Ku-I!) to determine I!or ro3. Calculate As4. Check Asminand Asmax

As is given, d required1. Calculate I! or ro2. Use the charts to determine

RlorRuorKu3. Calculate d4. Check Asminand Asmax

Note 4: If the both (As and d) are not given, consider assuming 1l=0.008-0.01and proceed as the previous procedure

Note 5: It should be noted that we have to check Asmin using Eq. 2.30 even ifll>!!min on the curve, because the curve tests only the value of 1.11J;,

Note 6: Since sometimes the beam depth is not known, a reasonableestimation for "d" can be concluded by assuming a=O.l d and .1l=0.01 and substituting in Eq. 2.34. gives: .

d = ll~:;)' , (2.36)

Note 2: It should be noted that beam depth needs to be increased if the pointis located outside the curve as shown in Fig. 2.16. .

Note 7: The design curves can be presented in a tabular form (Ru-ll) or (Ru-Ku) as given in appendix A

!!maxllmin

Rwnax

RI>Rl max

Fig. 2.16 Cases where the beam depth need to be increased

Rl max

As, d required1. Assume Rl=1/2RmaX (R:::::O.07)2. Use the charts to determine ro3. Calculate d, Asd 2 = M u A =to b d feu

feu b Rl S f4. Check Asminand Asmax

Note 1: Each curve terminates at the value of the maximum reinforcementratio !!max orromax. Thus, there is no need to check the maximummoment.or the maximum area of steel as long as the point is less thanthe maximum limit.

Note 3: For small values of Rl0.04), co can be approximated by ro=1.2 R1. Photo 2.8 Trammell Crow Center (209m ,50 stories)

Example 2.11A reinforced concrete cross-section is subjected to a bending moment of afactored value of 400 kN .m, The beam has a width of 200mm. It is required todesign the cross section using the (Rl-ro) curve, knowing that fcu=30 N/mm2and fy=280 N/mm2. .

Step 3: Check Asmin, Asmax

!O.22S-!T.: bd " O.225.J30 X2ooX95.0 = 836.JA,min = smaller of /" 280 .1.3A, = l.3x1899 = 2469 =836mm 2

Example 2.12A reinforced concrete cross-section is subjected to a bending moment of afactored value of 350 kN.m. The beam has a width of 250mm. It is required todesign the cross section using the (Ru-u) curve, knowing that fcu=25 N/mm2and fy=400 N/mm2 .

SolutionStep 1: Assume 11 and get Ru

Since both (As and d) is not given, then,Assume R=l!2 Rmax -l--+-~ Ru=2.4

From the curve 11= 0.8% (0.008)

Rumax

1l=0.80% IlmaxStep 2: Compute d, As

R = Muu b d 2

2.4 = 350 x 106

-l--l-d=763 mm250x d 2

As = J1 b d = 0.008 x 250 x 763 = 1526 mm' = 15.26 em"

Take d=800 mm, t=850 mmNote: it is more economical to use the calculated depth (763 rom) not the

chosen depth (800 mm) to compute AsStep 3: Check Asmin, Asmax

Example 2.13A reinforced concrete cross-section is subjected to a bending moment of afactored value of 290 kN.m. The beam has a width of 150mm. It is required todesign the cross section using the (Ru-ku) table, knowing that the material

22properties are fcu=40 N/mm and fy=360 N/mm .

SolutionStep1: Assume J.1 and get Ku

Assume J.1= 0.8% (0.008)From the table (Ru-ku) with fy=360 N/rrim2 determine Ku=0.655

Step 2: ComP':lte d, Asr-r-r-r-r-t-

d =K ~Mu ':"0.655 290xl06 =910.74mmu b 150

A, = J1 b d = 0.008x150 x910.74 = 1093 mm' = 10.93 em"

d=950 mm t=10QO mmNote: use the calculated depth (910 mm) to calculate As

Step 3: Check Asmin, Asmax

_ !0.225.JJ:: bd =; 0.225.JJ:: x150x950=563A,min smaller of I, 360 . =563cm 2

Example 2.14Redesign the beam in example 2.9 using the design aids (Ru-Jl) andassuming that d=600 mm

Example 2.15Determine the value of the ultimate load (Pu) that can be applied to thebeam shown in the figure using design aids.feu=25 N/mm2 and fy=360N/mm2 (Neglect own weight of the beam)

Since Ruis less than the Rumaxvalue in the curve thus A

3DOUBLY REINFORCE'D BEAMS AND TBEAMS

Photo 3.1 Edgar J. Kaufmann House (Falling-water), FrankLloyd Wright 1936, (6 m cantilever)

3.1 Doubly Reinforced Sections3.1.1 IntroductionDoubly reinforced sections ate those that include both tension and compressionsteel reinforcement. In most cases, they become necessary when architecturalrequirements restrict the beam depth.From the economic point of view, it is recommended to design the member as asingly reinforced section with tension reinforcement only. If the required area ofthe tension steel exceeds the maximum area of steel recommended by the code,

bFig. 3.2 Analysis of sections with compression reinforcement (steel yields)It will be ass~med that both the compression and the tension steel have yielded. Thestress block distance "a" is calculated utilizing the equilibrium of forces as follows

Compression force = Tension force (3.2)C, + Cs =T (3.3)

........................................'----------

(c

As..................................................................

es>e!1.l5

d

3.1.2 Analysis of DOUbly Reinforced SectionsCase A: compression steel yieldsThe. equilibrium of forces and strain compatibility shall be applied to analyze thesection. The strain distributions and the internal forces in beams withcompression reinforcement are shown in Fig. 3.2. The compressive force is thesum of two parts; i) concrete force Ce, and ii) compression steel force Cs.

....Jd:.

j?:.??.~ _ ~I~:= A's fylI.I5 I

..T :.... e's>e!1.15 ---~- d'.. --

0.67 t;,u b a................................: :~~:~~:::~~=~.:::=: _.............................................. Ce= 1.5

where a usually ranges from 0.1 to 0.4.

A; =a As ,............................................... (3.1)

14r----~------------'---------,

Despite all of the aforementioned benefits, adding compression steel inreinforced concrete beams will not increase the section moment capacitysignificantly. This is because the tension force is constant (T = Asf /1.15) andthe lever arm between the tension force and the resultant ofthe two compressiveforces (in concrete and in steel) is slightly affected by adding compressionreinforcement. This can be noticed by examining Fig. 3.1. In this figure, thevertical axis gives the percentage increase in the capacity of a doubly reinforcedsection as compared to an identical one without compression steel.The use of compression steel is more beneficial when the tension steel providedis near the maximum allowed percentage of steel Jinwx. Adding compressionreinforcement with a=0.6 will increase the beam capacity by 6 to 13 percent forbeams with 1l=0.8% and Il= J1max respectively.

compression steel should be added. Adding compression steel reinforcementmay change the mode of failure from compression failure to tension failure ormay change the section status from over-reinforced to under-reinforced section.Compression steel also reduces long-term deflection and increases beamductility. For economic considerations, the Egyptian code recommends limitingthe compression reinforcement amount to only 40% percent of the tension steel.

The compression area of steel A ~ is usually expressed as a ratio from the tensionarea of steel As as follows:

M - As I, (d a) A; t, (a ')u -T15 -2 +us- i:' (3.7)

0.67 feu b a A: f y' As J,1.5 +T15=T15 (3.4)

The compression steel stressf~ is checked using compatibility of strains as follows:e'. =0.003 c-d' =0003 a- 0.8 d'

s c' a .................................(3.5)

f' , . E 6 c -d' a-0.8d's =&s X s = 00--=600 (3.6)

c a

If the value oif, 'in Eq. 3.6 is less thanfJ1.l5, the analysis should be carried outacc~rding to the procedure outlined in case B. On the other hand, if,..'ln Eq.3.6. IS larger than fJ 1.15, the assumption of yielding of compression steel isvalid and the moment capacity can be determined by taking the moment of theforces around the concrete force as follows:

=0.8%

~=l.O%

0.5 0.60.40.30.20.1

h=360N/mm2

0-l""'-----"T----,.----,------.,--,----,.------1o

12

~ .8 +------------7'''-----,----=----'''=------1'-'

"E 6 t----7'~____=:::/"'----___.:::::::::::=====1i4+----_"----..~~--='--"'=----------___1

10 +----------------c~~----'----'----1

a=A)AsFig. 3.1 Effect of compression reinforcement on moment capacity

2+--7-:r-"C7""~-----------'------___j

Case B: compression steel does not yieldApplying the equilibrium condition and referring to Fig. 3.3, one gets:

Simplified Approach for the Analysis of Doubly ReinforcedSections

0.67(;u b a +A; 1,'= ~~ ~-., (3.8)

d:ax =1.25(1:'" ~~) (3.12)If the value of the actual d' /a is less than the value d'max/a, the compression steelwill yield and f~ equals to f/1.15. Table 3.1 lists the values of d'max/a thatensures yielding. .The ECP 203 presents a simplified approach for such an analysis. It permitsassuming that the compression steel yields if the ratio of the compression steeldepth d' to tension steel depth d is less than the values given in Table 3.1,otherwise a compatibility of strains (Eq.3.6) has to be utilized.

The previous procedure indicates. that the compression steel strain s', is affected- by the distance d' (refer to Fig.. 3.3). The compression steel yields if the distance

d' is small compared to the neutral axis distance as presented by Eq. 3.6.Setting f s = f/1.15, one can solve Eq. 3.6 for the maximum d' that ensuresyielding ofthe compression reinforcement.

0.67 feu b a1.5

___~-_--_.._.----r---,--

A'. s

b

As...........................................................,

. lOs> lOy/LIS

c

d

Fig. 3.3 Analysis of sections with compression reinforcement(steel does not vield) Table 3.1 Values of (d') to ensure yielding of compression steel

Substituting the expression off~ from Eq. 3.6 into Eq. 3.8 gives:. ,. (a-0.8d') Asf)' '(39)o4466j, b a + A 600 =-- .

'. cu .. s a 1.15

This can be reduced to a second order equation in terms of the stress blockdistance (a), given by

0.4466 feu b a' -(As f)'/1.l5-600x A; ) a-480A; d' =0 (3.10)

fiN/mmi) 240 280 360 400d'/d at c

The simplified approach for the analysis of a doubly reinforced section can besummarized in the following stepsGiven .:/cu,fy, b, d'; d, As and A ~Required iM;

Case A: check if d'/d< code limits (Table 3,1), then compression steel yields. Step 1 calculate "a" using Eq. 3.4.

Step 2 calculate M, using Eq. 3.7.

Case B: check if d'/d> code limits (Table 3.1), then compression steel does not yield Step 1 calculate "a" using Eq. 3.8 or Eq. 3.10.

Step 2 calculate Mu using Eq. 3.11.

3.1.3 Maximum Area of Steel for Doubly ReinforcedSections

To ensure ductile failure of a doubly reinforced section, the neutral axis distanceCmax is limited to that of the singly reinforced section as given in Table 4-1 in thecode or Table 2.1 given in Chapter 2. Thus, increasing the tension reinforcementabove A smax is allowed by the code only by adding compression reinforcementthat keeps the same neutral axis distance as shown in Fig. 3.4.1.A doubly reinforced section can be looked at as composed of a singly reinforcedconcrete section and a steel section. The singly reinforced section (Fig. 3.4.II)has an area of steel equal to As,max obtained from Table 2.1, and the steel sectionhas the same amount of top and bottom steel of area A ~ (Fig 3.4.III). Thus, themaximum area of steel for a doubly reinforced section Asd.max is given by

Asdmax =Asmax +A; (3.13)

Asd,max = A1 s,max (3.14a)-a

~J' = Jim., (for singlyrtl section(Table2.1) , (3.14b), .max I-a

IAsd,max = Jimax b d + A;I ~ (3.15)where J.!max is obtained from Table 2.1.The yielding of the compression reinforcement can be verified by comparing theratio d'/d and with the maximum allowed value given in Table 3.1. If thecompression steel does not yield, the maximum area of steel can be obtainedfrom:

A I = II bd+A' f.' (3.16)S( ,max rmax sf II 15

y

Maximum Moment CalculationReferring to Fig. 3.4, the maximum moment for doubly reinforced sectionsMud,max can be calculated using the following procedure

M udmax = M umax +M' (3.17)

M J. =M + A; fy (d -d') (3.18)uc .max u.max 1.15

b1- Doubly reinforced cross-section

T=Asd.max fyll.15_._ _.._. __ -'-~------I-

Es>Eyl1.l5

I 0.67 feu /1.5....................._ Q:'(.............._ r---,.. *?~ ?:??!.. ..

]' 1---.-----M' A's t N

.... .... .. ...i... .. .. .

111- Section with A'stop and bottom

(~s Mumax

A's.._ , _.. .. _ ..L..----!.__~

I' '1 E,>&/1.15 T2=A's fyl1.15b

As.max.....................,.................. . e .. --...................................... .. .L-__----l.__.

I_ 'I E,>E/ 1.15 Tt=As,lIlaxfylI.15b

11-Singly reinforced section with Asmax

d

d

Fig. 3.4 Maximum moment and area of steelfor doubly reinforced sections

[

f, = J;. /1.15J.l < J.lclmax

then .As < A sclmax

Mu < Mucl.ma.,

Photo 3.3 Peachtree Tower (1990), Atlanta USA (235m, 50 stories)

Example 3 .1 (compression steel yields)Find the moment capacity of the cross-section shown in Figure. Assume that d'

= 50 mm and the material properties are:feu = 25 Nzmrrr'fy = 400 N/mm2

oo\0

2 ald)min(O.I) ....okc/d = 217.6/550 = 0.396

Step 2: Check fsand f's

4

A's

Example 3.2Calculate the maximum area of steel and the maximum moment capacity that isallowed by the Egyptian Code (ECP 203) for the doubly reinforced section

sh~n in Example 3.1. The material properties are: feu = 25 N/mm2 and fy = 400N/mm2 .

~I- .j!,

Asd,max

Example 3.3 (compression steel does not yield)Find the moment capacity of the cross-section shown in figure.'feu = 30 Nzmrrr', and fy = 400 N/mm2 .

d'=100, 250 .1i :

500mm2

1900 mm'

d= 750 - 50 = 700 mm

Step 1: Compute a.d' 100 ..d = 700 = 0.143 > 0.10 (see table 3.1). .ccompression steel does not yield'We can use Eq 3.8 or Eq 3.10 to calculate a

0.67 leu b a + A' J: = A, I y.1.5 "1.15

Solution

0,4466 1cu ba' -(A, J;,/1.15-600xA; )a-480A; d'=O .- .'

Asf/1.1S

QR

As=1900

0.67 feu

2S0 1.511 '1 0.003 1--1

.!..~s:?_~t~-~~:~r~ ._~~.~~~~~~_,_._ ~~~~~s~:o 0V) 0t- t-

3350 a 2 - 360869 a - 24 X106 = 0

Solving for the only unknown "a"

a =154.2 mm-....~ c= 192.73 mm

~

IM ,= ASfJ,maxJ;'(d_amax)+ A;. k(amax. -d'). utl,max 1.15 2 . 1.15 2.

Mutl,max = [0.187 X25 X200x 5502 + 402x400 (550- 50)]1106 = 258.5 kN.m1.5 1.15

The same result can be obtained using Eq.3.20 as follows:amax= 0.8 cmax=0.8 x 0,42 x550 =184.8 mm

M " =1587X400(550_ 184.8) 402X400(184.8_50)~2585kN",I .max 1.15 ~ 2 + 1.15 2 ..m

,. 200 I

Solution

Step 1: .Calculate maximum area of steelFrom Table 2.1 and for fy =400 N/mm2 : ~max =4.31x 10-4 feu,since d'/d (0.09) < 0.176 (at c = cmx) , table 3.1, then compression steel has yieldedAsd,max = ~max b d +A'sAsd,m.x = (4.31 X 10-4 x 25) 200 x 550 + 402 = 1587 mrrr' > As(l520) o.k

Step 2: Calculate .max.imum moment capacityFrom Table 2-1: Rmax = 0.187, cmax/d = 0.42 for fy = 400 N/mm2

Using Eq.3 .19 to find Mud.max

R r b d? A' I,M = max J eu +_'_Y(d _ d')utl,ma, IS 1.15

Step 2: Check fs and f's.I' 192.73-100 2 400

J s= 600 288.6 N / mm < - compression steel does not yield192.73 1.15

Since c/d(0.275) < cJd = 0.63 then tension steel yields fs = f/1.15

Step 3: Compute Mu

Taking moment around concrete force Cc

M = AJy (d -::")-A' f'(d'-::") = AJy (d -::")+A'I' '(::"-d')n 1.15 2 s s 2 1.15 2 s . s 2

M = 1900x 400 (700- 154.2)+ 500x 288.6 (154.2 -100).= 408.3xlO = 408.3 kN.mII 1.15 2 . 2

!Final Resnlt: Mu = 408.3 kN.1IlI

00

3.1.4 Design of Doubly Reinforced Sections Using FirstPrinciples

The same procedure used in designing singly reinforced sections is used for thedesign of doubly reinforced sections. The unknowns in these types of problemsare the beam depth, area of steel, neutral axis position and the ratio of thecompression steel a. . .

Given .: feu,J;, u; h, d'Required : d, As and A'sUnknowns: a, d, As and A's

Since we have only two equilibrium equations,we have to limit the unknowns toonly two. If not given, the depth of the compression steel will be assumed 0.05-0.1 of the beam depth to ensure yielding of compressed bars for all steel grades.The design procedure can be summarized in the following steps:

1. Make the necessary assumptionsd' =0.05-0.10 d (compression steel yields for all fy)Assume As = f-lmax b d ---+(I..lmax for singly reinforced section (Table 4.1))Assume a=0.2-0A andEquilibrium of forces gives

0.67 lellb a (a x As) 1;. As I,-----"-''''--- + = --

1.5 1.15 1.15

Get a=J.d

Taking moment around the concreteforce gives

M= As 1;. (d _i!:-) + A; 1;. (!!--d')II 1.15 2 1.15 2

Solve the above equation to determine (a, d), then calculateAs = f-lmax b d

2. Check the minimum area of steel

3. Check the maximum area steel and the maximum moment by ensuring that

(c/d

f s = flUS

Since d' /d = 0.116

Example 3.5A reinforced concrete cross-section is subjected to 265 leN.m. Architecturalconsiderations require limiting the thickness of the section as much as possible.Economic considerations limited the value of a to 0.3. According to theseconstraints, design the cross section. Check the maximum area of steel and themaximum moments allowed by the code knowing that:b = 250 rom, feu = 30 Nzmm', and fy= 360 Nzrrmr'

Solution.Step 1: Assumptions

Given :!cu,J;" u; b, a (A'JRequired : d, d', AsUnknowns : a, d, d', AsWe have four unknowns, thus we shall assume two (d' and AS(Il))

1. Assume d' = 0.10 d2. Assume Il = Ilmax - (for singly reinforced section) .Il= 5x 10-4 feu =5 x l0-4x 30=0.015As = Jl b d =0.015x 250xd =3.75 dA; =a As = 0.3x3.75 d =1.125 d

Step 2: Apply the equilibrium equation T=C Eq.3.4Assume that compression and tension steel has yielded.

Step 3: Apply the second equilibrium equation, Eq. 3.7.Taking moment around concreteforce Cc

M= As 1;. (d _::..)+ A; f y (E.-d')II 1.15 2 1.15 2

265xl06 = 3;75 dx 360 (d- 0.2453d)+ 1.125 d X360(0.2453 d -0.1 d)1.15 2 1.15 2

. .

265 X 106=1037.9 d"

Step 5.1: Calculate maximum area of steelSince d'/d (0.1) < 0.21 (from Table 3.1) and even less than the code valueof 0.15, we can assume that compression steel yields.

. -4Asd.max = Jlmax b d + A ~ = 5x10 x 30 x 250 (550) + 568.45A.d,max = 2631 mm' > As(1895) .... o.k. I 250. I

. . A's=568 rnm2

= 470cm 2

= 556 em 2 < As ....... ,IJ.k

Example 3.6Design a doubly reinforced concrete cross-section to withstand an ultimate

. moment of 265 kN.m by assuming area of steel. Check the maximum area ofsteel and the maximum moments allowed by the code.b = 250 mm, feu = 30 Nzmm", and fy= 360 Nzmm'

SolutionStep 1: Assumptions

Given :!cu,J;', Mu, bRequired .: d, d', As, A ~Unknowns : a, d, d', As, A'sWe have five unknowns, thus we shall assume three (d' and As, A's(a))

1. Assume a=0.3, d'=50 mm2. use the approximate relation to assume As.

As =O.ll~Mu b =0.11 265 X106 x250 = 1492 mm'1;, 360

. 2Take As= 1500 mmA; =a As =0.3x1500=450mm 2

Step 2: Apply first equilibrium equation T =CAssume that compression and tension steel has yielded.0.67 x 30 x 250 x a 450x 360 1500 x 360------ + =---

1.5 1.15 1.15

a =98.12 mm andc =a/O.8= 122.65.mm

Step 3 : Apply second equilibrium equation !,M=OTaking moment around concrete force C,

Mu= As 1;. (d _!!-)+ A; I y (!!--d')

'US . 2 US 2

d = 614 mm. c/d = 122.65/614 = 0.2Since d'/d (50/614 = 0.081)< 0.15 (code limit for fy= 360 Nzmm"), thenour assumption that compression steel has yielded is correct.Take d = 650 mm and t = 700mm

Step 4: Calculate minimum area of steel

!0.225.JJ:: bd = 0.225.J30 x250x650=556

As min = smaller of' f, 360, ,l.3A s =L3xI500=1950 '

Step 5: Check maximum area of steel'Since c/d(0.2)

. As f/1.15As.. _._.

1'.>1'/1.15

Fig 3.5 Location of the neutral axis for doubly reinforced sections

_O._67_E.. = f.1 I)' (1-a)1.5 d 1.15I: .

where a.= A'JAs and l-t = As/(b . d)Define OJ = f.1 f)' (3.21)

feu0.67 x leu

1.5t--l

0.003A' T7_T A'!!!.:.!.5

s 0.-:.t c.-.-._._...._. ~._._._ ...... -'''''''-''''-' ._ .... -._.

M = As f)' (d _!!..) +- A; I)' (!!..-d')u 1.152 1.15 2

!!.. = 1.9468 cd{l- a) (3.22)d . .

Recalling the moment equation around the concrete compression force. .

0.67 feu ba A; 1;. As 1;.---"'"""--- +--=--

1.5 1.15 1.15Dividing by (feu b X d) and rearranging

3.1.5 Design of Doubly Reinforced Sections Using CurvesThe design of doubly reinforced sections from the first principles is complicated.Therefore, design curves were prepared to facilitate their design.In developing such curves, both compression steel and tension steel wereassumed to reach yield. In addition, two values for d' /d were used in developingthese curves and tables namely (0.05,0.1). These selected values were chosen tosatisfy