tribhuvan university -...

TRIBHUVAN UNIVERSITY

INSTITUTE OF ENGINEERING

PULCHOWK CAMPUS

DEPARTMENT OF CIVIL ENGINEERING

FINAL YEAR PROJECT REPORT

ON

EARTHQUAKE RESISTANT DESIGN OF

COMMERCIAL BUILDING

COURSE CODE (EG755CE)

PROJECT ADVISOR: PREPARED BY:

ER. DINESH GUPTA SAGAR KARKI CHHETRI(066BCE099)

SAMYOG SHRESTHA(066BCE104)

SANJEEMA BAJRACHARYA(066BCE107)

SANTOSH THAPA(066BCE111)

SUMAN RANA(066BCE133)

SUNIL MAHARJAN(066BCE135)

OCTOBER1 ,2013

ACKNOWLEDGEMENT

The trend of getting structures analyzed structurally and with seismic design for safety and economic

considerations is getting more and more popular nowadays, especially for commercial buildings that

house multiple facilities and immense flow of people. Keeping in view this need, the report

Earthquake resistant design of commercial building has been prepared as a part of B.E. final year

project as per the prescribed curriculum.

Thus, we would like to express our sincere gratitude to Institute Of Engineering, I.O.E., Pulchowk,

Lalitpur for providing us this opportunity to apply our engineering knowledge into actual practice. We

are extremely thankful to our project supervisor Er. Dinesh Gupta for providing us with careful

guidance with right logics, clearing up our confusions, providing incentives and regular examination

throughout the project. We also owe to Dr. Jishnu Subedi for his supervision on proper selection of

the building type and his guidance during the initial phase of the project.

We extend our heartfelt appreciation to our respected teacher Dr. Prof. Prem Nath Maskey, Er. Kamal

Thapa, DHOD, Institute of Engineering, Pulchowk, Lalitpur for their valuable suggestions.

We are also thankful to our seniors for their help regarding the general guidelines and sharing

valuable experiences of the project. Finally, we would like to acknowledge our gratitude towards each

other for such a united coordination amongst the group members during the project.

Project Members

Sagar Karki Chhetri (066BCE099)

Samyog Shrestha (066BCE104)

Sanjeema Bajracharya (066BCE107)

Santosh Thapa (066BCE111)

Suman Rana (066BCE133)

Sunil Maharjan (066BCE135)

TRIBHUVAN UNIVERSITY

INSTITUTE OF ENGINEERING

PULCHOWK CAMPUS

DEPARTMENT OF CIVIL ENGINEERING

The undersigned recommend to the Institute Of Engineering for acceptance, a project entitled,

“Earthquake Resistant Design of Commercial Building” submitted by Sagar Karki Chhetri,

Samyog Shrestha, Sanjeema Bajracharya, Santosh Thapa, Suman Rana and Sunil Maharjan in partial

fulfillment of requirement for the Bachelor Degree in Civil Engineering.

---------------------------- ---------------------------------- ------------------------------------

Dr.G.B.Motra Dr. Kamal Thapa Er. Dinesh Gupta

(External) (Internal) (Project Supervisor)

Lecturer, DHOD, I.O.E., I .O.E.,

I.O.E., Pulchowk Department of Civil Engineering Department of Civil Engineering

1st October, 2013

PREFACE

With the increasing number of commercial complex, in addition to the buildings being

designed using engineering knowledge, nowadays, seismic design is also gaining high

popularity. This report presents the in-depth seismic design of a commercial complex located

at Sundhara. The verification of adequacy of design has been done through computer analysis

with additional manual check on correctness of different design parameters.

This report includes the basic design concept of earthquake resistant design taking into account

various design considerations with respect to standards and finally the systematic design of

various members. Building analysis has been done using SAP2000 and calculations in excel.

The results of the calculations have been presented in tabular form and sample calculations

have been provided in detail. Various figures and sketches have been introduced to illustrate

the theory. References to the appropriate clauses of the standard codes of practices have been

made wherever necessary.

Every effort has been made to minimize the errors in the report. However, any error incurred

brought to the notice of project implementation team will be gladly mitigated.

Sagar Karki Chhetri

Samyog Shrestha

Sanjeema Bajracharya

Santosh Thapa

Suman Rana

Sunil Maharjan

Notations

Diameter of Bar

τc Shear Stress

γm Partial Safety Factor

Ab Area of Each Bar

Ag Gross Area of Concrete

Ah Horizontal Seismic Coefficient

Asc Area of Steel in Compression

Ast Area of Steel

Asv Area of Stirrups

bf Width of Flange

bw Width of Web

B Width

d Effective Depth

d′ Effective Cover

D Overall Depth

Df Depth of Flange

e Structure Eccentricity

E Young’s Modulus of Rigidity

Es Modulus of Elasticity of Steel

fck Characteristics Strength of Concrete

fy Characteristics Strength of Steel

fs Steel Stress of Service Load

h Height of building

I Importance Factor (For Base Shear Calculation)

I Moment of Inertia

Ip Polar Moment of Stiffness

k Lateral Stiffness

L Length of Member

Ld Development Length

M Bending Moment

Pc Percentage of Compression Reinforcement

Pt Percentage of Tension Reinforcement

Q Design Lateral Force

R Response Reduction Factor

Sa/g Average Response Acceleration Coefficient

Sv Spacing of Each Bar

T Torsional Moment due to Lateral Force

Ta Fundamental Natural Period of Vibrations

V′ Additional Shear

VB Design Seismic Base Shear

W Seismic Weight of Floor

Xu Actual Depth of Neutral Axis

Xul Ultimate Depth of Neutral Axis

Z Zone Factor

Abbreviations

CM Center of Mass

CR Center of Rigidity

D.L Dead Load

E.Q Earthquake Load

IS Indian Standard

L.L Live Load

RCC Reinforced Cement Concrete

SP Special Publication

Table of Content

Chapters Page no.

Chapter 1 INTRODUCTION

1.1 Background 1

1.2 Theme of project work 2

1.3 Objectives and Scope 2

1.4 Building description 3

1.5 Identification of loads 4

1.6 Method of analysis 4

1.7 Design 4

1.8 Detailing 5

1.9 Distribution of chapters 5

Architectural plan

SAP2000 layout

Chapter 2 STRUCTURAL SYSTEM AND LOADING

2.1 Structural Arrangement Plan 6

2.2 Vertical Load Calculation 6

2.3 Preliminary Design 8

2.4 Seismic Load 9

2.5 Load Combination 18

Chapter 3 STRUCTURAL ANALYSIS

3.1 Salient feature of SAP2000 20

3.2 Inputs and Outputs 21

3.3 Model verification 23

Chapter 4 SECTION DESIGN

4.1 Limit state Method 27

4.2 Design of structural elements 29

4.2.1 Design of slab 30

4.2.2 Design of Beam 65

4.2.3 Design of Column 123

4.2.4 Design of Staircase 129

4.2.5 Design of Basement Wall 136

4.2.6 Design of Lift Wall 142

4.2.7 Design of Mat Foundation 149

Chapter 5 DRAWINGS Sheet no.

Drawing of slab 1-5

Drawing of Beam 6-7

Drawing of Column 8-9

Drawing of Staircase 10-12

Drawing of Basement Wall 13

Drawing of Lift Wall 14

Drawing of Mat Foundation 15-16

ANNEX AND OUTPUT

Annex-I : Roof truss calculations

Annex-II : Elevator

BIBLIOGRAPHY

1

1.1 Background

Today, Kathmandu is a rapidly urbanizing city with building construction at just

about every corner of the city that one can see. Nowadays, with the awareness level of the

building owners increasing than in the past, the trend of having a building analyzed

scientifically before it is actually constructed is growing popular, especially in case of

medium to large commercial buildings, which is a good thing because such a practice helps

construction of more safer buildings which can eventually lead to avoidance of loss of lives

and property in case of a structural failure.

A designer has to deal with various structures ranging from simple ones like curtain

rods and electric poles to more complex ones like multistoried frame buildings, shell roofs

bridges etc. these structure are subjected to various load like concentrated loads uniformly

distributed loads, uniformly varying loads live loads, earthquake loads and dynamic forces.

The structure transfers the loads acting on it to the supports and ultimately to the ground.

While transferring the loads acting on the structure, the members of the structure are

subjected to the internal forces like axial forces, shearing forces, bending and torsional

moments.

Structural Analysis deals with analyzing these internal forces in the members of the

structures. Structural Design deals with sizing various members of the structures to resist the

internal forces to which they are subjected during their effective life span. Unless the proper

Structural Detailing method is adopted the structural design will be no more effective. The

Indian Standard Code of Practice should be thoroughly adopted for proper analysis, design

and detailing with respect to safety, economy, stability and strength.

The projected selected by our group is a medium commercial building located at

Sundhara, Kathamandu. According to IS 1893:2002, Kathmandu lies on Zone V, the

severest one. Hence the effect of earthquake is pre-dominant than the wind load. So, the

building is analyzed for Earthquake as lateral Load. The seismic coefficient design method

as stipulated in IS 1893:2002 is applied to analyze the building for earthquake. Special

reinforced concrete moment resisting frame is considered as the main structural system of

the building.

The project report has been prepared in complete conformity with various

stipulations in Indian Standards, Code of Practice for Plain and Reinforced Concrete IS 456-

Chapter 1

INTRODUCTION

2

2000, Design Aids for Reinforced Concrete to IS 456-2000(SP-16), Criteria Earthquake

Resistant Design Structures IS 1893-2000, Ductile Detailing of Reinforced Concrete

Structures Subjected to Seismic Forces- Code of Practice IS 13920-1993, Handbook on

Concrete Reinforcement and Detailing SP-34. Use of these codes have emphasized on

providing sufficient safety, economy, strength and ductility besides satisfactory

serviceability requirements of cracking and deflection in concrete structures. These codes are

based on principles of Limit State of Design.

This project work has been undertaken as a partial requirement for B.E. degree in

Civil Engineering. This project work contains structural analysis, design and detailing of a

commercial public building located in Kathmandu District. All the theoretical knowledge on

analysis and design acquired on the course work are utilized with practical application. The

main objective of the project is to acquaint in the practical aspects of Civil Engineering. We,

being the budding engineers of tomorrow, are interested in such analysis and design of

structures which will, we hope, help us in similar jobs that we might have in our hands in the

future.

1.2 Theme of Project work

This group under the project work has undertaken the structural analysis and design

of multi-storied commercial building. The main aim of the project work under the title is to

acquire knowledge and skill with an emphasis of practical application. Besides the utilization

of analytical methods and design approaches, exposure and application of various available

codes of practices is another aim of the work.

1.3 Objectives and Scopes

The specific objectives of the project work are

i. Identification of structural arrangement of plan.

ii. Modeling of the building for structural analysis.

iii. Detail structural analysis using structural analysis program.

iv. Sectional design of structural components.

v. Structural detailing of members and the system.

To achieve above objectives, the following scope or work is planned

i. Identification of the building and the requirement of the space.

3

ii. Determination of the structural system of the building to undertake the vertical and

horizontal loads.

iii. Estimation of loads including those due to earthquake

iv. Preliminary design for geometry of structural elements.

v. Determination of fundamental time period by free vibration analysis.

vi. Calculation of base shear and vertical distribution of equivalent earthquake load.

vii. Identification of load cases and load combination cases.

viii. Finite element modeling of the building and input analysis

ix. The structural analysis pf the building by SAP2000 for different cases of loads.

x. Review of analysis outputs for design of individual components

xi. Design of RC frame members, walls, mat foundation, staircase, and other by limit

state method of design

xii. Detailing of individual members and preparation of drawings as a part of working

construction document.

1.4 Building Description

Building Type : Medium Commercial Building, Located in Kathmandu

Structural System : RCC Space Frame

Plinth area covered : 1600 m2

Type of Foundation : Mat Foundation

No. of Storey : 6

Floor Height : 3.2m all floors

Type of Sub-Soil : Soft Soil (Type III)

According to IS 456-2000, Clause 27, structures in which changes in plan

dimensions take place abruptly shall be provided with expansion joints at the section where

such changes occur. Reinforcement shall not extend across an expansion joints and the break

between the sections shall be completed. Normally structure exceeding 45m in length is

designed with one or more expansion joints.

The design is intended to serve for the following facilities in the building:-

Basement for Parking ,

Other floors for rent for various commercial purposes like office,

departmental stores, banking, shop stall, etc.

4

1.5 Identification of loads

Dead loads are calculated as per IS 875 (Part 1) -1987

Seismic load according to IS 1893 (Part 1)-2002 considering Kathmandu

located at Zone V

Imposed loads according to IS 875(Part 2)-1987 has been taken as

For all the floor 4 KN/m2

For Staircase 5 KN/m2

For machine level live load has been taken 1.5 time 4 KN/m2 to

consider impact load due to elevator

1.6 Method of Analysis

The building is modeled as a space frame. SAP2000 is adopted as the basic tool for

the execution of analysis. SAP2000 program is based on Finite Element Method. Due to

possible actions in the building, the stresses, displacements and fundamental time periods are

obtained using SAP2000 which are used for the design of the members. Lift wall, mat

foundation, staircase, slabs are analyzed separately.

1.7 Design

The following materials are adopted for the design of the elements:

Concrete Grade: M20 and M25

M25 for the all columns

M20 for remaining structural elements

Reinforcement Steel – Fe415

Fe415 for the Slabs & Basement Wall

Fe415 HYSD for all the remaining elements

Fe415 for stirrups

Limit state method is used for the design of RC elements. The design is based on IS:

456-2000, SP-16, IS: 1893-2002, SP-34 and ‘Reinforced concrete design’ – Pillai and

Mennon and ‘Reinforced Concrete: Limit state design’- A.K. Jain. Were also used.

5

1.8 Detailing

The space frame is considered as a special moment resisting frame(SMRF) with a

special detailing to provide ductile behavior and comply with the requirements given in IS

13920-1993 and Hand book on Concrete Reinforcement and Detailing (SP-34).

1.9 Distribution of Chapter

This project has been broadly categorized into five chapters, Summary of each

chapter is mentioned below:

Chapter 1 : Introduction

Chapter 2 : Structural system and loading

In this chapter, briefing upon the structural arrangements is done with

necessary computations that are performed for the vertical load calculation,

preliminary design of the structure elements, seismic load calculation and the

different load combinations that are used.

Chapter 3 : Structural Analysis

This chapter deals with SAP2000 that is followed by the analysis of the

different structural members. This includes the inputs given and outputs

obtained in the process, the time period calculation and storey drift of the

building.

Chapter 4 : Structural Design

The design for the structural members is done by taking into account the

Limit State Method. This chapter includes the design procedures that are

followed for the design of slabs, beams, columns, staircases, basement wall,

lift wall and mat foundation.

Chapter 5 : Structural Detailing and Drawings

The various structural detailing and drawings of the different members as

obtained from their respective design are listed in this chapter.

ARCHITECTURAL PLAN

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SAP2000 LAYOUT

3D view

@ Z = 3.2m

@ Z = 6.4m

@ Z = 9.6m

@ Z = 12.8m

@ Z = 16m

@ Z = 19.2m

@ Z = 22.4m

6

2.1 Structural Arrangement Plan

The planning of the building has been done as per available land area, shape, space

according to building bylaws and requirement of commercial public building. The

positioning of columns, staircases, toilets, bathrooms, elevators etc are appropriately done

and accordingly beam arrangements is carried out so that the whole building will be

aesthetically, functionally and economically feasible.

The aim of design is the achievements of an acceptable probability that structures being

design will perform satisfactorily during their intended life. With an appropriate degree of

safety, they should sustain all the loads and deformations of normal construction and use and

have adequate durability and adequate resistance to the effect of misuse and fire.

The building consist of three blocks, namely First, Second and Third Block, separated by a

construction joint.

2.2 Vertical Load Calculation

a) Slab

Dead Load

Self-Weight of the slab= 150 mm x 25 KN/m3 = 3.75 KN/m2

Finishes = 50 mm x 20 KN/m3 = 1 KN/m2

Total = 4.75 KN/m2

Imposed Load

For All Floor = 4 KN/m2

Roof = 2 KN/m2

b) Wall

Masonry Wall

Thickness = 230 mm

Solid wall weight = 230 mm x 20 KN/m2 = 4.6 KN/m2

Solid wall weight after deducting opening 75 % of 4.6 = 3.45 KN/m2

Parapet Wall = 1.5 x 0.115 x 20 = 3.45 KN/m2

Chapter 2

STRUCTURAL SYSTEM AND

LOADING

7

Basement Wall

Thickness = 200 mm

Solid wall weight = 200 mm x 25 KN/m2 = 5.0 KN/m2

c) Column

Square = 0.55 x 0.55x 25 = 7.56 KN/m

Square = 0.65 x 0.65 x 25 = 10.56 KN/m

d) Beam

Straight Beam = 0.7 x 0.5 x 25 = 8.75 KN/m

Secondary Beam = 0.45 x 0.25 x 25 = 2.81 KN/m

Cinema hall Beam = 0.48 x 0.45 x 25 = 5.4 KN/m

e) Staircase

Dog Legged

Total thickness = 250 mm

Riser = 150 mm

Tread = 300 mm

Wt. of waist slab = 0.25 x 25 = 6.250 KN/m2

Wt. of each step = 0.50 x 0.15 x 0.3 x 25 = 0.56 KN/m

Wt. of landing = 0.25 x 25 = 6.250 KN/m2

Wt. of finishing = 1 KN/m

Imposed load = 5 KN/m2

Open Well

Total thickness = 220 mm

Riser = 180 mm

Tread = 250 mm

Wt. of waist slab = 0.22 x 25 = 5.50 KN/m2

Wt. of each step = 0.50 x 0.18 x 0.25 x 25 = 0.56 KN/m

Wt. of landing = 0.22 x 25 = 5.50 KN/m2

Wt. of finishing = 1 KN/m

Imposed load = 5 KN/m2

8

f) Lift

Thickness = 200 mm

Length = 10 m

Weight = 0.2 x 10 x 25 = 50 KN/m

2.3 Preliminary Design

Preliminary design is carried out to estimate approximate size of the structural

members before analysis of structure. Grid diagram is the basic factor for analysis in both

Approximate and Exact method and is presented below.

For column, M25 concrete; for rest M20 concrete. Fe415 steel for all

For Slab

Effective depth = span/αβγ & p=2.5% α = 26; β = 1: γ = 1.45

Slab (mm*mm) Ly/Lx Effective depth (mm)

1. (2500*6000) 2.4 66.31

2. (7000*3000) 2.33 79.57

3. (6000*5000) 1.2 132.62

Take overall depth of slab =150 mm

For beam

Main beam

D=span/15

Length (mm) D (mm)

7000 466.66

6000 400

5000 333

2500 166.67

Take Beam size =700mm*500mm

Secondary Beam

Say, D = 450 mm

Width of Beam B = 250 mm

For Column

For each floor level above column C-49,

9

Top slab area = (3+3)*(2.5+3.5) = 36 m2

Dead load of slab = 36*0.15*25 = 135 KN

Live load on slab = 36*4 =144 KN

Beam load = (3+3+3.5+2.5)*0.7*0.5*25 = 105 KN

Total load = 384 KN

As per code IS 875(PART 2) APPENDIX A (clause 3.2.1.2)

Total load on column C-49 = 384*7*(1-0.4)*1.5= 2419.2 KN

Assume percentage of steel = 2 %

As per IS 456 clause 39.3

Pu=0.4 fck AC+ 0.67 fy ASC

Or, 2419200= (0.4*25*(1-0.02) + 0.67*415*0.02) Ag

Ag = 157489.74 mm2

Assume square column section =√157489.74 = 396.849 mm

Take column of size 550mm*550mm

For Staircase

Doglegged Staircase

Depth of waist slab mmx

Spand 250

2.120

6000

20

Say Overall Depth = 250 mm

Similarly for open well Staircase

Overall Depth = 220 mm

2.4 Seismic Load

Seismic weight is the total dead load plus appropriate amount of specified imposed

load. While computing the seismic load weight of each floor, the weight of columns and

walls in any story shall be equally distributed to the floors above and below the storey. The

seismic weight of the whole building is the sum of the seismic weights of all the floors. It

has been calculated according to IS: 1893(Part I) – 2002.

IS: 1893(Part I) – 2002 states that for the calculation of the design seismic forces of

the structure the imposed load on roof need not be considered

The seismic weights and the base shear have been computed as below:

10

Seismic Weight Calculation:

Description

Specific

Weight

Length

Breadth

Thickness

or Height

KN/m3 m m m

Beam 25.00 7,6,5,2.5,1.5 0.5 0.7

Wall 20.00 6 0.23 3.2

Retaining Wall 25.00 6 0.2 3.2

Column 1 25.00 0.550 0.550 3.2

Column 2 25.00 0.650 0.650 3.2

Cantilever Beam 25.00 1.5 0.5 0.7

Parapet Wall 20.00 6,2.5.1.5 0.115 1.2

Partition wall + Floor Finish 1 KN/m2

Floor Finish For Staircase 1 KN/m2

Live Load Slab 4 KN/m2

Live Load Staircase 5 KN/m2

Sample calculation of seismic weight

Slab

Total slab area= 612 m2

Slab thickness= 0.15 m

Slab load= 612*0.15*25+612*1= 2907 KN

Column

Total no. of Columns = 28

Column load = 28*(0.55*0.55*3.2/2*25) = 338.8 KN

Beam

Total span of primary beam = 274m

Total span of secondary beam = 42m

Beam load = 274*0.5*0.7*25+42*0.45*0.25*25 = 2515.625 KN

Live load = 0

Total load = 5761.425 KN

11

Floor Type of load Total load(KN)

6th slab load 2907

5761.425

beam load 2515.625

column load 338.8

live load 0

5th slab load 4631.25

12251.975

beam load 4440.625

column load 980.1

live load 1950

steel roof 250

4th slab load 5510.133

13975.51963

beam load 4862.731

column load 1282.6

live load 2320.056

3rd slab load 7619.133

17946.76963

beam load 5788.581

column load 1331

live load 3208.056

2nd slab load 6797.25

16651.775

beam load 5613.125

column load 1379.4

live load 2862

1st slab load 6857.813

16764.0875

beam load 5639.375

column load 1379.4

live load 2887.5

12

Base Shear Calculation

According to IS 1893 (Part I): 2002 Cl. No. 6.4.2 the design horizontal seismic

coefficient Ah for a structure shall be determined by the following expression:

gR2

SIZA a

h

Where,

Z = Zone factor given by IS 1893 (Part I): 2002 Table 2, Here for Zone V, Z = 0.36

I = Importance Factor, I = 1 for commercial building

R = Response reduction factor given by IS 1893 (Part I): 2002 Table 7, R = 5.0

Sa/g = Average response acceleration coefficient which depends on Fundamental

natural period of vibration (Ta).

According to IS 1893 (Part I): 2002 Cl. No. 7.4.2

75.0075.0 hTa

Where,

h = height of building in m, h = 19.2 m

sec688.02.19*075.0 75.0 aT

For Ta= 0.688 and soil type III (Soft Soil) Sa/g = 2.428

Now,

131.052

5.2136.0

x

xxAh

According to IS 1893 (Part I) : 2002 Cl. No. 7.5.3 the total design lateral force or

design seismic base shear (VB) along any principle direction is given by

VB = Ah x W

Where, W = Seismic weight of the building

According to IS 1893 (Part I): 2002 Cl. No. 7.7.1 the design base shear (VB)

computed above shall be distributed along the height of the building as per the following

expression:

13

2

jj

n

1j

2

iiBi

hW

hWVQ

Where,

Qi = Design lateral force at floor i

Wi = Seismic weight of floor i

hi = Height of floor I measured from base

n = No. of storeys in the building

VB = 0.131*83351.6 =10926.94 KN

Storey Shear

FLOOR Wi hi (m) Wi hi2 Qi=VB Wi Hi

2/∑Wi hi2 Vi( KN )

6th 5761.425 19.2 2123892 2307.354424 2307.3544

5th 12251.98 16 3136506 3407.43835 5714.7928

4th 13975.52 12.8 2289749 2487.538685 8202.3315

3rd 17946.77 9.6 1653974 1796.845324 9999.1768

2nd 16651.78 6.4 682056.7 740.9730657 10740.15

1st 16764.09 3.2 171664.3 186.4926908 10926.643

10057842

Additional Shear Calculation Due to Torsion in Building

Center of Rigidity (CR) - A point through which a horizontal force is applied resulting in

translation of the floor without any rotation

Center of Mass (CM) - Center of gravity of all the floor masses.

Structural eccentricity (e)

e = CMCR

The eccentricity in building is calculated by

beeda

beedb

Where,

eda& edb = static eccentricity at floor a & b define as the distance between

center of mass and center of rigidity.

14

b = maximum dimension of the building perpendicular to the direction of

earthquake under consideration

and Dynamic magnification factors

Accidental eccentricity factor

From IS 1893 – 2002

1and05.0,5.1

Calculation by Simplified Analysis

The location of the center of rigidity is determined by

y

y

rk

xkx And

x

x

rk

yky

3

12L

EIk x And

3yL

EI12k

Where kx and ky are lateral stiffness of a particular element along the x and y axes.

E= Young’s Modulus of rigidity

I= Moment of Inertia

L= Length of the Member

The total torsional stiffness of a storey Ip about the center of rigidity is given by

)( 22 xkykI yxp

Where,

x , y = coordinates of the centroid of a particular element in plan from

the center of rigidity.

Ip = polar moment of stiffness

The additional shear on any frame on column line to a horizontal torsional moment T is

given by

xx

p

x'

x kI

yTV

yy

p

y'

y kI

xTV

15

Where,

'

xV Additional shear on any frame or column line in the x-direction due to torsional

moment

Vx = initial storey shear in x-direction due to lateral forces

Tx = yxeV , torsional moment due to lateral force in x-direction only

Kxx = total stiffness of the column line under consideration in the x-direction.

The subscript y represents y-direction

Center of stiffness calculation

Moment of inertia(I)

12

550*550 3

0.007625521 m4

modulus of elasticity(E) 255000 2.5E+11 N/m2

effective length(L) 0.65*3.2 2.08 m

stiffness of each column 3

12L

EI

2542147595 N/m

FIRST FLOOR

along y direction along x direction

no of

column

K Xn KXn no of

column

K Yn Kyn

8 2542147595 0 0 9 2.54E+09 0 0

8 2542147595 6 1.22023E+11 9 2.54E+09 7 1.6E+11

8 2542147595 12 2.44046E+11 9 2.54E+09 12 2.75E+11

7 2542147595 18 3.20311E+11 6 2.54E+09 17 2.59E+11

7 2542147595 24 4.27081E+11 8 2.54E+09 22 4.47E+11

8 2542147595 30 6.10115E+11 8 2.54E+09 27 5.49E+11

8 2542147595 36 7.32139E+11 8 2.54E+09 34 6.91E+11

7 2542147595 42 7.47391E+11 7 2.54E+09 36.5 6.5E+11

3 2542147595 44.5 3.39377E+11

TOTAL 3.54248E+12 TOTAL 3.03E+12

16

SECOND FLOOR


no of

column

K Xn KXn no of

column

K Yn Kyn

8 2542147595 0 0 9 2.54E+09 0 0

8 2542147595 6 1.22023E+11 9 2.54E+09 7 1.6E+11

8 2542147595 12 2.44046E+11 9 2.54E+09 12 2.75E+11

7 2542147595 18 3.20311E+11 6 2.54E+09 17 2.59E+11

7 2542147595 24 4.27081E+11 8 2.54E+09 22 4.47E+11

8 2542147595 30 6.10115E+11 8 2.54E+09 27 5.49E+11

8 2542147595 36 7.32139E+11 8 2.54E+09 34 6.91E+11

7 2542147595 42 7.47391E+11 7 2.54E+09 36.5 6.5E+11

3 2542147595 44.5 3.39377E+11

TOTAL 3.54248E+12 TOTAL 3.03E+12

THIRD FLOOR


no of

column

K Xn KXn no of

column

K Yn Kyn

3 254214759.5 44.5 3.3938E+10 9 2.54E+08 1.5 3431899253

7 254214759.5 42 7.4739E+10 9 2.54E+08 8.5 19447429103

7 254214759.5 36 6.4062E+10 9 2.54E+08 13.5 30887093281

7 254214759.5 30 5.3385E+10 6 2.54E+08 18.5 28217838306

6 254214759.5 24 3.6607E+10 8 2.54E+08 23.5 47792374789

6 254214759.5 18 2.7455E+10 8 2.54E+08 28.5 57960965170

7 254214759.5 12 2.1354E+10 8 2.54E+08 35.5 72196991703

7 254214759.5 6 1.0677E+10

7 254214759.5 0 0

Total 3.22217E+11 Total 2.59935E+11

17

FOURTH FLOOR


no of

column

K Xn KXn no of

column

K Yn Kyn

3 254214759.5 44.5 3.3938E+10 8 2.54E+08 35.5 72196991703

7 254214759.5 42 7.4739E+10 7 2.54E+08 28.5 50715844524

7 254214759.5 36 6.4062E+10 7 2.54E+08 23.5 41818327941

7 254214759.5 30 5.3385E+10 6 2.54E+08 18.5 28217838306

6 254214759.5 24 3.6607E+10 8 2.54E+08 13.5 27455194028

6 254214759.5 18 2.7455E+10 8 2.54E+08 8.5 17286603647

7 254214759.5 12 2.1354E+10 9 2.54E+08 1.5 3431899253

3 254214759.5 6 4575865671

7 254214759.5 0 0

TOTAL 3.1612E+11 TOTAL 2.41123E+11

FIFTH FLOOR


no of

column

K Xn KXn no of

column

K Yn Kyn

7 254214759.5 0 0 8 2.54E+08 34 6.91E+10

3 254214759.5 6 4575865671 7 2.54E+08 27 4.8E+10

7 254214759.5 12 21354039799 7 2.54E+08 22 3.91E+10

6 254214759.5 18 27455194028 6 2.54E+08 17 2.59E+10

6 254214759.5 24 36606925371 8 2.54E+08 12 2.44E+10

7 254214759.5 30 53385099499 8 2.54E+08 7 1.42E+10

7 254214759.5 36 64062119398 9 2.54E+08 0 0

7 254214759.5 42 74739139298 53

3 254214759.5 44.5 33937670396

TOTAL 3.16116E+11 TOTAL 2.21E+11

18

SIXTH FLOOR


no of

column

K Xn KXn no of

column

K Yn Kyn

3 254214759.5 42 32031059699 5 2.54E+08 34 4.32E+10

3 254214759.5 36 27455194028 5 2.54E+08 27 3.43E+10

7 254214759.5 30 53385099499 6 2.54E+08 22 3.36E+10

7 254214759.5 24 42708079599 4 2.54E+08 17 1.73E+10

7 254214759.5 18 32031059699 4 2.54E+08 12 1.22E+10

3 254214759.5 12 9151731343 3 2.54E+08 7 5.34E+09

3 2.54E+08 0 0

TOTAL 1.96762E+11 30 1.46E+11

Floor Center of mass Center of stiffness

X Y X Y

Top 26.036 19.589 25.800 19.133

Fifth 26.299 16.396 23.462 16.396

Fourth 23.824 18.282 23.462 17.896

Third 20.811 18.310 22.237 17.939

Second 21.268 17.405 21.773 18.633

First 21.391 17.293 21.773 18.633

Storey Design eccentricity(m) Storey

force(KN)

Torsional moment(KN-m)

ex ey Mty Mtx

6 -2.579 2.054 2307.354 -5950.458 4739.097

5 -6.480 2.408 3407.438 -22079.798 8203.668

4 -2.767 2.279 2487.539 -6882.921 5669.297

3 4.364 2.257 1796.845 7842.176 4056.074

2 2.984 -3.541 740.973 2210.983 -2623.851

1 2.799 -3.710 186.493 522.045 -691.814

19

2.5 Load Combination:

Different load cases and load combination cases are considered to obtain most

critical element stresses in the structure in the course of analysis.

There are together four load cases considered for the structural analysis and are

mentioned as below:

i.) Dead Load (D.L.)

ii.) Live Load (L.L)

iii.) Earthquake load in X-direction (E.Qx)

iv.) Earthquake load in Y-direction (E.Qy)

Following Load Combination are adopted as per IS 1893 (Part I): 2002 Cl. 6.3.1.2

i.) 1.5 (D.L + L.L)

ii.) 1.5 (D.L + E.Qx)

iii.) 1.5 (D.L - E.Qx)

iv.) 1.5 (D.L + E.Qy)

v.) 1.5 (D.L - E.Qy)

vi.) 1.2 (D.L + L.L + E.Qx)

vii.) 1.2 (D.L + L.L - E.Qx)

viii.) 1.2 (D.L + L.L + E.Qy)

ix.) 1.2 (D.L + L.L - E.Qy)

x.) 0.9 D.L + 1.5 E.Qx

xi.) 0.9 D.L -1.5 E.Qx

xii.) 0.9 D.L + 1.5 E.Qy

xiii.) 0.9 D.L -1.5 E.Qy

After checking the results, it was found that the stresses developed are most critical

for the following load combinations:

i.) 1.5 (D.L + L.L)

ii.) 1.2 (D.L + L.L + E.Qx)

iii.) 1.2 (D.L + L.L - E.Qx)

iv.) 1.2 (D.L + L.L + E.Qy)

v.) 1.2 (D.L + L.L - E.Qy)

The characteristic loads considered in the design of foundation are:

i.) Dead Load plus Live Load

ii.) Dead Load plus Earthquake Load

iii.) Dead Load minus Earthquake Load

To find the stress at the various points of the foundation, depth of footing and

reinforcements most critical factored loads are taken into account

20

3.1 Salient feature of SAP2000

SAP2000 represents the most sophisticated and user-friendly release of SAP series

of computer programs. Creation and modification of the model, execution of the analysis,

and checking and optimization of the design are all done through this single interface.

Graphical displays of the results, including real-time display of time-history displacements

are easily produced.

The finite element library consists of different elements out of which the three

dimensional FRAME element was used in this analysis. The Frame element uses a general,

three-dimensional, beam-column formulation which includes the effects of biaxial bending,

torsion, axial deformation, and biaxial shear deformations.

Structures that can be modeled with this element include:

• Three-dimensional frames

• Three-dimensional trusses

• Planar frames

• Planar grillages

• Planar trusses

A Frame element is modeled as a straight line connecting two joints. Each element

has its own local coordinate system for defining section properties and loads, and for

interpreting output.

Each Frame element may be loaded by self-weight, multiple concentrated loads, and

multiple distributed loads. End offsets are available to account for the finite size of beam and

column intersections. End releases are also available to model different fixity conditions at

the ends of the element. Element internal forces are produced at the ends of each element

and at a user-specified number of equally-spaced output stations along the length of the

element.

Loading options allow for gravity, thermal and pre-stress conditions in addition to

the usual nodal loading with specified forces and or displacements. Dynamic loading can be

in the form of a base acceleration response spectrum, or varying loads and base

accelerations.

Chapter 3

STRUCTURAL ANALYSIS

21

3.2 Inputs and Outputs

The design of earthquake resistant structure should aim at providing appropriate

dynamic and structural characteristics so that acceptable response level results under the

design earthquake. The aim of design is the achievement of an acceptable probability that

structures being designed will perform satisfactorily during their intended life. With an

appropriate degree of safety, they should sustain all the loads and deformations of normal

construction and use and have adequate durability and adequate resistance to the effects of

misuse and fire.

For the purpose of seismic analysis of our building we used the structural analysis

program SAP2000. SAP2000 has a special option for modeling horizontal rigid floor

diaphragm system.

A floor diaphragm is modeled as a rigid horizontal plane parallel to global X-Y

plane, so that all points on any floor diaphragm cannot displace relative to each other in X-Y

plane.

This type of modeling is very useful in the lateral dynamic analysis of building. The

base shear and earthquake lateral force are calculated as per code IS 1893(part1)2002 and

are applied at each master joint located on every storey of the building.

General description and assumptions for modelling

1. The building consists of a basement, a ground floor, first floor through sixth floor.

2. All the columns directly rest over the mat foundation and they are modelled as columns

with fixed support at their base.

3. The secondary beams are also included in the SAP model whereas the staircases are not

included, only their loads are distributed to appropriate beams.

4. The main beams rest centrally on columns to avoid local eccentricity.

5. The floor diaphragm is assumed to be rigid.

6. Centre line dimensions are followed for analysis and design.

7. For analysis purpose, the beams are assumed to be rectangular so as to distribute slightly

larger moment in columns.

8. Seismic loads will be considered acting in the horizontal direction (along either of the two

principal direcions) and not along the vertical direction.

9. The general layout of the building along with beam, column and slab numbering is

attached in the Annex.

22

10. The columns, beams and slabs are numbered from left to right and then from lower to

upper part of the plan. For instance, the first column of ground floor is CG-1, the first beam

of first floor is B1-1 and similarly the third slab of the fourth floor is S4-3.

11. For design purpose, the beams are grouped on the basis of their span; from the envelope

in SAP2000, the maximum moments (hogging and sagging) at three stations- left, middle

and right for each span are taken and designed.

12. For design purpose, the columns are grouped on the basis of their longitudinal rebar

percentage as obtained from SAP2000. The most critical column in compression (the one

with maximum longitudinal rebar) is designed and the same column is designed for shear.

13. The vertical forces and the moments in two directions on all columns (for the required

load comination) are extracted from the analysis result of SAP2000.

Sample output for Beam

TABLE: Element Forces - Frames

Frame Station Output Case

Case Type

P V2 V3 T M2 M3

Text m Text Text KN KN KN KN-m KN-m KN-m B1-1 0 Dead LinStatic 0 -1.32 8.882E-16 0.1815 0 -0.689

B1-1 0.75 Dead LinStatic 0 11.39 8.882E-16 0.1815 -7E-16 -4.466

B1-1 1.5 Dead LinStatic 0 24.1 8.882E-16 0.1815 -1E-15 -17.78

B1-1 0 Live LinStatic 0 -0.45 4.441E-16 0.0857 0 -0.238

B1-1 0.75 Live LinStatic 0 2.549 4.441E-16 0.0857 -3E-16 -1.024

B1-1 1.5 Live LinStatic 0 5.549 4.441E-16 0.0857 -7E-16 -4.061

B1-1 0 EqX LinStatic 0 -1.85 9.095E-13 6.103 -5E-13 -2.567

B1-1 0.75 EqX LinStatic 0 -1.85 9.095E-13 6.103 -1E-12 -1.184

B1-1 1.5 EqX LinStatic 0 -1.85 9.095E-13 6.103 -2E-12 0.2001

B1-1 0 EqY LinStatic 0 0.12 0 -0.0844 0 0.05

B1-1 0.75 EqY LinStatic 0 0.12 0 -0.0844 0 -0.04

B1-1 1.5 EqY LinStatic 0 0.12 0 -0.0844 0 -0.13

B1-1 0 DCON1 Combination 0 -1.98 1.332E-15 0.2723 0 -1.034

B1-1 0.75 DCON1 Combination 0 17.09 1.332E-15 0.2723 -1E-15 -6.698

B1-1 1.5 DCON1 Combination 0 36.16 1.332E-15 0.2723 -2E-15 -26.66

B1-2 0 Dead LinStatic 0 -2.4 8.882E-16 -0.0689 0 -1.094

B1-2 0.75 Dead LinStatic 0 11.29 8.882E-16 -0.0689 -7E-16 -4.431

B1-2 1.5 Dead LinStatic 0 24.98 8.882E-16 -0.0689 -1E-15 -18.03

B1-2 0 Live LinStatic 0 -0.93 4.441E-16 -0.0401 0 -0.456

B1-2 0.75 Live LinStatic 0 5.069 4.441E-16 -0.0401 -3E-16 -2.008

B1-2 1.5 Live LinStatic 0 11.07 4.441E-16 -0.0401 -7E-16 -8.06

B1-2 0 EqX LinStatic 0 0.324 9.095E-13 10.1221 -5E-13 -0.083

B1-2 0.75 EqX LinStatic 0 0.324 9.095E-13 10.1221 -1E-12 -0.326

23

3.3Model Verification

i) Shear force diagram ( Load Combination: 1.2DL+1.2LL+1.2EQX )

Figure 1: SFD of X-Z plane @ y = 1.5m

Figure 2: SFD of Y-Z plane @ x = 18m

24

ii) Bending moment diagram ( Load Combination: 1.2DL+1.2LL+1.2EQX )

Figure 3: BMD of X-Z plane @ y = 1.5m

Figure 4: BMD of Y-Z plane @ x = 18m

25

iii) Time period

Figure 5: Deformed shape of X-Z plane @ y = 1.5m (for mode 1)

Figure 6: Deformed shape of X-Z plane @ y = 1.5m (for mode 2)

26

iv) Storey drift

As per Clause no. 7.11.1 of IS 1893 (Part 1):2002, the storey drift in any storey due to

specified design lateral force with partial load factor of 1.0, shall not exceed 0.004 times the

storey height. From the SAP analysis, the displacements of the mass centres of various floors

are obtained and hence, the storey drift is computed.

EQX

Floor Displacement (mm) Storey drift (mm) Drift ratio

Ground 3.8 3.8 0.00120

First 9.4 5.6 0.00175

Second 15 5.6 0.00175

Third 20.3 5.3 0.00166

Fourth 23 2.7 0.00084

Fifth 26.6 3.6 0.00113

Sixth 29.4 2.8 0.00088

<0.004, OK

EQY

Floor Displacement (mm) Storey drift (mm) Drift ratio

Ground 3.6 3.6 0.00113

First 9 5.4 0.00169

Second 14.4 5.4 0.00169

Third 19.4 5.1 0.00159

Fourth 21.7 2.3 0.00072

Fifth 25 3.3 0.00103

Sixth 27.8 2.8 0.00088

<0.004, OK

27

Chapter 4

SECTION DESIGN

4.1 Limit state Method:

In the method if design based on limit state concept, the structure shall be designed

to withstand safely all loads liable to act on it throughout its life; it shall also satisfy the

serviceability requirements, such as limitations on deflection and cracking. The acceptable

limit for the safety and serviceability requirements before failure occurs is called a ‘limit state’.

The aim of design is to achieve acceptable probabilistic that the structure will not become unfit

for the use for which it is intended, that is, that it will not reach a limit state.

Assumptions for flexural member:

Plane sections normal to the axis of the member remain plane after bending.

The maximum strain in concrete at the outermost compression fiber is 0.0035.

The relationship between the compressive stress distribution in concrete and the strain in

concrete may be assumed to be rectangle, trapezoidal, parabola or any other shape which

results in prediction of strength in substantial agreement with the result of test. For design

purposes, the compressive strength of concrete in the structure shall be assumed to be 0.67

times the characteristic strength. The partial safety factor γm = 1.5 shall be applied in addition

to this.

The tensile strength of concrete is ignored.

The design stresses in reinforcement are derived from representative stress-strain curve for the

type of steel used. For the design purposes the partial safety factor γm = 1.15 shall be applied.

The maximum strain in the tension reinforcement in the section at failure shall not be less

than: 002.0E15.1

f

s

y

Where,

fy = characteristic strength of steel

Es = modulus of elasticity of steel

28

Limit state of collapse for compression:

Assumption:

In addition to the assumptions given above from i) to v), the following shall be assumed:

The maximum compressive strain in concrete in axial compression is taken

as 0.002.

The maximum compressive strain at highly compressed extreme fiber in concrete subjected

to axial compressive 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 fiber.

The limiting values of the depth of neutral axis for different grades of steel based on the

assumptions are as follows:

Fy xu,max

250 0.53

415 0.48

500 0.46

Materials adopted in our design:

M20 (1:1.5:3)

M25 (1:1:2)

Fe415

Use of SP16, IS456-2000, IS1893-2002, IS13920-1993, SP34:

After analyzing the given structure using the software SAP2000 the structural elements are

designed by Limit state Method. Account should be taken of accepted theories, experiment,

experience as well as durability.

The codes we use for the design are IS456-2000; IS1893-2002, IS13920-1993 and

Design aids are SP16 and SP34. Suitable material, quality control, adequate detailing and good

supervision are equally important during implementation of the project.

Use of different handbook for the design:

The structural elements (special staircases, lift wall, basement wall) which are not described

by the above mentioned codes and design aids were handled with the help of the handbooks

viz. Reinforced concrete Designer’s Handbook – Charles E. Reynolds and James C.

Steedmann, Reinforced Concrete Detailer’s Manual – Brian W. Boughton. For the design of

29

mat foundation, though there are several methods in practice, here conventional method of

mat footing design is adopted.

4.2 Design of structural elements:

The design includes design for durability, construction and use in service should be

considered as a whole. The realization of design objectives requires compliance with clearly

defined standards for materials, workmanship, and also maintenance and use of structure in

service.

This chapter includes all the design process of sample calculation for a single element as slab,

beam, column, staircases, basement wall, lift wall, ribbed slab and mat foundation.

4.2.1 Design of slab

4.2.2 Design of Beam

4.2.3 Design of Column

4.2.4 Design of Staircase

4.2.5 Design of Basement Wall

4.2.6 Design of Lift Wall

4.2.7 Design of Mat Foundation

30

4.2.1 Design of Slab

Design of 1.5m X 7m cantilever slab

Grade of Concrete M20 Grade of Steel Fe415(HYSD)

Ref Step Calculations Output

1.

2.

3.

4.

(i)

Thickness of slab and durability consideration

Centre-centre Spans

Lx=1.5 m

Ly=7 m

Assume,Depth

SpanRatio = 7 x 1.65 = 11.55 (a=7 for cantilever

beam; Clause 23.2.1)

d = 55.11

1500 129.87 mm≈ 130mm

Provide , d = 131 mm

Assuming clear span cover=15 mm

Providing 8 mm Ø bar

Total depth of slab, D= 131+ 15 + 8/2 = 150 mm,

Taking depth of slab=150mm

Design Load

Self-weight of slab = 0.15 x 25 = 3.75 KN/m2

Finishing load = 1.00 KN/m2

Total dead load = 3.75 + 1.00 = 4.75 KN/m2

Live load =4.0 KN/m2

Design load , w = 1.5(DL+LL) = 13.125KN/m2

Considering unit width of slab , wu= 13.125 KN/m

Moment Calculation

Mmax=𝑤𝑢∗𝑙2

2=14.766KNm

Design of slab:

Check for depth from Moment Consideration

Depth of Slab, d = mmx

bfck

M14.73

1000*20*138.0

10766.14

**138.0

6

max

d =131 mm

D=150 mm

31

IS 456-

2000

Annex

G.1.1

IS 456-

2000

cl.26.5.2.1

IS 456-

2000

Annex G

G-1.1.b)

IS 456-

2000

Clause

40.1

(ii)

(iii)

5.

6.

,which is less than provided depth,i.e 131 mm. (O.K.)

Calculation of Area of Steel

Min Ast = 0.12 % of bD = 0.12 x 1000 x 150 = 180 mm2

Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1311000)131100020

10766.146.411(

415

205.0

2

6

xxxx

xxx

=329.55mm2 >Min Ast (O.K.)

Number and spacing of bars:

Providing 8 mm Ø bars

Ab = 50.26 mm2

No. of bars= 56.626.50

55.329

b

st

A

A

Spacing of Bars, Sv = 1000xA

A

sd

b = 100055.329

26.50x = 152.56

mm

Provide 8 mm Ø @ 150 mm c/c

Act. Ast = 1000xS

A

v

b= 07.3351000

150

26.50x mm2

Pt = 0.256 %

Check for Shear

Shear force at the face of the support, Vu = wu* l =

13.125*1.5=19.6875 KN.

Nominal shear stress: v =Vu/(b*d)= 19.6875∗1000

1000∗131=0.150 N/mm2

Here, tension reinforcement of slab contribute in shear

For pt = 0.255 %

222.0c N/mm2

k c = 1.3*0.222=0.288

Hence, v <<k* c .(Safe).

Check for Development Length

Ld = bd

s

x4x6.1

=

47

2.1x4x6.1

415x87.0x

Ld ≤47 x 10 = 470 mm

Min Ast = 180

mm2

Spacing O.K.

8mm Ø@ 150

mm c/c

Act. Ast =

335.07mm2

(provided at

upper face,in

the tensile

zone of the

cantilever

slab)

Vu = 19.6875

KN

c 0.222

N/mm2

32

IS 456-

2000

cl.26.2.1

IS 456-

2000

cl.26.5.2.1

7.

8.

The code requires that the reinforcement should extend

by a length equal to Ld/3= 156.67mm beyond the face of the

support. This suggests that the width of the support

>(156.67+25)=181.667 mm.Here,support width =

450mm.Hence,anchorage condition is satisfied.

Also, at the simple support like this, where compressive

reaction confines the ends of bar,

o

u

d LVu

ML 1*3.1

Vu=19.6875KN

xu= bfck

Astfy

**36.0

**87.0=16.8mm.

M1u=0.87*fy*Ast*(d-0.416*xu)=15 KNm.

Providing 90° bend and a clear cover of 25mm at the side end,

Lo=ls/2-x’+3 Ø

Where, ls= width of support=450mm

x’=25mm

Ø=8mm

o

u LVu

M1*3.1 = 1214.48 >> 470mm. Hence,safe

Distribution Bars

Asd = 0.12 % of bD = 0.12 x 1000 x 150 =180 mm2


A

sd

b = 1000180

26.50x = 279.25 mm


Act. Ast = 1000xS

A

v

b= 06.2011000

250

26.50x mm2

Curtailment of Reinforcement and Detailing

Curtailment is done by as per simplified method and detailing is

done by as per Indian Standard Code IS SP 34

Hence, Safe

Hence, Safe

Asd=180 mm2

Act. Asd =

201.06 mm2

33

Design of 5m X 2.5m one long edge discontinuous slab



IS 456-

2000

Table 26

1

2


Clear Spans

Lx=2.5 m

Ly=5 m




Total depth of slab, D= 131 + 15 + 8/2 =150 mm

Since 22500

5000

x

y

l

l Design as Two Way Slab

Design Load

Self load of slab = 0.15 x 25 = 3.75 KN/m2


Dead load = 3.75 + 1.00 = 4.75 KN/m2

Live load = 4.0 KN/m2

Design load , w = 1.5(DL+LL) = 13.125 KN/m2

Considering unit width of slab , w= 13.125 KN/m

Moment Calculation

-ve Bending moment coefficient at continuous edge

αx= -0.085, αy= -0.037

+ve Bending moment coefficient at mid span

αx= 0.065, αy= 0.028

For Short Span

Support moment , Ms = - αxwlx2 = -0.085x 13.125 x 2.52 = -6.97

KN-m

Mid span moment , Mm = αywlx2 = 0.065 x 13.125 x 2.52 = 5.33

KN-m

For Long Span

d =131 mm

D=150 mm

lx=5000 mm

ly=6000 mm

34

IS 456-

2000

Annex

D.1.1

IS 456-

2000

Annex

G.1.1

IS 456-

2000

cl.26.5.2.1

IS 456-

2000

Annex G

G-1.1.b)

3

4

Support moment , Ms = - αxwlx2 = -0.037 x 13.125 x 2.52 = -

3.035 KN-m

Mid span moment , Mm = αywlx2 = 0.028 x 13.125 x 2.52 =

2.297KN-m



x

bx

M25.50

1000759.2

1097.6

759.2

6

max


Min Ast = 0.12 % of bD = 0.12 x 1000 x 150 = 180 mm2

Area of Steel along short span

Area of Steel at support ( Top Bars)

Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1311000)131100020

1097.66.411(

415

205.0

2

6

xxxx

xxx

= 150.97 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 1000790.150

26.50x = 332.91 mm


Actual ,Ast = 04.2011000250

26.501000 xx

S

A

v

b mm2

Pt = 0.153 %

Area of Steel at mid span (Bottom Bars)

Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1311000)131100020

1033.56.411(

415

205.0

2

6

xxxx

xxx

= 114.78 mm2 > Min Ast


d=50.25

mm<131mm

Min Ast = 180

mm2

Solved Eq.

for Ast

Spacing O.K.

8mm Ø@

250mm c/c

Act. Ast

=201.04mm2

Solved Eq.

for Ast

35

IS 456-

2000

Annex G

G-1.1.b

IS 456-

2000

Annex G

G-1.1.b

Ab = 50.26mm2


A

st

b = 100078.114

26.50x = 437.88 mm


Actual ,Ast = 04.2011000250

26.501000 xx

S

A

v

b mm2

Pt = 0.153 %

Area of Steel Along Long Span


Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1231000)123100020

10035.36.411(

415

205.0

2

6

xxxx

xxx

= 69.15 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 100015.69

26.50x = 726.84 mm


Actual ,Ast = 5.1791000280

26.501000 xx

S

A

v

b mm2

Pt = 0.146 %

Area of Steel at Mid Span ( Bottom Bars)

Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1231000)123100020

10297.26.411(

415

205.0

2

6

xxxx

xxx

= 52.2 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 10002.52

26.50x = 962.84 mm


Spacing O.K.

8mm Ø@ 250

mm c/c

Act. Ast

=201.04mm2

Solved Eq.

for Ast

Spacing O.K.

8 mm Ø@

280 mm c/c

Act. Ast =

179.5 mm2

Solved Eq.

for Ast

Spacing O.K.

8mm Ø@ 280

mm c/c

36

IS 456-

2000

Table 19

cl.40.2

5

6

Actual ,Ast = 5.1791000280

26.501000 xx

S

A

v

b mm2

Pt = 0.146 %

Check for Shear

For x-direction i.e. short span

Shear force at the face of the support, V = w lx = 13.125 x 2

.5.2

=16.40KN

Shear at critical section

16.40 KN

15.54 KN

d=0.131 m

2.5 m

131.05.25.2

40.16

uV

Vu = 15.54 KN


For pt = 0.153 %

28.0c N/mm2

k c bd = 1000

131100028.03.1 xxx = 47.68 KN > Vu

Check for Deflection

Along short Span

Since both ends are continuous, the basic value may be taken as

26

Act. Ast =

179.5 mm2

Vu = 16.04

KN

c 0.28

N/mm2

Hence, Safe

37

IS 456-

2000

cl.23.2.1

IS 456-

2000

Fig: 4

IS 456-

2000

cl.26.2.1

7

8

9

fs= ovidedPrSteelofArea

quiredReSteelofAreaf58.0 y = 0.58 x 415 x

04.201

78.114

= 137.42 N/mm2

Pt= 0.153 %

Modification factor (M.F.) = 2

dper = 07.48262

2500

..

xValueBasicxFM

lx mm < d provided.


Ld = bd

s

x4x6.1

=

47

2.1x4x6.1

415x87.0x

Ld = 47 x 8 = 376 mm

o1

d LV

ML

For Short Span

o1 L

V

M = 438.0275.0

40.16

235.5

m = 438 mm

For Long Span

o1 L

V

M = m31.0275.0

81.32

2297.2

= 310 mm

Distribution Bars

Asd = 0.12 % of bD = 0.12 x 1000 x 150 =180 mm2


A

sd

b = 1000180

26.50x = 279.22

mm


Act. Ast = 1000xS

A

v

b = 04.2011000250

26.50x mm2

Torsion Reinforcement

Torsion reinforcement is provided as per IS 456-2000

Cl.D.1.8,Cl. D.1.9

dper < d ,

Hence Safe

Hence, Safe

Hence, Safe

Asd=180 mm2

Act. Asd =

201.04 mm2

38

Design of 5m X 6m one long edge discontinuous slab



IS 456-

2000

Table 26

IS 456-

2000

Annex

D.1.1

1.

2.

3.

4.


Clear Spans

Lx=5 m

Ly=6 m





Since 22.15000

6000

x

y

l


Design Load



Dead load = 3.75 + 1.00 = 4.75 KN/m2




Moment Calculation


αx= -0.0520, αy= -0.037


αx= 0.039, αy= 0.028

For Short Span

Support moment , Ms = - αxwlx2 = -0.052x 13.125 x 52 = -17.07

KN-m

Mid span moment , Mm = αywlx2 = 0.039 x 13.125 x 52 = 12.80

KN-m

For Long Span

Support moment , Ms = - αxwlx2 = -0.037 x 13.125 x 52 = -12.15

KN-m

Mid span moment , Mm = αywlx2 = 0.028 x 13.125 x 52 =

9.31KN-m



x

bx

M64.78

1000759.2

1007.17

759.2

6

max


Min Ast = 0.12 % of bD = 0.12 x 1000 x 150 = 180 mm2



d =131 mm

D=150 mm

lx=5000 mm

ly=6000 mm

d=78.64

mm<131mm

Min Ast =

180 mm2

39

IS 456-

2000

Annex

G.1.1

IS 456-

2000

cl.26.5.2.1

IS 456-

2000

Annex G

G-1.1.b)

IS 456-

2000

Annex G

G-1.1.b

Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1311000)131100020

1007.176.411(

415

205.0

2

6

xxxx

xxx

= 384.3 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 10003.384

26.50x = 130.78 mm


Actual ,Ast = 08.4021000125

26.501000 xx

S

A

v

b mm2

Pt = 0.268 %


Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1311000)131100020

1080.126.411(

415

205.0

2

6

xxxx

xxx

= 283.34 mm2 > Min Ast


Ab = 50.26mm2


A

st

b = 100034.283

26.50x = 177.38 mm


Actual ,Ast = 08.4021000125

26.501000 xx

S

A

v

b mm2

Pt = 0.268 %



Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1231000)123100020

1015.126.411(

415

205.0

2

6

xxxx

xxx

= 298.38 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 100038.398

26.50x = 168.44 mm


Actual ,Ast = 3591000140

26.501000 xx

S

A

v

b mm2

Pt = 0.292 %


Solved Eq.

for Ast

Spacing

O.K.

8mm Ø@

125mm c/c

Act. Ast

=402.8mm2

Solved Eq.

for Ast

Spacing

O.K.

8mm Ø@

125 mm c/c

Act. Ast

=402.08mm2

Solved Eq.

for Ast

Spacing

O.K.

8 mm Ø@

140 mm c/c

Act. Ast =

359mm2

40

IS 456-

2000

Annex G

G-1.1.b

IS 456-

2000

Table 19

cl.40.2

IS 456-

2000

cl.23.2.1

5.

6.

Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1231000)123100020

1031.96.411(

415

205.0

2

6

xxxx

xxx

= 217.63 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 100063.217

26.50x = 230.94 mm


Actual ,Ast = 3591000140

26.501000 xx

S

A

v

b mm2

Pt = 0.292 %

Check for Shear



.5

=32.81KN


32.81 KN

31.09 KN

d=0.131 m

2.5 m

131.05.25.2

81.32

uV

Vu = 31.09 KN


For pt = 0.268 %

37.0c N/mm2

k c bd = 1000

131100037.03.1 xxx = 63.01 KN > Vu


Along short Span


26



08.402

34.283

= 169.62 N/mm2

Solved Eq.

for Ast

Spacing

O.K.

8mm Ø@

140 mm c/c

Act. Ast =

359mm2

Vu = 31.09

KN

c 0.37

N/mm2

Hence, Safe

41

IS 456-

2000

Fig: 4

IS 456-

2000

cl.26.2.1

7.

8.

9.

10.

Pt= 0.268 %


dper = 15.96262

5000

..

xValueBasicxFM

lx mm < d provided.


Ld = bd

s

x4x6.1

=

47

2.1x4x6.1

415x87.0x

Ld = 47 x 8 = 376 mm

o1

d LV

ML

For Short Span

o1 L

V

M = 47.0275.0

81.32

280.12

m = 470 mm

For Long Span

o1 L

V

M = 393.275.0

375.39

231.9

= 393 mm

Distribution Bars

Asd = 0.12 % of bD = 0.12 x 1000 x 150 =180 mm2


A

sd

b = 1000180

26.50x = 279.22

mm


Act. Ast = 1000xS

A

v

b = 04.2011000250

26.50x mm2



Cl.D.1.8,Cl. D.1.9




dper < d ,

Hence Safe

Hence, Safe

Hence, Safe

Asd=180 mm2

Act. Asd =

201.04 mm2

42

Design of 5m X 6m one short edge discontinuous



IS 456-

2000

Table 26

IS 456-

2000

Annex

D.1.1

IS 456-

2000

cl.26.5.2.1

1.

2.

3.

4.


Clear Spans

Lx=5 m

Ly=6 m





Since 22.15000

6000

x

y

l


Design Load



Dead load = 3.75 + 1.00 = 4.75 KN/m2




Moment Calculation


αx= -0.036, αy= -0.037


αx= 0.048, αy= 0.028

For Short Span


KN-m

Mid span moment , Mm = αywlx2 = 0.048 x 13.125 x 52 = 11.81

KN-m

For Long Span


KN-m


9.19KN-m



x

bx

M54.75

1000759.2

1075.15

759.2

6

max


Min Ast = 0.12 % of bD = 0.12 x 1000 x 150 = 180 mm2



d =131 mm

D=150 mm

lx=5000 mm

ly=6000 mm

d=75.54

mm<131mm

Min Ast =

180 mm2

43

IS 456-

2000

Annex

G.1.1

IS 456-

2000

Annex G

G-1.1.b)

IS 456-

2000

Annex G

G-1.1.b

Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1311000)131100020

1075.156.411(

415

205.0

2

6

xxxx

xxx

= 352.70 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 100070.352

26.50x = 142.5 mm


Actual ,Ast = 08.4021000125

26.501000 xx

S

A

v

b mm2

Pt = 0.268 %


Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1311000)131100020

1081.116.411(

415

205.0

2

6

xxxx

xxx

= 260.44 mm2 > Min Ast


Ab = 50.26mm2


A

st

b = 100044.260

26.50x = 192.98 mm


Actual ,Ast = 08.4021000125

26.501000 xx

S

A

v

b mm2

Pt = 0.268 %



Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1231000)123100020

1014.126.411(

415

205.0

2

6

xxxx

xxx

= 287.3 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 10003.287

26.50x = 174.94 mm


Actual ,Ast = 3591000140

26.501000 xx

S

A

v

b mm2

Pt = 0.292 %


Solved Eq.

for Ast

Spacing

O.K.

8mm Ø@

125mm c/c

Act. Ast

=402.08mm2

Solved Eq.

for Ast

Spacing

O.K.

8mm Ø@

125 mm c/c

Act. Ast

=402.08mm2

Solved Eq.

for Ast

Spacing

O.K.

8 mm Ø@

140 mm c/c

Act. Ast =

359 mm2

44

IS 456-

2000

Annex G

G-1.1.b

IS 456-

2000

Table 19

cl.40.2

IS 456-

2000

cl.23.2.1

5.

6.

Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1231000)123100020

1019.96.411(

415

205.0

2

6

xxxx

xxx

= 214.71 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 100071.214

26.50x = 234.08 mm


Actual ,Ast = 3591000140

26.501000 xx

S

A

v

b mm2

Pt = 0.292 %

Check for Shear



.5

=32.81KN


32.81 KN

31.09 KN

d=0.131 m

2.5 m

131.05.25.2

81.32

uV

Vu = 31.09 KN


For pt = 0.268 %

37.0c N/mm2

k c bd = 1000

131100037.03.1 xxx = 63.01 KN > Vu


Along short Span


26



08.402

34.283

= 169.62 N/mm2

Solved Eq.

for Ast

Spacing

O.K.

8mm Ø@

140 mm c/c

Act. Ast =

359mm2

Vu = 31.09

KN

c 0.37

N/mm2

Hence, Safe

45

IS 456-

2000

Fig: 4

IS 456-

2000

cl.26.2.1

7.

8.

9.

10.

Pt= 0.268 %


dper = 15.96262

5000

..

xValueBasicxFM

lx mm < d provided.


Ld = bd

s

x4x6.1

=

47

2.1x4x6.1

415x87.0x

Ld = 47 x 8 = 376 mm

o1

d LV

ML

For Short Span

o1 L

V

M = 46.0275.0

81.32

281.11

m = 460 mm

For Long Span

o1 L

V

M = m393.0275.0

375.39

219.9

= 393 mm

Distribution Bars

Asd = 0.12 % of bD = 0.12 x 1000 x 150 =180 mm2


A

sd

b = 1000180

26.50x = 279.22

mm


Act. Ast = 1000xS

A

v

b = 04.2011000250

26.50x mm2



Cl.D.1.8,Cl. D.1.9




dper < d ,

Hence Safe

Hence, Safe

Hence, Safe

Asd=180 mm2

Act. Asd =

201.04 mm2

46

Design of 5m X 6m two adjacent edges discontinuous slab



IS 456-

2000

Table 26

1.

2.

3.


Clear Spans

Lx=5 m

Ly=6 m





Since 22.15000

6000

x

y

l


Design Load



Dead load = 3.75 + 1.00 = 4.75 KN/m2




Moment Calculation


αx= -0.060, αy= -0.047


αx= 0.045, αy= 0.035

For Short Span


KN-m


14.76 KN-m

For Long Span

d =131 mm

D=150 mm

lx=5000 mm

ly=6000 mm

47

IS 456-

2000

Annex

D.1.1

IS 456-

2000

Annex

G.1.1

IS 456-

2000

Annex G

G-1.1.b)

4.

Support moment , Ms = - αxwlx2 = -0.047 x 13.125 x 52 = -

15.42 KN-m


11.484KN-m



x

bx

M45.84

1000759.2

1068.19

759.2

6

max


Min Ast = 0.12 % of bD = 0.12 x 1000 x 150 = 180 mm2



Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1311000)131100020

1068.196.411(

415

205.0

2

6

xxxx

xxx

= 448.10 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 100010.448

26.50x = 112.16 mm


Actual ,Ast = 6.5021000100

26.501000 xx

S

A

v

b mm2

Pt = 0.383 %


Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1311000)131100020

1076.146.411(

415

205.0

2

6

xxxx

xxx

= 329.41 mm2 > Min Ast


Ab = 50.26mm2

d=84.45

mm<131mm

Min Ast =

180 mm2

Solved Eq.

for Ast

Spacing

O.K.

8mm Ø@

100 mm c/c

Act. Ast

=502.6mm2

Solved Eq.

for Ast

48

IS 456-

2000

Annex G

G-1.1.b

IS 456-

2000

Annex G

G-1.1.b


A

st

b = 100041.329

26.50x = 152.57 mm


Actual ,Ast = 08.4021000125

26.501000 xx

S

A

v

b mm2

Pt = 0.307 %



Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1231000)123100020

1042.156.411(

415

205.0

2

6

xxxx

xxx

= 370.56 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 100056.370

26.50x = 135.63 mm


Actual ,Ast = 615.3861000130

26.501000 xx

S

A

v

b mm2

Pt = 0.295 %


Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1231000)123100020

10484.116.411(

415

205.0

2

6

xxxx

xxx

= 271.12 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 100012.271

26.50x = 185.37 mm


Spacing

O.K.

8mm Ø@

125 mm c/c

Act. Ast

=402.08mm2

Solved Eq.

for Ast

Spacing

O.K.

8 mm Ø@

130 mm c/c

Act. Ast =

386.615 mm2

Solved Eq.

for Ast

49

IS 456-

2000

Table 19

cl.40.2

IS 456-

2000

cl.23.2.1

5.

6.

Actual ,Ast = 22.2791000180

26.501000 xx

S

A

v

b mm2

Pt = 0.213 %

Check for Shear



.5

=32.81KN


32.81 KN

31.09 KN

d=0.131 m

2.5 m

131.05.25.2

81.32

uV

Vu = 31.09 KN


For pt = 0.383 %

423.0c N/mm2

k c bd = 1000

1311000423.03.1 xxx = 72.03 KN > Vu


Along short Span

Since both ends are continuous, the basic value may be taken

as 26



06.335

41.329

= 236.641 N/mm2

Spacing

O.K.

8mm Ø@

180 mm c/c

Act. Ast =

279.22 mm2

Vu = 31.09

KN

c 0.423

N/mm2

Hence, Safe

50

IS 456-

2000

Fig: 4

IS 456-

2000

cl.26.2.1

7.

8.

9.

10.

Pt= 0.255 %

Modification factor (M.F.) = 1.65

dper = 55.1162665.1

5000

..

xValueBasicxFM

lx mm < dprovided


Ld = bd

s

x4x6.1

=

47

2.1x4x6.1

415x87.0x

Ld = 47 x 8 = 376 mm

o1

d LV

ML

For Short Span

o1 L

V

M = 499.0275.0

81.32

276.14

m = 598 mm

For Long Span

o1 L

V

M = 420.0275.0

375.39

2485.11

= 476 mm

Distribution Bars

Asd = 0.12 % of bD = 0.12 x 1000 x 150 =180 mm2


A

sd

b = 1000180

26.50x = 279.22 mm


Act. Ast = 1000xS

A

v

b = 04.2011000250

26.50x mm2



Cl.D.1.8,Cl. D.1.9


Curtailment is done by as per simplified method and detailing

is done by as per Indian Standard Code IS SP 34

dper < d ,

Hence Safe

Hence, Safe

Hence, Safe

Asd=180 mm2

Act. Asd =

201.04 mm2

51

Design of 5m X 6m two long edges discontinuous slabs



IS 456-

2000

Table 26

IS 456-

2000

Annex

D.1.1

IS 456-

2000

cl.26.5.2.1

IS 456-

2000

Annex G

G-1.1.b)

1.

2.

3.

4.


Clear Spans

Lx=5 m

Ly=6 m





Since 22.15000

6000

x

y

l


Design Load



Dead load = 3.75 + 1.00 = 4.75 KN/m2




Moment Calculation


αx= 0.0, αy= -0.045


αx= 0.051, αy= 0.035

For Short Span

Mid span moment , Mm = αywlx2 = 0.051x 13.125 x 52 = 16.73

KN-m

For Long Span


KN-m

Mid span moment , Mm = αywlx2 = 0.035x 13.125 x 52 =

11.487KN-m



x

bx

M87.77

1000759.2

1073.16

759.2

6

max


Min Ast = 0.12 % of bD = 0.12 x 1000 x 150 = 180 mm2



Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

d =131 mm

D=150 mm

lx=5000 mm

ly=6000 mm

d=77.87

mm<131mm

Min Ast =

180 mm2

52

IS 456-

2000

Annex G

G-1.1.b

IS 456-

2000

Annex G

G-1.1.b

5.

= 1311000)131100020

1073.166.411(

415

205.0

2

6

xxxx

xxx

= 376.332 mm2 > Min Ast


Ab = 50.26mm2


A

st

b = 1000332.376

26.50x = 133.55 mm


Actual ,Ast = 08.4021000125

26.501000 xx

S

A

v

b mm2

Pt = 0.268 %



Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1231000)123100020

1076.146.411(

415

205.0

2

6

xxxx

xxx

= 353.62 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 100062.353

26.50x = 142.12 mm


Actual ,Ast = 3591000140

26.501000 xx

S

A

v

b mm2

Pt = 0.274 %


Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1231000)123100020

10478.116.411(

415

205.0

2

6

xxxx

xxx

= 270.83 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 10002

26.50x = 185.57 mm


Actual ,Ast = 3591000140

26.501000 xx

S

A

v

b mm2

Pt = 0.274 %

Check for Shear


Solved Eq.

for Ast

Spacing

O.K.

8mm Ø@

125mm c/c

Act. Ast

=402.8mm2

Solved Eq.

for Ast

Spacing

O.K.

8mm Ø@

140 mm c/c

Act. Ast

=359mm2

Solved Eq.

for Ast

Spacing

O.K.

8 mm Ø@

140 mm c/c

Act. Ast =

359 mm2

53

IS 456-

2000

Table 19

cl.40.2

IS 456-

2000

cl.23.2.1

IS 456-

2000

Fig: 4

IS 456-

2000

cl.26.2.1

6.

7.


.5

=32.81KN


32.81 KN

31.09 KN

d=0.131 m

2.5 m

131.05.25.2

81.32

uV

Vu = 31.09 KN


For pt = 0.268 %

37.0c N/mm2

k c bd = 1000

131100037.03.1 xxx = 63.01 KN > Vu


Along short Span


26



08.402

34.283

= 169.62 N/mm2

Pt= 0.268 %


dper = 15.96262

5000

..

xValueBasicxFM

lx mm < d provided.


Ld = bd

s

x4x6.1

=

47

2.1x4x6.1

415x87.0x

Ld = 47 x 8 = 376 mm

o1

d LV

ML

For Short Span

o1 L

V

M = 378.0 m = 378mm

For Long Span

o1 L

V

M = 462.275.0

375.39

276.14

= 462 mm

Vu = 31.09

KN

c 0.37

N/mm2

Hence, Safe

dper < d ,

Hence Safe

Hence, Safe

Hence, Safe

54

8.

9.

10.

Distribution Bars

Asd = 0.12 % of bD = 0.12 x 1000 x 150 =180 mm2


A

sd

b = 1000180

26.50x = 279.22

mm


Act. Ast = 1000xS

A

v

b = 04.2011000250

26.50x mm2



Cl.D.1.8,Cl. D.1.9




Asd=180 mm2

Act. Asd =

201.04 mm2

55

Design of 5m X 6m interior panel slab



IS 456-

2000

Table 26

1.

2.

3.


Clear Spans

Lx=5000mm

Ly=6000mm




Total depth of slab, D= 131 + 15 + 8/2 = 150 m

Since 22.15000

6000

x

y

l


Design Load



Dead load = 3.75 + 1.00 = 4.75 KN/m2



Considering unit width of slab , w=14.625 KN/m

Moment Calculation


αx = -0.043, αy= -0.032


αx= 0.032, αy= 0.024

For Short Span

Support moment , Ms = - αxwlx2 = -0.043x 14.625 x 5.02 = -

15.72 KN-m

Mid span moment , Mm = αywlx2 = 0.032x 14.625 x 5.02

= 11.7 KN-m

For Long Span

d =131 mm

D=150 mm

lx=5000 mm

ly=6000 mm

56

IS 456-

2000

Annex

D.1.1

IS 456-

2000

Annex

G.1.1

IS 456-

2000

Annex G

G-1.1.b)

4.

Support moment , Ms = - αxwlx2 = -0.032 x 14.625 x 5.02

= -11.7KN-m

Mid span moment , Mm = αywlx2 = 0.024 x 14.625 x 5.02 =

8.775 KN-m



x

bx

M47.75

1000759.2

1072.15

759.2

6

max


Min Ast = 0.12 % of bD = 0.12 x 1000 x 150 = 180 mm2



Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1311000)131100020

1072.156.411(

415

205.0

2

6

xxxx

xxx

= 351.98 mm2 > Min Ast


Ab = 50.26 mm2


A

st

b = 100098.351

26.50x = 142.8 mm


Actual ,Ast = 4021000125

26.501000 xx

S

A

v

b mm2

Pt = 0.268 %


Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1311000)131100020

107.116.411(

415

205.0

2

6

xxxx

xxx

= 257.9 mm2 > Min Ast


d=75.47

mm<131mm

Min Ast =

180 mm2

Spacing

O.K.

8mm Ø@

125 mm c/c

Act. Ast =

402 mm2

57

IS 456-

2000

Annex G

G-1.1.b

IS 456-

2000

Annex G

G-1.1.b

Ab = 50.26 mm2


A

st

b = 10009.257

26.50x = 194.88 mm


Actual ,Ast = 08.4021000125

26.501000 xx

S

A

v

b mm2

Pt = 0.268 %


Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1231000)123100020

107.116.411(

415

205.0

2

6

xxxx

xxx

= 276.34 mm2 > Min Ast


Ab = 50.26mm2


A

st

b = 100034.276

26.50x = 181.87 mm


Actual ,Ast = 3591000140

26.501000 xx

S

A

v

b mm2

Pt = 0.292 %


Ast = bd)bdf

M6.411(

f

f5.0

2ck

u

y

ck

= 1231000)123100020

10775.86.411(

415

205.0

2

6

xxxx

xxx

= 204.7 mm2 > Min Ast


Ab = 50.26mm2


A

st

b = 10007.204

26.50x = 245.58 mm

Spacing

O.K.

8mm Ø@

125 mm c/c

Act. As =

402.08 mm2

Spacing

O.K.

8mm Ø@

140 mm c/c

Act. Ast =

359

mm2

Spacing

O.K.

8mm Ø@

140 mm c/c

58

IS 456-

2000

Table 19

cl.40.2

5.

6.


Actual ,Ast = 3591000140

26.501000 xx

S

A

v

b mm2

Pt = 0.292 %

Check for Shear



0.5

=36.56 KN


36.56 KN

34.64 KN

d = 0.131m

2.5m

131.050.250.2

56.36

uV

Vu = 34.64 KN


For pt = 0.268 %

𝜏(max) = 2.8 N/mm2

V(max)=2.8*1000*131=366.8 KN

37.0c N/mm2

k c bd = 1000

131100037.033.1 xxx = 64.46 KN > Vu


Along short Span


26

Act. Ast =

359mm2

Vu = 34.64

KN

Vu<V(max)

c 0.370

N/mm2

Hence, Safe

59

IS 456-

2000

cl.23.2.1

IS 456-

2000

Fig: 4

IS 456-

2000

cl.26.2.1

7.

8.

9.

10.



08.402

9.257

= 154.4 N/mm2

Pt= 0.268 %


dper = 15.96262

5000

..

xValueBasicxFM

lx mm < d provided.


Ld = bd

s

x4x6.1

=

47

2.1x4x6.1

415x87.0x

Ld = 47 x 8 = 376 mm

o1

d LV

ML

For Short Span

o1 L

V

M = 538.0)25.001.0*8(

56.36

272.15

= 538 mm

For Long Span

o1 L

V

M = )25.001.0*8(

875.43

2775.8

*3.1 = 0.46 m = 460 mm

Provide Ld=540 mm

Distribution Bars

Asd = 0.12 % of bD = 0.12 x 1000 x 150 =180 mm2


A

sd

b = 1000180

26.50x = 279 mm


Act. Ast = 1000xS

A

v

b = 15.1861000270

26.50x mm2


Since this is interior panel no need of torsion reinforcement

Curtailment of Reinforcement

Curtailment is done by as per simplified method and Indian

Standard Code IS SP 34

dper < d ,

Hence

Safe

Hence, safe

Hence, safe

Asd=180 mm2

Act. Asd =

186.15 mm2

60

ONE WAY SLAB (2.5m X 6m)

1

2

3

4

5


Effective Spans

Lx=2.5 m

Ly=6 m

So, it is a one way slab.

Assume,Depth

Span Ratio = 26 x 1.45 = 37.7

d 7.37

250066.31 mm

Provide , D= 150 mm



Effective depth of slab, d= 150 - 15 - 8/2 = 131 mm

Design Load



Dead load = 3.75 + 1.00 = 4.75 KN/m2


Design load , wd = 1.5 DL = 7.13 KN/m2

wl = 1.5 LL = 6 KN/m2

Moment Calculation

Mu = −

𝑤𝑑∗𝑙2

10−

𝑤𝑙∗𝑙2

9

= -8.62KNm

Area of main rebars

Maximum bending moment = Mu = 0.87 fy Ast (d −𝐴𝑠𝑡 𝑓𝑦

𝑓𝑐𝑘 𝑏)

8.62 * 106 = 0.87 ∗ 415 ∗ Ast (131 −𝐴𝑠𝑡 415

20∗1000)

Ast = 187.83 mm2

Spacing of bars = ∗82

4∗187.83∗ 1000 = 276.67 𝑚𝑚

Provide 8 @250 c/c

Check for Shear


Shear force at the face of the support, Vu = 0.6*(wd + wl)* l = 0.6*(7.13+6)*2.5 = 19.70 KN

For pt = ∗82

4∗250∗131∗ 100 = 0.153%

61

6

7

8

9

10

28.0c N/mm2

k c bd = 1000

131*1000*28.0*3.1 = 47.68 KN > Vu


Along short Span

Since both ends are continuous, the basic value may be taken as 26

fs = ovidedPrSteelofArea

quiredReSteelofAreaf58.0 y

= 0.58 x 415 x 06.201

83.187

= 224.86 N/mm2

Pt= 0.153 %

Modification factor (M.F.) = 1.9

dper = 61.5026*9.1

2500

..

ValueBasicxFM

lx mm < d provided.


Ld = bd

s

x4x6.1

=

472.1x4x6.1

415x87.0x

Ld = 47 x 8 = 376 mm

o1

d LV

ML

For Short Span,

o1 L

V

M = 494.0275.0

7.19

262.8

m = 494 mm

Distribution Bars

Asd = 0.12 % of bD = 0.12 x 1000 x 150=180 mm2

Spacing of Bars, Sv = ∗82

4∗180∗ 1000 = 279 mm


Act. Ast = 1000xS

A

v

b = 04.2011000x250

26.50 mm2


Torsion reinforcement is provided as per IS 456-2000 Cl.D.1.8,Cl. D.1.9


Curtailment is done by as per simplified method and detailing is done by as per Indian

Standard Code IS SP 34

62


Grade of concrete M20 Grade of Steel Fe415 (HYSD)


1.

2.

For two adjacent edge discontinuous with Lx = 2.5 m

Shorter span of slab (Lx) = 2.5 m

Distance of Torsional Reinforcement from edge of slab

mL

L x

stT 5.05

Area of bar required for maximum mid-span moment Ast=

161 mm2 / m length

21224

3mmAxA ststT


Ab = 50.26 mm2


A

stT

b = 100061

26.50x = 823 mm


No. of bars required = barsS

L

V

stT2

300

500

2618

3mmAxA ststT


Ab = 50.26 mm2


A

stT

b = 1000122

26.50x = 411 mm



L

V

stT2

300

500

For one edge discontinuous with Lx = 3 m

Shorter span of slab (Lx) = 3 m

Lx = 2.5 m

LT st = 0.5 m

AT st = 122

mm2

Spacing

O.K.

8mm Ø@

300 mm c/c

AT st = 61

mm2

Spacing O.K.

8mm Ø@

300 mm c/c

63

3.


mL

L x

stT 6.05


359 mm2 / m length

224.2694

3mmAxA ststT


Ab = 50.26 mm2


A

stT

b = 100024.269

26.50x = 186.67

mm



L

V

stT4

180

600

262.1348

3mmAxA ststT


Ab = 50.26 mm2


A

stT

b = 100062.134

26.50x = 373.34

mm



L

V

stT2

300

600

For one short edge discontinuous with Lx = 5 m



mL

L x

stT 0.15


395 mm2 / m length

Lx = 3 m

LT st = 0.6 m

AT st =

269.24mm2

Spacing O.K.

8mm Ø@

180 mm c/c

AT st =

134.62mm2

Spacing O.K.

8mm Ø@

300 mm c/c

Lx = 5 m

LT st = 1 m

64

4.

213.1488

3mmAxA ststT


Ab = 50.26 mm2


A

stT

b = 100013.148

26.50x = 339.3 mm

Provide 8 mm Ø @300 mm c/c


L

V

stT4

300

1000

For two adjacent edge discontinuous with Lx = 5 m



mL

L x

stT 0.15


329.41 mm2 / m length

206.2474

3mmAxA ststT


Ab = 50.26 mm2

Spacing, Sv = 1000xA

A

stT

b = 100006.247

26.50x = 203.43 mm



L

V

stT5

200

1000

253.1238

3mmAxA ststT


Ab = 50.26 mm2


A

stT

b = 100053.123

26.50x = 406.86 mm



L

V

stT4

300

1000

AT st = 148.13

mm2

Spacing O.K.

8mm Ø@

300 mm c/c

Lx = 5 m

LT st = 1 m

AT st = 247.06

mm2

Spacing

O.K.

8mm Ø@

200 mm c/c

LT st = 1 m

AT st = 123.53

mm2

Spacing O.K.

8mm Ø@

300 mm c/c

65

4.2.2 Design of Beam Ductile Design of CantileverBeam

Concrete Grade = M20 Steel Grade = Fe415(HYSD)


IS13920:1993

cl. 6.1.1

IS13920:1993

cl.6.1.3

IS13920:1993

cl.6.1.2

IS13920:1993

cl.6.2.1b

IS13920:1993

cl.6.2.2

IS456-2000

cl.41.4.2

1.

2.

2.1

2.2

3.

i)

ii)

a)

Known Data

Overall Depth of Beam, D=700mm

Width of Beam, B=500mm

Considering 1 layer of 32mm bar at the top,i.e.

tension zone

Taking clear cover=25mm.

Effective depth, d=700-25-32/2 = 659 mm

Check for Axial Stress

Factored Axial Stress = 1.22N/mm2

Axial Stress = 1.22N/mm2 < 0.1 fck

Hence, design as flexural member.

Check for member size

Width of beam, B=500mm > 200mm

Depth of beam, D=700mm

B/D = 500/700 = 0.714 > 0.3

Hence, OK

Check for Limiting Longitudinal Reinforcement

Min. tension reinforcement

rstmin = 0.24y

ck

f

f = 0.24

415

20

= 2.586 x 10-3 = 0.26%

Ast,min = 0.26 x 500 x 659 / 100 = 856.7mm2

Max. steel reinforcement, As,max = 0.025bd

=0.025 x 500 x 659 = 8237.5mm2

Calculation of Longitudinal Steel Reinforcement

Mulim = 2.76bd2 = 2.76x500x6592 = 599.31KN-m

At left end:

For maximum -ve moment: (hogging moment)

Mu =129.19 KNm

Torsional Moment, Tu = 29.68 KNm

The longitudinal reinforcement shall be designed to

resist an equivalent bending moment, Me1

Me1 = Mu + Mt

The torsional moment,

Mt = Tu x7.1

b/D1 = 29.68x 7.1

500/7001 = 41.9 KNm

Since Mt<Mu, no longitudinal reinforcement shall be

provided on the flexural compression face.

Hence, Me1 = Mu+Mt = 129.19+ 41.9= 171.09 KNm

D=700mm

B=500mm

d=659mm

d’=25+32/2=41mm

d’/d=0.062

Flexural Member

Astmin=856.7mm2

Ast,max=8237.5mm2

Mulim =599.31KN-m

Singly Reinforced

Section

66

IS456-2000

Annex G

G-1.1b

IS13920:1993

cl.6.2.3

IS456-2000

cl.41.4.2

IS456-2000

Annex G

G-1.1b

IS13920:1993

cl.6.2.3

b)

Ast = bdbdf

M

f

f

ck

u

y

ck

2

6.411

5.0

=563.218mm2 < Ast,min

Providing minimum area of steel=856.7mm2

Provide 2-28mm

Act.Ast=1231.504mm2 (Top bars)

Pt=0.37%

Providing two longitudinal bars at the bottom as well

Asc must be at least 50% of Ast

50% of Ast =0.185% <0.26%(rmin)

So,providing minimum area of steel,

Asc=856.7mm2

Ast(required) = 856.7mm2(Top)

Asc(required) = 856.7mm2 (Bottom)

For maximum +ve moment: (sagging moment)

Mu = 221.25 KNm

Torsional Moment, Tu = 29.68KNm



Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 29.68x 7.1

500/7001 = 41.9 KNm



Hence, Me1 = Mu+Mt = 221.25 + 41.9=263.151 KNm

Here, Mulim>Mu Hence, The section is to be singly-

reinforced.

Ast = bdbdf

M

f

f

ck

u

y

ck

2

6.411

5.0

=992.37mm2 > Ast,min O.K. Provide 2-28mm


Pt=0.37%



50% of Ast =0.185% <0.26%(rmin)


Asc=856.7mm2

Ast(required) = 992.37mm2(Bottom)

Asc(required) = 856.7mm2 (Top)

For As(bottom), provide 2-28mm bars

Act.As(bottom)= 1231.5 mm2 (bottom)

For As(top), provide 2-28mm bars

Act.As(top)=1231.5 mm2 (top)

Singly Reinforced

Section

As(top)= 856.7mm2

As(bottom)=

992.37mm2

Act.As(top)=

1231.5mm2

Act.As(bottom)=1231.5

mm2

67

IS456-2000

cl.41.4.2

IS456-2000

Annex G

G-1.1b

IS13920:1993

cl.6.2.3

IS456-2000

cl.41.4.2

IS456-2000

cl.41.4.2.1

IS456-2000

iii)

a)

b)

At mid span


Mu = 27.33 KNm

Torsional Moment, Tu =17.263 KNm



Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 17.263x 7.1

500/7001 = 24.37 KNm



Hence, Me1 = Mu+Mt = 27.33 + 24.37= 51.7 KNm

Here,

Mulim>Mu Hence, The section is to be singly-

reinforced.

Ast = bdbdf

M

f

f

ck

u

y

ck

2

6.411

5.0

=117.77mm2 < Ast,min


Provide 2-28mm


Pt=0.37%



50% of Ast =0.185% <0.26%(rmin)


Asc=856.7mm2

Ast = 856.7mm2(Top)

Asc = 856.7mm2 (Bottom)


Mu = 1.51 KNm




Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 17.263x 7.1

500/7001 = 24.37 KNm

Since Mt>Mu, so longitudinal reinforcement shall be


Hence, Me1 = Mu+Mt = 1.51+ 24.37= 25.88 KNm

And,

Me2 =Mt-Mu=24.37-1.51=22.86KNm.


reinforced.

Singly Reinforced

Section

Singly Reinforced

Section

68

Annex G

G-1.1b

IS13920:1993

cl.6.2.3

IS13920:1993

cl.6.3.3

IS456:2000

cl.41.3.1

IS456:2000

Table 20

IS456:2000

Table 19

4.

4.1

Ast = bdbdf

M6.411

f

f5.0

2ck

u

y

ck

=128.02mm2 (Bottom bars)<Ast,min(856.7mm2)


Provide 2-28mm

Act.Ast=1231.504mm2 (Bottom bars)

Pt=0.37%



50% of Ast =0.185% <0.26%(rmin)


Asc=856.7mm2

Ast = 856.7mm2(Bottom)

Asc = 856.7mm2 (Top)





As,top/bottom=1/2*Max –ve steel at top of either joint

=0.5*1231.5

=615.75mm2 <1231.5mm2

Hence, O.K.


Act.As(bottom)= 1231.5mm2 (bottom)


Act.As(top)= 1231.5mm2 (top)

Check for shear

d=700-25-28/2=661mm

At left end

(sway to right):

Vu=502.733KN(+ve value from SAP)

(sway to left):

Vu=536.04KN(-ve value from SAP)

Design shear at left end=higher of the two values

=536.04KN

Equivalent shear,

Ve = Vu + 1.6b

Tu = 536.04+ 1.6 x

5.0

43.36 = 652.62KN

Equivalent nominal shear stress,

τve= 661500

1062.652 3

x

x= 1.97 N/mm2 < τcmax(=2.8N/mm2)

pt=3694.513/(500*661)*100%=1.12%

As(top)= 856.7mm2

As(bottom)= 856.7mm2

Act.As(top)=

1231.5mm2


mm2

τc=0.644 N/mm2

69

IS456:2000

cl.41.3.3

IS13920:1993

cl.6.3.5

IS13920:1993

cl.6.3.3

IS456:2000

Table 19

IS456:2000

cl.41.3.1

IS456:2000

Table 20

4.2

τc=0.644 N/mm2

Here, τve>τc, hence transverse reinforcement shall be

provided as follows:

Transverse reinforcement

Asv = )87.0(5.2)87.0( 111 y

vu

y

vu

fd

SV

fdb

ST

Let us use 8mm-4legged vertical stirrups,

Asv=201.06mm2

A clear cover of 25mm is assumed all around.

b1=b-2clear cover - = 500 – 2x25 – 28 = 422mm

d1=D-2clear cover - = 700 – 2x25 – 28= 622mm

201.062=

41587.06225.2

1004.536

41587.0622422

1043.36 36

xxx

x

xxx

x Sv

Sv=150.14mm

Also, Asv>=

y

cve

f87.0b Sv

201.062>=41587.0

644.097.1

x

x 500 x Sv

Sv<=108.75mm

Sv=100mm

Also,

Spacing of stirrups:least of

i)Sv=100mm

ii)d/4=661/4=165.25mm

ii)8* small=8*28=224mm

but >=100mm

Hence, Provide 8mm 4legged vertical

stirrups@100mmc/c over a length of 2d=1322mm.

At mid-span:

(sway to right):

Vu=45.4KN(+ve)

(sway to left):

Vu=45.33KN(-ve)


=519.13KN

Pt= %100*661*500

5.1231 =0.373%

τc=0.42 N/mm2

Equivalent shear,

Ve = Vu + 1.6b

Tu = 519.13 + 1.6 x

5.0

43.36 = 635.71KN

τve= 661500

1071.635 3

x

x = 1.923 N/mm2 < τcmax(=2.8N/mm2)


provided as below.

b1=422mm

d1=622mm

Asv=201.06mm2

Sv=100mm

8mm 4LVS@100m

mc/c

τc=0.42 N/mm2

70

IS456:2000

cl.41.3.3

IS13920:1993

cl.6.3.5

IS13920:1993

cl.6.3.3

IS456:2000

cl.41.3.1

IS456:2000

Table 20

IS456:2000

Table 19

IS456:2000

cl.41.3.3

4.3


Asv = )f87.0(d5.2

SV

)f87.0(db

S

y1

vu

y11

vu

Let us use 12mm -2legged vertical stirrups,

Asv=226.2mm2

100=

41587.06225.2

1013.519

41587.0622422

1043.36 36

xxx

x

xxx

x Sv

Sv=172.8mm

Also, Asv=

y

cve

f87.0b Sv

100=41587.0

42.0923.1

x

x 500 x Sv

Sv=108.67mm

Sv=100mm

Also,


i)Sv=100mm

ii)d/4=661/4=165.25mm


but >=100mm


stirrups@100mmc/c

At right end

(sway to right):


(sway to left):



=536.03KN

Equivalent shear,

Ve = Vu + 1.6b

Tu = 536.03+ 1.6 x

5.0

72.37 = 656.73KN


τve= 661500

1073.656 3

x

x= 1.987 N/mm2 < τcmax(=2.8N/mm2)

pt=3694.513/(500*661)*100%=1.12%

τc=0.644 N/mm2




Asv = )87.0(5.2)87.0( 111 y

vu

y

vu

fd

SV

fdb

ST


Asv=201.06mm2


b1=422mm

d1=622mm

Asv=201.06mm2

Sv=100mm

12mm 2LVS@100

mmc/c

τc=0.644 N/mm2

b1=422mm

71

IS13920:1993

cl.6.3.5

IS456:2000

26.5.1.7a)

4.4

4.5

4.6



201.062=

41587.06225.2

1003.536

41587.0622422

1072.37 36

xxx

x

xxx

x Sv

Sv=148.63mm

Also, Asv=

y

cve

f87.0b Sv

201.062=41587.0

644.0987.1

x

x 500 x Sv

Sv=108.11mm

Sv=100mm

Also,


i)Sv=100mm

ii)d/4=661/4=165.25mm


but >=100mm



Check for Shear Reinforcement Spacing

The spacing of stirrups for torsion shall not exceed the

least of the following

i) x1= short dimension of the stirrups

=500-2x25+8/2+8/2=458 450mm

ii)(x1+y1)/4

y1=long dimension of the stirrups

=700-2x25 + 8/2 + 8/2 = 658mm

Sv=(458+658)/4=279 270mm

iii)300mm

i.e.Sv,max= 270mm

Provision of Shear Reinforcement

Since the span is 1500 mm c/c only,

Provide 2-legged 12mm stirrups @ 100mm c/c

throughout.

Design of Side Reinforcement

Since, d>450mm, provide 0.1% of bD steel

reinforcement along both vertical faces

Arebar=0.001x500x700 = 350mm2

Hence, provide 2-12mm bar along each vertical side.

Act. Arebar=452.4mm2

d1=622mm

Asv=201.06mm2

Sv=100mm

8mm 4LVS@100m

mc/c

Sv,max= 270mm

2-12mm bar along

each vertical side.

72

Ductile Design of 6m span beam

Concrete Grade = M20 Steel Grade = Fe415 (HYSD)


IS13920:1993

cl. 6.1.1

IS13920:1993

cl.6.1.3

IS13920:1993

cl.6.1.2

IS13920:1993

cl.6.1.4

IS13920:1993

cl.6.2.1b

IS13920:1993

cl.6.2.2

1.

2.

i)

ii)

3.

i)

ii)

Known Data



Considering 1 layer of 32mm bar with spacer

bar to be used




Factored Axial Stress = 1.22N/mm2

Axial Stress = 1.22N/mm2 < 0.1 fck





B/D = 500/700 = 0.714 > 0.3

Hence, OK

Span Length, L=6m

L/D = 5.45/.7 = 7.79 > 4 OK



rstmin = 0.24y

ck

f

f = 0.24

415

20

= 2.586 x 10-3 = 0.26%

Ast,min = 0.26 x 500 x 659 / 100 = 856.7mm2


=0.025 x 500 x 659 = 8237.5mm2


Mulim = 2.76bd2 = 2.76x500x6592 = 599.31KN-m

At left support

D=700mm

B=500mm

d=659mm

d’=25+32/2=41mm

d’/d=0.062

Flexural Member

Astmin=856.7mm2

Ast,max=8237.5mm2

Mulim =599.31KN-m

73

IS456-2000

cl.41.4

IS456-2000

cl.41.4.2.1

SP16, Table 50

IS13920:1993

cl.6.2.3

a)

b)


Mu = 715.02 KNm


The longitudinal reinforcement shall be designed

to resist an equivalent bending moment, Me1

Me1 = Mu + Mt


Mt = Tu 7.1

b/D1 = 100.74 * 7.1

500/7001 =

142.22 KNm

Since Mt<Mu, no longitudinal reinforcement shall

be provided on the flexural compression face.

Hence, Me1 = Mu+Mt + = 715.02 + 142.22 =

857.24 KNm

Here,

Mulim<Mu Hence, The section is to be doubly-

reinforced.

Me2=Me1-Mulim = 857.24 – 599.31 =257.93 KNm

d’ = 25 + 32/2 = 41mm (Taking 32mm at

compression zone)

d’/d = 0.0622

2bd

Mu 2

6

659*500

10*24.8573.948

Pt=1.321%

Pc=0.385%


50% of Ast =0.661% > 0.385%

Ast = 1.321/100 x 500*659 = 4352.70mm2(Top)

Asc = 0.661/100 x 500*659 = 2176.35mm2

(Bottom)


Mu = 506.42 KNm



Doubly Reinforced

Section

d’ = 41mm

d’/d 0.1

74

IS456-2000

cl.41.4

IS456-2000

cl.41.4.2.1

SP16, Table 50

IS13920:1993

cl.6.2.3

iii)

a)


Me1 = Mu + Mt


Mt = Tu *7.1

b/D1 = 100.74* 7.1

500/7001 = 142.22

KNm



Hence, Me1 = Mu+Mt = 506.42+ 142.22= 648.64

KNm

Here,


reinforced.

Me2=Me1-Mulim = 648.64 – 599.31 = 49.33 KNm

d’ = 25 + 32/2 = 41mm (Taking 32mm at

compression zone)

d’/d = 0.0622

We have,

2bd

Mu 2

6

659*500

10*64.6482.987

Pt=1.029%

Pc=0.077%


50% of Ast =0.515% > 0.077%

Ast = 1.029/100 x 500*659 =

3390.56mm2(Bottom)

Asc = 0.515/100 x 500*659 = 1695.28mm2 (Top)

For As(top), provide 4-32mm and 2-28mm bars

Act.As(top)= 4448.50mm2 (top)

For As(bottom), provide 4-32mm and 2-16 bars

Act.As(bottom)=3619.115mm2 (bottom)

At mid span


Mu = 168.61 KNm

Doubly Reinforced

Section

d’ = 41mm

d’/d 0.1

As(top)= 4352.70mm2

As(bottom)=

3390.56mm2

Act.As(top)=

4448.50mm2

Act.As(bottom)

=3619.115m2

75

IS456-2000

cl.41.4

IS456-2000

cl.41.4.2.1

IS456-2000

Annex G

G-1.1b

IS13920:1993

cl.6.2.3

IS456-2000

cl.41.4

b)




Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 100.74* 7.1

500/7001 = 142.22

KNm



Hence, Me1 = Mu+Mt = 168.61 + 142.22= 310.83

KNm

Here,


reinforced.

Ast = bdbdf

M

f

f

ck

u

y

ck

2

6.411

5.0

=1437.09mm2 > Ast,min

Pt=0.436%

Providing two longitudinal bars at the bottom as

well


50% of Ast =0.218% <0.26%(rmin)


Asc=856.7mm2

Ast =1437.09mm2(Top)



Mu = 236.48 KNm




Me1 = Mu + Mt

Singly Reinforced

Section

76

IS456-2000

cl.41.4.2.1

IS456-2000

Annex G

G-1.1b

IS13920:1993

cl.6.2.3

IS456-2000

cl.41.4

IS456-2000

cl.41.4.2.1

iv)

a)


Mt = Tu x7.1

b/D1 = 100.74* 7.1

500/7001 = 142.22

KNm



Hence, Me1 = Mu+Mt = 236.48+142.22= 378.70

KNm

Here, Mulim>Mu Hence, The section is to be

singly-reinforced.

Ast = bdbdf

M6.411

f

f5.0

2ck

u

y

ck

=1795.43mm2 (Bottom bars) >Ast,min(856.7mm2)

Pt=0.545%

Providing two longitudinal bars at the bottom as

well


50% of Ast =0.272% >0.26%(rmin)


Asc = 897.72mm2 (Top)

For As(bottom), provide 2-32mm and 2-16mm bars




At right end


Mu =712.44 KNm




Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 97.62* 7.1

500/7001 = 137.82

As(top)= 1437.09mm2

As(bottom)=

1795.43mm2

Act.As(top)=

1608.50mm2

Act.As(bottom)

=2010.62mm2

77

SP16, Table 50

IS13920:1993

cl.6.2.3

IS456-2000

cl.41.4.2

IS456-2000

cl.41.4.2.1

b)

KNm



Hence, Me1 = Mu+Mt + = 712.44+ 137.82= 850.26

KNm

Here, Mulim<Mu Hence, The section is to be

doubly-reinforced.

Me2=Me1-Mulim = 850.26– 599.31 =250.95 KNm

d’ = 25 + 32/2 = 41mm (Taking 32mm at

compression zone)

d’/d = 0.0622

2bd

Mu 2

6

659*500

10*26.850 3.916

Pt=1.306%

Pc=0.369%


50% of Ast =0.653% > 0.369%

Ast = 1.306/100 x 500*659 = 4303.27mm2(Top)

Asc = 0.653/100 x 500*659 = 2151.64mm2

(Bottom)


Mu = 464.41 KNm




Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 97.62* 7.1

500/7001 = 137.82

KNm



Hence, Me1 = Mu+Mt + = 464.41+ 137.82= 602.23

KNm

78

SP16, Table 50

IS13920:1993

cl.6.2.3

IS13920:1993

cl.6.2.4

4.

i)

Here, Mulim<Mu Hence, The section is to be

doubly-reinforced.

Me2=Me1-Mulim = 602.23 – 599.31 = 2.92 KNm

d’ = 25 + 32/2 = 41mm (Taking 32mm at

compression zone)

d’/d = 0.0622

2bd

Mu 2

6

659*500

10*23.6022.773

Pt=0.958%

Pc=0.002%


50% of Ast =0.479% > 0.002%

Ast = 0.958/100 x 500*659 =

3156.61mm2(Bottom)

Asc = 0.479 /100 x 500*659 = 1578.31mm2 (Top)

For As(bottom), provide 4-32mm and 2-16mm bars



Act.As(top)=4448.50mm2 (top)

Minimum As,top/bottom=1/4*Max –ve steel at top of

either joint

=0.32%*500*659

=1054.4mm2 < As top/bottom at any

section

Check for shear

d=700-25-28/2=661mm

At left end

(sway to right):


(sway to left):



=339.348KN

As(top)=

4303.27mm2

As(bottom)=

3156.61mm2

Act.As(top)=

4448.50mm2

Act.As(bot)=3619.12

mm2

79

IS456:2000

cl.41.3.1

IS456:2000

Table 20

IS456:2000

Table 19

IS456:2000

cl.41.3.3

IS13920:1993

cl.6.3.5

Equivalent shear,

Ve = Vu + 1.6b

Tu = 339.348+ 1.6 x

5.0

74.100 =

661.716KN


τve= 659500

10716.661 3

x

x= 2 N/mm2 < τcmax(=2.8N/mm2)

pt=4448.50/(500*659)*100%=1.35%

τc=0.7 N/mm2

Here, τve>τc, hence transverse reinforcement shall

be provided as follows:


Asv = )87.0(5.2)87.0( 111 y

vu

y

vu

fd

SV

fdb

ST


Asv=201.06mm2




201.062=

41587.06185.2

10348.339

41587.0618418

1074.100 36

xxx

x

xxx

x Sv

Sv=119.08mm

Also, Asv=

y

cve

f87.0b Sv

201.062=41587.0

644.02

x

x 500 x Sv

Sv=107.07mm

Sv=100mm

Also,


i)Sv=100mm

ii)d/4=661/4=165.25mm


but >=100mm

Asv=201.06mm2

Sv=100mm

80

IS13920:1993

cl.6.3.3

IS456:2000

Table 19

IS456:2000

cl.41.3.1

IS456:2000

Table 20

IS456:2000

cl.41.3.3

4.2


stirrups@100mmc/c over a length of

2d=1322mm.

At mid-span:

(sway to right):

Vu=231.68KN(+ve)

(sway to left):

Vu=219.125KN(-ve)


=231.68KN

Pt= %100*659*500

43.1795 =0.545%

τc=0.50 N/mm2

Equivalent shear,

Ve = Vu + 1.6b

Tu = 231.68 + 1.6 x

5.0

74.100 =

554.05KN

τve= 661500

1005.554 3

x

x = 1.676 N/mm2

<τcmax(=2.8N/mm2)

Here, τve>τc, hence transverse reinforcement shall

be provided as below.


Asv = )f87.0(d5.2

SV

)f87.0(db

S

y1

vu

y11

vu


Asv=226.2mm2

226.2=

41587.06185.2

1068.231

41587.0618418

1074.100 36

xxx

x

xxx

x Sv

Sv=153mm

Also, Asv=

y

cve

f87.0b Sv

226.2=41587.0

50.0676.1

x

x 500 x Sv

Sv=138.89mm

8mm 4LVS

@100mmc/c

81

IS13920:1993

cl.6.3.5

IS456:2000

26.5.1.7a)

4.3

4.4

4.5

Sv=130mm

Also,


i)Sv=100mm

ii)d/4=661/4=165.25mm


but >=100mm


stirrups@125mmc/c

At right end

Provided as in left support


The spacing of stirrups for torsion shall not

exceed the least of the following


=500-2x25+8/2+8/2=458 450mm

ii)(x1+y1)/4


=700-2x25 + 8/2 + 8/2 = 658mm

Sv=(458+658)/4=279 270mm

iii)300mm

i.e.Sv,max= 270mm


Provide 4-legged 8mm stirrups @ 100mm c/c up

to length

2d=1320mm from left end

Provide 2-legged 12mm stirrups @ 125mm c/c

at mid span

Provide 4-legged 8mm stirrups @ 100mm c/c up

to length

2d=1320mm from right end

82

IS456:2000

26.5.1.3

4.6




Arebar=0.001x500x700 = 350mm2

Hence, provide 2-12mm bar along each vertical

side.


2-12mm bar along

each vertical side.

83

Ductile Design of 7m span inclined beam



IS13920:1993

cl. 6.1.1

IS13920:1993

cl.6.1.3

IS13920:1993

cl.6.1.2

IS13920:1993

cl.6.1.4

IS13920:1993

cl.6.2.1b

IS13920:1993

cl.6.2.2

1.

2.

i)

ii)

3.

Known Data



Considering 1 layer of 32mm bar with spacer bar to

be used




Factored Axial Stress = 0 N/mm2

Axial Stress = 0 N/mm2 < 0.1 fck





B/D = 500/700 = 0.714 > 0.3

Hence, OK

Span Length, L=6.35m

L/D = 6.35/.7 = 10 > 4 OK



rstmin = 0.24y

ck

f

f = 0.24

415

20

= 2.586 x 10-3 = 0.26%

Ast,min = 0.26 x 500 x 659 / 100 = 856.7mm2


=0.025 x 500 x 659 = 8237.5mm2


D=700mm

B=500mm

d=659mm

d’=25+32/2=41mm

d’/d=0.062

Flexural Member

Astmin=856.7mm2

Ast,max=8237.5mm2

84

IS456-2000

cl.41.4

IS456-2000

cl.41.4.2.1

IS13920:1993

cl.6.2.3

i)

ii)

a)

b)

Mulim = 2.76bd2 = 2.76x500x6592 = 599.31KN-m

At left support


Mu = 222.45 KNm




Me1 = Mu + Mt


Mt = Tu 7.1

b/D1 = 21.32 * 7.1

500/7001 = 30.10 KNm



Hence, Me1 = Mu+Mt + = 222.45 + 30.10 =252.55 KNm

Here,


reinforced.

Ast = bdbdf

M

f

f

ck

u

y

ck

2

6.411

5.0

=1145 > Ast,min

Pt=0.35%



50% of Ast =0.18% <0.26%(rmin)


Asc=856.7mm2

Ast =1145.00mm2(Top)



Mu = 65.46 KNm



Mulim =599.31KN-m

Singly Reinforced

Section

d’ = 41mm

d’/d 0.1

85

IS456-2000

cl.41.4

IS456-2000

cl.41.4.2.1

IS456-2000

Annex G

G-1.1b

IS456-2000

cl.41.4

iii)

a)


Me1 = Mu + Mt


Mt = Tu *7.1

b/D1 = 21.32* 7.1

500/7001 = 30.10 KNm



Hence, Me1 = Mu+Mt = 65.46+ 30.10=95.56 KNm

Here,


reinforced.

Ast = bdbdf

M

f

f

ck

u

y

ck

2

6.411

5.0

=413mm2 <Ast,min


Ast =856.70mm2(Bottom)

Asc = 856.70mm2 (Top)


Act.As(top)= 1182mm2 (top)


Act.As(bottom)=981mm2 (bottom)

At mid span


Mu = 0 KNm




Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 13.05* 7.1

500/7001 = 18.42 KNm

Singly Reinforced

Section

As(top)= 1145mm2

As(bottom)=

856.70mm2

Act.As(top)=

1182mm2

Act.As(bottom)

=981mm2

86

IS456-2000

cl.41.4.2.1

IS456-2000

Annex G

G-1.1b

IS456-2000

cl.41.4

IS456-2000

cl.41.4.2.1

IS456-2000

Annex G

G-1.1b

b)



Hence, Me1 = Mu+Mt = 0 + 18.42= 18.42 KNm

Here,


reinforced.

Ast = bdbdf

M

f

f

ck

u

y

ck

2

6.411

5.0

=78mm2 < Ast,min


Ast =856.70mm2(Top)



Mu = 111.42 KNm




Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 13.05* 7.1

500/7001 = 18.42 KNm



Hence, Me1 = Mu+Mt = 111.42+18.42= 129.84 KNm


reinforced.

Ast = bdbdf

M6.411

f

f5.0

2ck

u

y

ck

=566.16mm2 (Bottom bars) <Ast,min(856.7mm2)


Ast =856.70mm2(Bottom)

Asc = 856.70mm2 (Top

Singly Reinforced

Section

As(top)= 856.70mm2

As(bottom)=

87

IS456-2000

cl.41.4

IS456-2000

cl.41.4.2.1

IS456-2000

cl.41.4.2

iv)

a)

b)


Act.As(bottom)= 981 mm2 (bottom)


Act.As(top)=981mm2 (top)

At right end


Mu =180.15 KNm




Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 13.05* 7.1

500/7001 = 18.42 KNm



Hence, Me1 = Mu+Mt = 180.15+18.42= 198.57 KNm


reinforced.

Ast = bdbdf

M6.411

f

f5.0

2ck

u

y

ck

=885mm2 (Bottom bars) >Ast,min(856.7mm2)

Ast =885mm2(Top)

Asc =856.70mm2 (Bottom)


Mu = 77.28 KNm




Me1 = Mu + Mt


856.70mm2

Act.As(top)= 981mm2

Act.As(bottom)

=981mm2

88

IS456-2000

cl.41.4.2.1

IS13920:1993

cl.6.2.4

4.

i)

Mt = Tu x7.1

b/D1 = 13.05* 7.1

500/7001 = 18.42 KNm



Hence, Me1 = Mu+Mt = 95.70 KNm


reinforced.

Ast = bdbdf

M6.411

f

f5.0

2ck

u

y

ck

=414mm2 (Bottom bars) <Ast,min(856.7mm2)


Asc = 856.70mm2 (Top)


Act.As(top)= 1182mm2 (top)


Act.As(bottom)=981mm2 (bottom)

Minimum As,top/bottom=1/4*Max –ve steel at top of

either joint

=0.25*1182

=295mm2 < As top/bottom at any section

Check for shear

d=700-25-25/2=662mm

At left end

(sway to right):

Vu=0KN(+ve value from SAP)

(sway to left):



=143.63KN

Equivalent shear,

As(top)=856.7mm2

As(bottom)=

856.70mm2

Act.As(top)=

1182mm2

Act.As(bottom)

=981mm2

89

IS456:2000

cl.41.3.1

IS456:2000

Table 20

IS456:2000

Table 19

IS456:2000

cl.41.3.3

IS13920:1993

cl.6.3.5

Ve = Vu + 1.6b

Tu = 143.63+ 1.6 x

5.0

32.21 = 221.85KN


τve= 662500

1085.221 3

x

x= 0.64 N/mm2 < τcmax(=2.8N/mm2)

pt=1182/(500*662)*100%=0.35%

τc=0.4 N/mm2




Asv = )87.0(5.2)87.0( 111 y

vu

y

vu

fd

SV

fdb

ST


Asv=100.53mm2




100.53=

41587.06255.2

1063.143

41587.0625425

1032.21 36

xxx

x

xxx

x Sv

Sv=210mm

Also, Asv=

y

cve

f87.0b Sv

100.53=41587.0

4.064.0

x

x 500 x Sv

Sv=302mm

Sv=200mm

Also,


i)Sv=100mm

ii)d/4=662/4=165.5mm


but >=100mm

90

IS13920:1993

cl.6.3.3

IS456:2000

Table 19

IS456:2000

cl.41.3.1

IS456:2000

Table 20

IS13920:1993

cl.6.3.5

IS456:2000

cl.41.3.1

IS456:2000

Table 20

4.2

4.3



At mid-span:

(sway to right):

Vu=25.61KN(+ve)

(sway to left):

Vu=37.82KN(-ve)


=37.82KN

Pt= %100*662*500

7.856 =0.259%

τc=0.36 N/mm2

Equivalent shear,

Ve = Vu + 1.6b

Tu = 37.82 + 1.6 x

5.0

05.13 = 79.58KN

τve= 662500

1058.79 3

x

x = 0.24 N/mm2 < τcmax(=2.8N/mm2)

Here, τve<τc, hence nominal transverse reinforcement

shall be provided as below.


stirrups@300mmc/c

At right end

(sway to right):


(sway to left):

Vu=0KN(-ve value from SAP)


=125.51KN

Equivalent shear,

Ve = Vu + 1.6b

Tu = 125.51+ 1.6 x

5.0

05.13 = 167.27KN


τve= 662500

1027.167 3

x

x= 0.51 N/mm2 < τcmax(=2.8N/mm2)

91

IS456:2000

Table 19

IS456:2000

cl.41.4.3

IS13920:1993

cl.6.3.5

4.4

pt=1182/(500*662)*100%=0.357%

τc=0.4 N/mm2




Asv = )87.0(5.2)87.0( 111 y

vu

y

vu

fd

SV

fdb

ST


Asv=100.53mm2




100.53=

41587.06255.2

1051.125

41587.0625425

1005.13 36

xxx

x

xxx

x Sv

Sv=280mm

Also, Asv=

y

cve

f87.0b Sv

100.53=41587.0

40.051.0

x

x 500 x Sv

Sv=659mm

Also,


i)Sv=100mm

ii)d/4=662/4=165.5mm


but >=100mm







=500-2x25+8/2+8/2=458 450mm

Asv=100.53mm2

Sv=300mm

92

IS456:2000

26.5.1.7a)

IS456:2000

26.5.1.3

4.5

4.6

ii)(x1+y1)/4


=700-2x25 + 8/2 + 8/2 = 658mm

Sv=(458+658)/4=279 270mm

iii)300mm

i.e.Sv,max= 270mm


Provide 2-legged 8mm stirrups @ 200mm c/c up to

length 2d=1325mm from left end

Provide 2-legged 8mm stirrups @ 270mm c/c at mid

span


length 2d=1325mm from right end




Arebar=0.001x500x700 = 350mm2



2-12mm bar along

each vertical side.

93

Ductile Design of 5m span beam



IS13920:1993

cl. 6.1.1

IS13920:1993

cl.6.1.3

IS13920:1993

cl.6.1.2

IS13920:1993

cl.6.1.4

IS13920:1993

cl.6.2.1b

IS13920:1993

cl.6.2.2

1.

2.

2.1

2.2

3.

i)

ii)

Known Data




be used










B/D = 500/700 = 0.714 > 0.3

Hence, OK

Span Length, L=4.45m

L/D = 4.45/.7 = 6.36 > 4 OK



rstmin = 0.24y

ck

f

f = 0.24

415

20

= 2.586 x 10-3 = 0.26%

Ast,min = 0.26 x 500 x 659 / 100 = 856.7mm2


=0.025 x 500 x 659 = 8237.5mm2


Mulim = 2.76bd2 = 2.76x500x6592 = 599.31KN-m

At left support

D=700mm

B=500mm

d=659mm

d’=25+32/2=41mm

d’/d=0.062

Flexural Member

Astmin=856.7mm2

Ast,max=8237.5mm2

Mulim =599.31KN-m

94

IS456-2000

cl.41.4

IS456-2000

cl.41.4.2.1

SP16, Table

50

IS13920:1993

cl.6.2.3

a)

b)


Mu = 615.911 KNm




Me1 = Mu + Mt


Mt = Tu 7.1

b/D1 = 30.037x 7.1

500/7001 =

42.405KNm



Hence, Me1 = Mu+Mt = 615.91+ 42.41= 658.32 KNm

Here,

Mulim<Mu, hence, the section is to be doubly-

reinforced.

Me2=Me1-Mulim = 658.316 – 599.31 =59.006 KNm

d’ = 25 + 32/2 = 41mm (Taking 32mm at

compression zone)

d’/d = 0.0622

2bd

Mu 2

6

659*500

10*316.6583.03

Pt=1.029%

Pc=0.077%


50% of Ast =0.5145% > 0.045%

Ast = 1.029/100 x 500*659 = 3390.55mm2(Top)

Asc = 0.5145/100 x 500*659 = 1695.28mm2 (Bottom)

Provide at least 6-28 =3694.51mm2 (Top)

Provide at least 2-28 =1231.5mm2 (Bottom)


Mu = 453.95 KNm


Doubly Reinforced

Section

d’ = 41mm

d’/d 0.1

95

IS456-2000

cl.41.4

IS456-2000

cl.41.4.2.1

IS13920:1993

cl.6.2.3

iii)

a)



Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 36.43x 7.1

500/7001 = 53.544 KNm



Hence, Me1 = Mu+Mt = 453.95 + 53.544= 507.49 KNm

Here,

Mulim<Mu Hence, The section is to be singly-

reinforced.

Ast = bdbdf

M

f

f

ck

u

y

ck

2

6.411

5.0

=2540.40mm2 > Ast,min


Provide 4-32mm

Act.Ast=3217mm2 (Bottom bars)

Pt=0.98%

Providing two longitudinal bars at the top as well


50% of Ast =0.49% >0.26%(rmin)

So,providing area of steel,

Asc=1614.55mm2


Asc = 1614.55mm2 (Top)




Act.As(top)= 3078.76 mm2 (top)

At mid span


Mu = 96.039 KNm

Singly Reinforced

Section

As(top)= 3390.55mm2

As(bottom)=

2540.40mm2

Act.As(top)=

3694.513mm2


mm2

96

IS456-2000

cl.41.4

IS456-2000

cl.41.4.2.1

IS456-2000

Annex G

G-1.1b

IS13920:1993

cl.6.2.3

b)




Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 37.927x 7.1

500/7001 = 53.544

KNm



Hence, Me1 = Mu+Mt = 90.039 + 53.544= 143.583

KNm

Here,


reinforced.

Ast = bdbdf

M

f

f

ck

u

y

ck

2

6.411

5.0

=628.65mm2 < Ast,min


Provide 2-28mm


Pt=0.37%



50% of Ast =0.185% <0.26%(rmin)


Asc=856.7mm2

Ast = 856.7mm2(Top)



Mu = 64.33 KNm




Singly Reinforced

Section

97

IS456-2000

cl.41.4

IS456-2000

cl.41.4.2.1

IS456-2000

Annex G

G-1.1b

IS13920:1993

cl.6.2.3

iv)

a)

Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 39.06x 7.1

500/7001 = 55.143 KNm



Hence, Me1 = Mu+Mt = 64.33+ 55.143= 119.47 KNm


reinforced.

Ast = bdbdf

M6.411

f

f5.0

2ck

u

y

ck

=519.36mm2 (Top Bar)<Ast,min(856.7mm2)


Provide 2-28mm


Pt=0.37%



50% of Ast =0.185% <0.26%(rmin)


Asc=856.7mm2







At right end


Mu =576.88 KNm




98

IS456-2000

cl.41.4

IS456-2000

cl.41.4.2.1

SP16, Table

50

IS13920:1993

cl.6.2.3

IS456-2000

cl.41.4.2

IS456-2000

cl.41.4.2.1

b)

Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 39.06 x 7.1

500/7001 = 55.143

KNm



Hence, Me1 = Mu+Mt + = 576.88+ 55.143= 632.023

KNm

Here, Mulim<Mu Hence, The section is to be doubly-

reinforced.

Me2=Me1-Mulim = 632.023– 599.31 =32.713KNm

d’ = 25 + 32/2 = 41mm (Taking 32mm at

compression zone)

d’/d = 0.0622

2bd

Mu 2

6

659*500

10*023.632 2.91

Pt=.998%

Pc=0.045%


50% of Ast =0.499% > 0.248%

Ast = .998/100 x 500*659 = 3288.41mm2(Top)

Asc = 0.499/100 x 500*659 = 1644.21mm2 (Bottom)


Mu = 485.44 KNm




Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 37.92x 7.1

500/7001 = 53.53 KNm



99

IS13920:1993

cl.6.2.3

IS13920:1993

cl.6.2.2

IS456:2000

cl.41.3.1

4.

4.1

Hence, Me1 = Mu+Mt = 485.44 + 53.25=538.69 KNm

Here, Mulim<Mu Hence, The section is to be singly-

reinforced.

Ast = bdbdf

M

f

f

ck

u

y

ck

2

6.411

5.0

=2736.90mm2 >Ast,min

Providing area of steel=2736.90mm2

Provide 6-28mm


Pt=1.12%



50% of Ast =0.46% <0.26%(rmin)


Asc=1847mm2

Ast = 1847mm2(Top)



=0.309%*500*659

=1018.2mm2

Check for shear

d=700-25-28/2=661mm

At left end

(sway to right):

Vu=144.621(+ve value from SAP)

(sway to left):



=292.593KN

Equivalent shear,

Ve = Vu + 1.6b

Tu = 292.593+ 1.6 x

5.0

93.37 = 413.97KN


Act.As(top)=

3694.512mm2

Act.As(bottom)=

1847mm2

100

IS456:2000

Table 20

IS456:2000

Table 19

IS456:2000

cl.41.3.3

IS13920:1993

cl.6.3.5

4.2

τve= 661500

1097.413 3

x

x= 1.25 N/mm2 < τcmax(=2.8N/mm2)

pt=3694.513/(500*661)*100%=1.12%

τc=0.644 N/mm2




Asv = )87.0(5.2)87.0( 111 y

vu

y

vu

fd

SV

fdb

ST


Asv=226.19mm2




226.19=

41587.06225.2

10593.292

41587.0622422

1093.37 36

xxx

x

xxx

x Sv

Sv=109.11mm

Also,


i)Sv=100mm

ii)d/4=661/4=165.25mm


but >=100mm



At mid-span:

(sway to right):

Vu=188.604KN(+ve)

(sway to left):

Vu=213.941KN(-ve)


=213.941KN

Asv=100.53mm2

Sv=100mm

101

IS456:2000

Table 19

IS456:2000

cl.41.3.1

IS456:2000

Table 20

IS456:2000

cl.41.3.3

IS13920:1993

cl.6.3.5

4.3

Pt= %100*661*500

1847 =0.558%

τc=0.50N/mm2

Equivalent shear,

Ve = Vu + 1.6b

Tu = 213.914+ 1.6 x

5.0

927.37 = 335.31KN

τve= 661500

1031.335 3

x

x = 1.014 N/mm2 < τcmax(=2.8N/mm2)


provided as below.


Asv = )f87.0(d5.2

SV

)f87.0(db

S

y1

vu

y11

vu


Asv=100.53mm2

100.53=

41587.06225.2

1094.213

41587.0622422

1093.37 36

xxx

x

xxx

x Sv

Sv=128.67mm

Also, Asv=

y

cve

f87.0b Sv

100.530=41587.0

5.0014.1

x

x 500 x Sv

Sv=141.23mm

Sv=125mm

Also,


i)Sv=100mm

ii)d/4=661/4=165.25mm


but >=100mm


stirrups@125mmc/c

At right end

(sway to right):

102

IS13920:1993

cl.6.3.3

IS456:2000

cl.41.3.1

IS456:2000

Table 20

IS456:2000

Table 19

IS456:2000

cl.41.3.3


(sway to left):



=275.23KN

Equivalent shear,

Ve = Vu + 1.6b

Tu = 275.23+ 1.6 x

5.0

06.39 = 400.22KN


τve= 661500

1022.400 3

x

x= 1.211 N/mm2 < τcmax(=2.8N/mm2)

pt=3694.513/(500*661)*100%=1.12%

τc=0.644 N/mm2




Asv = )87.0(5.2)87.0( 111 y

vu

y

vu

fd

SV

fdb

ST


Asv=100.53mm2




100.53=

41587.06225.2

1023.275

41587.0622422

1006.39 36

xxx

x

xxx

x Sv

Sv=110.87mm

Also, Asv=

y

cve

f87.0b Sv

100.53=41587.0

644.012.1

x

x 500 x Sv

Sv=152.51mm

Sv=100mm

Also,

103

IS13920:1993

cl.6.3.5

IS456:2000

26.5.1.7a)

IS456:2000

26.5.1.3

4.4

4.5

4.6


i)Sv=100mm

ii)d/4=661/4=165.25mm


but >=100mm







=500-2x25+8/2+8/2=458 450mm

ii)(x1+y1)/4


=700-2x25 + 8/2 + 8/2 = 658mm

Sv=(458+658)/4=279 270mm

iii)300mm

i.e.Sv,max= 270mm



length

2d=1322mm from left end


span


length

2d=1322mm from right end




Arebar=0.001x500x700 = 350mm2



8mm 2LVS@125m

mc/c

2-12mm bar along

each vertical side.

104

Ductile Design of 2.5m span beam



IS13920:1993

cl. 6.1.1

IS13920:1993

cl.6.1.3

IS13920:1993

cl.6.1.2

IS13920:1993

cl.6.1.4

IS13920:1993

cl.6.2.1b

1.

2.

2.1

2.2

3.

i)

a)

Known Data




be used










B/D = 500/700 = 0.714 > 0.3

Hence, OK



rstmin = 0.24y

ck

f

f = 0.24

415

20

= 2.586 x 10-3 = 0.26%

Ast,min = 0.26 x 500 x 659 / 100 = 856.7mm2


=0.025 x 500 x 659 = 8237.5mm2


Mulim = 2.76bd2 = 2.76x500x6592 = 599.31KN-m

At left support

D=700mm

B=500mm

d=659mm

d’=25+32/2=41mm

d’/d=0.062

Flexural Member

Astmin=856.7mm2

Ast,max=8237.5mm2

Mulim =599.31KN-m

105

IS13920:1993

cl.6.2.2

IS456-2000

cl.41.4

IS456-2000

cl.41.4.2.1

SP16, Table

50

IS13920:1993

cl.6.2.3

IS456-2000

cl.41.4

b)


Mu = 579.6 KNm




Me1 = Mu + Mt


Mt = Tu 7.1

b/D1 = 36.43x 7.1

500/7001 = 51.43 KNm



Hence, Me1 = Mu+Mt = 579.6 + 51.43= 631.03 KNm

Here,


reinforced.

Me2=Me1-Mulim = 631.03 – 599.31 =31.72 KNm

d’ = 25 + 32/2 = 41mm (Taking 32mm at

compression zone)

d’/d = 0.0622

2bd

Mu 2

6

659*500

10*03.6312.906

Pt=0.999%

Pc=0.047%


50% of Ast =0.499% > 0.047%

Ast = 0.999/100 x 500*659 = 3294.63mm2(Top)

Asc = 0.499/100 x 500*659 = 1647.32mm2 (Bottom)


Mu = 553.93 KNm




Me1 = Mu + Mt

Doubly Reinforced

Section

d’ = 41mm

d’/d 0.1

106

IS456-2000

cl.41.4.2.1

SP16, Table

50

IS13920:1993

cl.6.2.3

IS456-2000

iii)

a)


Mt = Tu x7.1

b/D1 = 36.43x 7.1

500/7001 = 51.43 KNm



Hence, Me1 = Mu+Mt = 553.93 + 51.43= 605.36 KNm

Here,


reinforced.

Me2=Me1-Mulim = 605.36 – 599.31 = 6.05 KNm

d’ = 25 + 32/2 = 41mm (Taking 32mm at

compression zone)

d’/d = 0.0622

We have,

2bd

Mu 2

6

659*500

10*36.6052.788

Pt=0.964%

Pc=0.008%


50% of Ast =0.482% > 0.008%

Ast = 0.964/100 x 500*659 = 3176.38mm2(Bottom)

Asc = 0.482/100 x 500*659 = 1588.2mm2 (Top)





At mid span


Mu = 90 KNm




Me1 = Mu + Mt

Doubly Reinforced

Section

d’ = 41mm

d’/d 0.1

As(top)= 3294.63mm2

As(bottom)=3176.38mm2

Act.As(top)=3694.513

mm2

Act.As(bottom)=

3694.513mm2

107

cl.41.4

IS456-2000

cl.41.4.2.1

IS456-2000

Annex G

G-1.1b

IS13920:1993

cl.6.2.3

IS456-2000

cl.41.4.2

b)


Mt = Tu x7.1

b/D1 = 36.43x 7.1

500/7001 = 51.43 KNm



Hence, Me1 = Mu+Mt = 90 + 51.43= 141.43 KNm

Here,


reinforced.

Ast = bdbdf

M

f

f

ck

u

y

ck

2

6.411

5.0

=387.925mm2 < Ast,min


Provide 2-28mm


Pt=0.37%



50% of Ast =0.185% <0.26%(rmin)


Asc=856.7mm2

Ast = 856.7mm2(Top)



Mu = 71.49 KNm




Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 36.43x 7.1

500/7001 = 51.43 KNm


Singly Reinforced

Section

108

IS456-2000

Annex G

G-1.1b

IS13920:1993

cl.6.2.3

IS456-2000

cl.41.4

IS456-2000

iv)

a)


Hence, Me1 = Mu+Mt = 71.49+ 51.43= 122.92 KNm


reinforced.

Ast = bdbdf

M6.411

f

f5.0

2ck

u

y

ck

=306.53mm2 (Bottom bars)<Ast,min(856.7mm2)


Provide 2-28mm


Pt=0.37%

Providing two longitudinal bars at the bottom as well.


50% of Ast =0.185% <0.26%(rmin)


Asc=856.7mm2







At right end


Mu =743.91 KNm




Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 37.72x 7.1

500/7001 = 53.25 KNm

Singly Reinforced

Section

As(top)= 856.7mm2

As(bottom)= 856.7mm2

Act.As(top)=

1231.5mm2


mm2

109

cl.41.4.2.1

SP16, Table

50

IS13920:1993

cl.6.2.3

IS456-2000

cl.41.4.2

IS456-2000

cl.41.4.2

b)



Hence, Me1 = Mu+Mt + = 743.91+ 53.25= 797.16 KNm


reinforced.

Me2=Me1-Mulim = 797.16– 599.31 =197.85 KNm

d’ = 25 + 32/2 = 41mm (Taking 32mm at

compression zone)

d’/d = 0.0622

2bd

Mu 2

6

659*500

10*16.797 3.67

Pt=1.236%

Pc=0.248%


50% of Ast =0.618% > 0.248%

Ast = 1.236/100 x 500*659 = 4072.62mm2(Top)

Asc = 0.618/100 x 500*659 = 2036.31mm2 (Bottom)


Mu = 646.23 KNm




Me1 = Mu + Mt


Mt = Tu x7.1

b/D1 = 37.72x 7.1

500/7001 = 53.25 KNm



Hence, Me1 = Mu+Mt = 646.23 + 53.25= 699.48 KNm


reinforced.

Me2=Me1-Mulim = 699.48 – 599.31 = 100.17 KNm

d’ = 25 + 32/2 = 41mm (Taking 32mm at

Doubly Reinforced

Section

d’ = 41mm

d’/d 0.1

Doubly Reinforced

Section

110

SP16, Table

50

IS13920:1993

cl.6.2.3

IS13920:1993

cl.6.2.3

IS13920:1993

cl.6.3.3

IS456:2000

cl.41.3.1

4.

4.1

compression zone)

d’/d = 0.0622

2bd

Mu 2

6

659*500

10*48.6993.22

Pt=1.098%

Pc=0.549%


50% of Ast =0.549% > 0.149%

Ast = 1.098/100 x 500*659 = 3616.62mm2(Bottom)

Asc = 0.549 /100 x 500*659 = 1808.31mm2 (Top)






=0.5*3694.51

=1847.255mm2 <4310.26mm2

Hence, O.K.

Check for shear

d=700-25-28/2=661mm

At left end

(sway to right):


(sway to left):



=536.04KN

Equivalent shear,

Ve = Vu + 1.6b

Tu = 536.04+ 1.6 x

5.0

43.36 = 652.62KN


d’ = 41mm

d’/d 0.1

As(top)= 4072.62mm2

As(bottom)=

3616.62mm2

Act.As(top)=

4310.26mm2

Act.As(bottom)=

3694.513mm2

111

IS456:2000

Table 20

IS456:2000

Table 19

IS456:2000

cl.41.3.3

IS13920:1993

cl.6.3.5

τve= 661500

1062.652 3

x

x= 1.97 N/mm2 < τcmax(=2.8N/mm2)

pt=3694.513/(500*661)*100%=1.12%

τc=0.644 N/mm2




Asv = )87.0(5.2)87.0( 111 y

vu

y

vu

fd

SV

fdb

ST


Asv=201.06mm2




201.062=

41587.06225.2

1004.536

41587.0622422

1043.36 36

xxx

x

xxx

x Sv

Sv=150.14mm

Also, Asv >=

y

cve

f87.0b Sv

201.062>=41587.0

644.097.1

x

x 500 x Sv

Sv<=108.75mm

Sv=100mm

Also,


i)Sv=100mm

ii)d/4=661/4=165.25mm


but >=100mm



τc=0.644 N/mm2

b1=422mm

d1=622mm

Asv=201.06mm2

Sv=100mm

8mm 4LVS@100m

mc/c

112

IS13920:1993

cl.6.3.3

IS456:2000

Table 19

IS456:2000

cl.41.3.1

IS456:2000

Table 20

IS456:2000

cl.41.3.3

4.2

At mid-span:

(sway to right):

Vu=519.13KN(+ve)

(sway to left):

Vu=451.88KN(-ve)


=519.13KN

Pt= %100*661*500

5.1231 =0.373%

τc=0.42 N/mm2

Equivalent shear,

Ve = Vu + 1.6b

Tu = 519.13 + 1.6 x

5.0

43.36 = 635.71KN

τve= 661500

1071.635 3

x

x = 1.923 N/mm2 < τcmax(=2.8N/mm2)


provided as below.


Asv = )f87.0(d5.2

SV

)f87.0(db

S

y1

vu

y11

vu


Asv=226.2mm2

100=

41587.06225.2

1013.519

41587.0622422

1043.36 36

xxx

x

xxx

x Sv

Sv=172.8mm

Also, Asv>=

y

cve

f87.0b Sv

100>=41587.0

42.0923.1

x

x 500 x Sv

Sv<=108.67mm

Sv=100mm

Also,


τc=0.42 N/mm2

Asv=226.2mm2

Sv=100mm

113

IS13920:1993

cl.6.3.5

IS13920:1993

cl.6.3.3

IS456:2000

cl.41.3.1

IS456:2000

Table 20

IS456:2000

Table 19

IS456:2000

cl.41.3.3

4.3

i)Sv=100mm

ii)d/4=661/4=165.25mm


but >=100mm


stirrups@100mmc/c

At right end

(sway to right):


(sway to left):



=536.03KN

Equivalent shear,

Ve = Vu + 1.6b

Tu = 536.03+ 1.6 x

5.0

72.37 = 656.73KN


τve= 661500

1073.656 3

x

x= 1.987 N/mm2 < τcmax(=2.8N/mm2)

pt=3694.513/(500*661)*100%=1.12%

τc=0.644 N/mm2




Asv = )87.0(5.2)87.0( 111 y

vu

y

vu

fd

SV

fdb

ST


Asv=201.06mm2




201.062=

41587.06225.2

1003.536

41587.0622422

1072.37 36

xxx

x

xxx

x Sv

12mm 2LVS@100m

mc/c

τc=0.644 N/mm2

b1=422mm

d1=622mm

114

IS13920:1993

cl.6.3.5

IS456:2000

26.5.1.7a)

4.4

4.5

Sv=148.63mm

Also, Asv=

y

cve

f87.0b Sv

201.062=41587.0

644.0987.1

x

x 500 x Sv

Sv=108.11mm

Sv=100mm

Also,


i)Sv=100mm

ii)d/4=661/4=165.25mm


but >=100mm







=500-2x25+8/2+8/2=458 450mm

ii)(x1+y1)/4


=700-2x25 + 8/2 + 8/2 = 658mm

Sv=(458+658)/4=279 270mm

iii)300mm

i.e.Sv,max= 270mm





span



Asv=201.06mm2

Sv=100mm

8mm 4LVS@

100mmc/c

Sv,max= 270mm

115

IS456:2000

26.5.1.3

4.6




Arebar=0.001x500x700 = 350mm2



2-12mm bar along

each vertical side.

116

Ductile Design of 7m span Secondary beam



IS13920:1993

cl. 6.1.1

IS13920:1993

cl.6.1.3

IS13920:1993

cl.6.1.2

IS13920:1993

cl.6.1.4

IS13920:1993

cl.6.2.1b

IS13920:1993

cl.6.2.

1.

2.

2.1

2.2

3.

i)

ii)

Known Data




be used










B/D = 250/450 = 0.556 > 0.3

Hence, OK

Span Length, L=7-0.5/2-0.5/2 =6.5m

L/D = 6.5/.45 = 14.44 > 4 OK



rstmin = 0.24y

ck

f

f = 0.24

415

20

= 2.586 x 10-3 = 0.26%

Ast,min = 0.26 x 250 x 411 / 100 = 267.15mm2


=0.025 x 250 x 411 = 2568.75mm2


Mulim = 2.76bd2 = 2.76x250x4112 = 116.56KN-m

At left support

D=450mm

B=250mm

d=411mm

d’=25+28/2=39mm

d’/d=0.059

Flexural Member

Main beam = 0.5m

wide

Astmin=267.15mm2

Ast,max=2568.75mm2

Mulim =599.31KN-m

117

SP16, Table

50

IS13920:1993

cl.6.2.3

a)

b)

iii)

a)


Mu = 160.26 KNm


Since, torsional moment is too small we neglect torsion

in this beam.

Here,

Mulim<Mu, hence, the section is to be doubly-

reinforced.

d’/d = 0.059

Let us take the ratio as 0.1

2bd

Mu 2

6

411*250

10*26.1603.79

Pt=1.276%

Pc=0.336%


50% of Ast =0.638% > 0.336%

Ast = 1.276/100 x 250*411 = 1312mm2(Top)

Asc = 656mm2 (Bottom)


Since , the value is negative as extracted from SAP

Mu =0 KNm


So, it is governed by hogging moment.


Act.As(top)= 1432.56 mm2 (bottom)


Act.As(bottom)= 1231 mm2 (top)

At mid span


Mu = 117.20 KNm

Torsional Moment, Tu = 0.76 KNm (negligible)

Here,

Mulim>Mu Hence, The section is to be doubly-

reinforced.

Doubly Reinforced

Section

d’ = 39mm

d’/d 0.1

Doubly Reinforced

Section

118

SP16, Table

50

IS13920:1993

cl.6.2.3

b)

iv)

a)

b)

d’/d = 0.059

Let us take the ratio as 0.1

2bd

Mu 2

6

411*250

10*20.1172.78

Pt=0.968%

Pc=0.012%


50% of Ast =0.484% <0.012%

Ast = 0.968*250*411/100= 994.62mm2(Bottom)

Asc = 497.31mm2 (Top)


Mu = 0 KNm (no hogging moment)


So, design in the mid span is governed by sagging

moment.





At right end


Mu =160.15 KNm


which is same as the left end hogging moment.

So, Ast = 1312mm2(Top)

And, Asc = 656mm2 (Bottom)


Mu = 22.53 KNm


Here, Mulim<Mu Hence, The section is to be singly-

reinforced.

Ast = bdbdf

M

f

f

ck

u

y

ck

2

6.411

5.0

Act.As(top)=

1231.5mm2

Act.As(bottom)=

1231.5mm2

Singly Reinforced

Section

119

IS13920:1993

cl.6.3.3

IS456:2000

Table 20

IS456:2000

Table 19

IS456:2000

cl.41.3.3

4.

4.1

=156mm2 <Ast,min


Ast = 267.15mm2 (Top)



Act.As(top)= 1432.56 mm2 (bottom)


Act.As(bottom)= 1231 mm2 (top)


=0.25*1.276%*250*411

=327.78mm2(lesser than area at any section)

Check for shear

d=411mm

At left end

Design shear at left end=154.34KN

Tu = 0.76 KNm

Equivalent shear,

Ve = Vu + 1.6b

Tu = 154.34+ 1.6 x

5.0

76.0 = 156.77KN


τve= 411250

1077.156 3

x

x= 1.53 N/mm2 < τcmax(=2.8N/mm2)

pt=1432.56/(250*411)*100%=1.39%

τc=0.68 N/mm2




Asv = )87.0(5.2)87.0( 111 y

vu

y

vu

fd

SV

fdb

ST


120

IS13920:1993

cl.6.3.5

4.2

4.3

Asv=100.53mm2




100.53=

41587.03725.2

1077.156

41587.0372172

1076.0 36

xxx

x

xxx

x Sv

Sv=201.15mm

Also,


i)Sv=100mm

ii)d/4=411/4=102.75mm


but >=100mm



At mid-span:

(sway to right):


Pt= %100*411*250

1231 =1.2%

τc=0.65N/mm2

Ve = Vu + 1.6b

Tu = 9.92+ 1.6 x

5.0

76.0 = 12.35KN


τve= 411250

1035.12 3

x

x= 0.12 N/mm2 < τcmax(=2.8N/mm2)

Here, τve<τc, hence transverse reinforcement shall be

nominal


Asv=100.53mm2


stirrups@300mmc/c.

At right end


121

4.4

Tu = 0.76 KNm

Equivalent shear,

Ve = Vu + 1.6b

Tu = 154.34+ 1.6 x

5.0

76.0 = 156.77KN


τve= 411250

1077.156 3

x

x= 1.53 N/mm2 < τcmax(=2.8N/mm2)

pt=1432.56/(250*411)*100%=1.39%

τc=0.68 N/mm2




Asv = )87.0(5.2)87.0( 111 y

vu

y

vu

fd

SV

fdb

ST


Asv=100.53mm2




100.53=

41587.03725.2

1077.156

41587.0372172

1076.0 36

xxx

x

xxx

x Sv

Sv=201.15mm

Also,


i)Sv=100mm

ii)d/4=411/4=102.75mm


but >=100mm






122

4.5


=250-2x25+8/2+8/2=208

ii)(x1+y1)/4


=450-2x25 + 8/2 + 8/2 = 408mm

Sv=(208+408)/4=308

iii)300mm

i.e.Sv,max= 300mm





span



123

4.2.3 DESIGN OF COLUMN

Design of Square Column


Reference Step Calculations Output

IS 3920:1993

cl.7.1.1

1.

2.

Column ID: CG-14 (ground floor)

`

Known data:

Overall Depth of Column, D = 550mm

Width of Column, B = 550mm

Height, L = 3.2m

Clear height, l = 2.65m

Assume following data:

Clear cover, d= 40mm

diameter of longitudinal reinforcement, ø = 32mm

So, effective cover, d’= 40+32/2 = 56mm

Check for Axial Stress:

Lowest factored Axial Load = 4989.90 KN

Factored Axial Stress = Mpax

x49.16

550550

100090.4989

Axial Stress = 16.49> 0.1fck(2.5)

Hence, design as Column Member.

D =550mm

B =550mm

L = 3.2m

effective cover

d’= 56mm

Lowest among all

load combination

550mm

534.42KN/m

16.404KN/m

550mm

124

IS13920:1993

cl.7.1.2

IS13920:1993

cl.7.1.3

IS 456 : 2000

table 28

IS 456:2000

cl.25.1.2

IS 456:2000,

cl.26.5.3.1

IS 456:2000

cl.25.4

3.

4.

Check for Member Size:

Width of Column, B = 550mm > 200mm

Depth of Column, D = 550mm

B/D = 550/550 = 1 > 0.4

Hence, OK

Effective Length, le = 0.65 x l = 0.65 x 3.20 = 2.08m

Check for Short and Slender Column:

le/D = (2.08x1000)/550 = 3.78<12,(short column),ok

Limiting Longitudinal Reinforcement:

Min. Reinforcement,

= 0.8% of BD

= 0.8 x 550 x 550/100 = 2420 mm2

Max. Reinforcement, Max. Asc = 4% of BD

= 0.04 x 550 x 550 = 12100 mm2

But in extreme case, Max. Asc = 6% of BD

= 0.06 x 550 x 550 = 18150 mm2

Design for section:

Design of column for Max. Moment:

Data from sap analysis,

Pmax= -3731.833 KN

Mx = 16.404 KNm

My = -534.42 KNm

Min. eccentricity:

emin= l/500 + D/30 ≥ 20 mm

where,

l= unsupported length of the column

D= lateral dimension in plane of bending

emin= mmx

2073.2430

550

500

100020.3

Moment due to minimum eccentricity:

Min. Moment = Pu x emin = 3731.833 x 24.73/1000

=92.28KNm

le = 2.08m

Min.

Asc=2420mm2

Max. Asc =

12100mm2

extreme case,

Asc=18150mm2

125

SP16

chart 44

IS456:2000

cl.39.6

4.

Me = 92.28KNm > Mx

Mx= 92.28KNm

Me = 92.28KNm < MY

MY= 534.42KNm

Mu=1.15 22

yx MM = 623.668 KNm

493.055055025

0003731.833x1

xxBDf

P

ck

u

15.055055025

10668.6232

6

2

xx

x

BDf

M

ck

u

1.0550

56'

D

d

Assume reinforcement is uniformly distributed on

four sides,

ckf

p0.15

p = 0.15x25 = 3.75%

Hence adopt p= 3.4%

Now,

Check for Biaxial Moment

For

p/fck= 0.136 and Pu/fckBD=0.493

135.02

BDf

M

ck

u

Muxl=0.135x25x550x5502=561.52 KN m

scckygckuz A)f45.0f75.0(Af45.0p

= 6488.625 KN

αn = 0.667+1.667xPu/Puz

= 1.625

okM

M

M

Mn

uy

uy

n

uxl

ux ,197.01

Me = 92.28KNm

Mu=623.668KNm

Muxl

=561.52 KN m

Muyl =

561.52 KN m

Puz =6488.625 KN

αn = 1.625

126

IS 456:2000

table 19

IS 456:2000

cl. 40.2.2

IS 456:2000

cl

26.5.3.2.C.2

5.

6.

Design for Shear:

Percentage of steel provided= 3.4%

Design Shear Strength of concrete,

2/92.0 mmNc

Considering lowest, Pu = 4989.90 KN

For members subjected to axial compression Pu , the

design shear strength of concrete τc , shall be

multiplied by the following factor:

δ = 1+ ckg

u

fxA

Px3≤ 1.5

=1+ 97.2550*550*25

1000*90.49893

x> 1.5

Multiplying factor, δ = 1.5

Actual, Mpaxc 38.192.05.1

Shear capacity of the section,

Vc= 1.38*550*550/1000 = 417.45 KN

Shear force as per sap analysis

Vux = -286.396 KN

Vuy = 272.847 KN

Hence, the shear capacity of the column section

exceeds the induced shear force. So, shear

reinforcement is not required.

Design of Lateral Ties:

Diameter of ties:

øt ≥ not less than 6mm

≥ 0.25*maximum diameter of longitudinal

reinforcement

= 0.25*32 =8mm

Hence, adopt ties of 8mmø

Spacing of the ties:

Sv ≤ half the least lateral dimension of the

δ = 1.5

Act. τc=0.92 Mpa

Max. among all

load combination

127

IS

13920:1993

cl.7.3.3

IS

13920:1993

cl.7.4.8

IS

13920:1993

cl.7.4.6

IS

13920:1993

cl.7.4.1

compression member=275mm

Thus, provide 8mm ø lateral ties @ 200 c/c in central

part.

Area of cross-section of bar forming rectangular hoop

to be used as confining links

1

A

A

f

fhS18.0A

k

g

y

cksh

Ak = (550-2 x 40 + 2 x 8)2 =236196 mm2

h= Max of

mm

mm

5.1174

)4040550(

5.1174

)4040550(

where 3 is no. of bars in each face of column section

= 176.67 mm

Area of 8 mm ø bar = 50.26 mm2

Therefore,

1

236196

550550

415

255.11718.026.50

xxSx

or, S= 140.52 mm

Spacing of hoop should be least of

mm

DimensionLateralimumof

100

5.1374

550min4

1

but need not be less than 75 mm

Provide 8 mm ø links @ 100 mm c/c for a distance Lo

which shall not be less than

mm

mmSpanClearof

mmDimensionLateralerL

450

33.5336

32006

1

550arg

Hence, Provide 8 mm ø links @ 100 mm c/c for a

distance

Lo = 550 mm on either side from the joint.

øt = 8 mm

200mm c/c

Lo = 550 mm

8 mm ø links @

100 mm c/c

128

7. Splicing vertical bars

Maximum 50% to be spliced at one section.

Splicing in the middle half of the column height.

Clear length of lap = Development length (Ld)

Ld mm10326.1*25.1*4.1*4

32*415*87.0 =

So, lap length = 1050 mm

129

4.2.4 Staircase

Dog Legged Staircase

Concrete Grade=M20 Steel Grade = Fe415 (HYSD)

Ref. Step Calculations Output

1.

2.

i.

Known Data

Riser Height, R=150 mm

Tread Length=300 mm

Floor Height=3.2m

Flight Width, W=2.5m

No. of Risers in the flights=11

No. of Treads in the flights=10

Length of the flights=3.0m

Load Calculation

Flight,

Assuming Slab Thickness,D=250 mm

Self Wt.of Soffit Slab=0.335*0.25*25=2.1 KN/m per

step

1500mm 1500mm 3000mm

130

ii.

3.

4.

Wt. of Steps=0.5*0.15*0.3*25=0.56 KN/m per step

Total load per step=2.66 KN/m

Total load per m2=2.66/0.3=8.87 KN

Considering 1m Width of Slab,

Floor Finishing= 1.0 KN/m

Live Load=5 KN/m

Total Characteristics Load=14.87 KN/m

Design Load=1.5 x 14.87=22.305 KN/m

Landing


Self Wt. of Slab= 25 x .25=6.25 KN/m

Floor Finishing=1.0 KN/m

Live Load=5 KN/m


Design Load=1.5 x 12.25=18.375KN/m

Analysis

For Upper and Lower Flight,

Moment at ,

About Mid span,

Mmid=95.96 KN/m

Clear Cover=20 mm,16 mm dia. bars

Effecrtive Depth= 250-20-8= 222 mm

Design For Main Reinforcement,

For Mid Span

202221000

415122241587.01096.95 6

xx

xAxxAxxx st

st

Ast=1373.55 mm2 >Amin. (.0012 x 1000 x 250)

Required spacing of 16 mm Bars,

C/C Spacing=1000/1373.55 x 200.96=146.31 mm

Provide 16 mmØ @140 mm

Distributors Reinforcement,

Astmin=.0012 x 1000 x 250=300 mm2

Wf =

22.305 KN/ m

Wl =

18.375 KN/ m

Astreqd=1373.55

mm2

Astprod=1435.4

mm2

131

IS456-2000

Cl 26.5.2.1

IS456-2000

Cl 23.2.1.c

Fig.4

5.

6.

7.


C/C Spacing=1000/300.00 x 78.546=261.82 mm


Development Length

bd

s

dL

4

Ld = 752.18 mm

Provide Development Length 755 mm

Checking for depth of slab,

D=l/(20 x mt)

Percentage of steel,Pt=0.574

For fs=230.33Mpa

mt=1.25

D=6000/(20 x 1.25)=240 mm< 250 mm(O.K)

132

Open Well Staircase



1.

Fig:-Open Well Staircase

Known Data

Riser Height, R=180 mm

Tread Height,T=250 mm

Floor Height=3.2m

Flight Width, W=1.5m

No. of Treads in the flights=4

Length of the flights=1m

133

2.

i.

ii.

3.

4.

Load Calculation

Flight

Assuming Slab Thickness,D= 220 mm

Considering 1m Width of Slab

Self Wt.of Soffit Slab=0.308*0.22*25=1.694 KN/m per

step

Wt. of Steps=0.5*0.18*0.25*25=0.56 KN/m per step

Total load per step=2.254 KN/m

Total load per m2=2.254/0.25=9.02 KN


Floor Finishing= 1 KN/m

Live Load=5 KN/m


Design Load=1.5 x 15.02 =22.53 KN/m

Landing

Self Wt. of Slab= 25 x .22=5.5 KN/m

Floor Finishing= 1 KN/m

Live Load=5 KN/m


Design Load=1.5 x 11.5 =17.25KN/m

Analysis

For Upper and Lower Flight,

Maximum Bending Moment=14.77 KNm

Clear Cover=20 mm,10 mm dia.bars

Effecrtive Depth= 220-20-5=195 mm

For Intermediate Flight,

Maximum Bending Moment=65.49 KNm

Clear Cover=20 mm,16 mm dia.bars

Effecrtive Depth= 220-20-8=192 mm

Design For Main Reinforcement,

Wf = 22.53

KN/m

Wl = 17.25

KN/m

134

IS456-2000

Cl. G.1.1.b

Cl 26.5.2.1

5.

6.

7.

For Intermediate flight,

201921000

415119241587.01049.65 6

xx

xAxxAxxx st

st

Ast=1067.99 mm2 >Amin. (.0012 x 1000 x 220)


C/C Spacing=1000/1419.03 x 200.96=188.16 mm


For Upper and Landing,

201951000

415119541587.01077.14 6

xx

xAxxAxxx st

st

Ast = 214.69 mm2 >Amin. (.0012 x 1000 x 220)

C/C Spacing = 1000 x 78.546/214.69 = 365.86 mm


Distributors Reinforcement,

Astmin = .0012 x 1000 x 220= 264 mm2


C/C Spacing = 1000/264 x 78.546 = 183.88 mm


Devlopment Length


bd

s

4

ØLd

= 752 mm

Provide Devlopment Length 755 mm


bd

s

4

ØLd

= 470.11 mm

Provide Devlopment Length 475 mm

Checking for depth of slab,


Astreqd =

1067.9mm2

Astprod

=1086.3mm

2

Astreqd

=214.69

mm2

Astprod

=261.82

mm2

Astreqd =264

mm2

Astprod

=269.7mm2

135

IS456-2000

Cl 23.2.1.c

Fig.4

D = l/(20 x mt)

Percentage of steel,Pt = 0.49

For fs = 236.63Mpa

mt = 1.3

D = 5000/(20 x 1.3) = 192.31 mm< 220 mm(O.K)


D = l/(20 x mt)

Percentage of steel,Pt = 0.123

For fs = 235.62Mpa

mt = 2

D = 2500/(20 x 2) = 62.5 mm< 220 mm(O.K)

136

4.2.5 Design of basement wall

Introduction

Basement wall is constructed to retain the earth and to prevent moisture from seeping

into the building. Since the basement wall is supported by the mat foundation, the stability is

ensured and the design of the basement wall is limited to the safe design of vertical stem.

Basement walls are exterior walls of underground structures (tunnels and other earth

sheltered buildings), or retaining walls must resist lateral earth pressure as well as additional

pressure due to other type of loading. Basement walls carry lateral earth pressure generally

as vertical slabs supported by floor framing at the basement level and upper floor level. The

axial forces in the floor structures are , in turn, either resisted by shear walls or balanced by

the lateral earth pressure coming from the opposite side of the building.

Although basement walls act as vertical slabs supported by the horizontal floor

framing, keep in mind that during the early construction stage when the upper floor has not

yet been built the wall may have to be designed as a cantilever.

Design of vertical stem

The basement wall is designed as the cantilever wall with the fixity provided by the

mat foundation.

14.02 KN/m2

3.33 KN/m2

Due to Surcharge

(Rear or outer face)

Soil

Pressure

Basement Wall

(Front / Inner face)

Mat Footing

Fig: Basement Wall

137

4.2.5.1 Design of basement wall


Ref. Step Calculation Output

IS456:2000

Cl.32. 3.4

1

2

3

Design Constants

c/c of floor = 3.2m

Let clear height between the floor (h) =2.75 m

Unit weight of soil, γ = 17 KN/m3

Angle of internal friction of the soil, ө = 300

Surcharge produced due to vehicular movement is,

Ws = 10 KN/m2

Safe bearing capacity of soil, qs = 150.0 KN/m2

Moment calculation

Ka 333.030sin1

30sin1

sin1

sin1

Lateral load due to soil pressure,

Pa = Ka x γ x h2/2

= 0.333x17x2.752/2

= 21.40 KN/m

Lateral Load due to surcharge load,

Ps = Ka x Ws x h

= 0.333x10x2.75

= 9.15 KN/m

Characteristic Bending moment at the base of wall

Since weight of wall gives insignificant moment, so

this can be neglected in the design.

Mc = Pa x h/3 + Ps x h/2

= 21.40x2.75/3 + 9.15x2.75/2

= 32.2 KN-m

Design moment, M = 1.5Mc = 48.3 KN-m

Approximate design of section

Let effective depth of wall = d

BM = 0.138ƒckbd2

Pa =21.40KN/m

Ps = 9.15KN/m

M=48.3KN-m

138

IS456:2000

Cl.32.5.a

IS456:2000

26.5.2.2

IS456:2000

Cl.32.5.b

4

5

48.3x106 = 0.138x20x1000xd2

d = 132.28 mm

Let Clear cover is 30mm & bar is 20mm-Ф

Overall depth of wall, D = 132.28+30+10

= 172.28 mm

Take D = 200 mm

So , d = 200 – 30- 10 = 160 mm

Calculation of Main Steel Reinforcement

Ast=

2cky

ck

bdf

M6.411

xf2

bdf

Ast=

2

6

160100020

103.486.411

4152

201601000

xx

xx

x

xx

Ast = 954 mm2

Min. Ast = 0.0012xbxD = 0.0012x1000x200

= 240 mm2 < Ast

Max. Dia. of bar = D/8 = 200/8 = 25 mm

Providing 20mm-Ф bar , spacing of bar is

S=9544

1000202

x

xx=329 mm/m

Provide 20mm-Ф bar @300 mm c/c at the outer

face.

So, Provided Ast = 314.16x1000/300

= 1047.2 mm2

Pt = 1047.2x100/(1000x160) = 0.6545 %

Max. Spacing = 3d or 450mm

3d = 3x160 = 480 mm

Provide 8mmФ@450mm(nominal) c/c at the front

face.

Check for Shear

The critical section for shear strength is taken at a

distance of ‘d ’ from the face of support. Thus,

D = 200 mm

d = 160 mm

Ast = 954 mm2

S = 300 mm

Pt = 0.6545 %

139

IS456:2000

Cl.31.6.2.1

IS456:2000

Table-19

IS456:2000

Cl.23.2.a

IS456:2000

Cl.32.5.c.1

6

7

critical section is at d = 0.160 m from the top of

mat foundation.

i.e. at (2.75- 0.16) = 2.59 m below the top edge of

wall.

Shear force at critical section is,

Vu = 1.5x(Ka x Ws x Z + Ka x γ x Z2/2)

= 1.5x(0.333x10x2.59 + 0.333x17x2.592/2)

= 41.42 KN

Nominal shear stress , bd

Vuu

= 41.42x1000/(1000x160)

= 0.258 N/mm2

Permissible shear stress , τc = 0.529 N/mm2

τc > τu , Hence safe.


Leff = 2.75+d = 2.75+.16 = 2.91 m

Allowable deflection = leff/250 = 2910/250

= 11.64 mm

Actual Deflection = EI30

lp

EI8

lp eff4

aeff4

s

=

30

40.21

8

15.9

2050001601000

1229103

4

xx

x= 4.47 mm

which is less than allowable deflection, hence safe.

Calculation of Horizontal Reinforcement steel

bar

Area of Horizontal. Reinforcement = 0.002Dh

= 0.002x200x2750 = 1100 mm2

As the temperature change occurs at outer face of

basement wall, 2/3 of horizontal reinforcement is

provided at front face and 1/3 of horizontal

reinforcement is provided in inner face.

Outer face Horizontal Reinforcement steel,

Vu = 41.42 KN

τu = 0.258 N/mm2

τc = 0.529 N/mm2

140

IS456:2000

,

Cl.32.5.d

IS456:2000

Cl.26.2.1

8

= 2/3x1100= 733.33 mm2

Providing 12mm-Ф bar

No. of bar required, N = 733.33/113 = 7 nos.

Spacing = (h-clear cover at both sides- Ф)/(N-1)

= (2750-30-12)/(7-1) = 451 mm

Provide 12mm-Ф bar @ 450 mm c/c

Inner face Horizontal Reinforcement steel,

= 1/3x1100= 366.66 mm2

Providing 8mm-Ф bar

No. of bar required, N = 366.66/50.27 = 8 nos.

Spacing = (h-clear cover at both sides- Ф)/(N-1)

= (2750-30-12)/(8-1) = 386 mm

Provide 8mm-Ф bar @ 380 mm c/c

Max. spacing = 3d = 3x160 = 480 mm or 450 mm

Hence, spacing provided for Horizontal Steel is

OK.

Curtailment of Reinforcement

No bars can be curtailed in less than Ld distance

from the bottom of stem ,

Ld = bd

s

x4x6.1

=

2.146.1

2041587.0

xx

xx= 940 mm

The curtailment of bars can be done in two layers

1/3 and 2/3 heights of the stem above the base.

Let us curtail bars at 940mm (since, 1/3 distance =

916.66 mm is lesser than 940mm) from base, i.e.

1810mm from top.

Lateral load due to soil pressure ,

Pa = Ka x γ x h2/2

= 0.333x17x1.812/2

= 9.27 KN/m

Lateral load due to surcharge load,

Ps = Ka x Ws x h

141

= 0.333x10x1.81

= 6.03 KN/m

Characteristic Bending moment at the base of wall

is, Mc = Pa x h/3 + Ps x h/2

= 9.27x1.81/3 + 6.03x1.81/2

= 11.05 KN-m

Design Moment , M = 1.5Mc = 1.5x11.05 = 16.58

KN-m

Since this moment is less than half of the moment

at base of stem, spacing of vertical reinforcement is

doubled above 940 mm from the base of the wall.

But, minimum spacing is 450mm.

Provide 20mm-Ф bar @450 mm c/c above 916mm

from base at the outer face and no curtailment in

the inner face.

142

4.2.6 Design of Lift Wall

Load Calculation for Lift Wall Design

1. Ground Floor to fourth floor :

a) Lift Wall

Length = 10 m

Characteristic Load = 25 x 10 x 0.2 x 3.2 =160 KN

2. Machine Floor:

a) Lift Wall

Length =10 m

Characteristic Load =25 x 10 x 0.2 x 3.2 =160 KN

b) Slab

Dead Load =25 x 0.150 x 2.5 x 2.5 =23.43 KN

Live Load = 0.5 x 3 x 2.5 x 2.5 =9.375 KN

Total load =192.805 KN

3. Top Level:

a) Lift wall

Length =10 m

Characteristic Load =0.5x25 x 10 x 0.2 x 3.2 =80 KN

a) Slab

Dead Load =25 x 0.15 x 2.5 x 2.5 =23.43 KN

Total load =103.43 KN

Total Seismic Weight of the Lift =160x5+192.805+103.43 =1096.235 KN

143

Lateral Load Calculation for Lift

Total Seismic Load =1096.235 KN

h =22.4m

Ah = (Z x I x Sa) / (2 x R x g)

(IS 1893 -2002) Cl.6.4.1

Ta = 0.075 x h0.75 = 0.075 x 22.6570.75 = 0.772 Sec

(IS 1893 -2002) Cl.7.6

Sa/g = 1.67/Ta = 2.163

Z = 0.36

I = 1

R = 5

Ah = 0.36 x 1 x 2.163 / (2 x 5) = 0.116

VB = Ah x W

VB = 0.077 x 1096.235 = 127.16 KN

Qi = VB x (Wi x hi2)/∑ (Wi x hi

2)

Table for Lateral Load Calculation

FLOOR Wi hi Wi hi

2 Qi=VB Wi

Hi2/∑Wi hi

2 Vi KN Moment(KNm)

top 103.43 22.4 51897.0368 30.9699761 30.9699761

5th 192.805 19.2 71075.6352 42.41495968 73.38493579 99.10392353

4th 160 16 40960 24.44321101 97.82814679 333.935718

3rd 160 12.8 26214.4 15.64365504 113.4718018 646.9857878

2nd 160 9.6 14745.6 8.799555962 122.2713578 1010.095554

1st 160 6.4 6553.6 3.910913761 126.1822716 1401.363899

Ground 160 3.2 1638.4 0.97772844 127.16 1805.147168

Basement 0 0 0 0 127.16 2212.059168

213084.672

W6=103.43 KN

W1=160KN

W2=160 KN

W3=160 KN

W4=160 KN

W5=192.805KN

Lumped mass

144

Design of Lift Wall

1.20 m

2.5 m



IS 456-

2000

Cl.32.2.4 a

IS 456-

2000

Cl.32.2.3

IS 456-

2000

Cl.32.2.2

1

2

3

4

Known Data

Length of Lift Wall = 2.5 m

Breadth of Lift Wall = 2.5 m

Floor Height , H = 3.2 m

Assume, Wall Thickness, t = 200 mm

Check for Slenderness Ratio

Effective Height of the Wall , Hwe = 0.75H = 0.75 x 3.2 =

2.4 m

Slenderness Ratio , 30122.0

4.2

t

H ew

Minimum eccentricity

emin= 0.05t = 0.05 x 200 = 10 mm

Additional eccentricity

ea = mmxt

H we52.11

2002500

2400

2500

22

H = 3.2 m

t = 200 mm

emin= 10 mm

ea=20.48 mm

2.5 m

X

Y

145

IS 456-

2000

Cl.32.2.5

SP-16,

Chart 35

IS 456-

2000

Cl.32.5 a

5

6

a)

Ultimate load carrying capacity

Ultimate load carrying capacity per unit length of the

wall is ,

Puw = 0.3(t-1.2e-2ea)fck

= 0.3(200-1.2 x 10 -2 x 11.52) x 20 = 989.76 N/mm

. .

. Total Capacity of wall = 2243.53 KN

Calculation for Main Vertical Reinforcement

Assume, Clear cover = 20 mm

Using 12 mm Ø bar , Effective cover , d’ = 26 mm

When lateral load is acting along X- direction

Mu = m-KN 025.11062

2212.05

Vu = KN 58.632

127.16

Pu = KN 548.125 2

1096.25

2500

26'

D

d0.010

0442.0250020020

10025.11062

6

2

xx

x

bdf

M

ck

u

0548.0250020020

1000125.548

xx

x

bdf

P

ck

u

016.0ckf

p

p = 0.016x 20 = 0.32 %

Min, Ast = 0.12 % of bD < 0.32 %

. .

. Ast = 0.0032 x 200 x 2500= 1600 mm2

Area of 12 mm Ø = 113.09 mm2

No. of Bars = 09.113

1600 = 14.145 15 nos

Puw

=989.76N/m

m

d’ = 26 mm

p= 0.32 %

146

IS 456-

2000

Cl.32.5 b

SP-

16,Chart

31

IS 456-

2000

Cl.32.5 a

b)

. .

. Spacing of Bars, Sv = mm875.174115

12402500

Check for Spacing

Spacing of vertical steel reinforcement should be least of

3t = 3 x 200 = 600 mm

450 mm

To take account of the reversal effect, Provide 12 mm Ø

bars @ 170 mm c/c on both faces of the wall

When Lateral Load is acting along Y-direction

Mu = m-KN 025.11062

2212.05

Vu = KN 58.632

127.16

Pu = KN 548.125 2

1096.25

2500

26'

D

d0.010

0442.0250020020

10025.11062

6

2

xx

x

bdf

M

ck

u

0548.0250020020

1000125.548

xx

x

bdf

P

ck

u

016.0ckf

p

p = 0.016x 20 = 0.32 %

Min, Ast = 0.12 % of bD < 0.32 %

. .

. Ast = 0.0032 x 200 x 2500= 1600 mm2

Area of 12 mm Ø = 113.09 mm2

No. of Bars = 09.113

1600 = 14.145 15 nos

. .

. Spacing of Bars, Sv = mm875.174115

12402500

12 Ø @ 170

mm

p= 0.32 %

147

IS 456-

2000

Cl.32.5 b

IS 456-

2000

Cl.32.5 c

IS 456-

2000

Cl.32.4.2

IS 456-

2000

Cl.32.4.2.1

7

8

a)

Check for Spacing

Spacing of vertical steel reinforcement should be least of

3t = 3 x 200 = 600 mm

450 mm



Calculation of Horizontal Steel Reinforcement

Area of horizontal steel reinforcement = 0.2 % bH

= 0.002 x 200 x

3200

= 1280 mm2


No. of Bars = nos1231.1109.113

1280

. . . Spacing of Bars , Sv = mm90.290

112

3200



Check for Shear

When Lateral Load is acting along X- direction

Nominal Shear Stress,

23

/158.025008.0200

1058.63

8.0mmN

xx

x

Lxt

V

td

V

w

uu

v

Allowable Shear Stress,

2ckallowable mm/N4.320x17.0f17.0 τ > vτ

)(128.12500

3200WallHigh

L

H

w

w

cwτ should be lesser of

538.1202.0)28.13(3 1

xxfK

L

Hck

w

w

cw N/

mm2

12 Ø@ 170

mm c/c

12Ø @ 250

mm c/c

148

b)

2

2 /63.1128.1

128.120045.0

1

1

mmN

LH

LH

fK

w

w

w

w

ckcw

But not less than 2ck mm/N671.02015.0f15.0

. .

. cwτ =1.538 N/mm2 > vτ

When Lateral Load is acting along Y- direction

Nominal Shear Stress,

23

/158.025008.0200

1058.63

8.0mmN

xx

x

Lxt

V

td

V

w

uu

v

Allowable Shear Stress,

2ckallowable mm/N4.320x17.0f17.0 τ > vτ

)(128.12500

3200WallHigh

L

H

w

w

cwτ should be lesser of

538.1202.0)28.13(3 1

xxfK

L

Hck

w

w

cw N/

mm2

2

2 /63.1128.1

128.120045.0

1

1

mmN

LH

LH

fK

w

w

w

w

ckcw

But not less than 2ck mm/N671.02015.0f15.0

. .

. cwτ =1.538 N/mm2 > vτ

cw =1.538

N/mm2

Hence, Safe

cwτ =1.538

N/mm2

Hence, Safe

149

4.2.5 Design of Mat Foundation

It is necessary to provide a continuous footing under all the columns and walls if the loads

transmitted by the columns in a structure are so heavy or the allowable soil bearing pressure

small. Such a footing is called a raft or Mat Foundation. The raft is divided into series of

continuous strips centered on the appropriate column rows in the both directions as shown in

figure below. The shear and bending moment diagrams may be drawn using continuous

beam analysis or coefficients for each strip. The depth is selected to satisfy shear

requirements. The steel requirements will vary from strip. This method generally gives a

conservative design since the interaction of adjacent strips is neglected.

Design of Mat Foundation:

Required Data:

Case Considered = Case I = 1.5*(DL+LL)

Total Vertical Load = 210447.2KN

For Earthquake consideration also Soil Bearing Capacity (factored) = 150KN/m2

(From Soil investigation report)

Grade of concrete = 25KN/m3

Grade of steel = 415N/mm2

150

Calculation of Centre of gravity of plan area

mXiPi

582.22P

*X

mYiPi

4.17P

*Y

Geometrical Centre of the Mat Foundation:

X = 21.95m, Y = 18.508m

m 1.13221.955.058.22XXex

m608.0508.185.04.17YYey

Calculation of Moment of Inertia:

About X-X axis:

Ixx=194,203.23m4

I𝑦𝑦 = 264,964.76𝑚4

Area coverage of Mat:

A = 1645 m2

Mex = ΣP × ey = 210,447.2×-.608

= -127951.898KNm

Mey = ΣP × ex = 210447.2×1.132

= 238,226.23 KNm

Total Moment ( Mxx = Mex + ΣMx ) =-127951.898-202.918

=-128154.81KNm

Total Moment ( Myy = Mey + ΣMy) = 238226.23-4.328

=238221.902KNm

2KN/m931.1271645

210447.2

A

P

Soil Pressure calculation at different points:

yI

Mx

I

M

A

Pσ

xx

xx

yy

yy

151

Column No. x y Ϭ(x,y)

A-1 -21.95 18.992 95.6457627

A-2 -15.45 18.992 101.49189

A-3 -9.45 18.992 106.888315

A-4 -3.45 18.992 112.28474

A-5 2.55 18.992 117.681165

A-6 8.55 18.992 123.07759

A-7 14.55 18.992 128.474015

A-8 21.05 18.992 134.320142

B-1 -21.95 15.992 97.6271206

B-8 21.05 15.992 136.3015

C-1 -21.95 8.992 102.250289

C-8 21.05 8.992 140.924668

D-1 -21.95 3.992 105.552552

D-8 21.05 3.992 144.226931

E-1 -21.95 -1.008 108.854815

E-8 21.05 -1.008 147.529194

F-1 -21.95 -6.008 112.157078

F-8 21.05 -6.008 150.831457

F-9 23.55 -6.008 153.079967

G-1 -21.95 -11.008 115.459341

G-9 21.05 -11.008 154.13372

H-1 -21.95 -18.508 120.412735

H-2 -15.45 -18.508 126.258862

H-3 -9.45 -18.508 131.655287

H-4 -3.45 -18.508 137.051712

H-5 2.55 -18.508 142.448137

H-6 8.55 -18.508 147.844562

152

H-7 14.55 -18.508 153.240987

H-8 20.55 -18.508 158.637412

H-9 23.55 -18.508 161.335625

Note: Here the maximum upward soil pressure (161.336 KN/m2) is slightly greater than

safe bearing capacity (150 KN/m2) of foundation soil so it is necessary to increase the

strength of the foundation soil by using geotechnical soil stabilizing process like certain

depth of granular material packing.In the X-direction the raft is divided into eight strips i.e.

eight equivalent beams:

i. Beam A-A with 1.75m width and soil pressure of = 134.32KN/m2

ii. Beam B-B with 4.75m width and soil pressure of = 136.302KN/m2

iii. Beam C-C with 6m width and soil pressure of = 140.925KN/m2

iv. Beam D-D with 5m width and soil pressure of = 144.227KN/m2

v. Beam E-E with 5m width and soil pressure of = 147.53KN/m2

vi. Beam F-F with 5m width and soil pressure of = 153.08KN/m2

vii. Beam G-G with 6m width and soil pressure of = 154.134KN/m2

viii. Beam H-H with 4m width and soil pressure of = 161.34KN/m2

In the Y-direction the raft is divided into 9 strips i.e. nine equivalent beams:

Beam 1-1 with 3.5m width and soil pressure of and soil pressure of=120.413KN/m2

Beam 2-2 with 6m width and soil pressure of and soil pressure of=126.26KN/m2








From IS-456 Clause 22.5.1

Bending moment is Obtained by Coefficient 1/10 & ‘L’ as center to center column distance.

+M=-M=wl2/10

153

X direction

For strip A-A

Maximum Moment =134.32∗1.752

10=41.136KNm/m

For strip B-B


10=307.531KNm/m

For strip C-C

Maximum Moment =140.925∗62

10=507.33KNm/m

For strip D-D

Maximum Moment=144.227∗52

10=360.57KNm/m

For strip E-E


10=368.825KNm/m

For strip F-F


10=382.7KNm/m

For strip G-G


10=554.88KNm/m

For strip H-H


10=258.144KNm/m

Y-direction

For strip 1-1


10=147.51KNm/m

For strip 2-2


10=454.536KNm/m

For strip 3-3


10=473.98KNm/m

For strip 4-4


10=493.38KNm/m

For strip 5-5

154


10=512.82KNm/m

For strip 6-6


10=629.424KNm/m

For strip 7-7


10=551.66KNm/m

For strip 8-8

Maximum Moment=158.64∗3.752

10=223.09KNm/m

For strip 9-9

Maximum Moment=161.34∗1.752

10=49.41KNm/m

Calculation of depth of foundation

i. Calculation of depth from Moment Criterion

Maximum Moment,M=629.424KNm

Therefore,d=√𝑀

0.138𝑓𝑐𝑘∗𝑏=√

629.424∗106

3.45∗1000=427.132mm

ii. Calculation of Depth from Two way Shear

The depth of the raft will be governed by two way shear at one of the exterior columns.

In case location of critical shear is not obvious, it may be necessary to check all possible

locations. When Shear reinforcement is not provided, the calculated shear stress at

critical section shall not exceed Ksτc i.e. τv ≤ Ksτc

Where,

Ks = 0.5+βc but not greater than 1

βc= 1

Ks=1+ 0.5 =1.5>1

Hence Ks = 1

Shear strength of concrete τc ckf0.25

2N/mm25.1250.25 = τv

For Corner Column say H-1:

Column Load = 2516.645 KN

155

Perimeter, po = 2 ∗ (d

2+

550

2+ 500) = d + 1550

τv d1550d

102516.645

dp

V 3

o

u

d1550d

102516.64525.1

3

d = 841.77mm

For Side Column say H-6:

Column Load = 4528.58KN

Perimeter, po =

(. 5d +550

2+ 500) ∗ 2 + d + 550 = 2d + 2100

τv d21002

104528.58

dp

V 3

o

u

d

d21002

104528.58

dp

V25.1

3

o

u

d

d = 919.66mm

For side Column say C-8:

Column Load = 3218.834KN

Perimeter, po = (. 5d +550

2+ 500) ∗ 2 + d + 550 = 2d + 2100

τv d21002d

103218.834

dp

V 3

o

u

d21002d

103218.83425.1

3

d = 725.26mm

Hence depth is governed by 2 way shear.

Adopt effective depth=1000mm

Diameter of steel used=32Ø bars

Clear cover adopted=50mm

Overall depth=750+Clear cover+Ø/2=1066mm ~ 1070mm

156

Calculation of Reinforcement in the Foundation

Along X-direction

Minimum Reinforcement

Ast,min=0.12%bD=0.0012*1000*1070=1284mm2

Maximum negative moment = 629.424 𝑘𝑁𝑚

M = .87 ∗ fyAst ∗ d ∗ (1 −

Astfy

bdfck)

or, 554.88 ∗ 106 = .87 ∗ 415 ∗ Ast ∗ 1000 ∗ (1 −415Ast

1000∗1000∗25)

Therefore, Ast = 1578.2mm2/m> Ast,min

Provide 20mmφ bars at 160 mm c/c

Ast = 1963.5mm2/m> 1578.2mm2/m O.K.

Along Y-direction

Minimum Reinforcement

Ast,min=0.12%bD=0.0012*1000*1070=1284mm2

Maximum negative moment = 629.424 𝑘𝑁𝑚

M = .87 ∗ fyAst ∗ d ∗ (1 −

Astfy

bdfck)

or, 629.424 ∗ 106 = .87 ∗ 415 ∗ Ast ∗ 1000 ∗ (1 −415Ast

1000 ∗ 1000 ∗ 25)

Therefore, Ast = 1796.92mm2/m> Ast,min

Provide 20mmφ bars at 160 mm c/c

Ast = 1963.5mm2/m> 1796.92mm2/m O.K.

iii. Check for one-way shear:

Shear at critical section:

Vu=135.501KN

Pt=0.378%

157

From IS456:2000 Table 19,

Design shear strength, τc=0.426N/mm2

τv db

Vu

=0.135N/mm2 > τc. Hence,safe.

Result:

Along X-direction

Provide 20 mmφ bars at 160 mm c/c at the bottom and the top.

Along Y-direction

Provide 20 mmφ bars at 160 mm c/c at the bottom and the top

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Title

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Title

:

Re

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rce

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b

Pro

ject:

CO

MM

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Title

:

Re

info

rce

me

nt D

eta

il of sla

b

Pro

ject:

CO

MM

ER

CIA

L B

UIL

DIN

G

TR

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, SA

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Title

:

Re

info

rce

me

nt D

eta

il of sla

b

Pro

ject:

CO

MM

ER

CIA

L B

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DIN

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AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

,SU

MA

N, S

UN

IL

Title

:

Re

info

rce

me

nt D

eta

il of sla

b

Pro

ject:

CO

MM

ER

CIA

L B

UIL

DIN

G

TR

IBH

UV

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IVE

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ING

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Sta

ircase

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en

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S2

S2

S2

S2

S2

S2

S2

S2

S2

S2

S2

S2

S2

S2

S3

S3

S3

S3

S3

S3

S4

S4

S5

S5

S6

S6

S7

S7

S8

S8

S8

S8

S8

S8

S8

S8

S8

S8

S8

S8

S8

S8

S8

S8

S8

GE

NE

RA

L N

OT

ES

1. AL

L D

IMN

ES

ION

S A

RE

IN m

m, U

NL

ES

S O

TH

ER

WIS

E N

OT

ED

.2

. GR

AD

E O

F C

ON

CR

ET

E S

HA

LL

BE

M20 C

ON

FIR

MIN

G T

O IS

:456-20003

. RE

INF

OR

CE

ME

NT

SH

AL

L B

E H

IGH

ST

RE

NG

TH

DE

FO

RM

ED

BA

RS

OF

GR

AD

E F

e415 C

ON

FIR

MIN

G T

O IS

:1786-1985

SC

AL

E : N

ot T

o S

cale

DR

G N

o. :

SH

EE

T N

o. : 4

-A

SU

PE

RV

ISO

R:

Er.D

ine

sh G

upta

DA

TE

D : A

SH

OJ , 2

070

DR

AW

N B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

,SU

MA

N, S

UN

IL

DE

SIG

NE

D B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

,SU

MA

N, S

UN

IL

Title

:

Re

info

rce

me

nt D

eta

il of sla

b

Pro

ject:

CO

MM

ER

CIA

L B

UIL

DIN

G

TR

IBH

UV

AN

UN

IVE

RS

ITY

INS

TIT

UT

E O

F E

NG

INE

ER

ING

PU

LC

HO

WK

CA

MP

US

To

p R

ein

force

me

nt@

22

.4

8m

m@12

5m

mc/c

8m

m

@12

5m

mc/c

30

00

30

00

30

00

30

00

30

00

70

00

50

00

50

00

50

00

50

00

70

00

60

00

8m

m@

250

mm

c/c

8m

m@

250

mm

c/c

30

00

S2

S2

S2

S2

S2

S2

S2

S2

S2

S2

S2

S2

S3

S3

S3

S3

S3

S4

S6 S6

S6

S7

S7

S7

GE

NE

RA

L N

OT

ES

1. AL

L D

IMN

ES

ION

S A

RE

IN m

m, U

NL

ES

S O

TH

ER

WIS

E N

OT

ED

.2

. GR

AD

E O

F C

ON

CR

ET

E S

HA

LL

BE

M20 C

ON

FIR

MIN

G T

O IS

:456-20003

. RE

INF

OR

CE

ME

NT

SH

AL

L B

E H

IGH

ST

RE

NG

TH

DE

FO

RM

ED

BA

RS

OF

GR

AD

E F

e415 C

ON

FIR

MIN

G T

O IS

:1786-1985

SC

AL

E : N

ot T

o S

cale

DR

G N

o. :

SH

EE

T N

o. : 4

-B

SU

PE

RV

ISO

R:

Er.D

ine

sh G

upta

DA

TE

D : A

SH

OJ , 2

070

DR

AW

N B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

,SU

MA

N, S

UN

IL

DE

SIG

NE

D B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

,SU

MA

N, S

UN

IL

Title

:

Re

info

rce

me

nt D

eta

il of sla

b

Pro

ject:

CO

MM

ER

CIA

L B

UIL

DIN

G

Bo

ttom

Re

info

rce

men

t@2

2.4

30

00

30

00

30

00

30

00

30

00

30

00

60

00

70

00

50

00

50

00

50

00

50

00

70

00

8m

m@

12

5m

m c/c

8m

m@

250

mm

c/c

8m

m@

250

mm

c/c

8m

m@

250

mm

c/c

TR

IBH

UV

AN

UN

IVE

RS

ITY

INS

TIT

UT

E O

F E

NG

INE

ER

ING

PU

LC

HO

WK

CA

MP

US

S2

S2

S2

S2

S2

S2

S2

S2

S2

S2

S2

S2

S3

S3

S3S

3S

3

S4

S6

S6

S6

S7

S7

S7

2500

625

17501750

1250

7000

1250

5000

12 Ø

2LV

S @

100 m

m c/c

8 Ø

2L

VS

@ 100

mm

c/c8

Ø @

200 mm

c/c8

Ø @

100 mm

c/c8

Ø @

100 mm

c/c8

Ø @

125 mm

c/c8

Ø @

100 mm

c/c

6-28 Ø

6-28 Ø

6-28 Ø

2-28 Ø

3-28 Ø

3-28 Ø

3-28 Ø

5-28 Ø

6-28 Ø

2-28 Ø

2-28 Ø

6-28 Ø

6-28 Ø

L-S

ection

along

Grid G

-G of F

irst floor

X-S

ections

TR

IBH

UV

AN

UN

IVE

RS

ITY

INS

TIT

UT

E O

F E

NG

INE

ER

ING

PU

LC

HO

WK

CA

MP

US

Pro

ject:

CO

MM

ER

CIA

L B

UIL

DIN

G T

itle:

Beam

Detailing

DA

TE

D : A

SH

OJ , 2070

DR

AW

N B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

, SU

MA

N, S

UN

IL

DE

SIG

NE

D B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

, SU

MA

N, S

UN

IL

SU

PE

RV

ISO

R:

Er. D

inesh

Gupta

SC

AL

E : N

ot To S

cale

DR

G N

o. :

SH

EE

T N

o. : 6

GE

NE

RA

L N

OT

ES

1. AL

L D

IMN

ES

ION

S A

RE

IN m

m, U

NL

ES

S O

TH

ER

WIS

E N

OT

ED

.2

. GR

AD

E O

F C

ON

CR

ET

E S

HA

LL

BE

M2

0 CO

NF

IRM

ING

TO

IS:456-2000

3. R

EIN

FO

RC

EM

EN

T S

HA

LL

BE

HIG

H S

TR

EN

GT

H D

EF

OR

ME

D B

AR

S O

F G

RA

DE

Fe415 C

ON

FIR

MIN

G T

O IS

:1786-1985

L-S

ection

along

Grid

7-7 of Ground floor

15001500

15001500

15001500

1500

15001500

15001500

15001500

1500

1500

1320

28101320

13202810

13201320

28101320

28101320

1320

12m

mØ

2L

VS

@ 1

30mm

c/c8

mm

f 2

LV

S @

100m

m c/c

8m

mf

4LV

S@

100m

m c/c

550550

550550

550

60006000

60006000

4-32

Ø +

2-16Ø4

-32Ø

+ 2-16Ø

4-32

Ø +

2-28Ø4

-32Ø

+ 2-28Ø700

500

AT

E-E

AT

D-D

AT

C-C

AT

B-B

AT

A-A

4-3

2Ø

+ 2

-16Ø

4-3

2 Ø+

2-28 Ø

2-12Ø

ED

CB

A

ED

CB

A

8m

mf

2L

VS

@1

00mm

c/c8

mm

f 4L

VS

@1

00mm

c/c

4-32

Ø +

2-16Ø4

-32Ø

+ 2-16Ø

4-32

Ø +

2-28Ø4

-32Ø

+ 2-28Ø

8m

mf

2L

VS

@1

00mm

c/c8

mm

f 4L

VS

@1

00mm

c/c

4-32

Ø +

2-16Ø4

-32Ø

+ 2-16Ø

4-32

Ø +

2-28Ø4

-32Ø

+ 2-28Ø

12m

mØ

2L

VS

@ 1

30mm

c/c8

mm

f 2

LV

S @

100m

m c/c

8m

mf

4LV

S@

100m

m c/c

4-32

Ø +

2-16Ø4

-32Ø

+ 2-16Ø

4-32

Ø +

2-28Ø4

-32Ø

+ 2-28Ø

700

500

4-3

2Ø

+ 2

-16Ø

4-3

2 Ø+

2-28 Ø

2-12Ø

500

2-3

2Ø

+ 2

-16Ø

2-32 Ø

2-12Ø

700

500

4-3

2Ø

+ 2-16Ø

4-3

2 Ø+

2-28 Ø

2-12Ø

X-S

ections

16

00

1500

550550

550550

2-28 Ø

2-28 Ø

7-28 Ø

6-28 Ø 625

13201320

1320

700

500

AT

A-A

6-2

8Ø

6-28 Ø

2-12Ø

700

500

AT

C-C

3-2

8Ø

6-28 Ø

2-12Ø

700

500

AT

E-E

2-2

8Ø

2-28 Ø

2-12Ø

AT

B-B

700

500

2-2

8Ø

2-28 Ø

2-12Ø

AT

D-D

700

500

2-28 Ø

2-12Ø

8 Ø

@ 150 m

m c/c

450S

econdary

beamM

ain b

eam

2-26

Ø +

1-16Ø

700

6-28 Ø

3-2

8Ø

700

500

4-3

2Ø

+ 2-16Ø

4-3

2 Ø+

2-28 Ø

2-12Ø

12 Ø

2LV

S @

100 m

m c/c

2-28 Ø

2-28 Ø

2-32

Ø +

2-16Ø

2-32Ø

990990

12m

m 2L

VS

@ 1

30 c/c

2-32

Ø +

2-16Ø

2-32Ø

12m

m 2L

VS

@ 1

00 c/c

1500

8m

m Ø

2 L

VS

@ 20

0 mm

c/c

A

A

D D

CC

BB

EE

15

0

2-28Ø

SC

AL

E : N

ot To S

cale

DR

G N

o. :

SH

EE

T N

o. : 7

SU

PE

RV

ISO

R:

Er. D

inesh

Gupta

DA

TE

D : A

SH

OJ , 2070

DR

AW

N B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

, SU

MA

N, S

UN

IL

DE

SIG

NE

D B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

, SU

MA

N, S

UN

IL

GE

NE

RA

L N

OT

ES

1. AL

L D

IMN

ES

ION

S A

RE

IN m

m, U

NL

ES

S O

TH

ER

WIS

E N

OT

ED

.2

. GR

AD

E O

F C

ON

CR

ET

E S

HA

LL

BE

M2

0 CO

NF

IRM

ING

TO

IS:456-2000

3. R

EIN

FO

RC

EM

EN

T S

HA

LL

BE

HIG

H S

TR

EN

GT

H D

EF

OR

ME

D B

AR

S O

F G

RA

DE

Fe415 C

ON

FIR

MIN

G T

O IS

:1786-1985

8m

mØ

@ 20

0mm

c/c

2-25 Ø

A

B

C

2-25 Ø

2-25

Ø+

1-16Ø

2-25 Ø

2-25

Ø+

1-16 Ø

L-S

ection of th

e inclined beam

1750

1750

1325

C

B

A

Title:

Beam

Detailing

Pro

ject:

CO

MM

ER

CIA

L B

UIL

DIN

G

TR

IBH

UV

AN

UN

IVE

RS

ITY

INS

TIT

UT

E O

F E

NG

INE

ER

ING

PU

LC

HO

WK

CA

MP

US

1325

2-25 Ø

8m

mØ

@ 27

0mm

c/c

700

500

AT

C-C

AT

B-B

AT

A-A

2-2

5Ø

2-2

5 Ø+

1-16 Ø

2-12Ø

X-S

ections

1600

8m

mØ

@ 27

0mm

c/c2

-25 Ø

+1-16Ø

2-25

Ø+

1-16Ø

2-25 Ø

2-25 Ø

2-25 Ø

2-25 Ø

1250

1325

7000

5000

D

D

E

E

AT

D-D

AT

E-E

700

500

2-2

5Ø

2-25 Ø

2-12Ø

700

500

2-2

5Ø

2-2

5 Ø+

1-16 Ø

2-12Ø

700

500

2-2

5Ø

2-2

5 Ø+

1-16 Ø

2-12Ø

700

500

2-2

5Ø

2-25 Ø

2-12Ø

8m

mØ

@ 25

0mm

c/c8

mm

Ø @

200m

m c/c

8m

mØ

@ 25

0mm

c/c

550550

550

TR

IBH

UV

AN

UN

IVE

RS

ITY

INS

TIT

UT

E O

F E

NG

INE

ER

ING

PU

LC

HO

WK

CA

MP

US

Pro

ject:

CO

MM

ER

CIA

L B

UIL

DIN

G T

itle:

Reinforcem

ent Details of C

olum

ns

DA

TE

D : A

SH

OJ , 2070

DR

AW

N B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

, SU

MA

N, S

UN

IL

DE

SIG

NE

D B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

, SU

MA

N, S

UN

IL

SU

PE

RV

ISO

R:

Er. D

inesh Gupta

SC

AL

E : N

ot To S

cale

DR

G N

o. :

SH

EE

T N

o. : 8

S.N

.C

OL

UM

NS

IZE

RE

INF

OR

CE

ME

NT

LO

NG

ITU

DIN

AL

TIE

S L

o(C

onfined

Length)

SA

MP

LE

OF

TIE

S

1.

2.

3.

5.

4.

6.

7.

550 X 550

4 # 32 f

+12

# 28 f

Lo

H - 2 L

oS

EC

TIO

N

12 #

32 f

8f

@ 1

008

f @

140

550 mm

8f

@ 1

00

8f

@ 1

00

8f

@ 1

00

8f

@ 1

00

650 mm

650 mm

Tab

le of C

olum

n Reinforcem

ent

135°

Detail of S

tirrup

12f

C-1

C-2

C-3

C-4

C-5

550 X 550

550 X 550

550 X 550

550 X 550

12 #

28 f

12 #

25 f

8 # 25 f

8f

@ 1

40550 m

m

8f

@ 1

40550 m

m

8f

@ 1

40550 m

m

8f

@ 1

40550 m

m

Cinem

ahall -1

Cinem

ahall -2

650 X

650

650 X 6

50

8 # 32 f +8 # 28 f

16

# 28

f

8f

@ 1

008

f @

140

8f

@ 1

008

f @

140

GE

NE

RA

L N

OT

ES

1. A

LL

DIM

NE

SIO

NS

AR

E IN

mm

, UN

LE

SS

OT

HE

RW

ISE

NO

TE

D.

2. G

RA

DE

OF

CO

NC

RE

TE

SH

AL

L B

E M

25 C

ON

FIR

MIN

G T

O IS

:456-20003

. RE

INF

OR

CE

ME

NT

SH

AL

L B

E H

IGH

ST

RE

NG

TH

DE

FO

RM

ED

BA

RS

OF

GR

AD

E F

e500 CO

NF

IRM

ING

TO

IS:1786-1985

5-28f3-28f

8mmf

@15

0c/c

8mmf

@10

0c/c

TR

IBH

UV

AN

UN

IVE

RS

ITY

INS

TIT

UT

E O

F E

NG

INE

ER

ING

PU

LC

HO

WK

CA

MP

US

Pro

ject:

CO

MM

ER

CIA

L B

UIL

DIN

G T

itle:

Reinforcem

ent Detail of S

taircase

DA

TE

D : A

SH

OJ , 2070

DR

AW

N B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

, SU

MA

N, S

UN

IL

DE

SIG

NE

D B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

, SU

MA

N, S

UN

IL

SU

PE

RV

ISO

R:

Er.D

inesh G

upta

SC

AL

E : N

ot To S

cale

DR

G N

o. :

SH

EE

T N

o. : 10

GE

NE

RA

L N

OT

ES

1. A

LL

DIM

NE

SIO

NS

AR

E IN

mm

, UN

LE

SS

OT

HE

RW

ISE

NO

TE

D.

2. G

RA

DE

OF

CO

NC

RE

TE

SH

AL

L B

E M

20

CO

NF

IRM

ING

TO

IS:456-2000

3. R

EIN

FO

RC

EM

EN

T S

HA

LL

BE

HIG

H S

TR

EN

GT

H D

EF

OR

ME

D B

AR

S O

F G

RA

DE

Fe500 C

ON

FIR

MIN

G T

O IS

:1786-1985

15

00

30001500

5000

2500

10001500

1500

2000

1500

6000

XX

12

1

2

3 3

Dog

Leg

ged StairC

ase

180m

m

300m

m16m

Dia@

140mm

10m

m D

ia@250m

m

16m

m D

ia@140m

m

16m

m D

ia@140m

m

15

00

mm

TR

IBH

UV

AN

UN

IVE

RS

ITY

INS

TIT

UT

E O

F E

NG

INE

ER

ING

PU

LC

HO

WK

CA

MP

US

Pro

ject:

CO

MM

ER

CIA

L B

UIL

DIN

G T

itle:

Reinforcem

ent Detail of S

taircase

DA

TE

D : A

SH

OJ , 2070

DR

AW

N B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

, SU

MA

N, S

UN

IL

DE

SIG

NE

D B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

, SU

MA

N, S

UN

IL

SU

PE

RV

ISO

R:

Er.D

inesh G

upta

SC

AL

E : N

ot To S

cale

DR

G N

o. :

SH

EE

T N

o. : 11

GE

NE

RA

L N

OT

ES

1. A

LL

DIM

NE

SIO

NS

AR

E IN

mm

, UN

LE

SS

OT

HE

RW

ISE

NO

TE

D.

2. G

RA

DE

OF

CO

NC

RE

TE

SH

AL

L B

E M

20

CO

NF

IRM

ING

TO

IS:456-2000

3. R

EIN

FO

RC

EM

EN

T S

HA

LL

BE

HIG

H S

TR

EN

GT

H D

EF

OR

ME

D B

AR

S O

F G

RA

DE

Fe500 C

ON

FIR

MIN

G T

O IS

:1786-1985

Section

A-A

2-16m

m B

ars

10m

m D

ia@250m

m10

mm

Dia@

250mm

Section

B-B

16

mm

Dia@

140m

m1

0mm

Dia@

250mm

10m

m D

ia@250m

m

10m

m D

ia@250m

m

8mm

dia nosing

10m

m dia nosing

A A

B B

220m

m

10m

m D

ia@250m

m

16m

m D

ia@280m

m

10

mm

Dia@

250mm

16

mm

Dia@

140m

m16

mm

Dia@

140m

m

16m

m D

ia@280m

m

Beam

Wall

SC

AL

E : N

ot To

Scale

DR

G N

o. :

SH

EE

T N

o. : 12

SU

PE

RV

ISO

R:

Er. D

inesh Gupta

DA

TE

D : A

SH

OJ , 2070

DR

AW

N B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

, SU

MA

N, S

UN

IL

DE

SIG

NE

D B

Y :S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

, SU

MA

N, S

UN

IL

Title:

Reinforcem

ent Detail of S

taircase P

roject:

CO

MM

ER

CIA

L B

UIL

DIN

G

TR

IBH

UV

AN

UN

IVE

RS

ITY

INS

TIT

UT

E O

F E

NG

INE

ER

ING

PU

LC

HO

WK

CA

MP

US

1500m

m

15

00

mm

180m

m

22

0mm

250m

m12m

m D

ia@100m

m

10m

m D

ia@250

mm

10m

m D

ia@250m

m10

mm

Dia@

250m

m

10m

m D

ia@250m

m

10m

m D

ia@250m

m

10m

m D

ia @250m

m

10m

m D

ia@250m

m

10m

m D

ia@250m

m

1500m

m

1500

180

mm

22

0mm

250m

m10

mm

Dia@

250mm

10m

m D

ia@250m

m

10

mm

Dia@

250mm

10

mm

Dia@

250mm

10

mm

Dia@

250m

m

10m

m D

ia@250m

m

12m

m D

ia @100m

m

10m

m D

ia@250m

m

10m

m D

ia@2

50mm

12m

m D

ia@100m

m

10m

m D

ia@250m

m

10m

m D

ia @250m

m

12m

m D

ia@100m

m10m

m D

ia@250m

m

12m

m D

ia@100m

m1

0mm

Dia@

250mm

10m

m D

ia@250m

m

10m

m D

ia@250m

m

250m

m

180m

m2

20mm

10

mm

Dia@

250mm

Open W

ell Interm

ediate Flig

ht

Open W

ell low

er Flig

hts

Op

en W

ell Upp

er Flights

10m

m D

ia@250m

m

(sec at 1-1)

(sec. at 2-2)

(sec. at 3-3

)

GE

NE

RA

L N

OT

ES

1. A

LL

DIM

NE

SIO

NS

AR

E IN

mm

, UN

LE

SS

OT

HE

RW

ISE

NO

TE

D.

2. G

RA

DE

OF

CO

NC

RE

TE

SH

AL

L B

E M

20

CO

NF

IRM

ING

TO

IS:456-2000

3. R

EIN

FO

RC

EM

EN

T S

HA

LL

BE

HIG

H S

TR

EN

GT

H D

EF

OR

ME

D B

AR

S O

F G

RA

DE

Fe500 C

ON

FIR

MIN

G T

O IS

:1786-1985

GE

NE

RA

L N

OT

ES

1. AL

L D

IMN

ES

ION

S A

RE

IN m

m, U

NL

ES

S O

TH

ER

WIS

E N

OT

ED

.2

. GR

AD

E O

F C

ON

CR

ET

E S

HA

LL

BE

M20 C

ON

FIR

MIN

G T

O IS

:456-20003

. RE

INF

OR

CE

ME

NT

SH

AL

L B

E H

IGH

ST

RE

NG

TH

DE

FO

RM

ED

BA

RS

OF

GR

AD

E F

e415 C

ON

FIR

MIN

G T

O IS

:1786-1985

SC

AL

E : N

ot To S

cale

DR

G N

o. :

SH

EE

T N

o. : 13

SU

PE

RV

ISO

R:

Er.D

inesh G

upta

DA

TE

D : B

HA

DR

A , 2070

DR

AW

N B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

,SU

MA

N, S

UN

IL

DE

SIG

NE

D B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

,SU

MA

N, S

UN

IL

Title:

Reinforcem

ent Detail of B

asement W

all P

roject:

CO

MM

ER

CIA

L B

UIL

DIN

G

TR

IBH

UV

AN

UN

IVE

RS

ITY

INS

TIT

UT

E O

F E

NG

INE

ER

ING

PU

LC

HO

WK

CA

MP

US

0.20

Sectio

n at 1-1

8m

m- f

@300m

m c/c

12

mm

-f@

45

0m

m c/c

20

mm

-f@

22

5mm

c/c

1070

1810940

Colum

n

Colum

n

SO

IL

12m

m-f

@450m

m c/c

8m

m-f

@380m

m c/c

11

Reinfo

rcemen

t Details o

f Basem

ent Wall

32m

m- f

@330m

m c/c

32m

m-f

@500m

m c/c

8m

m-f

@380m

m c/c

20

mm

-f@

45

0mm

c/c

20m

m-f

@300m

m c/c

8m

m-f

@300m

m c/c

Plan

of b

asement w

all

PA

RK

ING

2500

2500

650

6501200

12 m

m Ø

@ 1

70 m

m c/c

Vertical B

ars

12 m

m Ø

@ 2

50m

m c/c

Horizon

tal Bars

12 m

m Ø

@ 1

70m

m c/c

Vertical B

ars

12 m

m Ø

@ 2

50 m

m c/c

Ho

rizon

al Bars

32m

m Ø

@ 50

0 mm

c/c

32 m

m Ø

@ 5

00 m

m c/c

PL

AN

SE

CT

ION

OF

LIF

T W

AL

LS

EC

TIO

N 1-1

1

1

GE

NE

RA

L N

OT

ES

1. A

LL

DIM

NE

SIO

NS

AR

E IN

mm

, UN

LE

SS

OT

HE

RW

ISE

NO

TE

D.

2. GR

AD

E O

F C

ON

CR

ET

E S

HA

LL

BE

M20 C

ON

FIR

MIN

G T

O IS

:456-2000

3. R

EIN

FO

RC

EM

EN

T S

HA

LL

BE

HIG

H S

TR

EN

GT

H D

EF

OR

ME

D B

AR

S O

F G

RA

DE

Fe41

5 CO

NF

IRM

ING

TO

IS:1

786-1985

32 m

m Ø

@ 3

30 m

m c/c

32 m

m Ø

@ 3

30 m

m c/c

1070

SC

AL

E : N

ot To S

cale

DR

G N

o. :

SH

EE

T N

o. : 14

SU

PE

RV

ISO

R:

Er. D

inesh Gupta

DA

TE

D : A

SH

OJ , 2070

DR

AW

N B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

, SU

MA

N, S

UN

IL

DE

SIG

NE

D B

Y : S

AG

AR

, SA

MY

OG

, SA

NJE

EM

A, S

AN

TO

SH

, SU

MA

N, S

UN

IL

Title:

Lift W

all Reinforcem

ent P

roject:

CO

MM

ER

CIA

L B

UIL

DIN

G

TR

IBH

UV

AN

UN

IVE

RS

ITY

INS

TIT

UT

E O

F E

NG

INE

ER

ING

PU

LC

HO

WK

CA

MP

US

Lift W

all Reinforcem

ent

1.1320

0.6080

H G

E D C B A

78

65

43

21

6000

6000

6000

34502550

6000

6000

6000

500

6000

500

2500

7000

5000

5000

2500

5000

7000

500

AA

BB

GE

NE

RA

L N

OT

ES

:

1.A

LL

DIM

EN

SIO

NS

AR

E IN

MM

UN

LE

SS

OT

HE

RW

ISE

SP

EC

IFIE

D

2.G

RA

DE

OF

CO

NC

RE

TE

SH

AL

L B

E M

25

CO

NF

OR

MIN

G T

O IS

456:2000

3.R

EIN

FO

RC

EM

EN

T S

HA

LL

BE

HY

SD

BA

RS

OF

GR

AD

E F

E4

15 C

ON

FO

RM

ING

TO

IS 1786:1985

TR

IBH

UV

AN

UN

IVE

RS

ITY

INS

TIT

UT

E O

F E

NG

INE

ER

ING

PU

LC

HO

WK

CA

MP

US

PR

OJE

CT

:C

OM

ME

RC

IAL

BU

ILD

ING

TIT

LE

:M

AT

FO

UN

DA

TIO

N R

EIN

FO

RC

EM

EN

T D

ET

AIL

SP

RO

JEC

T S

UP

ER

VIS

OR

:E

R. D

INE

SH

GU

PT

A

SC

AL

E: N

OT

DR

AW

N T

O S

CA

LE

DR

AW

ING

NO

.:

SH

EE

T N

O.: 15

DA

TE

D: A

SW

IN,2070

DR

AW

N B

Y: S

AG

AR

,SA

MY

OG

,SA

NJE

EM

A,S

AN

TO

SH

,SU

MA

N,S

UN

IL

DE

SIG

NE

D B

Y: S

AG

AR

,SA

MY

OG

,SA

NJE

EM

A,S

AN

TO

SH

,SU

MA

N,S

UN

IL

XX

YY

500

20m

m dia distribution

bars @

160m

m c/c(top)

20m

m dia m

ain bars

@ 1

60m

m c/c(top)

20m

m dia distribution

bars @

16

0mm

c/c(bottom)

20m

m dia m

ain bars

@ 1

60mm

c/c(bottom)

5000

550m

m*5

50mm

column

20m

m dia @

160m

m c/c

20m

m dia @

160m

m c/c

650m

m*6

50mm

cinema hall colum

n

SE

CT

ION

AT

A-A

550

mm

*55

0m

m colum

n

SE

CT

ION

AT

B-B

1070

1040

1070

1040

GE

NE

RA

L N

OT

ES

:

1.A

LL

DIM

EN

SIO

NS

AR

E IN

MM

UN

LE

SS

OT

HE

RW

ISE

SP

EC

IFIE

D

2.G

RA

DE

OF

CO

NC

RE

TE

SH

AL

L B

E M

25

CO

NF

OR

MIN

G T

O IS

456:2000

3.R

EIN

FO

RC

EM

EN

T S

HA

LL

BE

HY

SD

BA

RS

OF

GR

AD

E F

E4

15 C

ON

FO

RM

ING

TO

IS 1786:1985

TR

IBH

UV

AN

UN

IVE

RS

ITY

INS

TIT

UT

E O

F E

NG

INE

ER

ING

PU

LC

HO

WK

CA

MP

US

PR

OJE

CT

:C

OM

ME

RC

IAL

BU

ILD

ING

TIT

LE

:M

AT

FO

UN

DA

TIO

N R

EIN

FO

RC

EM

EN

T D

ET

AIL

SP

RO

JEC

T S

UP

ER

VIS

OR

:E

R. D

INE

SH

GU

PT

A

SC

AL

E: N

OT

DR

AW

N T

O S

CA

LE

DR

AW

ING

NO

.:

SH

EE

T N

O.: 16

DA

TE

D: A

SW

IN,2070

DR

AW

N B

Y: S

AG

AR

,SA

MY

OG

,SA

NJE

EM

A,S

AN

TO

SH

,SU

MA

N,S

UN

IL

DE

SIG

NE

D B

Y: S

AG

AR

,SA

MY

OG

,SA

NJE

EM

A,S

AN

TO

SH

,SU

MA

N,S

UN

IL

650m

m*650

mm

cinema hall colum

n

20m

m dia @

160m

m c/c

20m

m dia @

160m

m c/c

20m

m dia @

160m

m c/c

20m

m dia @

160m

m c/c

20mm

dia @ 160m

m c/c

20m

m dia @

160mm

c/c

65

0mm

*650

mm

cinema hall co

lum

n

20m

m dia @

16

0m

m c/c

DE

TA

IL B

DE

TA

IL A

DE

TA

IL B

20

mm

dia @

160m

m c/c

20m

m d

ia @ 16

0m

m c/c

20

mm

dia @ 16

0mm

c/c

DE

TA

IL A

20m

m dia ch

air @ 1000m

m c/c

20

mm

dia ch

air @ 1

000m

m c/c

ANNEX-I

ROOF TRUSS

Dead load

GI sheeting = 0.084 𝐾𝑁/𝑚2

Fixing = 0.025 𝐾𝑁/𝑚2

Services = 0.1 𝐾𝑁/𝑚2

Self wt of roof truss=( 𝑠𝑝𝑎𝑛

3+ 1) ∗ 10 𝐾𝑁/𝑚2

=( 12

3+ 5) ∗ 10

= 0.09 𝐾𝑁/𝑚2

Assume self wt of purlin =200 𝑁 𝑚⁄

Total dead load =(0.084+0.025+0.1+0.009)*2+0.2

= 0.8 𝐾𝑁 𝑚⁄

False ceiling wt =0.16 𝐾𝑁/𝑚2

=0.16*4.635*2

= 1.48 KN

Other fittings =0.1 KN

Total joint load =1.6 KN

Live load

Slope angle=18.43⁰

Live load = 750-20*(18.43⁰-10⁰)

=581.4 𝑁/𝑚2

=0.6 𝐾𝑁/𝑚2

Total live load = 0.6*2 =1.2 𝐾𝑁 𝑚⁄

Wind load

Wind speed = 47 𝑚 𝑠𝑒𝑐⁄

k1= 1 TABLE 1- LIFE=50 YR,V=47 M/SEC

k2= 0.992 TABLE 2 - class B ,category3,HT=22.4

K3= 1.36 ANNEX-C.2

𝑉𝑏= k1*k2*k3*47 = 63.40864 m/sec

Pz=0.6*𝑉𝑏2 = 2412.393376 𝑁/𝑚2

= 2.5 𝐾𝑁/𝑚2

𝐶𝑝𝑖= ±0.2 (considering the normal permeability of building )

Finally roof load on per meter length of purlin

Joint load on bottom member of truss

wind angle windward leeward

h/w=1.86

0⁰ -0.78 -0.6 TABLE 5

90⁰ -0.8 -0.8

external pressure coefficent

windward leeward windward leeward windward leeward

0⁰ -0.78 -0.6 -0.2 -0.58 -0.4 5.27 -3.0566 -2.108

0.2 -0.98 -0.8 5.27 -5.1646 -4.216

90⁰ -0.8 -0.8 -0.2 -0.6 -0.6 5.27 -3.162 -3.162

0.2 -1 -1 5.27 -5.27 -5.27

2.108pd

(KN/m)

Wind load F(KN/m)pressure coefficent on roof

wind load on roof

Cpewind

angle Cpi

(Cpe-Cpi)

DEAD 0.8 KN/m 0.4 KN/m

LIVE 1.2 KN/m 0.6 KN/m

WIND 5.27 KN/m 2.635 KN/m

LOAD

TYPE

INTERMEDIATE

PURLIN

END

PURLIN

1.6 KN 0.74 KN

INTERMEDIATE

TRUSS

END

TRUSS

LOAD

TYPE

CEILING

FINISH

ANNEX-II

ELEVATOR

Bibliography

Books

1. Reinforced Concrete Design S. U. Pillai & D. Menon

2. Reinforced Concrete Limit State Design A.K. Jain

3. Reinforced Concrete Detailer’s Manual Brian W. Boughton

4. Advanced Structure Analysis A.K. Jain

5. Dynamics of Structure Anil K. Chopra

6. Reinforced Concrete Design S.N. Sinha

7. Reinforced concrete Designer’s Handbook Charles E. Reynolds and James C.

Steedmann

Codes

1. Plain & Reinforced Concrete Code of Practice - IS 456:2000

2. Criteria for Earthquake Resistant Design of Structure - IS 1893(Part I):2000

3. Design Aids for Reinforced Concrete - SP 16

4. Handbook on Concrete Reinforcement & Detailing - SP 34(S & T):1987

5. Design and Construction of Raft Foundation – IS 2950(Part I)-1981

6. Ductile Detailing of Reinforced Concrete Structures IS13920:1993

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