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
BA
SE
ME
NT
PL
AN
60006000
60006000
60006000
60002500
50005000500050007000
LIFT
LIFT
PLA
NT R
OO
M
ELE
CTR
ICR
OO
M
7000 2500
AB
CD
EF
GH
I
12345678
AB
CD
EF
GH
I
12345678
1500
GR
OU
ND
FL
OO
R P
LA
N
2500
700050005000500050007000
60006000
60006000
60006000
60002500
550
550
15001500
30001500
21001800
2100
LIFT
LIFT
SHOP
101
SHOP
102SHOP
103SHOP
104
SHOP
105
SHOP
106SHOP
107
SHOP
108SHOP
109
SHOP
110
SHOP
111
SHOP
112
SHOP
125
SHOP
125A
SHOP
126
SHOP
128
SHOP
127A
SHOP
127
SHOP
113
SHOP
114SHOP
115
SHOP
116SHOP
117
SHOP
118
SHOP
119
SHOP
120
SHOP
121
SHOP
122
SHOP
123
SHOP
124
1500
1000
150020001500
AB
CD
EF
GH
I
AB
CD
EF
GH
I
12345678
2345678 1
2500
FIR
ST
FL
OO
R P
LA
N
700050005000
60006000
60006000
60006000
60002500
50005000500050007000
H
2500
LIFT
LIFT
1500
SHOP
201SHOP
202SHOP
203SHOP
204SHOP
205SHOP
206SHOP
207SHOP
208SHOP
209
SHOP
210
SHOP
211SHOP
212
SHOP
225SHOP
225ASHOP
226
SHOP
228SHOP
227ASHOP
227
SHOP
229
SHOP
229A
SHOP
230A
SHOP
230
SHOP
213
SHOP
214SHOP
215
SHOP
216SHOP
217
SHOP
218SHOP
219
SHOP
220
SHOP
221
SHOP
222SHOP
223
SHOP
224
AB
CD
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AB
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2345678 1
SE
CO
ND
FL
OO
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LA
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LIFT
LIFT
OFFICE
OFFICE
AB
CD
EF
GH
I
AB
CD
EF
GH
I
12345678
60006000
60006000
700050005000500050007000
45678
2500
23 1
OFFICE
OFFICE
OFFICE
OFFICE
OFFICE
OFFICE
OFFICE
OFFICE
OFFICE
OFFICE OFFICE
OFFICE
60006000
60006000
2500
15001500
60006000
60006000
60006000
6000
LIFT
LIFT
CEN
TRA
LA
TRIU
M
OFFIC
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WLIN
GA
LLEY
CIN
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AH
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TH
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FL
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LA
N
2500
AB
CD
EF
GH
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12345678
AB
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700050005000500050007000
45678 23 1
15001500
6000
60006000
60006000
60006000
6000
LIFT
LIFT
CEN
TRA
LA
TRIU
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A H
ALL A
CIN
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A H
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OFFIC
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FO
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Re
fresh
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Pro
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Pro
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CIN
EM
AH
ALL 'A
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BB
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EM
AH
ALL 'B
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BY
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TSTO
ILET
15001000
AB
CD
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CD
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12345678
700050005000500050007000
45678 23 1
1500
LIFT
LIFT
STEEL ROOF
FIF
TH
FL
OO
R P
LA
N
AB
CD
EF
GH
I
AB
CD
EF
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60006000
60006000
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45678 23 1
CENTRAL
ATRIUM
LIFT
LIFT
STEEL ROOF
SIX
TH
FL
OO
R P
LA
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AB
CD
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AB
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CENTRAL
ATRIUM
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
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
THIRD FLOOR
along y direction along x direction
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
along y direction along x direction
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
along y direction along x direction
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
along y direction along x direction
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
Spacing of Bars, Sv = 1000xA
A
sd
b = 1000180
26.50x = 279.25 mm
Provide 8 mm Ø @ 250 mm c/c
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
Grade of Concrete M20 Grade of Steel Fe415(HYSD)
Ref Step Calculations Output
IS 456-
2000
Table 26
1
2
Thickness of slab and durability consideration
Clear Spans
Lx=2.5 m
Ly=5 m
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
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
Finishing load = 1.00 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
Check for depth from Moment Consideration
Depth of Slab, d = mmx
x
bx
M25.50
1000759.2
1097.6
759.2
6
max
Calculation of Area of Steel
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 1000790.150
26.50x = 332.91 mm
Provide 8 mm Ø @ 250 mm c/c
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
Providing 8 mm Ø bars
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
Spacing of Bars, Sv = 1000xA
A
st
b = 100078.114
26.50x = 437.88 mm
Provide 8 mm Ø @ 250 mm c/c
Actual ,Ast = 04.2011000250
26.501000 xx
S
A
v
b mm2
Pt = 0.153 %
Area of Steel Along Long Span
Area of Steel at support ( Top Bars)
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 100015.69
26.50x = 726.84 mm
Provide 8 mm Ø @ 280 mm c/c
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 10002.52
26.50x = 962.84 mm
Provide 8 mm Ø @ 280 mm c/c
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
Here, tension reinforcement of slab contribute in shear
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.
Check for Development Length
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
Spacing of Bars, Sv = 1000xA
A
sd
b = 1000180
26.50x = 279.22
mm
Provide 8 mm Ø @ 250 mm c/c
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
Grade of Concrete M20 Grade of Steel Fe415(HYSD)
Ref Step Calculations Output
IS 456-
2000
Table 26
IS 456-
2000
Annex
D.1.1
1.
2.
3.
4.
Thickness of slab and durability consideration
Clear Spans
Lx=5 m
Ly=6 m
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
Since 22.15000
6000
x
y
l
l Design as Two Way Slab
Design Load
Self load of slab = 0.15 x 25 = 3.75 KN/m2
Finishing load = 1.00 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= -0.0520, αy= -0.037
+ve Bending moment coefficient at mid span
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
Check for depth from Moment Consideration
Depth of Slab, d = mmx
x
bx
M64.78
1000759.2
1007.17
759.2
6
max
Calculation of Area of Steel
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)
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 10003.384
26.50x = 130.78 mm
Provide 8 mm Ø @ 125 mm c/c
Actual ,Ast = 08.4021000125
26.501000 xx
S
A
v
b mm2
Pt = 0.268 %
Area of Steel at mid span (Bottom Bars)
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
Providing 8 mm Ø bars
Ab = 50.26mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 100034.283
26.50x = 177.38 mm
Provide 8 mm Ø @ 125 mm c/c
Actual ,Ast = 08.4021000125
26.501000 xx
S
A
v
b mm2
Pt = 0.268 %
Area of Steel Along Long Span
Area of Steel at support ( Top Bars)
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 100038.398
26.50x = 168.44 mm
Provide 8 mm Ø @ 140 mm c/c
Actual ,Ast = 3591000140
26.501000 xx
S
A
v
b mm2
Pt = 0.292 %
Area of Steel at Mid Span ( Bottom Bars)
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 100063.217
26.50x = 230.94 mm
Provide 8 mm Ø @ 140 mm c/c
Actual ,Ast = 3591000140
26.501000 xx
S
A
v
b mm2
Pt = 0.292 %
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
=32.81KN
Shear at critical section
32.81 KN
31.09 KN
d=0.131 m
2.5 m
131.05.25.2
81.32
uV
Vu = 31.09 KN
Here, tension reinforcement of slab contribute in shear
For pt = 0.268 %
37.0c N/mm2
k c bd = 1000
131100037.03.1 xxx = 63.01 KN > Vu
Check for Deflection
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
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 %
Modification factor (M.F.) = 2
dper = 15.96262
5000
..
xValueBasicxFM
lx mm < d provided.
Check for Development Length
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
Spacing of Bars, Sv = 1000xA
A
sd
b = 1000180
26.50x = 279.22
mm
Provide 8 mm Ø @ 250 mm c/c
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
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
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
Grade of Concrete M20 Grade of Steel Fe415(HYSD)
Ref Step Calculations Output
IS 456-
2000
Table 26
IS 456-
2000
Annex
D.1.1
IS 456-
2000
cl.26.5.2.1
1.
2.
3.
4.
Thickness of slab and durability consideration
Clear Spans
Lx=5 m
Ly=6 m
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
Since 22.15000
6000
x
y
l
l Design as Two Way Slab
Design Load
Self load of slab = 0.15 x 25 = 3.75 KN/m2
Finishing load = 1.00 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= -0.036, αy= -0.037
+ve Bending moment coefficient at mid span
αx= 0.048, αy= 0.028
For Short Span
Support moment , Ms = - αxwlx2 = -0.036x 13.125 x 52 = -15.75
KN-m
Mid span moment , Mm = αywlx2 = 0.048 x 13.125 x 52 = 11.81
KN-m
For Long Span
Support moment , Ms = - αxwlx2 = -0.037 x 13.125 x 52 = -12.14
KN-m
Mid span moment , Mm = αywlx2 = 0.028 x 13.125 x 52 =
9.19KN-m
Check for depth from Moment Consideration
Depth of Slab, d = mmx
x
bx
M54.75
1000759.2
1075.15
759.2
6
max
Calculation of Area of Steel
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)
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 100070.352
26.50x = 142.5 mm
Provide 8 mm Ø @ 125 mm c/c
Actual ,Ast = 08.4021000125
26.501000 xx
S
A
v
b mm2
Pt = 0.268 %
Area of Steel at mid span (Bottom Bars)
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
Providing 8 mm Ø bars
Ab = 50.26mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 100044.260
26.50x = 192.98 mm
Provide 8 mm Ø @ 125 mm c/c
Actual ,Ast = 08.4021000125
26.501000 xx
S
A
v
b mm2
Pt = 0.268 %
Area of Steel Along Long Span
Area of Steel at support ( Top Bars)
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 10003.287
26.50x = 174.94 mm
Provide 8 mm Ø @ 140 mm c/c
Actual ,Ast = 3591000140
26.501000 xx
S
A
v
b mm2
Pt = 0.292 %
Area of Steel at Mid Span ( Bottom Bars)
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 100071.214
26.50x = 234.08 mm
Provide 8 mm Ø @ 140 mm c/c
Actual ,Ast = 3591000140
26.501000 xx
S
A
v
b mm2
Pt = 0.292 %
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
=32.81KN
Shear at critical section
32.81 KN
31.09 KN
d=0.131 m
2.5 m
131.05.25.2
81.32
uV
Vu = 31.09 KN
Here, tension reinforcement of slab contribute in shear
For pt = 0.268 %
37.0c N/mm2
k c bd = 1000
131100037.03.1 xxx = 63.01 KN > Vu
Check for Deflection
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
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 %
Modification factor (M.F.) = 2
dper = 15.96262
5000
..
xValueBasicxFM
lx mm < d provided.
Check for Development Length
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
Spacing of Bars, Sv = 1000xA
A
sd
b = 1000180
26.50x = 279.22
mm
Provide 8 mm Ø @ 250 mm c/c
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
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
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
Grade of Concrete M20 Grade of Steel Fe415(HYSD)
Ref Step Calculations Output
IS 456-
2000
Table 26
1.
2.
3.
Thickness of slab and durability consideration
Clear Spans
Lx=5 m
Ly=6 m
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
Since 22.15000
6000
x
y
l
l Design as Two Way Slab
Design Load
Self load of slab = 0.15 x 25 = 3.75 KN/m2
Finishing load = 1.00 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= -0.060, αy= -0.047
+ve Bending moment coefficient at mid span
αx= 0.045, αy= 0.035
For Short Span
Support moment , Ms = - αxwlx2 = -0.06x 13.125 x 52 = -19.68
KN-m
Mid span moment , Mm = αywlx2 = 0.045 x 13.125 x 52 =
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
Mid span moment , Mm = αywlx2 = 0.035 x 13.125 x 52 =
11.484KN-m
Check for depth from Moment Consideration
Depth of Slab, d = mmx
x
bx
M45.84
1000759.2
1068.19
759.2
6
max
Calculation of Area of Steel
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
1068.196.411(
415
205.0
2
6
xxxx
xxx
= 448.10 mm2 > Min Ast
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 100010.448
26.50x = 112.16 mm
Provide 8 mm Ø @ 100 mm c/c
Actual ,Ast = 6.5021000100
26.501000 xx
S
A
v
b mm2
Pt = 0.383 %
Area of Steel at mid span (Bottom Bars)
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
Providing 8 mm Ø bars
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
Spacing of Bars, Sv = 1000xA
A
st
b = 100041.329
26.50x = 152.57 mm
Provide 8 mm Ø @ 125 mm c/c
Actual ,Ast = 08.4021000125
26.501000 xx
S
A
v
b mm2
Pt = 0.307 %
Area of Steel Along Long Span
Area of Steel at support ( Top Bars)
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 100056.370
26.50x = 135.63 mm
Provide 8 mm Ø @ 130 mm c/c
Actual ,Ast = 615.3861000130
26.501000 xx
S
A
v
b mm2
Pt = 0.295 %
Area of Steel at Mid Span ( Bottom Bars)
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 100012.271
26.50x = 185.37 mm
Provide 8 mm Ø @ 180 mm c/c
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
For x-direction i.e. short span
Shear force at the face of the support, V = w lx = 13.125 x 2
.5
=32.81KN
Shear at critical section
32.81 KN
31.09 KN
d=0.131 m
2.5 m
131.05.25.2
81.32
uV
Vu = 31.09 KN
Here, tension reinforcement of slab contribute in shear
For pt = 0.383 %
423.0c N/mm2
k c bd = 1000
1311000423.03.1 xxx = 72.03 KN > Vu
Check for Deflection
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
06.335
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
Check for Development Length
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
Spacing of Bars, Sv = 1000xA
A
sd
b = 1000180
26.50x = 279.22 mm
Provide 8 mm Ø @ 250 mm c/c
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
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
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
Grade of Concrete M20 Grade of Steel Fe415(HYSD)
Ref Step Calculations Output
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.
Thickness of slab and durability consideration
Clear Spans
Lx=5 m
Ly=6 m
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
Since 22.15000
6000
x
y
l
l Design as Two Way Slab
Design Load
Self load of slab = 0.15 x 25 = 3.75 KN/m2
Finishing load = 1.00 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= 0.0, αy= -0.045
+ve Bending moment coefficient at mid span
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
Support moment , Ms = - αxwlx2 = -0.045 x 13.125 x 52 = -14.76
KN-m
Mid span moment , Mm = αywlx2 = 0.035x 13.125 x 52 =
11.487KN-m
Check for depth from Moment Consideration
Depth of Slab, d = mmx
x
bx
M87.77
1000759.2
1073.16
759.2
6
max
Calculation of Area of Steel
Min Ast = 0.12 % of bD = 0.12 x 1000 x 150 = 180 mm2
Area of Steel along short span
Area of Steel at mid span (Bottom Bars)
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
Providing 8 mm Ø bars
Ab = 50.26mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 1000332.376
26.50x = 133.55 mm
Provide 8 mm Ø @ 125 mm c/c
Actual ,Ast = 08.4021000125
26.501000 xx
S
A
v
b mm2
Pt = 0.268 %
Area of Steel Along Long Span
Area of Steel at support ( Top Bars)
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 100062.353
26.50x = 142.12 mm
Provide 8 mm Ø @ 140 mm c/c
Actual ,Ast = 3591000140
26.501000 xx
S
A
v
b mm2
Pt = 0.274 %
Area of Steel at Mid Span ( Bottom Bars)
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 10002
26.50x = 185.57 mm
Provide 8 mm Ø @ 140 mm c/c
Actual ,Ast = 3591000140
26.501000 xx
S
A
v
b mm2
Pt = 0.274 %
Check for Shear
For x-direction i.e. short span
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.
Shear force at the face of the support, V = w lx = 13.125 x 2
.5
=32.81KN
Shear at critical section
32.81 KN
31.09 KN
d=0.131 m
2.5 m
131.05.25.2
81.32
uV
Vu = 31.09 KN
Here, tension reinforcement of slab contribute in shear
For pt = 0.268 %
37.0c N/mm2
k c bd = 1000
131100037.03.1 xxx = 63.01 KN > Vu
Check for Deflection
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
08.402
34.283
= 169.62 N/mm2
Pt= 0.268 %
Modification factor (M.F.) = 2
dper = 15.96262
5000
..
xValueBasicxFM
lx mm < d provided.
Check for Development Length
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
Spacing of Bars, Sv = 1000xA
A
sd
b = 1000180
26.50x = 279.22
mm
Provide 8 mm Ø @ 250 mm c/c
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
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
Asd=180 mm2
Act. Asd =
201.04 mm2
55
Design of 5m X 6m interior panel slab
Grade of Concrete M20 Grade of Steel Fe415(HYSD)
Ref Step Calculations Output
IS 456-
2000
Table 26
1.
2.
3.
Thickness of slab and durability consideration
Clear Spans
Lx=5000mm
Ly=6000mm
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 m
Since 22.15000
6000
x
y
l
l Design as Two Way Slab
Design Load
Self load of slab = 0.15 x 25 = 3.75 KN/m2
Finishing load = 1.00 KN/m2
Dead load = 3.75 + 1.00 = 4.75 KN/m2
Live load = 5.0 KN/m2
Design load , w = 1.5(DL+LL) = 14.625 KN/m2
Considering unit width of slab , w=14.625 KN/m
Moment Calculation
-ve Bending moment coefficient at continuous edge
αx = -0.043, αy= -0.032
+ve Bending moment coefficient at mid span
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
Check for depth from Moment Consideration
Depth of Slab, d = mmx
x
bx
M47.75
1000759.2
1072.15
759.2
6
max
Calculation of Area of Steel
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
1072.156.411(
415
205.0
2
6
xxxx
xxx
= 351.98 mm2 > Min Ast
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 100098.351
26.50x = 142.8 mm
Provide 8 mm Ø @ 125 mm c/c
Actual ,Ast = 4021000125
26.501000 xx
S
A
v
b mm2
Pt = 0.268 %
Area of Steel at mid span (Bottom Bars)
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
Providing 8 mm Ø bars
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
Spacing of Bars, Sv = 1000xA
A
st
b = 10009.257
26.50x = 194.88 mm
Provide 8 mm Ø @ 125 mm c/c
Actual ,Ast = 08.4021000125
26.501000 xx
S
A
v
b mm2
Pt = 0.268 %
Area of Steel Along Long Span
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
Providing 8 mm Ø bars
Ab = 50.26mm2
Spacing of Bars, Sv = 1000xA
A
st
b = 100034.276
26.50x = 181.87 mm
Provide 8 mm Ø @ 140 mm c/c
Actual ,Ast = 3591000140
26.501000 xx
S
A
v
b mm2
Pt = 0.292 %
Area of Steel at Mid Span ( Bottom Bars)
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
Providing 8 mm Ø bars
Ab = 50.26mm2
Spacing of Bars, Sv = 1000xA
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.
Provide 8 mm Ø @ 140 mm c/c
Actual ,Ast = 3591000140
26.501000 xx
S
A
v
b mm2
Pt = 0.292 %
Check for Shear
For x-direction i.e. short span
Shear force at the face of the support, V = w lx = 14.625 x 2
0.5
=36.56 KN
Shear at critical section
36.56 KN
34.64 KN
d = 0.131m
2.5m
131.050.250.2
56.36
uV
Vu = 34.64 KN
Here, tension reinforcement of slab contribute in shear
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
Check for Deflection
Along short Span
Since both ends are continuous, the basic value may be taken as
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.
fs= ovidedPrSteelofArea
quiredReSteelofAreaf58.0 y = 0.58 x 415 x
08.402
9.257
= 154.4 N/mm2
Pt= 0.268 %
Modification factor (M.F.) = 2
dper = 15.96262
5000
..
xValueBasicxFM
lx mm < d provided.
Check for Development Length
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
Spacing of Bars, Sv = 1000xA
A
sd
b = 1000180
26.50x = 279 mm
Provide 8 mm Ø @ 270 mm c/c
Act. Ast = 1000xS
A
v
b = 15.1861000270
26.50x mm2
Torsion Reinforcement
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
Thickness of slab and durability consideration
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
Assuming clear span cover=15 mm
Providing 8 mm Ø bar
Effective depth of slab, d= 150 - 15 - 8/2 = 131 mm
Design Load
Self load of slab = 0.150 x 25 = 3.75 KN/m2
Finishing load = 1.00 KN/m2
Dead load = 3.75 + 1.00 = 4.75 KN/m2
Live load = 4.0 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
For x-direction i.e. short span
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
Check for Deflection
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.
Check for Development Length
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
Provide 8 mm Ø @ 250 mm c/c
Act. Ast = 1000xS
A
v
b = 04.2011000x250
26.50 mm2
Torsion Reinforcement
Torsion reinforcement is provided as per IS 456-2000 Cl.D.1.8,Cl. D.1.9
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
62
Torsion Reinforcement
Grade of concrete M20 Grade of Steel Fe415 (HYSD)
Ref Step Calculations Output
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
stT
b = 100061
26.50x = 823 mm
Provide 8 mm Ø @ 300 mm c/c
No. of bars required = barsS
L
V
stT2
300
500
2618
3mmAxA ststT
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
stT
b = 1000122
26.50x = 411 mm
Provide 8 mm Ø @ 300 mm c/c
No. of bars required = barsS
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.
Distance of Torsional Reinforcement from edge of slab
mL
L x
stT 6.05
Area of bar required for maximum mid-span moment Ast=
359 mm2 / m length
224.2694
3mmAxA ststT
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
stT
b = 100024.269
26.50x = 186.67
mm
Provide 8 mm Ø @ 180 mm c/c
No. of bars required = barsS
L
V
stT4
180
600
262.1348
3mmAxA ststT
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
stT
b = 100062.134
26.50x = 373.34
mm
Provide 8 mm Ø @ 300 mm c/c
No. of bars required = barsS
L
V
stT2
300
600
For one short edge discontinuous with Lx = 5 m
Shorter span of slab (Lx) = 5 m
Distance of Torsional Reinforcement from edge of slab
mL
L x
stT 0.15
Area of bar required for maximum mid-span moment Ast=
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
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
stT
b = 100013.148
26.50x = 339.3 mm
Provide 8 mm Ø @300 mm c/c
No. of bars required = barsS
L
V
stT4
300
1000
For two adjacent edge discontinuous with Lx = 5 m
Shorter span of slab (Lx) = 5 m
Distance of Torsional Reinforcement from edge of slab
mL
L x
stT 0.15
Area of bar required for maximum mid-span moment Ast=
329.41 mm2 / m length
206.2474
3mmAxA ststT
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing, Sv = 1000xA
A
stT
b = 100006.247
26.50x = 203.43 mm
Provide 8 mm Ø @200 mm c/c
No. of bars required = barsS
L
V
stT5
200
1000
253.1238
3mmAxA ststT
Providing 8 mm Ø bars
Ab = 50.26 mm2
Spacing of Bars, Sv = 1000xA
A
stT
b = 100053.123
26.50x = 406.86 mm
Provide 8 mm Ø @300 mm c/c
No. of bars required = barsS
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)
Ref Step Calculations Output
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
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 = 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
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) = 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
For maximum -ve moment: (hogging moment)
Mu = 27.33 KNm
Torsional Moment, Tu =17.263 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 = 17.263x 7.1
500/7001 = 24.37 KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
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
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 = 856.7mm2(Top)
Asc = 856.7mm2 (Bottom)
For maximum +ve moment: (sagging moment)
Mu = 1.51 KNm
Torsional Moment, Tu = 17.263 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 = 17.263x 7.1
500/7001 = 24.37 KNm
Since Mt>Mu, so longitudinal reinforcement shall be
provided on the flexural compression face.
Hence, Me1 = Mu+Mt = 1.51+ 24.37= 25.88 KNm
And,
Me2 =Mt-Mu=24.37-1.51=22.86KNm.
Here, Mulim>Mu Hence, The section is to be singly-
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)
Providing minimum area of steel=856.7mm2
Provide 2-28mm
Act.Ast=1231.504mm2 (Bottom 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 = 856.7mm2(Bottom)
Asc = 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)
As,top/bottom=1/2*Max –ve steel at top of either joint
=0.5*1231.5
=615.75mm2 <1231.5mm2
Hence, O.K.
For As(bottom), provide 2-28mm bars
Act.As(bottom)= 1231.5mm2 (bottom)
For As(top), provide 6-28mm bars
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
Act.As(bottom)=1231.5
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)
Design shear at left end=higher of the two values
=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)
Here, τve>τc, hence transverse reinforcement shall be
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
Transverse reinforcement
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,
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 12mm 2legged vertical
stirrups@100mmc/c
At right end
(sway to right):
Vu=181.215KN(+ve value from SAP)
(sway to left):
Vu=154.99KN(-ve value from SAP)
Design shear at left end=higher of the two values
=536.03KN
Equivalent shear,
Ve = Vu + 1.6b
Tu = 536.03+ 1.6 x
5.0
72.37 = 656.73KN
Equivalent nominal shear stress,
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
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=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
b1=b-2clear cover - = 500 – 2x25 – 28 = 422mm
d1=D-2clear cover - = 700 – 2x25 – 28= 622mm
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,
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.
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)
Ref Step Calculations Output
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
Overall Depth of Beam, D=700mm
Width of Beam, B=500mm
Considering 1 layer of 32mm bar with spacer
bar to be used
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
Span Length, L=6m
L/D = 5.45/.7 = 7.79 > 4 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 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)
For maximum -ve moment: (hogging moment)
Mu = 715.02 KNm
Torsional Moment, Tu = 100.74 KNm
The longitudinal reinforcement shall be designed
to resist an equivalent bending moment, Me1
Me1 = Mu + Mt
The torsional moment,
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%
Asc must be at least 50% of Ast
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)
For maximum +ve moment: (sagging moment)
Mu = 506.42 KNm
Torsional Moment, Tu = 100.74 KNm
The longitudinal reinforcement shall be designed
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)
to resist an equivalent bending moment, Me1
Me1 = Mu + Mt
The torsional moment,
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 = 506.42+ 142.22= 648.64
KNm
Here,
Mulim<Mu Hence, The section is to be doubly-
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%
Asc must be at least 50% of Ast
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
For maximum -ve moment: (hogging moment)
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)
Torsional Moment, Tu = 100.74 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 = 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 = 168.61 + 142.22= 310.83
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
=1437.09mm2 > Ast,min
Pt=0.436%
Providing two longitudinal bars at the bottom as
well
Asc must be at least 50% of Ast
50% of Ast =0.218% <0.26%(rmin)
So,providing minimum area of steel,
Asc=856.7mm2
Ast =1437.09mm2(Top)
Asc = 856.70mm2 (Bottom)
For maximum +ve moment: (sagging moment)
Mu = 236.48 KNm
Torsional Moment, Tu = 100.74 KNm
The longitudinal reinforcement shall be designed
to resist an equivalent bending moment, Me1
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)
The torsional moment,
Mt = Tu x7.1
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 = 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
Asc must be at least 50% of Ast
50% of Ast =0.272% >0.26%(rmin)
Ast = 1795.43mm2(Bottom)
Asc = 897.72mm2 (Top)
For As(bottom), provide 2-32mm and 2-16mm bars
Act.As(bottom)= 2010.62 mm2 (bottom)
For As(top), provide 2-32mm bars
Act.As(top)=1608.50 mm2 (top)
At right end
For maximum -ve moment: (hogging moment)
Mu =712.44 KNm
Torsional Moment, Tu = 97.62 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 = 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
Since Mt<Mu, no longitudinal reinforcement shall
be provided on the flexural compression face.
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%
Asc must be at least 50% of Ast
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)
For maximum +ve moment: (sagging moment)
Mu = 464.41 KNm
Torsional Moment, Tu = 97.62 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 = 97.62* 7.1
500/7001 = 137.82
KNm
Since Mt<Mu, no longitudinal reinforcement shall
be provided on the flexural compression face.
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%
Asc must be at least 50% of Ast
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(bottom)= 3619.115mm2 (bottom)
For As(top), provide 4-32mm and 2-28mm 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):
Vu=128.211KN(+ve value from SAP)
(sway to left):
Vu=339.348KN(-ve value from SAP)
Design shear at left end=higher of the two values
=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
Equivalent nominal shear stress,
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:
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 – 32 = 418mm
d1=D-2clear cover - = 700 – 2x25 – 32= 618mm
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,
Spacing of stirrups:least of
i)Sv=100mm
ii)d/4=661/4=165.25mm
ii)8* small=8*16=128mm
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
Hence, Provide 8mm 4legged vertical
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)
Design shear at left end=higher of the two values
=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.
Transverse reinforcement
Asv = )f87.0(d5.2
SV
)f87.0(db
S
y1
vu
y11
vu
Let us use 12mm -2legged vertical stirrups,
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,
Spacing of stirrups:least of
i)Sv=100mm
ii)d/4=661/4=165.25mm
ii)8* small=8*16=128mm
but >=100mm
Hence, Provide 12mm 2legged vertical
stirrups@125mmc/c
At right end
Provided as in left support
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
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
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
2-12mm bar along
each vertical side.
83
Ductile Design of 7m span inclined beam
Concrete Grade = M20 Steel Grade = Fe415 (HYSD)
Ref Step Calculations Output
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
Overall Depth of Beam, D=700mm
Width of Beam, B=500mm
Considering 1 layer of 32mm bar with spacer bar to
be used
Taking clear cover=25mm.
Effective depth, d=700-25-32/2 = 659 mm
Check for Axial Stress
Factored Axial Stress = 0 N/mm2
Axial Stress = 0 N/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
Span Length, L=6.35m
L/D = 6.35/.7 = 10 > 4 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
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
For maximum -ve moment: (hogging moment)
Mu = 222.45 KNm
Torsional Moment, Tu = 21.32 KNm
The longitudinal reinforcement shall be designed to
resist an equivalent bending moment, Me1
Me1 = Mu + Mt
The torsional moment,
Mt = Tu 7.1
b/D1 = 21.32 * 7.1
500/7001 = 30.10 KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
Hence, Me1 = Mu+Mt + = 222.45 + 30.10 =252.55 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
=1145 > Ast,min
Pt=0.35%
Providing two longitudinal bars at the bottom as well
Asc must be at least 50% of Ast
50% of Ast =0.18% <0.26%(rmin)
So,providing minimum area of steel,
Asc=856.7mm2
Ast =1145.00mm2(Top)
Asc = 856.70mm2 (Bottom)
For maximum +ve moment: (sagging moment)
Mu = 65.46 KNm
Torsional Moment, Tu = 21.32 KNm
The longitudinal reinforcement shall be designed to
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)
resist an equivalent bending moment, Me1
Me1 = Mu + Mt
The torsional moment,
Mt = Tu *7.1
b/D1 = 21.32* 7.1
500/7001 = 30.10 KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
Hence, Me1 = Mu+Mt = 65.46+ 30.10=95.56 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
=413mm2 <Ast,min
So,providing minimum area of steel,
Ast =856.70mm2(Bottom)
Asc = 856.70mm2 (Top)
For As(top), provide 2-25mm and 1-16mm bars
Act.As(top)= 1182mm2 (top)
For As(bottom), provide 2-25mm bars
Act.As(bottom)=981mm2 (bottom)
At mid span
For maximum -ve moment: (hogging moment)
Mu = 0 KNm
Torsional Moment, Tu = 13.05 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 = 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)
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
Hence, Me1 = Mu+Mt = 0 + 18.42= 18.42 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
=78mm2 < Ast,min
So,providing minimum area of steel,
Ast =856.70mm2(Top)
Asc = 856.70mm2 (Bottom)
For maximum +ve moment: (sagging moment)
Mu = 111.42 KNm
Torsional Moment, Tu = 13.05 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 = 13.05* 7.1
500/7001 = 18.42 KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
Hence, Me1 = Mu+Mt = 111.42+18.42= 129.84 KNm
Here, Mulim>Mu Hence, The section is to be singly-
reinforced.
Ast = bdbdf
M6.411
f
f5.0
2ck
u
y
ck
=566.16mm2 (Bottom bars) <Ast,min(856.7mm2)
So,providing minimum area of steel,
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)
For As(bottom), provide 2-25mm bars
Act.As(bottom)= 981 mm2 (bottom)
For As(top), provide 2-25mm bars
Act.As(top)=981mm2 (top)
At right end
For maximum -ve moment: (hogging moment)
Mu =180.15 KNm
Torsional Moment, Tu = 13.05 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 = 13.05* 7.1
500/7001 = 18.42 KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
Hence, Me1 = Mu+Mt = 180.15+18.42= 198.57 KNm
Here, Mulim>Mu Hence, The section is to be singly-
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)
For maximum +ve moment: (sagging moment)
Mu = 77.28 KNm
Torsional Moment, Tu = 13.05 KNm
The longitudinal reinforcement shall be designed to
resist an equivalent bending moment, Me1
Me1 = Mu + Mt
The torsional moment,
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
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
Hence, Me1 = Mu+Mt = 95.70 KNm
Here, Mulim>Mu Hence, The section is to be singly-
reinforced.
Ast = bdbdf
M6.411
f
f5.0
2ck
u
y
ck
=414mm2 (Bottom bars) <Ast,min(856.7mm2)
Ast = 856.70mm2(Bottom)
Asc = 856.70mm2 (Top)
For As(top), provide 2-25mm and 1-16mm bars
Act.As(top)= 1182mm2 (top)
For As(bottom), provide 2-25mm bars
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):
Vu=143.63KN(-ve value from SAP)
Design shear at left end=higher of the two values
=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
Equivalent nominal shear stress,
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
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-2legged vertical stirrups,
Asv=100.53mm2
A clear cover of 25mm is assumed all around.
b1=b-2clear cover - = 500 – 2x25 – 25 = 425mm
d1=D-2clear cover - = 700 – 2x25 – 25= 625mm
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,
Spacing of stirrups:least of
i)Sv=100mm
ii)d/4=662/4=165.5mm
ii)8* small=8*16=128mm
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
Hence, Provide 8mm 2legged vertical
stirrups@200mmc/c over a length of 2d=1325mm.
At mid-span:
(sway to right):
Vu=25.61KN(+ve)
(sway to left):
Vu=37.82KN(-ve)
Design shear at left end=higher of the two values
=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.
Hence, Provide 8mm 2legged vertical
stirrups@300mmc/c
At right end
(sway to right):
Vu=125.51KN(+ve value from SAP)
(sway to left):
Vu=0KN(-ve value from SAP)
Design shear at left end=higher of the two values
=125.51KN
Equivalent shear,
Ve = Vu + 1.6b
Tu = 125.51+ 1.6 x
5.0
05.13 = 167.27KN
Equivalent nominal shear stress,
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
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-2legged vertical stirrups,
Asv=100.53mm2
A clear cover of 25mm is assumed all around.
b1=b-2clear cover - = 500 – 2x25 – 25 = 425mm
d1=D-2clear cover - = 700 – 2x25 – 25= 625mm
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,
Spacing of stirrups:least of
i)Sv=100mm
ii)d/4=662/4=165.5mm
ii)8* small=8*16=128mm
but >=100mm
Hence, Provide 8mm 2legged vertical
stirrups@250mmc/c over a length of 2d=1325mm.
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
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
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
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
Provide 2-legged 8mm stirrups @ 250mm c/c up to
length 2d=1325mm from right end
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
2-12mm bar along
each vertical side.
93
Ductile Design of 5m span beam
Concrete Grade = M20 Steel Grade = Fe415(HYSD)
Ref Step Calculations Output
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
Overall Depth of Beam, D=700mm
Width of Beam, B=500mm
Considering 1 layer of 32mm bar with spacer bar to
be used
Taking clear cover=25mm.
Effective depth, d=700-25-32/2 = 659 mm
Check for Axial Stress
Factored Axial Stress = 0 N/mm2
Axial Stress = 0 N/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
Span Length, L=4.45m
L/D = 4.45/.7 = 6.36 > 4 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 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)
For maximum -ve moment: (hogging moment)
Mu = 615.911 KNm
Torsional Moment, Tu = 30.037 KNm
The longitudinal reinforcement shall be designed to
resist an equivalent bending moment, Me1
Me1 = Mu + Mt
The torsional moment,
Mt = Tu 7.1
b/D1 = 30.037x 7.1
500/7001 =
42.405KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
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%
Asc must be at least 50% of Ast
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)
For maximum +ve moment: (sagging moment)
Mu = 453.95 KNm
Torsional Moment, Tu = 37.927 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)
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 = 36.43x 7.1
500/7001 = 53.544 KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
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
Providing minimum area of steel=2540.40mm2
Provide 4-32mm
Act.Ast=3217mm2 (Bottom bars)
Pt=0.98%
Providing two longitudinal bars at the top as well
Asc must be at least 50% of Ast
50% of Ast =0.49% >0.26%(rmin)
So,providing area of steel,
Asc=1614.55mm2
Ast = 2540.40mm2(Bottom)
Asc = 1614.55mm2 (Top)
For As(bottom), provide 6-28mm bars
Act.As(bottom)= 3694.51 mm2 (bottom)
For As(top), provide 5-28mm bars
Act.As(top)= 3078.76 mm2 (top)
At mid span
For maximum +ve moment: (sagging moment)
Mu = 96.039 KNm
Singly Reinforced
Section
As(top)= 3390.55mm2
As(bottom)=
2540.40mm2
Act.As(top)=
3694.513mm2
Act.As(bottom)=3078.76
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)
Torsional Moment, Tu = 37.927 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 = 37.927x 7.1
500/7001 = 53.544
KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
Hence, Me1 = Mu+Mt = 90.039 + 53.544= 143.583
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
=628.65mm2 < 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 top 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 = 856.7mm2(Top)
Asc = 856.7mm2 (Bottom)
For maximum -ve moment: (hogging moment)
Mu = 64.33 KNm
Torsional Moment, Tu = 39.06 KNm
The longitudinal reinforcement shall be designed to
resist an equivalent bending moment, Me1
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
The torsional moment,
Mt = Tu x7.1
b/D1 = 39.06x 7.1
500/7001 = 55.143 KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
Hence, Me1 = Mu+Mt = 64.33+ 55.143= 119.47 KNm
Here, Mulim>Mu Hence, The section is to be singly-
reinforced.
Ast = bdbdf
M6.411
f
f5.0
2ck
u
y
ck
=519.36mm2 (Top Bar)<Ast,min(856.7mm2)
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 = 856.7mm2(Bottom)
Asc = 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)
At right end
For maximum -ve moment: (hogging moment)
Mu =576.88 KNm
Torsional Moment, Tu = 39.06 KNm
The longitudinal reinforcement shall be designed to
resist an equivalent bending moment, Me1
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
The torsional moment,
Mt = Tu x7.1
b/D1 = 39.06 x 7.1
500/7001 = 55.143
KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
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%
Asc must be at least 50% of Ast
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)
For maximum +ve moment: (sagging moment)
Mu = 485.44 KNm
Torsional Moment, Tu = 37.92KNm
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 = 37.92x 7.1
500/7001 = 53.53 KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
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
Act.Ast=3694.512mm2 (Bottom bars)
Pt=1.12%
Providing two longitudinal bars at the top as well
Asc must be at least 50% of Ast
50% of Ast =0.46% <0.26%(rmin)
So,providing minimum area of steel,
Asc=1847mm2
Ast = 1847mm2(Top)
Asc = 3694.512mm2 (Bottom)
As,top/bottom=1/4*Max –ve steel at top of either joint
=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):
Vu=292.593KN(-ve value from SAP)
Design shear at left end=higher of the two values
=292.593KN
Equivalent shear,
Ve = Vu + 1.6b
Tu = 292.593+ 1.6 x
5.0
93.37 = 413.97KN
Equivalent nominal shear stress,
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
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-2legged vertical stirrups,
Asv=226.19mm2
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
226.19=
41587.06225.2
10593.292
41587.0622422
1093.37 36
xxx
x
xxx
x Sv
Sv=109.11mm
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 12mm 2legged vertical
stirrups@240mmc/c over a length of 2d=1322mm.
At mid-span:
(sway to right):
Vu=188.604KN(+ve)
(sway to left):
Vu=213.941KN(-ve)
Design shear at left end=higher of the two values
=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)
Here, τve>τc, hence transverse reinforcement shall be
provided as below.
Transverse reinforcement
Asv = )f87.0(d5.2
SV
)f87.0(db
S
y1
vu
y11
vu
Let us use 8mm -2legged vertical stirrups,
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,
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 2legged vertical
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
Vu=275.23KN(+ve value from SAP)
(sway to left):
Vu=153.16KN(-ve value from SAP)
Design shear at left end=higher of the two values
=275.23KN
Equivalent shear,
Ve = Vu + 1.6b
Tu = 275.23+ 1.6 x
5.0
06.39 = 400.22KN
Equivalent nominal shear stress,
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
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-2legged vertical stirrups,
Asv=100.53mm2
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
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
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 2legged vertical
stirrups@100mmc/c over a length of 2d=1322mm.
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
Provide 2-legged 8mm stirrups @ 100mm c/c up to
length
2d=1322mm from left end
Provide 2-legged 8mm stirrups @ 125mm c/c at mid
span
Provide 2-legged 8mm stirrups @ 100mm c/c up to
length
2d=1322mm from right end
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
8mm 2LVS@125m
mc/c
2-12mm bar along
each vertical side.
104
Ductile Design of 2.5m span beam
Concrete Grade = M20 Steel Grade = Fe415(HYSD)
Ref Step Calculations Output
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
Overall Depth of Beam, D=700mm
Width of Beam, B=500mm
Considering 1 layer of 32mm bar with spacer bar to
be used
Taking clear cover=25mm.
Effective depth, d=700-25-32/2 = 659 mm
Check for Axial Stress
Factored Axial Stress = 0 N/mm2
Axial Stress = 0 N/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 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)
For maximum -ve moment: (hogging moment)
Mu = 579.6 KNm
Torsional Moment, Tu = 36.43 KNm
The longitudinal reinforcement shall be designed to
resist an equivalent bending moment, Me1
Me1 = Mu + Mt
The torsional moment,
Mt = Tu 7.1
b/D1 = 36.43x 7.1
500/7001 = 51.43 KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
Hence, Me1 = Mu+Mt = 579.6 + 51.43= 631.03 KNm
Here,
Mulim<Mu Hence, The section is to be doubly-
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%
Asc must be at least 50% of Ast
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)
For maximum +ve moment: (sagging moment)
Mu = 553.93 KNm
Torsional Moment, Tu = 36.43 KNm
The longitudinal reinforcement shall be designed to
resist an equivalent bending moment, Me1
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)
The torsional moment,
Mt = Tu x7.1
b/D1 = 36.43x 7.1
500/7001 = 51.43 KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
Hence, Me1 = Mu+Mt = 553.93 + 51.43= 605.36 KNm
Here,
Mulim<Mu Hence, The section is to be doubly-
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%
Asc must be at least 50% of Ast
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)
For As(bottom), provide 6-28mm bars
Act.As(bottom)= 3694.513mm2 (bottom)
For As(top), provide 6-28mm bars
Act.As(top)=3694.513mm2 (top)
At mid span
For maximum -ve moment: (hogging moment)
Mu = 90 KNm
Torsional Moment, Tu = 36.43 KNm
The longitudinal reinforcement shall be designed to
resist an equivalent bending moment, Me1
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)
The torsional moment,
Mt = Tu x7.1
b/D1 = 36.43x 7.1
500/7001 = 51.43 KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
Hence, Me1 = Mu+Mt = 90 + 51.43= 141.43 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
=387.925mm2 < 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 = 856.7mm2(Top)
Asc = 856.7mm2 (Bottom)
For maximum +ve moment: (sagging moment)
Mu = 71.49 KNm
Torsional Moment, Tu = 36.43 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 = 36.43x 7.1
500/7001 = 51.43 KNm
Since Mt<Mu, no longitudinal reinforcement shall be
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)
provided on the flexural compression face.
Hence, Me1 = Mu+Mt = 71.49+ 51.43= 122.92 KNm
Here, Mulim>Mu Hence, The section is to be singly-
reinforced.
Ast = bdbdf
M6.411
f
f5.0
2ck
u
y
ck
=306.53mm2 (Bottom bars)<Ast,min(856.7mm2)
Providing minimum area of steel=856.7mm2
Provide 2-28mm
Act.Ast=1231.504mm2 (Bottom 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 = 856.7mm2(Bottom)
Asc = 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)
At right end
For maximum -ve moment: (hogging moment)
Mu =743.91 KNm
Torsional Moment, Tu = 37.72 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 = 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
Act.As(bottom)=1231.5
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)
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
Hence, Me1 = Mu+Mt + = 743.91+ 53.25= 797.16 KNm
Here, Mulim<Mu Hence, The section is to be doubly-
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%
Asc must be at least 50% of Ast
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)
For maximum +ve moment: (sagging moment)
Mu = 646.23 KNm
Torsional Moment, Tu = 37.72KNm
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 = 37.72x 7.1
500/7001 = 53.25 KNm
Since Mt<Mu, no longitudinal reinforcement shall be
provided on the flexural compression face.
Hence, Me1 = Mu+Mt = 646.23 + 53.25= 699.48 KNm
Here, Mulim<Mu Hence, The section is to be doubly-
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%
Asc must be at least 50% of Ast
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)
For As(bottom), provide 6-28mm bars
Act.As(bottom)= 3694.51mm2 (bottom)
For As(top), provide 7-28mm bars
Act.As(top)=4310.26mm2 (top)
As,top/bottom=1/2*Max –ve steel at top of either joint
=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):
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,
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
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.
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)
Design shear at left end=higher of the two values
=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)
Here, τve>τc, hence transverse reinforcement shall be
provided as below.
Transverse reinforcement
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,
Spacing of stirrups:least of
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
ii)8* small=8*28=224mm
but >=100mm
Hence, Provide 12mm 2legged vertical
stirrups@100mmc/c
At right end
(sway to right):
Vu=536.03KN(+ve value from SAP)
(sway to left):
Vu=451.88KN(-ve value from SAP)
Design shear at left end=higher of the two values
=536.03KN
Equivalent shear,
Ve = Vu + 1.6b
Tu = 536.03+ 1.6 x
5.0
72.37 = 656.73KN
Equivalent nominal shear stress,
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
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
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,
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.
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
Provide 4-legged 8mm stirrups @ 100mm c/c up to
length 2d=1322mm from left end
Provide 2-legged 12mm stirrups @ 100mm c/c at mid
span
Provide 4-legged 8mm stirrups @ 100mm c/c up to
length 2d=1322mm from right end
Asv=201.06mm2
Sv=100mm
8mm 4LVS@
100mmc/c
Sv,max= 270mm
115
IS456:2000
26.5.1.3
4.6
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
2-12mm bar along
each vertical side.
116
Ductile Design of 7m span Secondary beam
Concrete Grade = M20 Steel Grade = Fe415(HYSD)
Ref Step Calculations Output
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
Overall Depth of Beam, D=450mm
Width of Beam, B=250mm
Considering 1 layer of 28mm bar with spacer bar to
be used
Taking clear cover=25mm.
Effective depth, d=450-25-28/2 = 411 mm
Check for Axial Stress
Factored Axial Stress = 0 N/mm2
Axial Stress = 0 N/mm2 < 0.1 fck
Hence, design as flexural member.
Check for member size
Width of beam, B=250mm > 200mm
Depth of beam, D=450mm
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
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 250 x 411 / 100 = 267.15mm2
Max. steel reinforcement, As,max = 0.025bd
=0.025 x 250 x 411 = 2568.75mm2
Calculation of Longitudinal Steel Reinforcement
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)
For maximum -ve moment: (hogging moment)
Mu = 160.26 KNm
Torsional Moment, Tu = 0.76 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%
Asc must be at least 50% of Ast
50% of Ast =0.638% > 0.336%
Ast = 1.276/100 x 250*411 = 1312mm2(Top)
Asc = 656mm2 (Bottom)
For maximum +ve moment: (sagging moment)
Since , the value is negative as extracted from SAP
Mu =0 KNm
Torsional Moment, Tu = 0.76 KNm
So, it is governed by hogging moment.
For As(top), provide 2-28mm and 1-16mm bars
Act.As(top)= 1432.56 mm2 (bottom)
For As(bottom), provide 2-28mm bars
Act.As(bottom)= 1231 mm2 (top)
At mid span
For maximum +ve moment: (sagging moment)
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%
Asc must be at least 50% of Ast
50% of Ast =0.484% <0.012%
Ast = 0.968*250*411/100= 994.62mm2(Bottom)
Asc = 497.31mm2 (Top)
For maximum -ve moment: (hogging moment)
Mu = 0 KNm (no hogging moment)
Torsional Moment, Tu = 0.76 KNm
So, design in the mid span is governed by sagging
moment.
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)
At right end
For maximum -ve moment: (hogging moment)
Mu =160.15 KNm
Torsional Moment, Tu = 0.76 KNm
which is same as the left end hogging moment.
So, Ast = 1312mm2(Top)
And, Asc = 656mm2 (Bottom)
For maximum +ve moment: (sagging moment)
Mu = 22.53 KNm
Torsional Moment, Tu = 0.76KNm
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
So,providing minimum area of steel,
Ast = 267.15mm2 (Top)
Asc = 267.15mm2 (Bottom)
For As(top), provide 2-28mm and 1-16mm bars
Act.As(top)= 1432.56 mm2 (bottom)
For As(bottom), provide 2-28mm bars
Act.As(bottom)= 1231 mm2 (top)
As,top/bottom=1/4*Max –ve steel at top of either joint
=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
Equivalent nominal shear stress,
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
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-2legged vertical stirrups,
120
IS13920:1993
cl.6.3.5
4.2
4.3
Asv=100.53mm2
A clear cover of 25mm is assumed all around.
b1=b-2clear cover - = 250 – 2x25 – 28 = 172mm
d1=D-2clear cover - = 450 – 2x25 – 28= 372mm
100.53=
41587.03725.2
1077.156
41587.0372172
1076.0 36
xxx
x
xxx
x Sv
Sv=201.15mm
Also,
Spacing of stirrups:least of
i)Sv=100mm
ii)d/4=411/4=102.75mm
ii)8* small=8*16=128mm
but >=100mm
Hence, Provide 8mm 2legged vertical
stirrups@200mmc/c over a length of 2d=820mm.
At mid-span:
(sway to right):
Design shear at left end=9.92KN
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
Equivalent nominal shear stress,
τve= 411250
1035.12 3
x
x= 0.12 N/mm2 < τcmax(=2.8N/mm2)
Here, τve<τc, hence transverse reinforcement shall be
nominal
Let us use 8mm-2legged vertical stirrups,
Asv=100.53mm2
Hence, Provide 8mm 2legged vertical
stirrups@300mmc/c.
At right end
Design shear at left end=154.34KN
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
Equivalent nominal shear stress,
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
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-2legged vertical stirrups,
Asv=100.53mm2
A clear cover of 25mm is assumed all around.
b1=b-2clear cover - = 250 – 2x25 – 28 = 172mm
d1=D-2clear cover - = 450 – 2x25 – 28= 372mm
100.53=
41587.03725.2
1077.156
41587.0372172
1076.0 36
xxx
x
xxx
x Sv
Sv=201.15mm
Also,
Spacing of stirrups:least of
i)Sv=100mm
ii)d/4=411/4=102.75mm
ii)8* small=8*16=128mm
but >=100mm
Hence, Provide 8mm 2legged vertical
stirrups@200mmc/c over a length of 2d=820mm.
Check for Shear Reinforcement Spacing
The spacing of stirrups for torsion shall not exceed the
least of the following
122
4.5
i) x1= short dimension of the stirrups
=250-2x25+8/2+8/2=208
ii)(x1+y1)/4
y1=long dimension of the stirrups
=450-2x25 + 8/2 + 8/2 = 408mm
Sv=(208+408)/4=308
iii)300mm
i.e.Sv,max= 300mm
Provision of Shear Reinforcement
Provide 2-legged 8mm stirrups @ 200mm c/c up to
length 2d=820mm from left end
Provide 2-legged 8mm stirrups @ 300mm c/c at mid
span
Provide 2-legged 8mm stirrups @ 200mm c/c up to
length 2d=820mm from right end
123
4.2.3 DESIGN OF COLUMN
Design of Square Column
Concrete Grade = M25 Steel Grade = Fe415 (HYSD)
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
Considering 1m Width of Slab,
Self Wt. of Slab= 25 x .25=6.25 KN/m
Floor Finishing=1.0 KN/m
Live Load=5 KN/m
Total Characteristics Load=12.25 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.
Required spacing of 10 mm Bars,
C/C Spacing=1000/300.00 x 78.546=261.82 mm
Provide 10 mmØ @250 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
Concrete Grade = M20 Steel Grade = Fe415 (HYSD)
Ref. Step Calculations Output
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
Considering 1m Width of Slab,
Floor Finishing= 1 KN/m
Live Load=5 KN/m
Total Characteristics Load=15.02 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
Total Characteristics Load=11.5 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)
Required spacing of 10 mm Bars,
C/C Spacing=1000/1419.03 x 200.96=188.16 mm
Provide 16 mmØ @185 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
Provide 10 mmØ @300 mm
Distributors Reinforcement,
Astmin = .0012 x 1000 x 220= 264 mm2
Required spacing of 10 mm Bars,
C/C Spacing = 1000/264 x 78.546 = 183.88 mm
Provide 10 mmØ @180 mm
Devlopment Length
For Intermediate flight,
bd
s
4
ØLd
= 752 mm
Provide Devlopment Length 755 mm
For Upper and Landing,
bd
s
4
ØLd
= 470.11 mm
Provide Devlopment Length 475 mm
Checking for depth of slab,
For Intermediate flight,
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)
For Upper and Landing,
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
Concrete Grade = M20 Steel Grade = Fe415 (HYSD)
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.
Check for Deflection
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
Concrete Grade = M20 Steel Grade = Fe415 (HYSD)
Ref. Step Calculations Output
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
To take account of the reversal effect, Provide 12 mm Ø
bars @ 170 mm c/c on both faces of the wall
Calculation of Horizontal Steel Reinforcement
Area of horizontal steel reinforcement = 0.2 % bH
= 0.002 x 200 x
3200
= 1280 mm2
Providing 12 mm Ø bar
No. of Bars = nos1231.1109.113
1280
. . . Spacing of Bars , Sv = mm90.290
112
3200
To take account of the reversal effect, Provide 12 mm Ø
bars @ 250 mm c/c on both faces of the wall
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
Beam 3-3 with 6m width and soil pressure of and soil pressure of=131.66KN/m2
Beam 4-4 with 6m width and soil pressure of and soil pressure of=137.05KN/m2
Beam 5-5 with 6m width and soil pressure of and soil pressure of=142.45KN/m2
Beam 6-6 with 6m width and soil pressure of and soil pressure of=174.84KN/m2
Beam 7-7 with 6m width and soil pressure of and soil pressure of=153.24KN/m2
Beam 8-8 with 3.5m width and soil pressure of and soil pressure of=158.64KN/m2
Beam 9-9 with 1.75m width and soil pressure of and soil pressure of=161.34KN/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
Maximum Moment =136.302∗4.752
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
Maximum Moment=147.53∗52
10=368.825KNm/m
For strip F-F
Maximum Moment =153.08∗52
10=382.7KNm/m
For strip G-G
Maximum Moment=154.134∗62
10=554.88KNm/m
For strip H-H
Maximum Moment=161.34∗42
10=258.144KNm/m
Y-direction
For strip 1-1
Maximum Moment =120.413∗3.52
10=147.51KNm/m
For strip 2-2
Maximum Moment =126.26∗62
10=454.536KNm/m
For strip 3-3
Maximum Moment =131.66∗62
10=473.98KNm/m
For strip 4-4
Maximum Moment=137.05∗62
10=493.38KNm/m
For strip 5-5
154
Maximum Moment=142.45∗62
10=512.82KNm/m
For strip 6-6
Maximum Moment =174.84∗62
10=629.424KNm/m
For strip 7-7
Maximum Moment=153.24∗62
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
GE
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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. : 2
-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
Sta
ircase
Op
en
we
ll
Botto
m R
ein
forc
em
ent@
6.4
Sta
ircase
8m
m@125m
mc/c
8m
m@
500m
mc/c
8m
m@
140m
mc/c
8m
m@
125m
mc/c
8m
m@
140m
mc/c
8m
m@
125m
mc/c
8m
m@
140m
mc/c
8m
m@
140m
mc/c
25
0
60
00
60
00
60
00
60
00
60
00
60
00
25
00
95
00
50
00
50
00
50
00
50
00
70
00
15
00
8m
m@2
00m
mc/c
8m
m@2
50m
mc/c8
mm
@200m
m c
/c
8m
m@2
50m
mc/c
60
00
Ce
ntra
l Atriu
m
S1
S1
S1
S1
S1
S1
S1
S1
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S3S
3
S3S3
S3
S3
S3S
3S
3
S3
S3
S3
S4
S4
S4
S4
S5S5
S7
S7
S7
S7
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. : 2
-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
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
Sta
ircase
Op
en
we
ll
Top R
ein
forc
em
ent@
6.4
8m
m@140m
mc/c
8m
m@125
mm
c/c
8m
m@2
50m
mc/c
Sta
ircase
8m
m@2
50m
mc/c
8m
m@2
50m
mc/c
8m
m@2
50m
mc/c
8m
m@140m
mc/c
8m
m@140m
mc/c
8m
m@2
50m
mc/c
8m
m@
125
mm
c/c
8m
m@125m
mc/c
8m
m@2
50m
mc/c
8m
m@140m
mc/c
8m
m@125m
mc/c
8m
m@125m
mc/c
8m
m@125m
mc/c
8m
m@2
50m
mc/c
60
00
25
06
000
60
00
60
00
60
00
60
00
50
00
50
00
50
00
50
00
70
00
15
00
8m
m@2
00m
mc/c
8m
m@2
00m
mc/c
8m
m@2
00m
m c
/c
8m
m@2
50m
mc/c
70
00
60
00
25
00
S1
S1
S1
S1
S1
S1
S1
S1
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S3
S3
S3
S3
S3
S3
S3
S3
S3
S3
S3
S3
S4
S4
S4
S4
S5
S5
S7
S7
S7
S7
S8
S8
S8
S8
S8
S8
S8
S8
Ce
ntra
l Atriu
m
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. : 3
-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
Bo
ttom
Re
info
rce
me
nt@
16
Sta
ircase
Sta
ircase
60
00
60
00
60
00
60
00
60
00
60
00
60
00
15
00
70
00
50
00
50
00
50
00
50
00
70
00
15
00
8m
m@125m
mc/c
8m
m@125m
mc/c
Op
en
we
llC
INE
MA
HA
LL
CIN
EM
A H
AL
L
S1
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S2
S3S
3
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
Ce
ntra
l Atriu
m
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. : 3
-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
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
Sta
ircase
Op
en
we
ll
To
p R
ein
force
me
nt@
16
Sta
ircase
8m
m@140m
mc/c
8m
m@125m
mc/c
8m
m@2
50m
mc/c
8m
m@125m
mc/c
8m
m@125m
mc/c
8m
m@140m
mc/c
8m
m@125m
mc/c
8m
m@125m
mc/c
8m
m@
125
mm
c/c
8m
m@130m
mc/c
8m
m@2
50m
mc/c
8m
m@2
50m
mc/c
CIN
EM
A H
AL
L
Au
dito
rium
CIN
EM
A H
AL
L
S1
S2
S2
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