SP16-Design Aid For RC To IS456-1978

DESIGN AIDS
FOR
REINFORCED CONCRETE
TO IS : 456-l 978
As in the Original Standard, this Page is Intentionally Left Blank
DesignAids
For ReinforcedConcrete
to IS : 4564978
BUREAU OF INDIAN STANDARDS

BAHADUR SHAH ZAFAR MARC, NEW DLEHI 110 002
SP16:1980
FIRST PUBLISHED SEPTEMBER 1980
ELEVENTH REPRINT MARCH 1999
(Incorporatinp Amendment No. I)
0 BUREAU OF INDIAN STANDARDS
UDC 624.0 12.45.04 (026)
PRICE Rs. 500.00
I’KiNTED 1N INDIA AT
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AND PI II3LISHED BY
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FOREWORD
Users of various civil engineering codes have been feeling the need for explanatory hand-
books and other compilations based on Indian Standards. The need has been further emphasized
in view of the publication of the National Building Code of India 1970 and its implementation.
In 1972, the Department of Science and Technology set up an Expert Group on Housing and
Construction Technology under the Chairmanship of Maj-Gen Harkirat Singh. This Group
carried out in-depth studies in various areas of civil engineering and constr,uction practices.
During the preparation of the Fifth Five Year Plan in 1975, the Group was assigned the task
of producing a ,Science and Technology plan for research, development and extension work
in the sector of housing and construction technology. One of the items of this plan was the
production of design handbooks, explanatory handbooks and design aids based on the National
Building Code and various Indian Standards and other activities in the promotion of National
Building Code. The Expert Group gave high priority to this item and on the recommendation
of the Department of Science and Technology the. Planning Commission approved the follow-
ing two projects which were assigned to the Indian Standards Institution:
a) Development programme on Code implementation for building and civil engineering

construction, and
b) Typification for industrial buildings.
A Special Committee for Implementation of Science and Technology Projects (SCIP)

consisting of experts connected with different aspects (see page viii ) was set up in 1974 to advise
the IS1 Directorate General in identification and for guiding the development of the work under
the Chairmanship of Maj-Gen Harkirat Singh, Retired Engineer-in-Chief, Army Headquarters
and formerly Adviser ( Construction) Planning Commission, Government of India. The
Committee has so far identified subjects for several explanatory handbooks/compilations
covering appropriate Indian Standards/Codes/Specifications which include the following:
Functional Requirements of Buildings

Functional Requirements of Industrial Buildings
Summaries of Indian Standardsfor Building Materials
Building Construction Practices
Foundation of Buildings
Explanatory Handbook on Earthquake Resistant Design and Construction (IS : 1893
.
Des& %?for Reinforced Concrete to IS : 456- 1978
Explanatory Handbook on Masonry Code
Commentary on Concrete Code ( IS : 456 )
Concrete Mixes
Concrete Reinforcement
Form Work
Timber Engineering
Steel Code ( IS : 800 )
Loading Code
Fire Safety
Prefabrication ,
Tall Buildings
Design of Industrial Steel Structures
Inspection of Different Items of Building Work
Bulk Storage Structures in Steel
Bulk Storage Structures in Concrete
Liquid Retaining Structures
.
Construction Safety Practices

Commentaries on Finalized Building Bye-laws
Concrete Industrial Structures
One of the explanatory handbooks identified is on IS : 456-1978 Code of practice for

plain and reinforced concrete ( third revision). This explanatory handbook which is under
preparation would cover the basis/source of each clause; the interpretation of the clause and
worked out examples to illustrate the application of the clauses. However, it was felt that some
design aids would be of help in designing as a supplement to the explanatory handbook. The
objective of these design aids is to reduce design time in the use of certain clauses in the Code
for the design of beams, slabs and columns in general building structures.
For the preparation of the design aids a detailed examination of the following handbooks
was made :
4 CP : 110 : Part 2 : 1972 Code of practice for the structural use of concrete : Part 2
Design charts for singly reinforced beams, doubly reinforced beams and rectangular
columns. British Standards Institution.
‘4 AC1 Publication SP-17(73) Design Handbook in accordance with the strength design
methods of AC1 318-71, Volume 1 ( Second Edition). 1973. American Concrete
Institute.
cl Reynolds ( Charles E ) and Steadman ( James C ). Reinforced Concrete Designer’s
Handbook. 1974. Ed. 8. Cement and Concrete Association, UK.
4 Fintel ( Mark ), Ed. Handbook on Concrete Engineering. 1974. Published by Van
Nostrand Reinhold Company, New York.
The charts and tables included in the design aids were selected after consultation with
some users of the Code in India.
The design aids cover the following:
a) Material Strength and Stress-Strain Relationships;

b) Flexural Members ( Limit State Design);
c) Compression Members ( Limit State Design );
d) Shear and Torsion ( Limit State Design );
e) Development Length and Anchorage ( Limit State Design );
f) Working Stress Method;
g) Deflection Calculation; and
h) General Tables.
The format of these design aids is as follows:
a) Assumptions regarding material strength;

b) Explanation of the basis of preparation of individual sets of design aids as related
to the appropriate clauses in the Code; and
c) Worked example illustrating the use of the design aids.
Some important points to be noted in the use of the design aids are:
4 The design units are entirely in SI units as per the provisions of IS : 456-1978.
b) It is assumed that the user is well acquainted with the provisions of IS : 456-1978
before using these design aids.
4 Notations as per IS : 456-1978 are maintained here as far as possible.
d) Wherever the word ‘Code’ is used in this book, it refers to IS : 456-1978 Code of
practice for plain and reinforced concrete ( third revision ).
4 Both charts and tables are given for flexural members. The charts can be used con-
veniently for preliminary design and for final design where greater accuracy is needed,
tables may be used.
vi
f) Design of columns is based on uniform distribution of steel on two faces or on four
faces.
g) Charts and tables for flexural members do not take into consideration crack control
and are meant for strength calculations cnly. Detailing rules given in the Code should
be followed for crack control.
h) If the steel being used in the design has a strength which is slightly different from the
one used in the Charts and Tables, the Chart or Table for the nearest value may be
used and area of reinforcement thus obtained modified in proportion to the ratio of
the strength of steels.
j) In most of the charts and tables, colour identification is given on the right/left-hand
corner along with other salient values to indicate the type of steel; in other charts/
tables salient values have been given.
These design aids have been prepared on the basis of work done by Shri P. Padmanabhan,
Officer on Special Duty, ISI. Shri B. R. Narayanappa, Assistant Director, IS1 was also
associated with the work. The draft Handbook was circulated for review to Central Public
Works Department, New Delhi; Cement Research Institute of India, New Delhi; Metallurgical
and Engineering Consultants (India) Limited, Ranchi, Central Building Research Institute,
Roorkee; Structural Engineering Research Centre, Madras; M/s C. R. Narayana Rao, Madras;
and Shri K. K. Nambiar, Madras and the views received have been taken into consideration
while finalizing the Design Aids.
vii
.SPECIAL COMMIlTEE FOR IMPLEMENTATION OF SCIENCE AND
TECHNOLOGY PROJECTS (SCIP)
Chairman
MAJ-GEN HARKIRAT SINGH
W-51 Greater Kailash I, New Delhi 110048
Members
SEIR~A. K. BANERJEE Metallurgical and Engineering Consultants (India) Limited,
Ran&i
PROF DINESH MOHAN Central Building Research Institute, Roorkee
DR S. MAUDOAL Department of Science and Technology, New Delhi
DR M. RAMAIAH Structural Engineering Research Centre, Madras
SHRI T. K. SARAN Bureau of Public Enterprises, New Delhi
SHRI T. S. VEDAGIRI Central Public Works Department, New Delhi
DR ‘H. C. VISVESVARAYA Cement Research Institute of India, New Delhi
SHRI D. AJITHA SIMHA Indian Standards Institution, New Delhi
(Member Secrewv)
. ..
Vlll
CONTENTS
Page
LIST OF TABLES M THE EXPLANATORY -TEXT ... ... x

LIST OF CHARTS . .. . .. xi
LIST OF TABLES . .. . . . Xiv
SYMBOLS .. . . . . xvii
CONVERSK)N FACTORS . .. . . . xix
1. MATERIAL STRENGTH AND STRESS-STRAIN RELATIONSHIPS 3
1.1 Grades of Concrete .. . .. 3

1.2 Types and Grades of Reinforcement ... ... 3
1.3 Stress-strain Relationship for Concrete . .. .. . 4
1.4 Stress-strain Relationship for Steel . .. . .. 4
2. FLEXURAL MEMBERS ... .. . 9
2.1 Assumptions .. . . .. 9
2.2 Maximum Depth of Neutral Axis ... ... 9
2.3 Rectangular Sections . .. ... 9
2.3.1 Under-Reinforced Sections .. . . .. 10
2.3.2 Doubly Reinforced Sections ... ... 12
2.4 T-Sections . .. .. . 14
2.5 Control of Deflection . .. . .. 14
3. COMPRESSION MEMBERS ... . .. 99
3.1 Axially Loaded Compression Members . .. . . . 99

3.2 Combined Axial Load and Uniaxial BendIng ... . . . 99
3.2.1 Assumptions ..* . . . 100
3.2.2 Stress Block Parameters when the Neutral iAxisLies .. . . . . 101
Outside the Section
3.2.3 Construction of Interaction Diagram .. . . . . 101
3.3 Compression Members Subject to Biaxial Bending .. . . . . 104
3.4 Slender Compression Members . .. . . . 106
4. SHEAR AND TORSION . .. . .. 175

4.1 Design Shear Strength of Concrete . .. ..* 175
4.2 Nominal Shear Stress .. . . . . 175
4.3 Shear Reinforcement .. . . . . 175 .
4.4 Torsion . .. . . . 175
ix
Page
5. DEVELOPMENT LENGTH AND ANCHORAGE . .. .. . 183
5.1 Development Length of Bars . .. . .. 183

5.2 Anchorage Value of Hooks and Bends . .. . . . - 183
6. WORKING STRESS DESIGN .. . ... 189
6.1 Flexural Members . .. . .. 189

6.1.1 Balanced Section . .. ... 189
6.1.2 Under-Reinforced Section . .. .. . 189
6.1.3 Doubly Reinforced Section . .. ... 190
6.2 Compression Members ... . .. 190
6.3 Shear and Torsion .. . ... 191
6.4 Development Length and Anchorage ... 1.. 191
7. DEFLECTION CALCULATION .. . . .. 213
7.1 Effective Moment of Inertia . .. . .. 213

7.2 Shrinkage and Creepl)eflections . .. .. . 213
LIST OF TABLES IN THE EXPLANATORY TEXT
Table
A Salient Points on the Design Stress Strain Curve for Cold Worked
Bars ... . .. 6
B Values of F for Different Grades of Steel . .. . .. 9
C Limiting Moment of Resistance and Reinforcement Index for Singly

Reinforced Rectangular Sections . .. .. . 10
D Limiting Moment of Resistance Factor Mu,ii,/bd’, N/mm2 for Singly.
Reinforced Rectangular Sections .. . .. . 10
E Maximum Percentage of Tensile Reinforcement Pt,lim for Singly
Reinforced Rectangular Sections .. . .., 10
F Stress in Compression Reinforcement, fX N/mma in Doubly
Reinforced Beams with Cold Worked Bars ... ... 13
G Multiplying Factors for Use with Charts 19 and 20 .. . .. . 13
H Stress Block Parameters When the Neutral Axis Lies Outside the
Section ... . .. 101
I Additional Eccentricity for Slender Compression Members . . . . ,. 106
J Maximum Shear Stress rc,max ... . .. 175
K Moment of Resistance Factor M/bd’, N/mm” for Balanced
Rectangular Section .. . . .. I89
L Percentage of Tensile Reinforcement P1,b.i for Singly Reinforced
Balanced Section .*. ... 189
M Values of the Ratio A,/&, . .. . .. 190
X
LIST OF CHARTS
chart
No. PW
FLEXURE - Singly Reinforced Section
1 CL= 15 N/mm’, fy= 250 N/mm’ d- 5 to 30 cm ... 17

2 Lk - 15 N/mm*, fu= 250 N/mm* d = 30 to 55 cm .*. 18
3 f etr- 15 N/mm*, fr= 250 N/mms d - 55 to 80 cm .. . 19
4 f ck= 15 N/mm*, fy= 415 N/mm* d= 5 to 30 cm ... 21
5 f Ed - 15 N/mm*, fi - 415 N/mm* d I 30 to 55 cm . .. 22
6 fsk- 15 N/mm*, f, -415 N/mm* d-55 to 80 cm . .. 23
7 fck= 15 N/mm*, fi -500 N/mm* d== 5 to 30 cm . .. 25
8 f ctr= 15 N/mm*, fy -
500N/mm* d = 30 to 55 cm ... 26
9 f& - 15 N/mm*, ’ fv -
500N/mm* d-55 to 80 cm . .. 27
10 f& x 20 N/mm*, .fy=250 N/u& ,d P 5 to 30 cm ... 29
11 f& - 20 N/mm*, fy -250 N/mm% d - 30 to 55 cm ... 30
12 f ek= 20 N/mm’, f, =
2% N/mm’ d = 55 to 80 cm .. . 31
13 fh - 20 N/mm*, I;- 415 N/mm* d- 5 to 30 cm ... 33
14 fdrI 20 N/mm*, fv -415 N/mm’ d-30 to 55 cm .. . 34
15 f ck- 20 N/mm*, fy -
415 N/m’ d- 55 to 80 cm ... 35
16 f& - 20 N/mm’, fy -500 N/mm* d= 5 to 30 cm . .. 37
17 fd - 20 N/mm’, fr -500 N/mm* d-30 to 55 cm . .. 38
18 hk - 20 N/mm*, & = 500 N/mm* d I 55 to 80 cm ... 39
FLEXURE - Doubly Reinforced Section
19 fr I 250 N/mm’, d-d’ - 20 to 50 cm ... .. . 41

20 fr I 250 N/mm*, d-d’ - 50 to 80 cm . .. ... 42
CONTROL OF DEFLECTION
21 fr - 250 N/mm’ ... ... 43

22 fr I 415 N/mm’ . .. ,.. 44
23 fi - 500N/mm’ . .. .. . 45
AXIAL COMPRESSION
24 h - 250 N/mm’ . .. .“. 109

25 ft - 415 N/mm’ . .. ... 110
26 A-5OON/mm’ ... .. . 111
xi
Chart
No. Page
COMPRESSION WITH BENDING - Rectangular Section -
Reinforcement Distributed Equally on Two Sides
27 f, = 250 N/mm% d’/D = 0.05 . .. . .. 112
28 fi = 250 N/mms d’/D = 0.10 . .. ... 113
29 h I 250 N/mm9 d’/D = 0.15 . .. . .. 114
30 fv LI 250 N/mm9 d’/D = 020 . .. ... 115
31 fr = 415 N/mm9 d’/D = 0.05 . .. .. . 116
32 fu = 415 N/mm9 d’/D - 0.10 1.. . .. 117
33 fr PD415 N/mm9 d’/D P 0.15 . . . . .. 118
34 fy - 415 N/mm9 d’/D - 0.20 . . . ... 119
35 fr - 508 N/mm9 d’/D = 0.05 . . . ... 120
36 fr - 500 N/mm4 d’/D = 0.10 . . . ... 121
37 fr = 500 N/mm9 d’/D = 0.15 . . . ... 122
38 fy - 500 N/mm3 d’/D - 0.20 . . . . .. 123
COMPRESSION WITH BENDING - Rectangular Section -

Reinforcement Distributed Equally on Four Sides
39 J, I) 250 N/mm* d’/D - 0.05 . .. .. . 124
40 .fx I 250 N/mm” d’/D = 0.10 . .. . .. 125
41 fy= 250 N/mm9 d’/D = 0.15 .. . ... 126
42 fy - 250 N/mm2 d’lD = 0.20 . .. . .. 127
43 fy II 415 N/mm9 d’/D = 0.05 . .. ... 128
44 frP 415 N/mm9 d’/D c 0.10 . .. .. . 129
45 fr I 415 N/mm9 d’/D = 0.15 . .. . .. 130
46 fr - 415 N/mm9 d’/D = 0.20 . .. . .. 131
47 fy = 500 N/mm3 d’/D P 0.05 . .. . .. 132
48 f, - 500 N/mm9 d’/D = 0.10 .. . .. . 133
49 fv - 500 N/mm’ d’lD = 0.15 . .. . .. 134
50 fu = 500 N/mm9 d’/D = 0.20 . .. .. . 135
COMPRESSION WlTH BENDlNG - Circular Section

51 fx - 250
N/mm9 d’/D = 0.05 ... . .. 136
52 fv P 250
N/mm2 d’/D = 0.10 . .. . .. 137
53 fy = 250
N/mm’ d’/D = 0.15 ... . .. 138
54 fr = 250
N/mm’ d’/D = 0.20 . .. .. . 139
55 fyP 415N/mm9 d’/D = 0.05 . .. . .. 140
56 fr- 415N/mm9 d’!D = 0.10 . .. ... 141
57 fy = 415
N/mm’ d’/D = 0.15 ... . .. 142
58 fy - 415
N/mm9 d’/D I= 0.20 . .. i .. 143
59 fi - 500
N/mm9 d’/D = 0.05 ... . . .. 144
60 fu- 500N/mm” d’/D = 0.10 ... . .. 145
61 h-500 N/mm* d’/D = 0.15 . .. .. . 146
62 fv-500 N/mm* d’/D = 020 . .. . .. 147
63 Values of Puz for Compression Members . .. . .. 148
64 Biaxial Bending in Compression Members . .. . .. 149
65 Slender Compression Members - Multiplying Factor k for . . . 150
Additional Moments
xii
ClUWt
No. Page
TENSION WITH BENDING - Rectangular Section -
Reinfomment Distributed Equally on Two Sides
66 h - 250 N/mm’ d’/D = @l5 and 020 .. . . .. 151
67 fr - 250 N/mm’ d’/D -0.05 and 010 ... . .. 152
68 fr - 415 N/mm’ d’/D - @OS .. . . .. 153
69 & = 415 N/mm* d’/D a 0.10 .. . . .. 154
70 h - 415 N/mm’ d’/D P 0.15 . . ., .., 155
71 /r - 415 N/mm’ d’/D - 020 . .. . .. 156
72 II-=5OON/llltII’ d’/D - 0.05 ... . .. 157
73 h-5OON/IIUU’ d’/D = 010 ... . .. 158
74 h-SOON/mm’ d’/D - 0.15 .. . . .. 159
75 &-so0 N/lIlOI’ d’/D - O-20 . .. ... 160
TENSION WITH BENDING - Rectangular Section - Reinforcement
Distributed Equally on Four Sides
76 f, - 250 N/mm’ d’/D - 0.05 and 010 . .. . .. 161
77 fr - 250 N/mm* d’/D - 0.15 and 020 ... .,. 162
78 /r - 415 N/mm’ d’/D = 0.05 . .. .. . 163
79 /r I 415 N/mma d’/D P 0.10 .. . ... 164
80 fr - 415 N/mm’ al/D - 0.15 . .. . .. 165
81 fr = 415 N/mm’ d’/D - 090 .. . ... 166
82 h-5OON/mm’ d’/D = 0.05 . .. .. . l6i
83 Jt-5OON/lIllII’ d’/D = 0.10 . .. ... l68
a4 fr - 500N/mm* d’/D - 0.15 . .. . .. 169
85 A-5OON/IUlll’ d*/D= 020 . .. ... 170
86 Axial Compiession (Working Stress Design) 0, - 130 N/mm* . . . 193

87 Axial Compression (Working Stress Design) am - 190 N/mm* . . . 194
88 Moment of Inertia of T-Beams ... 215
89 Effective Moment of Inertia for Calculating Deflection . .. 216
90 Percentage, Area and Spacing of Bars in Slabs . .. 217
‘Pl Effective Length of Columns - Frame Restrained .Against Sway . . . 218
92 Effective Length of Columns - Frame Without Restraiht to Sway 219
. ..
xul
LIST OF TABLES
Table
No. Page
FLEXURE - Reinforcement Percentage, pI for Singly Reinforced Sections
1 fft -15 N/mm’ ...... 47

2 20 N/mm’
f CL = ...... 48
3 f CL
- 25. N/mm* ...... 49
4 fd = 30 N/mm’ ...... 50
FLEXURE - Moment of Resistance of Slabs, kN.m Per Metre Width
5 fd - 15 N/mm’ /r- 250 N/mm* Thickness - 10.0 cm ... 51

6 f ck- 15N/llIHl’fy- 250 N/mm* Thickness - 11.0 cm ... 51
7 fdr- 15 N/mm* fr I 250 N/mm* Thickness = 120 cm ... 52
8 fdr-15 N/mm’ fy- 250 N/mm* Thickness - 13.0 cm . .. 52
9 f & - 15 N/mm’. fy w 250 N/mm* Thickness = 14.0 cm . .. 53
10 Id-15 N/mm’ f,- 250 N/mm’ Thickness - 15.0 cm . .. 53
11 fd - 15 N/mm* fy- 250 N/mm* Thickness - 175 cm ... 54
12 fck- 15 N/mm’ fy- 250 N/mm* Thickness = 20.0 cm ... 55
13 fd - 15 N/mm* fi - 250 N/mm* Thickness - 22.5 cm ... 56
14 fck - ,15 N/mm* fy - 250 N/mm* Thickness = 25.0 cm . .. 57
15 fck- 15 N/mm* fy - 415 N/mm* Thickness - 10.0 cm ... 58
16 . .. 58
17 2: f :: :rGI i- - 415 N/mm* Thickness - 11.0
120 cm ... 59
18 fd - 15 N/mm* fy I 415 N/mm* Thickness .- 13.0 cm ... 59
19 fd - 15 N/mm* fv I 415 N/mm* Thcikness - 140 cm . .. 60
20 f ck- 15 N/mm* f, - 415 N/mm* Thickness - 15.0 cm . .. 61
21 fck= 15 N/mm* fv- 415 N/mm* Thickness - 17.5 cm . .. 62
22 fdt- 15 N/mm* fr - 415 N/mm* Thickness - 20.0 cm . .. 63
23 fsk- 15N/mm’ fy E 415 N/mm* Thickness I 225 cm .. . 64
24 fa- 15 N/m* fy = 415 N/mm* Thickness - 25.0 cm . .. 65
25 f clr-m N/mm*$, - 250 N/mm* Thickness - 10.0 cm ... 66
26 f ck - 20 N/mm’ f, - 250 N/mm* Thickness - 11.0 cm ... 66
27 f clr-mN/=’ &I- 250 N/mm* Thickness - 12.0 cm . .. 67
28 fck-2ON/mm* h - 250 N/mm* Thickness - 13.0 cm .. . 67
29 f: z $ :\=I 2 -I 250 N/mm* Thickness - 15.0
14.0 cm . .. 68
30 ... 68
31 fd - 20 N/m+ h I 250 N/mm* Thickness - 17.5 cm a*- 69
32 fck- 20 N/mm*fr - 250 N/mm* Thickness I 20.; cm . . . 70
33 f& - 20 N/mm’ fy - 250 N/mm* Thickness - 22.5 cm . . . 71
34 fck - 20 N/mm’ f, I 250 N/mm* Thickness - 25.0 cm, . . . 72
xiv
Table
NO.
35 f ck - 20 N/mm2 h - 415 N/mm2 Thickness - 100 cm . .. 73

36 fti- 20 N/mm2 h I 415 N/mm2 Thickness - 110 cm ... 73
37 1; - 20 N/mm2 h - 415 N/mm2 Thickness - 12.0 cm . .. 74
38 - 415 N/mm2 Thickness - 13.0 cm ... 74
39 - 415 N/mm2 Thickness - 14.0 cm . .. 75
40 f,+- 20 N/mm2 f, - 415 N/mm2 Thickness - 15.0 cm . .. 76
41 fck - 20 N/mm2 fy - 415 N/mm2 Thickness - 175 cm ... 77
42 - 415 N/mm2 Thickness - 200 cm ... 78
43 2: 1: i/z: i - 415 N/mm2 Thickness - 225 cm . .. 79
44 fck- 20 N/mm2 fy - 415 N/mm2 Thickness - 25.0 cm . .. 80
FLEXURE - Reinforcement Percentages for Doubly

Reinforced Sections
45 f& - 15 N/mm2 fr - 250 N/mm2 . .. . .. 81
46 fr - 250 N/mm2 . .. ... 82
47 &- 250 N/mm2 . .. . .. 83
48 f;k- 30N/mm2 fr P 250 N/mm2 . .. . .. 84
49 fek- 15N/mm2 fy I 415 N/mm’ .. . ... 85
50 fy I 415 N/mm2 . .. ... 86
51 fr - 415 N/mm2 . .. . .. 87
52 fck- 30 N/mm2 fu - 415 N/mm2 . .. ... 88
53 fdr- 15 N/mm2 fr- 500N/mm2 ... ... 89
54 fck \- 20 N/mm2 fY I: 500 N/mm2 . .. ... 90
55 f ck= 25 N/mm2 fy - 500N/mm2 ... . .. 91
56 fck- 30 N/mm2 fr-5OON/r.llm2 . .. . .. 92
FLEXURE - Limiting Moment of Resistance Factor, Mo,u,&,# /a, for

Singly Reinforced T-beams N/mm*
57 fv- 250 N/mm’ .. . ... 93
58 fy- 415 N/mm= . .. ..* 94
59 fy- 500N/mm’ . .. . .. 95
60 Slender Compression Members - Values of P, . .. . .. 171
61 Shear - Design Shear Strength of Concrete, rc, N/mm* . .. . .. 178
62 Shear - Vertical Stirrups ... . .. 179
63 Shear - Bent-up Bars . .. . .. 179
DEVELOPMENT LENGTH
64 Plain Bars ... . .. 184
65 Deformed bars, fuP 415 N/mm* .. . . .. 184
66 Deformed bars, fr - 500N/mm* ... . .. La5
67 Anchorage Value of Hooks and Bends . .. 2.. 186
WORKING STRESS METHOD - FLEXURE - Moment of

Resistance Factor, M/bd’, N/mm’ for Singly Reinforced Sections
a* - 5-O N/mm* . .. . .. 195
2 u,bc - 7.0 N/mm’ .. . . ..
70 uca - 8.5 N/mm’ .. . .*. ::
71 ucbc- 10.0 N/mm* ... . .. 198
xv
Table
No.
WORKING STRESS DESIGN - FLEXURE - Rchforccmcnt
Percentages for Doubly Reinforced S&ions
72 uck- 5.0 N/mm1 aa - 140 N/mm8 . .. . ..

73 acbs- 79 N/mms w I 140 N/mm’ . .. .. .
74 oti - 8.5 N}mn’ an I 140 N/mm’ ... ...
75 a&C- 10.0 N/mm* au - 140 .N/J& ... .. .
76 ati I 5.0 N/m& u,, - 230 N/mm’ . .. ...
77 aa - 7.0 N/mm* au P 230 N/inn+ . .. ...
78 oca - 8.5 v/mm’ a,( - 230 N/mm’ .. . ...
79 oek - 10.0 N/mm* ust- 230 N/mm’ .. . . . ..
WORKING STRJZSSMETHOD-SHEAR
80 Permiklble Shear Stress in Concrete rc, N/mm* .. . .. .

81 Vertical Stirrups . .. .. .
82 Bent-up Bars ... .. .
WORKING STRESS METHOD - DEVELOPMENT LENGTH
83 Plain Bars . ..
84 Deformed Bars - uat- 230 N/mm*, e - 190 N/mm’ .. .
85 Deformed Bars - u,, - 275 N/mm*, uc - 190 N/mm’ . ..
86 Moment of Inertia - Values of b@/l2 000 . ..
MOMENT OF INERTIA OF CRACKED SECTION-Values of Ir/
87 &Id - 095 ... . .. 221

88 d’ld - 090 ... ... 222
89 d’jd - 015 . .. . .. 223
90 d’ld - 020 . .. ..I 224
DEPTH OF NEUTRAL &US - Values of n/d by Elastic Theory
91 d’/d I 005 1.. ... 225

92 d’/d - 0.10 l .. . .. 226
93 d’/d - 0.15 ... .. . 227
94 d’jd = 020 . .. . .. 228
95 Areas of Given Numbers of Bars in cm* ... .. . 229

96 Areas of Bars at Given Spacings .. . . .. 230
97 Fixed End Moments for Prismatic Beams .. . .. . a31
98 Detlection Formulae for Prismatic Beams . .. . .. 232
xvi
SYMBOLS
AC = Area of concrete E Flexural tensik strength

I Gross area of section (modulus of rupture) of
4 concrete
A* = Area of steel in a column or in a
singly reinforced beam or slab = Stress in steel
AC - Area of compression steel - Compressive stress in steel
corresponding to a strain of
A I” = Area of stirrups 0402
A so DCArea of additional tensile = Stress in the reinforcement
reinforcement nearest to the tension face of a
act = Deflection due to creep member subjected to combined
6 = Deflection due to shrinkage axial load and bending
b = Breadth of beam or shorter = Cytrteristic yield strength of
dimensions of a rectangular
column P Design yield strength of steel
br = Effective width of flange in a = Effective moment of inertia
T-beam
P Moment of inertia of the gross
bw = Breadth of web in a T-beam section about centroidal axis,
b, = Centre-to-centre distance between neglecting reinforcement
corner bars in the direction of = Moment of inertia of cracked
width section
D I Overall depth of beam or slab or = Flexural stiffness of beam
diameter of column or large1
dimension in a rectangular := Fkxural stiffness of column
column or dimension of a = Constant or coefficient or factor
rectangular column in the
direction of bending = Development length of bar
Di LI Thickness of flange in a T-beam = Length of column or span of
d - Effective depth of a beam or slab beam
d’,d’ = distance of centroid of com- = Effective length of a column,
pression reinforcement from bending about xx-axis
the extreme compression fibre
of the concrete section = Effective length of a column,
bending about yy-axis
d, G Centre to centre distance between
comer bars in the direction of = Maximum moment under service
depth loads
EC = Modulus of elasticity of concrete - Cracking moment
ES = Modulus of elasticity of steel = Design moment for limit state
P Eccentricity with respect to major Design (factored moment)
ha
axis (xx-axis) M u3h-n - Limiting moment of resistance of
= Eccentricity with respect to a singly reinforced rectangular
C.Y
minor axis (yy-axis) beam
= Minimum eccentricity Mu, e Design moment about xx-axis
= Compressive stress in concrete at MUY a Design moment about &-axis

the level of centroid of M”l, = Maximum uniaxial moment
compression reinforcement capacity of the section with
fck = Charircteristic compressive axial load, bending about
strength of concrete xx-axis
xvii
&I - Maximum uniaxial moment Xl = Shorter dimension of the stirrup
capacity of the section with &I = Depth of neutral axis at the
axial load, bending about limit state of collapse
yy-axis
Xu,mox = Maximum depth of neutral axis
Mel - Equivalent bending moment in limit state design
MU, - Additional moment, MU - Mn,tim Yc = Distance from centroidal axis
in doubly reinforced beams of gross section, neglecting
Mu,timrr= Limiting moment of resistance reinforcement, to extreme fibre
of a T-beam in tension
m = Modular ratio Yl = Longer dimension of stirrup
P = Axial load z = Lever arm
pb - Axial load corresponding to the a P Angle
condition of maximum - Partial safety factor for load
compressive strain of 0903 5 in Yr
concrete and OQO2 in the Ym - Partial safety factor for material
outermost layer of tension strength
steel in a compression member t = Creep strain in concrete
P” = Design axial load for limit state ecbc - Permissible stress in concrete in
design (factored load) bending compression
P P Percentage of reinforcement 6X = Permjssible stress in concrete in
- Percentage of compression direct compression
PC
reinforcement, 100 A,,/bd 01 = Stress in steel bar
PC let Percentage of tension reinforce- es 3: Permissible stress in steel in
ment, -l,OOAst/bd compression
Ptr - Additional percentage of tensile 011 = Permissible stress in steel in
reinforcement ’ doubly tension
reinforced beams, ‘I”00A,t,/bd
es, I Permissible stress in shear
ST - Spacing of stirrups reinforcement
T” - Torsional moment due to P Nominal shear stress
factored loads 7Y
V - Shear force 7bd P Design bond stress
VS I Strength of shear reinforcement k - Shear stress in concrete
(working stress design) ‘5w - Equivalent shear stress
V&l = Sbear force due to factored loads
Q,mu - Maximum shear stress in concrete
VW = Stren of shear reinforcement with shear reinforcement
8himit state design) i Creep coefficient
8
x = Dept;: neutral axis at service
9 - Diameter of bar
...
XVlll
CONVERSION FACTORS
Conversely
To Convert into Mu&ply by Multiply
by
(1) (2) (3) (4)

I -~
Loads and Forces
Newton kilogram o-102 0 9.807
Kilonewton Tonne 0.102 0 9.807
Moments and Torques

Newton metre kilogram metre o-102 0 9.807
Kilonewton metre Tonne metre o-102 0 9.807
Stresses
Newton per mm* kilogram per mm’ o-102 0 9.807
Newton per mm’ kilogram per cm2 10.20 O-0981
xix
1. MATERIAL STRENGTHS AND
STRESS-STRAIN RELATIONSHIPS
I.1 GRADES OF CONCRETE 1.1.1 Generally. Grades ;ti IS and M 20 are

used for flexural members. Charts for flexural
The following six grades of concrete can members and tables for slabs are, therefore,
be used for reinforced concrete work as given for these two grades ordy. However,
specified in Table 2 of the Code (IS : 4% tables for design of flexural members are
1978*): given for Grades M 15, M 20, M 25 and M 30.
M 15, M 20, M 25, M 30, M 35 and M 40. 1.1.2 The charts for compression members
are applicable to all grades of concrete.
The number in the grade designation refers
to the characteristic compressive strength,
fti, of 15 cm cubes at 28 days, expressed in 1.2 TYPES AND GRADES OF
N/mmZ ; the characteristic strength being REINFORCEMENT BARS
defined as the strength below which not
more than 5 percent of the test results are The types of steel permitted for use as re-
expected to fall. inforcement bars in 4.6 of the Code and their
characteristic strengths (specified minimum
*Code d practice for plain and reinforced concrete yield stress or O-2 percent proof stress)
( third revision ). are as follows:
Type oj Steel Indian Standard Yield Stress or O-2 Percent

Proof Stress
Mild steel (plain bars) IS : 432 (Part I)-1966* 1 26 z$fm;rni,or bars up to
Mild steel (hot-rolled deform- IS : 1139-1966t r 24 kgf/mm* for bars over

ed bars) I- 20 mm dia
Medium tensile steel (plain IS : 432 (Part I)-1966*> 36 lkfe;i2’ . bars up to

bars)
I 34.5 kgf/mm* for bars over
Medium tensile steel (hot- IS : 1139-1966t 20 mm’dia up to 40 mm
rolled deformed bars) 1 iiia
33 kgf/mm” for bars over
40 mm dia
High yield strength steel (hot- IS : 1139-1966t 42.5 kgf/mm2 for all sizes
rolled deformed bars)
High yield strength steel IS : 1786-1979$ 7 4 15 N/mm2 for all bar sizes
(cold-twisted deformed 500 N/mm* for all bar sizes
bars)
Hard-drawn steel wire fabric IS : 1566-19674 and 49 kgf/mm*

IS : 432 (Part II)-19661
Nom-S1 units have been used in IS: 1786-19793; in other Indian Standards. SI units will be adopted
in their next revisions.
*Specification for mild steel and medium tensile steel bars and hard-drawn steel wire for concrete
reinforcement: Part I Mild steel and medium tensile steel bars (second revision).
tSpecification for hot rolled mild steel, medium tensile steel and high yield strength steel deformed
bars for concrete reinforcement (revised).
$Specificatiod for cold-worked steel high strength &formed bars for concrete reinforcement (second
WlSiO#).
&+eciiication for hard-drawn steel wire fabric for concrete reinforcement (#rsr revisfon ).
ijSpecification for mild steel and medium tensile steel bars and hard-drawn steel wire for
concrete reinforcement: Part II Hard drawn steel wire (second revision).
MATERIAL STRENGTHS AND STRESS-STRAIN RELATIONSHIPS 3

Taking the above values into consideration,
most of the charts and tables have been
prepared for three grades of steel having
characteristic strength& equal to 250 N/mm*,
415 N/mm2 and 500 N/mm2.
1.2.1 If the steel being used in a design has

a strength which is slightly diflerent from the
above values, the chart or table for the nearest
value may be used and the area ofreinforce-
ment thus obtained be modi$ed in proportion
to the ratio of the strengths.
I/ /
1.2.2 Five values of fY (includinglthe value .I
I a.002 0’001
for hard-drawn steel wire fabric) have been
included in the tables for singly reinforced STRAIN
sections.
FIG. 1 DESIGN STRKSS-STRAINCURVE FOR
CONCRETE
1.3 STRESS-STRAIN RELATIONSHIP

FOR CONCRETE
The Code permits the use of any appro-

priate curve for the relationship between the
compressive stress and strain distribution
in concrete, subject to the condition that it
results in the prediction of strength in subs-
tantial agreement with test results [37.2(c)
of the Code]. An acceptable stress-strain
curve (see Fig. 1) given in Fig. 20 of the Code
will form the basis for the design aids in this
publication. The compressive strength of con- . 200000 N/mm’
crete in the structure is assumed to be O-67fd.
With a value of l-5 for the partial safety
factor ym for material strength (35.4.2.1 of
the Code), the maximum compressive stress ? --
in concrete for design purpose is 0.446 fck STRAIN
(see Fig. I).
FIG. 2 STRESS-STRAINCURVE FOR MILD STEEL
be adopted. According to this, the stress
1.4 STRESS-STRAIN RELATIONSHIP is proportional to strain up to a stress of
FOR STEEL 0.8 fY. Thereafter, the stress-strain curve is
defined as given below:
The modulus of elasticity of steel, E,, is Stress hu#aslic~srrain
taken as 200 000 N/mm2 (4.6.2 of the Code). O*SOfy Nil
This value is applicable to all types of 0.85 fr OQOOl
reinforcing steels. 0*9ofy 0.0% 3
0*9sf, o*ooo 7
The design yield stress (or 0.2 percent proof 0.975 fy 0~0010
stress) of steel is equal to fr/ym. With a value l-O& 0.002 0
of l-15 for ym (3.5.4.2.2 of the Code), the The stress-strain curve for design purposes is
design yield stress fv becomes 0#87 f,. The obtained by substituting fYe for fY in the
stress-strain relations1.tp for steel in tension above. For two grades of cold-worked bars
and compression is assumed to be the same. with 0.2 percent proof stress values of
415 N/mms and 500 N/mm2 respectively,
For mild steel, the stress is proportional the values of total strains and design stresses
to strain up to yield point and thereafter the corresponding to the points defined above
strain increases at constant stress (see Fig. 2). are given in Table A (see page 6). The stress-
For cold-worked bars, the stress-strain strain curves for these two grades of cold-
relationship given in Fig. 22 of the Code will worked bars have been plotted in Fig. 3.
4 DESIGN AIDS FOR REINFORCED CONCRETE

1
500 so0
450
m2
soo/‘1.0
‘iv’ ‘mnl
400
UC/l1-n
350
T
< 300
2.
250
200
150
100
50
0
0 0.001 o-002 0.003 o-004 0*005
STRAIN
FIG. 3 STRESS-STRAIN
CURVESFOR COLD-WORKED STEELE
MATERIAL SrRENGTHS AND STRESS-STRAINRELATIONSHIPS 5

TABLE A SALIENT POINTS ON THE DESlGN STRESS-STRAIN CURVE GOR
COLD-WORKED BARS
( Chse 1.4 )
STRESS LEVEL f, 0 415 N/mm’ fy= 500 N/mm8

* ,_-.-k b
f-- >
Strain Stress Strain Stress
(1) (‘1 (3) (4) (5)
N/mm* N/mm*
0.80 fyd 090144 288.1, woo174 347.8

0.85 fyd 0031 63 306.7 0.001 95 369.6
0.90&l 0~00192 324.8 0.002 26 391.3
0’95 fyd 0032 4 I 342.8 0.002 77 413.0
0.975 fyd 0.002 76 351.8 0.003 12 423.9
l’ofyd 0.003 80 360.9 MO4 17 434.8
NOTE -- Linear interpolation may be done for intermediate values.
6
2. FLEXURAL MEMBERS
2.2 ASSUMPTJONS 2.2 MAXIMUM DEPTH OF NEUTRAL

. AXIS
The basic assumptions in the design of
flexur&lmembers for the limit state of col- Assumptions (b) and (f’)govern the maximum
lapse are fcivenbelow (see 37.2 of the Code): depth of neutral axis in flexural members.
T& strain distribution across a member
4 Plane sections normal to the axis of corresponding to those limiting conditions
the member remain plane after bending. is shown in Fig, 4. The maximum depth of
This means that the strain at any point neutral axis x,,, - is obtained directlyfrom
on the cross section is directly propor- the strain diagram by considering similar
tional to the distance from the neutral triangles.
RXiS. 0.003 5
x0,,_
W Ihe maximum strain in concrete at d (0.005 5 f 0.87 f,/&)
the outermost compression fibre is
0903 5. The values of * for three grades of
d The design stress-strain relationship reinforcing steel are given in Table B.
for concrete is taken as indicated in
Fig. 1.
d) The tensile strength of concrete is TABLE B VALUES OF F FOR
ignored.
DIFFERENT GRADES OF STEEL
e), Tbt design stresses
in reinforcement 2.2)
(Cfuu.re
are derived from the strains using
the stress-strain relationship given -in f,, N/mms 250 415 500
Fig. 2 and 3.
f) The strain in the tension reinforcement 7 0531 0.479 O-456
is to be not less than
2.3 RECTANGULAR SECTIONS

This assumption is intended to ensure The compressive stress block for concrete
ductile fail&e, that is, the tensile is represented by the design stress-strain
reinforcementhas to undergo a certain curve as in Fig. 1. It is. seen from this stress
degree of inelastic deformation before block (see Fig. 4) that the centroid of com-
the concrete fails in compression. pressive force in a rectangular section lies
0*0035
t
X u,m*a
!zzx + 0*002 O=87 f-,

E*
STRESS
STRAIN
DIAGRAM
OIAGRAM
FIQ. 4 SINOLY REINFQRCSDSECTION
FLEXURAL MRMM3R.S
at a distance or U-416 xu (wnlcn nas oecn
TABLE E MAXIMUM PERCENTAGE OF
rounded off to 0.42 xu in the code) from the TENSILE REINFORCEMENT pt,lim FOR
extreme compression fibre; and the total force SINGLY REINFStRmTNSRE!aANGW
of compression is 0.36 fck bxu. The lever arm,
that is, the distance between the centroid
of compressive force and centroid of tensile (c%u.w 2.3)
force is equal to (d - 0.416 x,). Hence the
upper limit for the moment of resistance of a fdr, /y, Nhm’
N/mm* r b
singly reinforced rectangular section is given 250 415 u)o
by the following equation: 1.32
15
M u,lim = O-36& bxu,,, 1.76 ;g “0%
x(d - 0.416 ~u,mu) 220
.
4 2% l.43 YE
Substituting for xu,- from Table B and
transposing fdr bd2, we get the values of
tie limiting moment of resistance factors for 2.3.1 Under-Reinforced Section
singly reinforced rectangular beams and
slabs. These values are given in Table C. Under-reinforced section means a singly
The tensile reinforcement percentage, pt,lim reitiorced section with reinforcement per-
corresponding to the limiting moment of centage not exceeding the appropriate value
resistance is obtained by equating the forces given in Table E. For such sections, the
of tension and compression. depth of neutral axis xu will be smaller than
x”,,,,~. The strain in steel at the limit state
of collapse will, therefore, be more than
0.87 fy
Substituting for xu,mPxfrom Table B, we get - + 0902 and, the design stress in
E.
the values of Pt,lim fYj& as given in Table C. steel will be 0937fy. The depth of neutral
axis is obtained by equating the forces of
tension and compression.
TABLE C LIMITING MOMENT OF
RESISTANCE AND REINFORCEMENT INDEX
FOR SINGLY REl;~&FOR~N~ RECTANGULAR ‘G (0.87 fr) - 0.36 fdrb xu
(Clause 2.3)
j& N/mm* 250 415 500

The moment of resistance of the section is
M*,lhl - 0.149 W138 0.133 equal to the prdduct of the tensile force
--
Lk bd’ and the lever arm.
Plrllrnfy 21.97 19.82 18.87
Mu = pG (@87f,) (d - 0,416 xu)
/ ck
=O*87fy & l- 0.4165 bd2

( )( )
The values of the limiting moment of resis-
tance factor Mu/bd2 for different grades of Substituting foi $ we get
concrete and steel are given in Table D. The
corresponding percentages of reinforcements
are given in Table E. These are the maximum _ _
permissible percentages for singly reinforced
sections. x 1- 1.005 &$]bda
C
2.3.Z.Z Charts 1 to 28 have been prepared
TABLE D LIMITING MOMENT OF
RESISTANCE FAVOR Mu,,im/bd’, N/mm’ FOR by assigning different values to Mu/b and
SINGLY REINFC);&yE$sECTANGULAR plotting d versus pt. The moment values in
the charts are in units of kN.m per metr$
(Clause 2.3) width. Charts are given for three grades of
steel and, two grades of concrete, namely
/CL, fy, N/-Y M 15 and M 20, which are most commonly
N/mm’ used for flexural members. Tables 1 to 4
rK------ 500
cover a wider range, that ‘is, five values of
15 2.24 2.00 fy and four grades of concrete up to M 30.
3: 2.98
3.73 Is:
3.45 2.66
3.33 In these tables, the values of percentage of
30 4.47 414 3.99 reinforcement pt have been tabulated against
Mu/bd2.
10 DESIGN AIDS FOR WNFORCED CONCRETE

2.3.2.2 The moment of resistance of slabs, Retbrring to C/r& 6, corresponding to
with bars of different diameters and spacings h&,/b - 567 kN.m and d = 5625 cm,
are given in Tables 5 to 44. Tables are given
for concrete grades M 15 and M 20, with Percentage of steel pt - lOOAs M = 0.6
two grades of steel. Ten different thicknesses
ranging from 10 cm to 25 cm, are included. 0.6 bd 0.6~30~5625 __O1 ,,*
These tables take into account 25.5.2.2 .*. A,= -jijiy 100
of the Code, that is, the maximum bar
diameter does not exceed one-eighth the thick-
ness of the slab. Clear cover for reinforce-
ment has been taken as 15 mm or the bar For referring to Tables, we need the value
diameter, whichever is greater [see 25.4.1(d)
Mu
of the Code]. Jn these tables, the zeros at
the top right hand comer indicate the region
of
w
where the reinforcement percentage would M” 170 x IO’
exceed pt,lim; and the zeros at the lower bd’ - -3m6.25 x 56.25 x IO’
left hand comer indicate the region where I 1.79 N/mm’
the reinforcement is less than the minimum
according to 25.5.2.1 of the Code. From Table 1,
Percentage of reinforcement, pt = 0.594
Example 1 Singly Reinforced Beam 0.594 x 30 x 56.25 _ ,omo2,,*
.*. As-
Determine the main tension reinforcement 100
required for a rectangular beam section
with the following data: Example 2 Slab
Sixe of beam 3ox6Ocm
Concrete mix M 15 Determine the main reinforcement re-
Characteristic strength 415 N/mm’ quired for a slab with the following data:
of reinforcement
*Factored moment 170kN.m Factored moment 9.60 kN.m
*Assuming 25 mm dia bars with 25 mm E%etre

clear cover, Depth of slab 10 cm
Concrete mix M 15
Effectivedepth I 60 - 2.5 -2;- 5625 cm Characteristic strength a) 4 15 N/mm2
of reinforcement b) 250 N/mm*
From Table D, for fr P 415 N/mm’ and
fcrc
- 15 N/mm* h&l-HODOF REPERRING
TO TABLESFOR SLABS
MWliUJM’ p 2.07 N/mm:
Referring to Table 15 (for fy - 415 N/mmz),
v$g$ x (1000)’ directly we get the following reinforcement
for a moment of resistance of 9.6 kN.m
e; 2.07 x 101kN/m* per metre width:
8 mm dia at 13 cm spacing
:. &am - 2.07 x 1O’W or 10 mm dia at 20 cm spacing
30
I 2-07 x 10’ x fa x Reinforcement given in the tables is based
on a cover of 15 mm or bar diameter which-
I 1965 kN.m ever is greater.
$%ua] moment. of. 170 kN.m is less *than
The sectton 1stherefore to be destgned MFXHOD OF RFNRRJNG TO FLBXURB CHART
as u’~mm’singly
reinforced (unde&einforced)
rectangular section. Assume 10 mm dia bars with 15 mm cover,
OF RBFQIRIN
fVfM’HOD GTOFU3XURECHART d - 10 - 1.5 - 9 =8cm
For referring to Chart, we need the value of a) For fy= 415 N/mm’
moment per metre width. From Table D, Mu,tidb# = 2.07 N/mm*
Mu/b-g = 567kN.m per metre width. :. J%lirn - 2.07 x lOa x z x (A)’

= 13.25kN.m ’ _’
*The term ‘factoredmoment’means the moment
due to characteristic loads multiplied by the appro- Actual bending moment of 960 kN.m is less
priate value of p&rtialsafety factor yf. than the limiting bending moment.
FLExuRALmMBERs 11
Referring to Chart 4, reinforcement per- The lever arm for the additional moment of
centage, pt 6 0.475 resistance is equal to the distance between
Referring to Chart 90, provide centroids of tension reinforcement and com-
8 mm dia at 13 cm spacing pression reinforcement, that is (d-d’) where
or 10 mm dia at 20 cm spacing. d’ is the distance from the extreme compres-
sion fibre to the centroid of compression
Alternately, reinforcement. Therefore, considering the
A, = O-475 x 100 x &J = 3.8 cm* per moment of resistance due to the additional
tensile reinforcement and the compression
metre width. reinforcement we get the following:
From Table %, we get the same reinforce-
ment as before. Mu, - Asts (0*87f,) (d - a,)
also, Mu, =&Us-fQC)(d-J’)
b) Forf, = 250 N/mm* where
From Table D, Mu&bd” = 2.24 N/mm2 A1t2 is the area of additional tensile rein-
forcement,
M u&m = 2.24 x 10’ x 1 x(h) AK is the area of compression reinforce-
= 14.336 kN.m ‘---’ ment,
Actual bending moment of 9.6 kN.m is less I= is the stress in compression reinforce-
ment, and
than the limiting bending moment.
Referring to Chart 2, reinforcement per- fee is the compressive stress in concrete at
the level of the centroid of compres-
centage, pt = 0.78 sion reinforcement.
Referring to Churf PO, provide 10 mm dia
at 13 cm spacing. Since the additional tensile force is balanced
by the additional compressive force,
2.3.2 Doubly Reinforced Sections - Doubly
reinforced sections are generally adopted A, (l;c - fee)= At, (0*87&j
when the dimensions of the beam have been Any two of the above three equations may
predetermined from other considerations be used for finding Alt, and A,. The total
and the design moment exceeds the moment tensile reinforcement Ast is given by,
of resistance of a singly reinforced section.
The additional moment of resistance needed Ast = Pblim bd$
m Asc,
is obtained by providing compression re-
inforcement and additional tensile reinforce- It will be noticed that we need the values of
ment. The moment of resistance of a doubly frc and J& before we can calculate Al.
reinforced section is thus the sum of the The approach, given here is meant for design
limiting moment of resistance Mu,lim of a of sections and not for analysing a given
singly reinforced .section and the additional section. The depth of neutral axis is, therefore,
moment of resistance Mu,. Given the values taken as equal to x,,,-. As shown in Fig. 5,
of Mu which is greater than M”,lim, the value strain at the level of the compression reinforce-
of Mu, can be calculated. d’
ment will be equal to O-003 5 1- -
Mu, = Mu - Muslim ( XU,UWZ
>
STRAIN OlAGRkM
FIG. 5 DOUBLY REINKIRCED SECI-ION
12 DESIGN AIDS FOR REINFORCED CONCRIXE

For values of d’/d up to 0.2, fee isequal to The values of pt and pc for four values of
0446 fck; and for mild steel reinforcement d’jd up to 0.2 have been tabulated against
fz would be equal to the design yield stress MU/bd2 in Tables 45 to 56. Tables are given
of 0.87 fY. When the reinforcement is cold- for three grades of steel and four grades
worked bars, the design stress in compression of concrete.
reinforcement fw for different values of
d’/d up to 0.2 will be as given in Table F.
Example 3 Doubly Reinforced Beam
Determine the main reinforcements re-
TABLE F STRESS IN COMPRESSION
REINFORCEMENT ftc, N/mm* IN DOUBLY quired for a rectangular beam section with
REINFORCED BEAMS WITH COLD- the following data:
WORKED BARS
(Clause 2.3 2) Size of beam 30 x 6Ocm
Concrete mix M 15
f Y9 d’ld
Characteristic strength of 415 N/mm2
N/mm’ -A , reinforcement
0.0s 0.10 O-15 0.20 320 kN.m
Factored moment
415 355 353 342 329
500 424 412 395 370 Assuming 25 mm dia bars with 25 mm
clear cover,
2.3.2.2 Astz has been plotted against (d -d’) d = fj0 - 2.5 - 225 = 56*25 cm
for different values of MU, in Charts 19 and
20. These charts have been prepared for From Table D, for fy = 415 N/mm2 and
fs = 217.5 N/mm2 and it is directly appli- fck = 15 N/mm2
cable. for mild steel reinforcement with yield Mu,linJbd”=2.07 N/mm2 = 2.07 x IO2kN/m”
stress of 250 N/mm*. Values of Aat? for other .*. Mu,lim-2.07 x 103bd2
grades of steel and also the values of A, can 30 56.25 56.25
be obtained by multiplying the value read -2.07 x 10”x loo x Ts- x -100-
from the chart by the factors given in Table G.
The multiplying factors for A=, given in = 196.5 kN.m
this Table, are based on a value of fee corres- Actual moment of 320 kN.m is greater
ponding to concrete grade M20, but it can than Mu,lim
be used for all grades of concrete with little .*. The section is to be designed as a doubly
error. reinforced section.
TABLE G MULTIPLYING FACTORS FOR Reinforcement from Tables

USE WITH CHARTS 19 AND 20
‘Clause 2.3.2.1) Mu 320
$$ = O-562 5)2 x 103~~~~~ N/mm2
f FACTOR FACTOR FOR A, FOR d’jd
N&P FOR c-- 2.5 + 1.25 i o,07
A at* 0.05 0.10 0.15 0.2
d’/d c
5625 >
250 1.00 1.04 1.04 1.04 1.04
Next higher value of d’/d = 0.1 will be used
415 0.60 0.63 0.63 0.65 0.68 for referring to Tables.
500 0.50 0.52 0.54 0.56 0.60
Referring to Table 49 corresponding to
2.3.2.2 The expression for the moment of MU/bd2 = 3.37 and $ = 0.1,
resistance of a doubly reinforced section may
also be written in the following manner: Pt = 1.117, pc = 0.418
.
.. At - 18.85 cm2, A, = 7.05 cm2
Mu = Mu,lim + %(0*87fy) (d-d’)
REINFORCEMENTFROM CHARTS
Mu Mu,lim
(d-d’) = (56.25 - 3.75) - 52.5 cm
bj2 = ___bd” + -&(0*87f,)( I- ;>
Mu2 - (320 - 196.5) = 123.5 kN.m
where
ptz is the additional percentage of tensile Chart is given only for fy = 250 N/mm2;
reinforcement. therefore use Chart 20 and modification
Pt = phlim + pt2 factors according to Table G.
Referring to Chart 20,
PC =P”[-L-]
Art2 (for fY = 250 N/mm2) = 10.7 cm2
FLEXURAL MEMBERS 13
usia Jl¶odibrion factors given in Table G distribution in the flanp would not be uni-
for BY= 415 N/nuns, form. The expression Bven in E-22.1 of the
Code is an approximation which makes allo-
I& - 10.7 x 0.60 r! 6-42 cm* wance for the variation of stress in the flange.
,& I 10.7 x 0.63 = 674cm’ This expression is obtained by substitutin#
JYfor &in the equation of E-2.2 of the CO&
Referring to ruble E, yf being equal to (0.15 X,,m&+ 065 or)
pt,nm - 072 but not greater than Dr. With this m&&a-
5625 x 30 tion,
.*. Ast,u,n -0.72 .x ,oo - 1215cm’
Mudin~~T9 Mu,lirn,web f 0446 f&
A*: E 12.15+ 642 = 18.57cm’
These values of At and AE are comparable Mr-WY+ - f )
to the values obtained from the table. Dividing both sides by&k bw P,
2.4 T-SECTIONS
The moment of resistanceof a T-beam can x(& l)$(l+$)
be considered as the sum of the moment of
resistance of the concrete in the web of width where
b, and the contribution due to ,flanges of xu;= + 0.65 !$
width br.
The maximum moment of resistance is ob- but .f < $
tained when the depth of neutral axis is x,,,~.
When the thickness of flange is small,
that is, less than about 0.2 d, the stress in the Using the above expression, the ~2:
flange will be uniform or nearly uniform of the moment of resistance
(see Fig. 6) and the centroid of the compres- Mu,lim,T~ck b,# for different values of h/b*
sive force in the flange can be taken at Df/2 and &/d have been worked out and given in
from the extreme compression fibre. There- Tables 57 to 59 for three grades of steel.
fore, the following expression is obtained for
the limiting moment of resistance of T-beams 2.5 CONTROL OF DEFLECTION
with small values of Dfjd.
2.5.2 The deffection of beams and slabs
would generally be, within permissible limits
x(br-bw)h( d-$) if the ratio of span to effective depth of the
whereMll,llltniiveb member does not exceed the values obtained
in accordance with 22.2.1 of the Code. The
30.36 fd bwxu,,,,.x (d-O.416 x0,,,,&. following basic values of span to effective
The equation given in E-2.2 of the code is the depth are given:
same as above, with the numericals rounded
off to two decimals. When the flange thick-, En\!;;;Eorted 20
ness is greater than about 0.2 d, the above
expression is not corre4ztbecause the stress Cantilever “4
0.0.
047 f” 0.87 f,
-_b -* 0.002
E,
STRAIN DIAGRAM STRESS DIAORAM
ho. 6 T-SECTION
14 DBSIGN AIDS FOR REINPORCED CDNCRETE

Further modifying factors are given in 25.3 The values read from these Charts
order to account for the effects of grade and are directly applicable for simply supported
percentage of tension reinforcement and members of rectangular cross section for
percentage of compression reinforcement. spans up to 10 m. For simply supported or
continuous spans larger than 10 m, the values
2.5.2 In normal designs where the reinforce- should be further multiplied by the factor
ment provided is equal to that required from (lo/span in me&es). For continuous spans
strength considerations, the basic values of or cantilevers, the values read from the charts
span to effective depth can be multiplied by are to be modified in proportion to the basic
the appropriate values of the modifying values of span to effective depth ratio. The
factors and given in a form suitable for direct tn.l~G$ing factors for this purpose are as
reference. Such charts have been prepared ..
as explained below : conned; spans
4 The basic span to effective depth ratio &
for simply supported members is multi-
plied by the modifying factor for ten- In the case of cantilevers which are longer
sion reinforcement (Fig. 3 of the Code) than 10 m the Code recommends that the
and plotted as the base curve in the deflections should be calculated in order to
chart. A separate chart is drawn for ensure that they do. not exceed permissible
each grade of steel. In the chart, span limits.
to effective depth ratio is plotted on
the vertical axis and the tensile 2.54 For flanged beams, the Code recom-
reinforcement percentage is dotted on mends that the values of span to effective
the horizontal axis. depth ratios may be determined as for rectan-
Eoeons, subject to the followmg modi-
b) When the tensile reinforcement ex- ..
.ceeds ~I,II,,, the section will be doubly
reinforced. The percentage of compres- 4 The reinforcement percentage should
sion reinforcement is proportional to be,bcz&zm the area brd while referrmg
the additional tensile reinforcement
@t - PM,,) as explained in 2.3.2. b) The value of span to effective depth
However, the value of Pt,lim and pc ratio obtained as explained earlier
will depend on the grade of concrete should be reduced by multiplying by the
also. Therefore, the values of span to following factors:
effective depth ratio according to base b&v Factor
curve is modified as follows for each
grade of concrete: >:::3
For intermediate values, linear interpola-
1) For values of pt greater than tion may be done.
the appropriate value of pt,lim,
the value of (pt - pt,lim) is cal- Nom --The above method for flanged beams
culated and then the percentage of alay sometimes give anomalous mwlts. If the fhges
compression reinforcement p= re- arc ignored and the beam is considered as a rectangular
quired is calculated. Thus, the section, the value of span to effective depth ratio thus
obtained (percen of rciaforcemcnt being based
value of pc corresponding to a value on the area l&) Ys ould always be oa the safe side.
of pt is obtained. (For this purpose
d’/d has been assumed as 0.10 but 2.5.5 In the case of tw way slabs supported
the chart, thus obtained can gene- on all four sides, the sPorter span should be
rally be used for all values of d’/d considered for the purpose of calculating the
in the normal range, without signi- span to effective depth ratio (see Note 1 below
ficant error in the value of maximum 23.2 of the Code).
span to effective depth ratio.)
2) The value of span to effective depth 2.5.6 In the case of flat slabs the longer span
ratio of the base curve is multiplied should be considered (30.2.2 of the Code).
by the modifying factor for com- When drop panels conforming to 30.2.2 of
pression reinforcement from Fig. 4 the Code are not provided, the values of span
of the Code. to effective depth ratio obtained from the
3) The value obtained above is plotted Charts should be multiplied by 0.9.
on the same Chart in which the base
curve was drawn earlier. Hence Example 4 Control of Deflection
the span to effective depth ratio for
doubly reinforced section is plotted Check whether the depth of the member
against the tensile reinforcement in the following cases is adequate for control-
percentage pt without specifically ling deflection :
indicating the value of pc on the a) Beam of Example 1, as a simply suppor-
Chart. ted beam over a span of 7.5 m
FLBXURAL MEMBERS 15
b) Beam of Example 3, as a cantilever beam .*. The section is satisfactory for control
over a span of 4.0 m of deflection.
Cl Slab .of Example 2, as a continuous Span
slab spanning in two directions the c) Actual ratio of
Effective depth
shorter and longer spans being, 2.5 m
and 3.5 m respectively. The moment =-=2.5 31.25
given in Example 2 corresponds to 0.08
shorter spa’n. (for slabs spanning in two directions,
the shorter of the two is to be con-
Span sidered)
a>Actual ratio of
Eflective depth (i) Forfv = 415 N/mm2
pt = 0,475
= (56.;5;,oo) = 13.33 Referring to Chart 22,
Percentage of tension reinforcement Max Span = 23.6
required, (-> d
pt = 0.6 For continuous slabs the factor
Span obtained from the Chart should be
Referring to Char1 22, value of Max T multiplied by 1.3.
( >
corresponding to Pt = 0.6, is 22.2. :. Max “7 for continuous slab
Actual ratio of span to effective depth is less
than the allowable value. Hence the depth = 23.6 x 1.3 F 30.68
provided is adequate for controlling deflec- Actual ratio of span to effective depth is
tion. slightly greater than the allowable. Therefore
Span the section may be slightly modified or actual
b) Actual ratio of deflection calculations may be made to as-
Etfective depth
certain whether it is within permissible limits.
‘(d&J = 7.11 (ii) F0r.j; = 250 N/mm2
pt = 0.78
Percentage of tensile reinforcement, Referring to Chart 21,
pr = 1.117
Referring to Churl 22, Max Span = 31.3
(-1 d
Max value of %!a? = 21.0 :. For continuous slab,
( Cl 1
Max %% = 31.3 x 1.3 = 40.69
For cantilevers, values read from the d
Chart are to be multiplied by 0.35. Actual ratio of span to effective depth is
:. Max value of 1 less than the allowable value. Hence the
I/d for ) =21.0x0*35=7.35 section provided is adequate for controlling
cantilever J deflection.

fY
240
250
415
TABLE 1 FLEXURE - REINFORCEMENT PERCENTAGE, pc FOR SINGLY 480
REINFORCED SECTIONS
500 :
7ck
25
N/mm= b
250 415 480
0.141 Oa5 0.074 O-398

E
040
O-166 @lOO OQ86 i’
. i;:
O-45 .
8’E .
8’::; o&E O-503
O-50 Q240 O-144 O-125 O-510 0.423
O-276 O-265 0.159 0.138 O-518 O-448 O-430

O-302 0.290 0.175 0.151 O-526 0.455 O-436
O-329 0.164 O-461 O-443
O-356 E i-E O-178 8:Z 0468
O-383 O-368 Oh 0’191 0’550 O-475 8:::
0.410 O-394 O-237 Q-205 0.558 0482 o-463

O-421 O-405 0.244 0.211 O-566 0489 O-469
0.433 O-415 O-250 O-216 0’574 0.496 O-476
O-426 O-257 o-222 O-582 0’503 0483
IZ O-437 0’263 O-227 O-590 o-510 0.490
0448 O-270 0993 O-517 O-497
8::;; O-458 0.276 l-007 0.525
0489 0469 0283 1*021 O-532 o”:?i
O-500 0.480 O-289 1935 0.539 0.518
O-512 0.491 0.2% l-049 0.546 0.525
0.523 O-502 0.303 0.262 0554 O-532

O-535 0.513 o-267 0.561 0.539
0.546 0.524 EJ O-273 0.569
0.558 O-323 O-279 O-576
O-570 8:::; O-329 O-285 O-584
1.10 O-581 0.558 O-291 l-136 0.685 O-592
l-12 0393 0370 Ei O-297 l-151 O-693
l-14 @605 0.581 O-350 O-303 l-166 O-703
l-16 0.592 O-357 O-309 1.181 0.712
l-18 .
8’:;; O-604 O-364 O-315 l-197
1.20 0.641 O-615 0.371 0.321 l-212
1.22 0.653 0.627 0.378 0.327 1228
O-665 O-639 0.385 0.333 1,243
::z O-678 O-650 0.392 O-339 1.259
l-28 O-690 O-662 O-399 O-345 1.275
1.30
l-32 8E I:G
1.34 O-727 O-698
O-740 O-710
I:: 0.752 O-722
0.765 O-734 0442 O-382

O-778 o-747 O-389
0.759 X:t:Y 0.395
x:z 0.771 O-465
0.816 O-784 0.472 x:z
NOTE-Blanks indicate inadmissible reinforcement percentage (see Table E).
FLeXURAL MEMBERS
240
250
415
--
480 TABLE 2 FLEXURE -REINFORCEMENT PERCENTAGE,

REINFORCED ‘SECTIONS
pt FOR SINGLY
f f& - zO.N/m'
ck
20
O-073 1.253 l-203 O-627
“0%
O-188 Efl
izi! :‘26:
l-216 O-633 .
x’%
O-615
O-213
O-237
0.111
0’123
FEJ
. 1%
LOZ
:z
1.256
I:: 0.621
0.628
O-131 l-323 0.661 O-635

O-143
E pj xzz
XE . 0.655
O-181 .
:‘:z O-690 0.662
O-201 O-193 l-338 o-697

.
:‘z l-352 0.704
::E Ffg l-423 1966 o-711
0.242 l-438 1380 o-719
0.255 O-245 l-452 1’394 O-726
l-05 0.538 0.517 0.258 1467 O-734

1.10 O-566 O-543 ::a:; 1~482 O-741
1’15 0.570 O-297 8:;:; l-497 O-748
OO’E 0.597 O-311 0.298 0.756
.
:‘z *
O-650 @624 0’325 0312 :::;; 0.764
O-678 O-651 0.339 1.542 1481
E O-707 O-679 O-354 l-558 1’495
E 0.736 0.707
O-765 0.735 .
EL! ::z :::1’:
l-50 0.795 0’763 0397 l-604 l-540
1.55 O-825
;:g 8$;:
.
l-70 O-916
l-75 0947
l-80 1.632
I-85 *
%! X:E ;:gj l-647
1’041 O-521 0.500 l-663
E 1.073 o-537 O-515 _g;;
200 lm6 0.553 o-531 .
l-119 0.537 1.782 l-711

3g 0’543
:::z O-550 .
:‘E .
:‘%
E l-159 O-556 l-833 l-760
210 l-172 O-562
l-185
.
f’:: 1*199
.
f’:; .
:‘E
220 l-239
Num -Blanks lndk$teinadmissible

reinforcement
pcmntagc (see Table J%
48 DESIGNAIDS FOR REINFORCEDCGNCRRI’E

fY
240
250
4'15
TABLE 3 FLEXURE - REINFORCEMENT PERCENTAGE,
REINFORCED SECTIONs
pt FOR SINGLY 480
500
/ck = 25 N/mms
7ck
Mlw2, fu,N/mm2
L
25
N/mms 7’
--Y
250 415 480 500. 250 415 500
0.30 0.146 0.140 0.084 O-070 1.415 1.358 O-818 O-679

0.35 O-171 0.164 O-099 1448 l-390 0.837
O-195 0188 0.113 :%z 1482 l-422 :z:
g O-211 Of27 O-106 1.515 l-455 ::::z O-727
:z 0.236 O-142 O-118 l-549 1487 O-896 O-744
O-271 O-260 O-156 0.130 l-584 l-520

O-296 0.171 0.142 % 1.618 ::;:z 8%
0?321 !% 0.186 O-154 290 1.653 :::z O-956 O-794
O-347 0.333 O-201 0167 2-95 l-689 l-621 o-977 O-811
O-373 O-358 O-216 0.179 3.00 l-724 l-655 0.997 0.828
0.80 0.399 0231 0191 3.05 l-760 1.018 0.845

O-85 t-I.425 O-204 3.10 1.797 ::% 1.039 O-863
0.451 X% 0.216 3.15 1.834 l-760 l-061 0.880
8C O-477 0.276 0229 l-871 l-796 l-082 0.898
lfl0 O-504 O-291 O-242 :::: 1909 1.832 1.104 O-916
l-05 O-530 0.509 0.307 0.255 1.947 l-869 0.935

1.10 6535 O-322 0.267 33% l-962 l-884 ::::; 0942
1.15 8% O-561 0.338 0.280 l-978 l-899
O-611 0353 0.293 :::t 1993 l-914 :::z
::z O-638 .
z-;;: 0.369 0.306 3-38 2QO9 l-929 1.162
l-30 0.666 0.639 O-385 0.333 0.320 2.025 l-944

l-35 0.693 O-401 0.347 O-333
0.721 X:% w417 0.346 Z:% .
:‘zc
:z 0.749 @719 0.433 8% O-359 2.072 l-989
1.50 O-777 0.746 O-449 O-388 0373 2-088 2.005
o-773 O-466 0403 O-387 2.104

O-482 2.120
8::: 0.499 x:::: x%Y
0.856 0.515 0.428 z:::
O-883 O-532 8:% O-442 2.170
1.80 0.949 O-911 O-549 O-415 0.456 3.60 2.186 2.099
l-83 0979 O-940 O-566 O-489 0.470 3.62 2.203 2.115
1.90 l-009 0968 O-583 0.484 2.219 2.131
1.95 l-038 O-601 fj:$ O-498 ::z 2.147
2-00 1,068 .
YZ O-618 o-513 3.68 ;:22:; 2.163
2-05 1.055 O-635 O-527 3.70 2.270 2.179
210 :z O-653 8% 0542 3-72 2.287 2196
2.15 l-160 : :% 0.671 O-580 O-557 3.74 2.304
l-191 l-143 0.689 0.596 O-572
$12”: 1.222 1.173 O-707 0.611 0.587
l-254 l-204 0.725 O-627 O-602
13283 l-234 O-743 O-643 0.617
l-317 l-265 0.762 O-632
l-350 0781 ZR 0.648
l-382 : :% o-799 O-691 0.663
NOTE- Blanks indicate inadmissible reinforcement percentage (see Table E).
m3xuRAL MEMBERS 49
'Y
240
250
415
480 TABLE 4 FLEXURE - REINFORCEMENT PERCENTAGE, pi FOR SINGLY
REINFORCED SECTIONS
500
fck fd = 30 N/mm2
30 MUW,
A. N/mm2
*
N/mm2 r240 -7
250 500 250 415 480 500
0.140 0.070 255 1.374 1.319 0.794 @687

0.082 0.812 0.702
.
8’::; 0.093 fZ ::zi ::zi 0.830 0.718
@211 0.105 270 1467 1.408 0.848 0.733
0235 0.117 275 1.498 1.438 0.866 0.749
1.530 0.734
c8”: 1.562 ~~~ is; 0.750
1.594 1.530 0.797 0.765
tz 1.626 1.561 0813 0.781
3.00 1.659 1.592 0.829 0.796
0380 3.05 1.691 1.624 0.978 0.846 0.812
0405 3.10 1.725 1.656 0.862 0.828
0.429 3.15 1.758 1.687 zz3 0.879 0.844
0.454 1.791 1.720 1.036 0.896 0.860
0.479 33:g 1.825 1.752 1.055 0.913 0.876
0.525 0.252 3.30 1.859 1.785 1.075 0.930
0552 0.265 3.35 1.893 1.818 1.095 0.947
0.578 Ct.277 1.851 1.115 0.964
0.290 :z zi I.884 1.135 0.981
.
x’z 0.303 3.50 1998 1.918 1.156 0.999
@631 0316 3.55 1.952 l-176 0976
~~i?i
0.712
0739
x:::
0.709
xz
0.355
3.60
3.65
3.70
2z
2105
2.142
1.986
2021
2056
1.197
1.218
1.239
KG;
1.053
1.071
zf:
1.028
0.766 0’735 0.368 3.75 2.178 2091 1.260 1.089 1.046
0.762 @381 3.80 2215 1.281 1.108
8’;1;: 0.788 0.394 3.85 2253 1.303 KS
oi49 2291 1.325 .
:‘:z 1.099
xz:: x:z it:z 2329 1.347 1.164 1.118
x:zi 0.868 0.434 4.00 2367 1.369 1.184
0.932 0.895 1.391
0961 0.922 oo:z f:iz i::z 1.414
0.989 0.950 0.475 2485 2386
1.018 0977 0488 2.525 2.424
1946 1.005 0502 2566 2463
1.075
:x.
1.163
1.192
1’173
::i!f
1.260
1289
NOTE- Blanksindicateinadmissible
reinforcement
pyceniage(see TableE).
so DESIGN AIDS FOR REmFORCJiD CONCREIB

3. COMPRESSION MEMBERS
3.1 A$MU\;,OADED COMPRESSION lower sectioqs would eliminate tho need for
any calculation. This is particularly useful
as an aid for deciding the sizes of columns
All compression members arc to be designed at the preliminary design stage of multi-
for a minimum eccentricity of load in two sforeyed buildings.
oribcioal directions. Clause 24.4 of the Code
I
specifiks the following minimum eccentri-

city, eminfor the design of columns: Example 5 Axially Loaded Column
1 Determine the cross section and the
emin=ggg i-D 3o subject to a minimum of
reinforcement required for an axially loaded
2 cm. columc with the following data:
where
Factored load 3000kN
1 is the unsupported length of the column Concrete grade M20
(see 24.1.3 of the Code for definition of Characteristic strength of 415 N/mm’
unsupported length), and reinforcement
D is the lateral dimension of the column Unyoyior;d length of 3.0 m
in the direction under consideration.
After determining the eccentricity, the section The cross-sectional dimensions required will
should be designed for combined axial load depend on the percentage of reinforcement.
and bending (see 3.2). However, as a simplifi- Assuming 1.0 percent reinforcement and
cation, when the value of the mininium referring to Chart 25,
eccentricity calculated as above is less than or
equal to 0*05D, 38.3 of the Code permits Required cross-sectional area of column,
the design of short axially loaded compression A, - 2 700 cm*
members by the following equation: Provide a section of 60 x 45 cm.
P,=@4f,k AC-i-0.67fY Ar Area of reinforcement, A, - 1.0 x m~$,j2
where 1: 27 cm8
PU is the axial load (ultimate), We have to check whether the minimum
A, is the area of concrete, and eccentricity to be considered is within 0.05
Asc is the afea of reinforcement. times the lateral dimensions of the column.
The above equation can be written as In the direction of longer dimension,
PA, --I D
P” = 0.4 f& A, - +$) t- 0.67fy loo &in -
500 +30
3*0x 102 60
where = + jg P 0.6 j-2.0 - 2.6 cm
500
As is the gross area of cross section, and or, e&D = 26160 = O-043
p is the percentage of reinforcement.
Dividing both sides by A,, In the direction of the shorter dimension,
3.0 x 102 45
PU = @4&( 1 - j&) +“‘67fy $j emrn =
500
+
30
= 0.6 + 1.5
= 2.1 cm
or, e,i,/b = 2*1/45 = @047
Charts 24 to 26 can be used for designing The minimum eccentricity ratio is less than
short columns in accordance with the above @05 in both directions. Hence the design of
equations. In the lower section of these the section by the simplified method of 38.3
charts, P./A, has been plotted against of the Code is valid.
reinforcement percentage p for different
grades of concrete. If the cross section of
the column is known, PU/Al can be calculated 3.2 COMBINED AXIAL LOAD AND
and the reinforcement percentage read from UNIAXIAL BENDING
the chart. In the upper section of the charts,
PU/As is plotted against PUfor various values As already mentioned in 3.1, all com-
of AS. The combined use of the upper and pression members should be designed for
COMPRESSION MEMBERS 99
minimum eccentricity of load. It should compressed edge to zero at the opposik
always be ensuredthat the scotionis designed ed~. For purely axial compression, the
for a moment which is not lessthan that due strain is assumed to be uniformly equal
to the prescribedtinimum eccentricity. 00002 acxossthe seotion[see 38.l(a) of the
Code]. The strairidistributionlines for these
3.2.1 Amanptio&Assumptiom (a), (c),
~’ oases intersecteaeh other at a depth of
(d) and (e) for flexural members (see 2.1)
are also applicable to members subjcoted ~ffom the highly compressed edge. This
to combined axial load and bending. ‘The
assumption (b) that the maximum strain point is assumedto act as a fulcrum for the
in concrete at the outermost eom ression strain distribution line when the neutral
fibre is 04N35 is also applicable wi en the axis lies outsidethe motion (see Fig. 7). This
neutralaxis k within the seotionand in the leads to the assumption that the strain at
limitingcase when the neutralaxis lies along the highly compresseded~ is 00035 minus
one edge of the section; in the latter oasc 0?5 times the strain at the least compressed
the strain varies from 0@035 at the highly edge [see 38.Z(b) of the Cole],
“-i t-
:
-
1
---1-”
I
I
● ‘* ● ● ●
*
● ●
i
c
● ●
b I :
0 !
0 * ●
t
● HIWilmY C6MPRE S SE II
I ● 00 I J: EOOE
CENTRO13AL AXIS
+’+ ikh ROW OF REINFORCEMENT
STRAIN DIAGRAMS
0035
Neutral axis
wlthln the scctlon
-30/7
-1
-— -----
Neutral axis
outside the sect ion
FIG. 7 Cmramm Am- LOAD AND UNIAXIAL BENDING
No DESIGNAIDSIK)RREINFORCED CONCR81E
3.2.2 Stress Block Parameters Wh&n the Area of stress block
Neutral Ax& Lies O&side the Section - When
the neutral axis lies outside the section, - 0446f,D-5 g 4,-D
the shape of the stress block will be as. indi- ( >
cated in Fig. 8. The stress is uniformly
= 04461&D +gD
0446fd for a distance of Ly from the highly
compressed edge because the strain is more - 0446fdr D
[l-&&J]
than 0402 and thereafter the stress diagram
is parabolic.
The centroid of the stress block will be
found by taking moments about the highly
compressed edge.
Moment about the highly compressed edge
D
pO1446fckD
(1 i -$ gD
The position of the centroid is obtained by

dividing the moment by the area. For diier-
ent values of k, the area of stress block and
STRAIN DIAORAM the position of its centroid are given in
Table H.
t
i
TABLE H STRESSBLOCKPARAhUTTERS
WHENTHE NETmmtA&mA?N LIES OUTSIDE
(Clause 3.2.2)
O-446 1,
BTRESS OIAORAW
FIG. 8 STRBSS
BLOCK WHEN THE NEUTRAL
Am h¶ oUT?3IDE THE SECTION
Let x0- kD and let g be the ditference Nom-Values of stress block parametershave
between the strxs at the highly compressed been tabulatedfor valuesof k upto4’00for infom-
edgo and the stress at the least compressed tion only. For constructionof interactiond@cams
edge. Considering the geometric properties it b merally adaquatato considervalues of k up to
about 1.2.
of a parabola,
33.3 Constructionof InteractionDiagram -

Design charts for combined axial compression
and bending are given in the form of inter-
action diagmms in which curyes for PJbDfd
versus MdbD* fb are plotted for different
1 values of p/f&, where p is the reinforcement
-o+Mf& & percentage.
( 1
COMPRESSlONMEMBERS 101
3.2.3.2 For the case of purely axial com- The above expression can be written as
pression, the points plotted on the y-axis
n
of the charts are obtained as follows:
P,= 0446f,rbd + ‘g (A - 0.446 fek)

Taking moment of the forces about the
centroid of the section,
where
fr is the compressive stress in steel corres-
ponding to a strain of 0.002. + xg (.Ai - fci).Yi
The second term within parenthesis repre- i- 1
sents the deduction for the concrete replaced
by the reinforcement bars. This term is where
usually neglected for convenience. However,
a9 a better approximation, a constant value C,D is the distance of the centroid of the
corresponding to concrete grade M20 has concrete stress block, measured from
been used in the present work so that the the highly compressed edge; and
error is negligibly small over ;he range of
concrete mixes normally used. An accurate Yi is the distance from the centroid of the
consideration of this term will necessitate section to the ith row of reinforce-
the preparation of separate Charts for each ment; positive towards the highly
grade of concrete, which is not considered
compressed edge and negative to-
wards the least compressed edge.
worthwhile.
Dividing both sides of the equation by
3.2.3.2 When bending moments are also
fck bD”,
acting in addition to axial load, the points
for plotting the Charts are obtained by
assuming different positions of neutral axis. c, (O-5-Cd
For each position of neutral axis, the strain
distribution across the section and the n
stress block parameters are determined as +X*(&i -&i)(s)
explained earlier. The stresses in the rein-
forcement are also calculated from the i- 1
known strains. Thereafter the resultant axial
force and the moment about the centroid b) When the neutral axis lies within the
of the section are calculated as follows: section
a) When the neutral axis lies outside the
section In this case, the stress block parameters
are simpler and they can be directly incorpora-
li ted into the expressions which are otherwise
same as for the earlier case. Thus we get the
following r;xpressions:
i-1
where =@36k+ c &j&&d

i-1
Cl - coefficient for the area of stress
block to be taken from Table H
(see 3.2.2);
Ad
Pi - where A,i is the area of rein- n
bx
forcement in the ith row;
fii - stress in the ith row of reinform
ment, compression being positive
and tension being negative; where
fci - stress in concrete at the level of
the ith row of reinforcement; and Depth of neutral axis
k-
n - number of rows of reinforcement. D

An approximation is made for the value only; they do not take into account crack
Offci for M20, as in the case of 3.2.3.1. For control which may be important for tension
circular sections the procedure is same as members.
above, except that the stress block para-
meters given earlier are not applicable; Example 6 Square Column with Uniaxial
hence the section is divided into strips and Bending
summation is done for determining the
forces and moments due to the stresses in Determine the reinforcement to be provided
concrete. in a. square column subjected to uniaxial
bending, with the following data:
3.2.3.3 Chartsfor compression with bending -
Charts for rectangular sections have been Size of column 45 x 45cm
given for reinforcement on two sides (Charts Concrete mix M 25
27 to 38) and for reinforcement on four Characteristic strength of 415 N/mm%
sides (Charts 39 to 50). The Charts for the reinforcement
latter case have been prepared for a section Factored load 2500kN
with 20 bars equally distributed on all sides, (characteristic load
but they can be used without significant. multiplied by yr)
error for any other number of bars (greater Factored moment 200 kN.m
than 8) provided the bars are distributed Arraugement of
equally on the four sides. The Charts for reinforcement: (a) On two sides
circular section (Charts 51 to 62) have been (b) On four sides
prepared for a section with 8 bars, but they
can generally be used for sections with any (Assume moment due to minimum eccentri-
number of bars but not less than 6. Charts city to be less than the actual moment).
have been given for three grades of steel
and four values of d’/D for each case men- Assuming 25 mm bars with 40 mm cover,
tioned above. d = 40 + 12.5 OP52.5 mm z 5.25 cm
d’/D = 5.25145 - 0.12
The dotted lines in these charts indicate Charts for d’/D = 0.15 will be used
the stress in the bars nearest to the tension f& = 25 x 2 45
500x x45103
x lo2
_ = 0.494
face of the member. The line for fs, I; 0
indicates that the neutral axis lies along the
outermost row of reinforcement. For points 200 x 106
25x45~45~45~103
_ - = 0.088
lying above this line on the Chart, all the
bars in the section will be in compression. a) Reinforcement on two sides,
The line for fSt = fYd indicates that the Referring to Chart 33,
outermost tension reinforcement reaches the p/fck = 0.09
design yield strength. For points below this Percentage of reinforcement,
line, the outermost tension reinforcement p = 0.09 x 25 - 2.25
undergoes inelastic deformation while succes- As = p bD/lOO = 2.25 x 45 x 45/100
sive inner rows may reach a stress of fyd. = 45.56 cm2
It should be noted that all these stress values b) Reinforcement on four sides
are at the failure condition corresponding from Chart 45,
to the limit state of collapse and not at work- p&k = 0.10
ing Ioads. p p. 0.10 x 25 = 2.5
As = 2.5 x 45 x 45/100 = 50.63 cm”
3.2.3.4 Charts for tension with bending -
These Charts are extensions of the Charts
for compression with bending. Points for Example 7 Circular Column with Uniaxial
plotting these Charts are obtained by assum- Bending
ing low values of k in the expressions given
earlier. For the case of purely axial tension, Determine the reinforcement to be pro-
vided in a circular column with the following
Pu - g (O-87fy) data:
Diameter of column 50 cm
hk (@87fy) Grade of concrete M20
Characteristic strength 250 N/mm2 for
Charts 66 to 75 are given for rectangular of reinforcement bars up to
sections with reinforcement on two sides 20 mm+
and Charts 76 to 85 are for reinforcement 240 N/mm2 for
on four sides. It should be noted that these bars over
charts are meant for strength calculations 20mm#

Factored load 16OOkN Core_diie;; = 50-2 (4-O - 0.8)
Factored moment 125 kN.m
Lateral reinforcement : AI/AC IP 5O’/43*6’ = 1.315
(a) Hoop reinforcement 0.36 (A,,& - 1)falJlr
(b) Helical reinforcement I egg; 0.315 x 201250
(Assume moment due to minimum eccentri-

city to be less than the actual moment). Volume of helical reinforcement
Volume of core
Assuming 25 mm bars with 40 mm cover, Aarc .(42*8)
--_------= 0.09 A,J,
d’ = 40 x 12.5 = 52.5 mm m 5.25 cm
d’/D - 5.25150 = 0.105 ;(43*6%) Q, a,
Charts for d’/D = 0.10 will be used. where, Ash is the area of the bar forming
the helix and sh is the pitch of the helix.
(a) Column with hoop reinforcement In order to satisfy the coda1 requirement,
1600 x 103 0.09 Art&k > O*OO91
20 x50 x 50 x ioa - o’32 For 8 mm dia bar, Ati = O-503 cm2
125 x 10 sh ( 0.09 x 0.503
20 x 50 x 50 x 50 x 103 - 0.05 ’ 0.0091
‘__ < 4.97 cm
Referring to Chart 52, for fy I 250 N/mm1
p/fck = o-87
= 0.87 x 20 = 1.74 3.3 COMPRESSION MEMBERS SUB-
A: = pnD2/400 JECT TO BIAXIAL BENDING
= 1.74 x nx50x50/400=34*16cm2
Exact design of members subject to axial
Forf, I 240 N/mm2, load and biaxial bending is extremely
AS = 34.16 x 2501240 = 35.58 cm2 laborious. Therefore, the Code permits the
design of such members by the following
(b) Column with Helical Reinforcement equation :
According to 38.4 of the Code, the $rength

of a compression member with hehcal re--
inforcement is 1.05 times the strength of a lhere
similar member with lateral ties. Therefore,
the, given load and moment should be divided M,,, M,, are the moments about x and y
by 1.05 before referring to the chart. axes respectively due to design loads,
M “Xl, MUYl are the maximum uniaxial
Hence, moment capacities with an axial load
P,, bending about x and y axes res-
pectively, and
ozn is an exponent whose value depends on
Pu/Puz (see table below) where
P uz = 0.45 fck A, + 0*75fy As:
From Chart 52, for fy = 250 N/mm2, PUIPUZ ‘an
p,$_k = 0.078
p = 0.078 x 20 = 1.56 go.2 1.0
As = 1.56 x x x 50 x 50/44X )0*8 2.0
= 30.63 cm2
For fy = 240 N/mm%, A, = 30.63 x 2501240 For intermediate values, linear interpo-
= 31.91 cm2 lation may be done. Chart 63 can be used
for evaluating Puz.
According to 38.4.1 of the Code the ratio
of the volume of helical reinforcement to the For different values of Pu/Puz, the appro-
volume of the core shall not be less than priate value of azn has been taken and curves
0.36 (A,/Ac - 1) fck Ify where A, is the for the. equation
gross area of the section and Ac is the area
of the core measured to the outside diameter (!$)“’ + (z)=” = 1.0 have been
of the helix. Assuming 8 mm dia bars for the
helix, plotted in Chart 64.
104 DESIGN AIDS FOR REINFORCED CONCRETE .

ExampIe 8 Rectangular colrmu, with Biaxial
Be?tdi?lg
DeWmine the reinforcement to be pro-
vided in a short column subjected to biaxial
bending, with the following data:
size of column Referring to Churn 64, the permissible v&a
Concrete mix EPcm
415 N/mm’ MBa
Characteristic strength
of reinforcement
Factored load, P,, 1600kN
ofns,, qrmsponding to the above v&see
Factored moment acting 120 kN qual to 0.58.

parallel to the larger
dimension, M,w The actual value of 0.617 is only sli&ly
Factored moment acting 90 kN higher than the value read from the Chart.
parallel to the shorter This can be made up by slight increase in
dimension, Mu, reinforcement.
Moments due to minimum eccentricity are A6 - 1-2x40x60 _2&8,.&
less than the values given above. 100

12 bars of 18 mm will give A.130.53 c&
Reinforcement is distributed equally on Reinforcement percentage provided,
four sides.
p _ 30.53 x 100
As a iirst trial assume the reinforcement 6o x40 - 1.27
percentage, p= 1.2
With this percentage, the section may be
P&k - 1*2/l 5 - 0.08 rechecked as follows:
Uniaxial moment capacity of the section p/f&- l-27115= 0.084 7
about xx-axis: Referring to Chart 44,
5.25
d’/D - 6. - 0.087 5 - 0,095
f$
Chart for d’/D = 0.1 will be used. Mw z 0.;9;. &li x 40 x 60’ x 10*/10*
1600 x 10s .
p&k bD = 15 x 40 x 60 x 10”- 0444 Referring to Chart 45
Referring to Chart 44, l
- 0.085
M&k bD= = 0.09 f+
M WI z W&354
x&52 60 x 40’ x 10a/lO’
:. MUX, - 0.09 x 15 x 40 x 60’ x loylo~
- 194.4 kN.m Referring to Char; 63,
Uniaxial moment capacity of the section
about yv_axis: PUZ= 10.4 N/mm2
Al
5.25
d’JD = 40- 0.131 Puz - 10.4 x 60 x 40 x lO’/lO~
- 2 496 kN
Chart for d’/D - 0.15 will be used.
Referring to Chart 45,
M&k bD’ - 0.083
:. MuYl - 0.083 x 15 x 60 x40*x 10a/lO’
I 119.52kN.m
Calculation of P,,: Referring to Chart 64,
Referring to Chart 63 corresponding to
p = 1.2,fu= 415 and fck= 15, Corresponding to the above values of
Muy PU
the permissible value of
MuY, and z’
MUX is 0.6.
-
10.3 x 40 x 60X MUX,
108/10SkN
2 472 kN Hence the section is O.K.
COMPRESSION, MEMBERS 105

3.4 SLENDER COMPRESSION In accordance with 38.7.1.1 of the Code,
MEMBERS the additional moments may be reduced by
the multiplying factor k given below:
When the slenderness ratio D&?. or # of
a compression member exceeds 12, it is
considered to be a slender compression
member (see 24.2.2 of the Code); In and i, where
being the effective lengths with respect to
the major axis and minor axis respectively. P,, = 0.45 &k Ac + 0.75 fy A, which
When a compression member is slender with may be obtained from Chart 63, and Pb is the
respect to the major axis, an additional axial load corresponding to the condition of
moment Mu given by the following equation maximum compressive strain of 0.003 5
(modified as indicated later) should be in concrete and tensile strain of O%Ml2in
taken into account in the design (see 38.7.1 outermost layer of tension steel.
of the Code) :
Though this modification is optional ac-
cording to the Code, it should always be
taken advantage of, since the value of k
could be substantially less than unity.
Similarly, if the column is slender about the The value of Pb will depend on arrangement
minor axis an additional moment M.,, should of reinforcement and the cover ratio d’/D,
be considered. in addition to the grades of concrete and
Pub &” steel. The values of the coefficients required
M =
ay 2000 (1
b
for evaluating Pb for various cases are given
in Table 60. The values given in Table 60
The expressions for the additional moments are based on the same assumptions as for
can be written in the form of eccentricities members with axial load and uniaxial bending.
of load, as follows: The expression for k can be written as
follows :
Mu - P, eu
where
Chart 65 can be used for finding the ratio

of k after calculating the ratios P,/Pu, and
pb/&z.
Example 9 Slender Column (with biaxial

Table 1 gives the values b or 3 for bending)
different values of slenderness ratio. Determine the reinforcement required for
a column which is restrained against sway,
with the following data:
TABLE I ADDITIONAL ECCENTRICITY FOR
SLENDER COMPRESSION MEMBERS Size of column 40 x 30 cm
(Chuxe 3.4) Concrete grade M 30
Characteristic strength 415 N/mm1
of reinforcement
Effective length for 6-O m
bending parallel to
larger dimension, Z,
Effective length for 5.0 m
bending parallel to
shorter dimension, ly
Unsupported length 70m
Factored load 1500kN
Factored.moment in the 40 kN.m at top
d!““;f larger and 22.5 kN.m
at bottom
DESIGN AIDS FOR RRINFORCED CONCRKI-E

Factored moment in the 30 kN.m at top O-028x 3
Pb (about yy-axis) i 0.184 -+ 30
direction of shorter ~tdJOcN.xn
dimension x40x30x30
x 10*/1os
The column is bent in double curvature. 672 kN
.pby -
Reinforcement will be distributed equally
on four sides. .*. k, I p -;b; I ‘;ym- 17r
6-o x 100
‘-Lx
D P 40 = 15.0 > 12 = oG5
Puz - Pu 2700-1500
k.y = p “7. -
cy=
I 5.0 x 100 = 16-7 > 12 pby - 2 700 - 672
b 30
= o-592
Therefore the column is slender about
both the axes. The additional moments calculated earlier,
From Table I, will now be multiplied by the above values
of k.
For Lz P 15, eJD = 0.113
= 67.8 x O-625 = 42.4 kN.m
:: = 63.0 x 0.592 - 37.3 kN.m
For blay = 167, e,/b = O-140
The additional moments due to slenderness
effects should be added to the initial moments
Additional moments:
after modifying the initial moments as
M 1x = Puex = 1 500 x0-1 13 x & -67.8kN.m follows (see Note 1 under 38.7.2 of the Code) :
M,,=(O*6 x 40 - 0.4 x 22.5) = 15-OkN.m
May = Pue, = 1 500 x0*14x &=63*0 kN.m KY= (0.6 x 30 - 0.4 x 20) = 10.0 kN.m
The above moments will have to be reduced The above actual moments should be com-
in accordance with 38.7.1.1 of the Code; pared with those calculated from minimum
but multiplication factors can be evaluated eccentricity consideration (see 24.4 of the
only if the reinforcement is known. Code) and greater value is to be taken as the
initial moment for adding the additional
For first trial, assumep p: 3.0 (with reinforce- moments.
ment equally on all the four sides).
40
&-=40x 30= 1200cm2
ex = - &+$-7g + 3o= 2*73cm
From Chart 63, Puz/As = 22.5 N/mm2 = 2.4 cm

.-. Pu = 22.5 x 1200 x 102/10s =2 700 kN Both e, and e, are greater than 20 cm.
Calculation of Pb : Moments due to minimum eccentricity:
Assuming 25 mm dia bars with 40 mm cover Mux = 1 500 x ‘g = 41.0 kN.m
d’/D (about xx-axis) cs g = 0.13 > 15.0 kN.m
Chart or Table for d’/d P O-15 will be 1500x2’4 = 36.0 kN.m
Muy -
used. 100
5.25 > 10.0 kN.m
d’/D (about yy-axis) = 3. = 0.17
Chart or Table for d’/d = 0.20 will be :. Total moments for which the column
used. is to be designed are:
From Table 60, MUX- 41.0 + 42.4 = 83.4 kN.m
iu Uy= 36.0 + 37.3 = 73.3 kN.m
Pb (about xx-axis) = (k, + k2 &rbD
The section is to be checked for biaxial
3 betiding.
0.196 + 0.203 X o
1500 x 10s
Pul& bD =
x 30 x 30 x 40 x 102/103 30x 30 x40 x 102
= -779 kN = 0.417

i
Plfck - .g = 0.10
MUY
M
-
73.3
Ic -103.7 = 0.71
UYl
PulPu, = - 1500 = 0.56

:. MUX1= 0.104 x 30 x 30 x 40 x 40 x 2700
103/10’
Referring to Chciit 64, the maximum allow-
= 149.8 kN.m able value of M,,/M,,, corresponding to the
Refep; to ihart 46 (d’/D P O-20), above values of M,,/M,,, and PuIPuz is 0.58
v cka 2 = 0,096 which is slightly higher than’ the actual value
:. M”Yl =0*096 x 30 x 40 x 30’~ 30 x of 0.56. The assumed reinforcement of 3.0
103/106 percent is therefore satisfactory.
= 103.7 kN.m A s = pbD/lOO - 3.0 x 30 x 40/100
M”,
- ic -
83.4
e 0.56 L= 36.0 cm2
M”,, 149.8
108
'Y
250
415
- 500 Chart 63 VALUES OF Puz for COMPRESSION MEMBERS
f
ck
15
20
25
30
ti i i i i i i i i i i i i i i i i i i IWi
148 ._ DESIGN AIDS FOR REINFORCED CONCRETE .a

-.. .
Chart64 BIAXIALBENDINGIN COMPRESSION MEMBERS
0 0.1 O-2 O-3 0.4 0.5 O-6 O-7 0.8 0.9 1.0
%/L
COMPRFSSIONMEMBERS
Chart 65 SLENDER COMPRESSION MEMBERS -
Multiplying Factor k for Additional Moments
k+
ur-pu
P
PUZ -Pe
a 04
04
O-3
0*2
o-1
150 DESIGN AIDS FOR REINFORCED CONCRl3-b

TABLE 60 SLENDER COMPRESSION MEMBERS-VALUES OF A
lktMgdW_:
Valora of k,
&ID
r 9
005 WlO 015 OQQ
0219 om7 01% 0184
0172 @MO 0149 0.138
V@mof4
#ID
fr
N,‘mrn~
r
O-05 0.10 &lS 020
\
COMPRESSION
MEMBERS 171
Y”
4. SHEAR AND TORSION

4.f DESIGN SHEAR STRENGTH OF For a series of inclined stirrups, the value
CONCRETE of Vup/d for vertical stirrups should be
multiplied by (since i- coscc) where cc is
*The design shear strength of concrete is the angle between the inclined stirrups and
given in Table 13 of the Code. The values the axis of the member. The multiplying
given in the Code are based on the following factor works out to 1.41 and 1.37 for 45”
equation: and 60” angles respectively.
For a bent up bar,

VuI= 0*87fY ASvsince
where
Values of V,,, for different sizes of bars,
g =0.8 fck/6*89pl, but not less than 1.0, bent up at 45” and 60” to the axis of the
and Pt = 100 A&&. member are given in Table 63 for two grades
of steel.
The value of ‘F~corresponding to pl varying
from 0.20 to 3.00 at intervals of 0.10 are given
4.4 TORSION
in Table 61 for different grades of concrete.
Separate Charts or Tables are not given
4.2 NOMINAL SHEAR STRESS for torsion. The method of design for torsion
is based on the calculation of an equivalent
The nominal shear stress 7” is calculated shear force and an equivalent bending
by the following equation: moment. After determining these, some of
VU the Charts and Tables for shear and flexure
7” = - can be used. The method of design for
bd torsion is illustrated in Example 11.
where
V,, is the shear force.
Example 10 Shear
When rv exceeds 7c, shear reinforcement
should be provided for carrying a shear Determine the shear reinforcement (vertical
equal to Vu- Q bd. The shear stress rv should stirrups) required for a beam section with
not in any case exceed the values of T~,~, the following data:
given in Table J. (If T”> T~,~~, the section
is to be redesigned.) Ream size ‘30 x 60 cm
Depth of beam acrn
Concrete grade M 15
TABLE J MAXIMUM SHEAR STRESS w,mu Characteristic strength 250 N/mma
of stirrup reinforcement
CON- GRADE Ml5 M20 M25 M30 M35 M40 Tensile reinforcement 0.8
Q., mu, N/mm’ 25 2% 3-l 3.5 3-l 40 percentage
Factored shear force, Vu 180 kN
4.3 SHEAR REINFORCEMENT Assuming 25 mm dia bars with 25 mm cover,
The design shear strength of vertical d = 60 -T - 2.5 = 56.25 cm

stirrups is given by the following equation:
v “I _ @87f,A,vd Shear stress, 7” =i g -30 ~8p,$o,“,,

-
sv
= l-07 N/mm*
where
A,” is the total cross sectional area of From Table J for M15, 7c,max= 2.5 N/mm2
the vertical legs of the stirrups, and T” is less than 7c,mu
sv is the spacing (pitch) of the stirrups.
From Table 61, for P1=0.8, ~~20.55 N/mm*
The shear strength expressed as Vu/d are given
in Table 62 for different diameters and Shear capacity of concrete section = Q bd
spacings of stirrups, for two grades of steel. = 0*55 x 30 x 56.25 x 102/103=92*8 kN
SHEAR AND TORSION 175

Shear to be carried by stirrups, VU,==V, -~bd Referring to Table I, corresponding to
= 180 - 92.8 = 87.2 kN Mujbdz .= 2.05
V”,
-EL-- 87.2 = 1.55 kN/cm PI = 0.708
d 5625
Referring to Table 62, for steelf, -250 N!mme. A,, = O-708 x 30 x 5625/100 = 1 l-95 cm*
Provide 8 mm diameter two legged vertical Provide 4 bars of 20 mm dia (A*= 12.56 cm*)
stirrups at 14 cm spacing. on the flexural tensile face. As Mt is less
than MU,we need not consider Me, according
to 40.4.2.1 of the Code. Therefore, provide
only two bars of 12 mm dia on the compres-
Example I I Torsion sion face, one bar being at each corner.
Determine the reinforcements required for As the depth of the beam is more than
a rectangular beam section with the following 45 cm, side face reinforcement of 0.05 percent
data : on each side is to be provided (see 25.5.1.7
and 25.5.1.3 of the Code). Providing one
Size of the beam 30 Y 6Ocm bar at the middle of each side,
Concrete grade M 15
Characteristic strength 415 N/mm2 Spacing of bar = 53.412 = 267 cm
of steel 0.05 x 30 x 26.7
Factored shear force 95 kN Area required for each bar= .- ,oo
Factored torsional 45 kN.m = 040 cm*
moment
Factored bending moment 11S kN.m Provide one bar of 12 mm dia on each side.
Transverse reinforcement (see 40.4.3 of the
Assuming 25 mm dia bars with 25 mm cover, Code) :
d = 60 - 2.5 - ‘G - 56.25 cm
Area of two legs of the stirrup should satisfy
the following:
Equivalent shear,
Vc = V-I- 1*6(f ,
45
595-t 1*6x m = 95-l-240 = 335 kN
Equivalent shear stress. W---30 cm-
V, 335 x 101
%e = Fd = 3. x 56.25 >rlo2 = 1.99 N/mm*
From Table J, for M 15, ‘F~,,,,.~= 2.5 N/mm”
b, = 23 cm
k ,-FLEXURAL
TENSION
FACE
1 I
~~~is less than sc.,,,-; hence the section does
not require revision.
From Table 61, for an assumed value of
pt = 0.5,
T. = 0.46 N/mm* c T”=.
Hence longitudinal and transverse reinforcc- 7
ments are to be designed Longitudinal 6C1 cm
reinforcement (see 40.4.2 of the Code): Y&6cm d, 953-4
Equivalent bending moment, cm
Me,.= M,C Mt *
-m
1
= 115-J-79.4 -
*
= 194.4 kN.m
194.4x 10”
M,Jbd2 = 30 x (56.292 x 103 = 2.05 N/mma
176 DESIGN AIDS FOR REINFORCED CONCREFE

Assuming diameter of stirrups as 10 mm Nom-It is only a coincidence that the values of
Aav (@87/rllSv cdcuhted by the hvo cqlm-
da = 60 - (2.5 + l-O)-(2*5+0%)-53.4 cm tions 8rc the srmc.
b1=30-2(25+1.0)=23cm
Aav(0*87&J 45 x 10’
S” -23 x 53.4 x lOa Referring T&e 62 (forfr - 415 N/mm’).
Provide 10 mm + two legged stirrups at
+25 x9553.4
x IO?
x 10-366.4-l-71.2 12.5 cm spacing.
= 437.6 N/mm
P 438 kN/cm According to 25.5.2.7(u) of the Code, the
Area of all the legs of the stirrup should spacing of stirrupa shall not exceed xl,
satisfy the condition that A& should not (x, C yJ4 and 300 mm, where x1 and 1
arc the short and long dimensions of tL
stirrup.
From Table 61, for tensile reinforcement
percentage of @71, the value of o is O-53 xl - 30 - 2(2.5 - O-5)= 26 cm
N/mm’ y,r60-2(25-05)=56cm
(xl f y&/4 - (26 i- 56)/4 - 20.5 cm
- (I.99 - 0.53)
30 x 10 10 mm + two legged stirru at 12.5 cm
II 438 N/mm -438 kN/cm spacing will satisfy all the cod$ requirements.
SHEAR AND TORSION 177

f
ck
15
20
25 -
30
TABLE 61 SHEAR - DESIGN SHEAR STRENGTH OF CONCRETE, TC)N/mm2
35
fck, N/mm*
40 pt I - --.
,
IS 20 25 30 35 40
032 0.33 0.33 0.33 0.34 034

0.38 0.39 0.39 0.40 0.40 0.41
0.43 8:Z 0.45 045 046 046
0.46 0.49 0.50 0.50 0.51
0.50 0.51 0.53 0.54 0.54 0.55

0.53 0.55 056 0.57 0.58 059
0.80 0.55 0.57 0.59 iE 0.61 062
0.57 8:Z 0.62 8:; 0.65
% O%O 0.64 066 0.68
1.10 0.62 ~~~ 0.66 0.68 069 0.70

1.20 0.63 0.69 0.70 0.72 0.73
1.30 065 0.68 0.71 0.72 0.74 0.75
1’40 0.67 0.70 072 0.74 076 0.77
1.50 068 072 0.74 076 0.78 0.79
1*60 0.69 073 076 0.78 0.80 0.81

1.70 0.71 0.75 0.77 0.80 0.81 0.83
0.71 076 0.79 0.81 0.83 0.85
:z 0.71 077 0.80 0.83 085 086
2.00 0.71 0.79 0.82 084 0.86 088
210 0.71 0.80 0.83 086 0.88 090

220 071 0.81 0.84 087 0.89 0.91
2.30 0.71 0.82 0.86 088 0.91 0.93
0.71 082 0.87 092 0.94
f% 0.71 0.82 0.88 ;I: 0.93 0.95
0.71 0.82 X:E 0.92 094 097

E 0.71 0.82 0.93 0.96 0.98
E
3.00
x:::
0.71
@82
0.82
0.82
091
0.92
0.92
0.94
0.95
0.96
097
0.98
0.99
0.99
l*oO
1.01
178 DESIGN AIDS FGR RHNFGRCBD CGNCmB

'v
250
415
TABLE 62 SHEAR -VERTICAL STIRRUPS
Values of VW/d for two legged stirrups, kN/cm.
J, = 250 N/mm2 /x = 415 N/mm*

STIRRUP DIAMEIZR,mm DIAMETER,
mm
SPmNO, , -7
em 6 8 10 12 ’ 6 8 10
5 4373 6833 4083 7.259 11.342

3644 3403
3.124 6E f :% 9452
8.102
1537 2733 4271 t:;:: 4537 7089
l-367 2429 3.796 2269 4.033 6302
-1.230 2186 3,416 2042 3.630 5.671
1.988 3.106 4472 I.856 3.299 5.156 7424
1.822 2847 4100 1.701 3.025 4726 6806
1682 2628 3.784 1.571 2792 4363
1.562 1.458 2593 4051 L%
1.458 ::z Et 1.361 2420 3.781 5445
2135 3.075 1.276 2269 3.545 5.104
::E 2010 2894 1.201 2135 3.336 4804
1.215 1.898 2733 1.134 3.151
1.151 1.798 2589 1.075 .
:8:8 2985 t:;:;:
1.093 1.708 2460 1.020 1.815 2.836 4083
0.875 1.367 0.817 l-452
0.729 1.139 0681 1.210 :z
0625 0.976 0’583 1.037 1.620
0.547 0.854 0.510 0907 1.418
0.486 0.759 0.454 0.807 1.260
TABLE 63 SHEAR- BENT-UP BARS
Values of Vu, for singal bar, kN
BAR /i - 250 N/mm* fy = 415 N/mm2

Dm, I , I >
mm a = 45’ a=60° Q i= 45” a=60°
:; 1739
1208 21.30
1479 28.87
20.05 2456
35.36
:B” 30.92
3914 47.93
37.87 51.33
64.97 %E
20 48.32 5918 SO.21 98.23
5846 7160 _. __
97.05 118.86
75.49 9246 125.32
94.70 115.98 15720 :;;*z.
123.69 151.49 205.32 251.47
15654 191.73 25986 318-27
NOTE- a is the angle between the bent-up bar and the axis of the member.
8HEAR AND TOIlsION 179

5. DEVELOPMENT LENGTH
AND ANCHORAGE
5.2 DEVELOPMENT LENGTH OF positive moment reinforcement [see 25.2.3.3(c)
BARS of the Code] and for determining the length
The development length Ld, is given by of lap splices (see 25.2.5.2 of the Code).
Values of this development length for diffe-
Ld = ~+ es rent grades of steel and concrete are given
4 Tbd in Tables 64 to 66. The tables contain the
where development length values for bars in tension
as well as compression.
4 is the diameter of the bar,
a, is the stress in the bar, and 5.2 ANCHORAGE VALUE OF HOOKS
7bd is the design bond stress given in AND BENDS
25.2.1.1 of the Code. In the case of bars in tension, a standard
hook has an anchorage value equivalent to a
The value of the development kngth straight length of 16# and a 90” bend has
corresponding to a stress of 0937 fY in the an anchorage value of 84. The anchorage
reinforcement, is required for determining values of standard hooks and bends for
the maximum permissible bar diameter for different bar diameters are given in Table 67.
DEVIXWMENT LRNGTH AND ANCHORAGE

183
'v
250
415
f
ck TABLE 64 DEVELOPMENT LENGTH FOR FULLY STRESSED PLAIN BARS
15 Jj = 250 N/mm* for bars up to 20 mm diameter.

= 240 K/mm’ for bars over 20 mm diameter.
20
25 Tabulated values are in ccatimatns.
30 TENSIONBARS COMPRESSION
BARS
BAR GRADE OF CONCRETE GRADE OF CONCRETE
DIAMETER, f-A----, ,
mm Ml5 M20 M25 M30 Ml5 M20 M25 -M301
x 43.5
326 % 23.3
31.1 21.8
290 261
348 21.8 18.6 174
23.2
10 544 453 38.8 363 43.5 z:: 290
12 65.3 544 46.6 435 522 435 348
:: 97.9
87.0 725
81.6 621
699 653
58.0 696
78.3 65.3
5&O 497
559 zt
z 108.8
114.8 z.76 820
777 725
766 91.9
870 766
725 65.6
621 58.0
61.2
25 1305 108.8 93.2 870 104.4 87.0 746 696
;; 146.2 1392
121.8 1193
lW4 111.4
974 133.6
1169 974 95.5
835 %
36 :z:x 1566 1342 125.3 150.3 f:::: 107.4 10&2
NOTE -The development lengths given above ara for a stress of @87/y in the bar.
_
TABLE 65 DEVELOPMENT LENGTH FOR FULLY STRESSED DEFORMED BARS
fy PI 415 N/mm*
Tabulated values are in cantimetrcs.
TENSIONBARS COMPRESSION BARS

DIAMETER, C- A- -- I -
Ml5 M20 M25 M30 Ml5 M20 M25 M30
33’8 28.2 242 22.6 271 193 18.1

45.1 37.6 322 391 361 25.8 241
564 47-o 37.6 45.1 322 30.1
677 564 45.1 542 38.7 361
75.2 72.2 51.6 48.1

1% 846 z:: 81.2 58.0 542
112.8 940 SC
124-l 103.4 827 z*: %Z %“z
141.0 117.5 100.7 940 112.8 94.0 806 75.2

1580 1128 105.3 126.4 105.3
1:A:: 128.9 120.3 144.4 120.3 1E 2::
180.5
203.1 169.3 145.0 135.4 162.5 135.4 1161 108.3
Nora-The development lengths given above are for a stress of 087fy in the bars.

TABLE 67 ANCHORAGE VALUE OF HOOKS AND BENDS
Tabulated values are in centimetres.
BAR DIAMETER,
mm 6 8 10 12 16 18 20 22 25 28 32 36
Anchorage Value of
hook 9.6 128 160 192 25.6 28.8 320 35.2 40.0 448 51.2 57-6
Anchorage Value of
90” bend 4.8 6.4 8-O 9.6 12.8 14.4 16.0 17.6 20-O 22.4 25.6 28.8
-I---
4 cb mir
_L_-
STANDARD HOOK STANDARD 90’ BEND
STANDARD HOOK AND BEND
Type of Steel Min Valye of k

Mild steel 2
Cold worked steel 4
NOTE 1 -Table is applicable to all grades of reinforcqment bars.
No,rli 2 - Hooks and binds shall conform to the details given above.

6. WORKING STRESS DESIGN
6.1 FLEXURAL MEMBERS Reinforcement percentage Pt,bal for balanced
section is determined by equating the com-
Design of flexural members by working pressive force and tensile force.
stress method is based on the well known
assumptions given in 43.3 of the Code. _ kdb _ PI,~I bd as1
a,h
The value of the modular ratio, m is given by 2 100
280 93.33 50 k.a,a
m E-E- hbal = a t
3 acbc acbc .
Therefore, for all values of acb we have
The value of pt,w for different values of
m acbc = 93.33 a,bc and a,t are given in Table L.
b;bc
TABLE L PERCENTAGEOF TENSILE
7
-. REINFORCEMENTP..,,., FOR SINGLY
REINFORCEDBALANCEDSECTION
.,“_.
(Clause 6.1.1)
esl N/mm*
$&ma L
,
Gil 230 275
l-k
5.0 0.71 0.31 O-23
T-1
7.0 l-00 0.44 0.32
FIG. 9 BALANCEDSECTION(WORKING 8.5 1.21 053 0.39
STRESSDESIGN)
10.0 1.43 0.63 O-46
6.1.1 Balanced Section (see Fig. 9)

6.1.2 Under Reinforced Section
Stress in steel = ast =maLbc(+-1)
The position of the neutral axis is found
by equating the moments of the equivalent
areas.
1 pt bd
-= bkdz = loo m (d - kd)
k 2
93.33
k= bd2 7 = bd2 ‘$ (1 - k)
as( f- 93.33
The value of k for balanced section depends
only on qt. It is independent of a,bc. Moment k2 = p$(l - k)
of resistance of a balanced section is given
by hfbal = yach k( 1 - f );The values k2 + !$- !!!=o.
of Mbal/bd2 for different values of U&c and The positive root of this equation is given by
asI are given in Table K. k = -- Ptm + p2,m2 + ptm
100 J F (100)” - 50
TABLEK MOMENTOF RESISTANCEFACTOR

M/b@,N/mm*FOR BALANCED This is the general expression for the depth
RECTANGULARSECTION of neutral axis of a singly reinforced section.
Moment of resistance of an under-reinforced
=cbc ‘%I,N/mm* section is given by
N/mm’ c +‘-
140 230 275
5.0 0.87 0.65 0.58
7-o 1.21 0.91 O-81
8.5 147 1.11 0.99
10-o 1.73 1.30 l-16 Values of the moment of resistance factor
M/bd2 have been tabulated against pt in
WORKINGSTRm DESIGN
189
x (1 -;)a&
To& tensile reinforcement A,l is given by

Ast = AM, -f- Astt
FIG. 10 DOUBLYR~r~~onci?n SECTION where Atu = pl,bPI &f

(WORKINGSmm.s D~SGN) 100
and A,Q =
Tables 68 to 71. The Tables cover four
grades of concrete and five values of uu.
The compression reinforcement can be ex-
6.1.3 Doubly Reinforced Section - Doubly pressed as a ratio of the additional tensile
reinforced sections are adopted when the reinforcement area Altp.
bending moment exceeds the moment of
resistance of a balanced section.
M=b&i-M’
USI 1
The additional moment M’ is resisted by = Qcbc (1.5 m - 1) (l-d’/kd)
providing compression reinforcement and
0 additional tensile reinforcement. The stress
in the compression reinforcement is taken Values of this ratio have been tabulated for
as I.5 m times the stress in the surrounding different values of d’/d and ucbcin Table M.
concrete. The table includes two values of ust. The
Taking moment about the centroid of values of pt and pc for fear values of d’/d
tensile reinforcement, have been tabulated against’ M/bd’ in
Tables 72 to 79. Tables are given for four
grades of concrete and two grades of steel.
M’ = PC bd(1.5 m - 1) ucbc
-ldc
TABLE M VALUESOF THE RATIO A&,,,

(Clause6.1.3)
%t “cbc d’ld
I +& (1.5m - 1) uck N/mm’ N/mm’ m--
0.15 0.20
5.0 1.19 1.38 l-66 2.07

x(1 -$)(I-;)bd’ 1.20 I.68 2.11
140 I ;:y l-22 % 1.70 213
i 10.0 1.23 l-44 1.72 2.15
Equating the additional tensile force and

dditional compressive force, ;:g 2.65
2.61 3.60
3.55 5.63
5.54
214
2.12 2.71
2.68 ;I$ .
:*:I:
k$! (1.5 m - l)U&( 6.2 COMPRESSION MEMBERS

Xi l-2)
Charts 86 and 87 are given for determining
the permissible axial load on a pedestal or
or (pt - pt,bd olt short column reinforced with longitudinal
bars and lateral ties. Charts are given for
=pc (1*5m-l)ucbc two vrdues of 0%. These charts have been
(l-ii) made in accordance with 45.1 of the Code.

According to 46.3 of the Code, members Tables 81 and 82 are given for design of
subject to combined axial load and bending shear reinforcement.
designed by methods based bn elastic theory
should be further checked for their strength 6.4 DEVELOPMENT LENGTH AND
under ultimate load conditions. Therefore
ANCHORAGE
it would be advisable to design such members
directly by the limit state method. Hence, The method of calculating development
no design aids are given for designing such length is the same as given under limit state
members-- by elastic theory. design. The difference is only in the values
of bond stresses. Development lenaths for
6.3 SHEAR AND TORSION plain bars and two grade; of deformed bars
are given in Tables 83 to 85.
The method of design for shear and torsion
by working stress method are similar to the Anchorage value of standard hoolcs and
limit state method. The values of Permissible bends as given in Table 67 are applicable
shear stress in concrete are given in Table 80. to working stress method also.
WORKlNC3 STRESS DESlCiN 191

%t
130
140
IWORKING STRESS METHOD
190
230
TABLE 68 FLEXURE - MOMENT OF RESISTANCE FACTOR,
SINGLY REINFORCED SECTIONS
iU/bda, N/mm* FOR
275
.
50
130
0,146
O-158 x:2:
190 230
O-258
275
pt
0.47
0.48
f
130
0.542
O-553
140
est. N/mma
0.583
0.595
190 230 L-
275
O-170 O-248 Ei. O-49 O-564 O-607

O-181 O-265 O-321 0.50 O-574 O-619
O-193 O-282 O-341 O-51 0.585 0.630
0.205 O-299 O-362 @52 0.5% O-642

0.216 O-316 O-383 O-53 O-607 O-654
O-228 O-333 O-54 0,618 O-665
.
Es 0.55 0.629 0x77
.
8’::: .
x’::; 0444 0.56 OTi40 O-689
O-262 O-383 O-650 O-700

0400 O-661 O-712
::z 0.417 0:672 0.724
o-297 0.433 O-683 0.735
O-308 O-450 O-693 O-747
O-319 0.467 0.62 0.704 O-758

O-331 O-483 0.63 O-715 O-770
O-342 Oao 0.726 O-781
O-353 @516 x:s O-736 O-793
0’364 @533 O-66 0.747 O-804
O-376 O-758
O-387 0.768 X:iE
O-398 O-779 0’839
O-850
:z X:E O-862
0431 0.631 O-811
o-443 0.647 O-821
O-663
ez O-679 .
:ii;
O-476 0.695 O-853
O-487 O-711 O-77 O-864
O-498 0.728
z; .
8’0;:
O+O 0.895
%t
130
140 1 WORKING STRESS METHOD 1
190
230
275
TABLE 69 FLEXURE - MOMENT OF RESISTANCE FACTOR, M/hP, N/mm* FOR
%bc
.
70 us:, N/mm*
pt
130 140 190 230 275- 130 140 190 230 27i
O-242 0.428 O-76 O-869 O-936
O-266 0.470 O-77 @880 0948
O-289 0.511 0.78 0.891 O-960
@552 O-79 0.971
.
:::: O-593 0.80 8ZZ O-983
O-358 O-523 O-633 O-757 O-81 o-994
O-381 O-557 O-674 O-806 O-82
:zi .
x’z!i z::
0.83 F%
1.029
O-449 O-657 O-795 8:: l-041
O-690 O-835 O-86 l-052
.
8Z O-723 O-87 -l-064
__.
0.739 x::;: O-88 l-075
iFg O-89 l-087
. .
X’Ei O-90 l-099
O-539 O-581 l-031 l-110
O-551 O-593 :z l*!Ml l-122
O-562 @605 O-821 l-052 l-133
O-573 O-617 0837 l-145
O-584 0.629 O-854 :iE l-156
O-641 O-96 l-084 1.16

O-653 O-97 1.179
O-98 ::z: l*l!bl
:zE 099 1.116 1.202
OS89 1.00 l-127
O-651 @701 O-951

O-713 O-967 ::::I
:z. O-983 1.158
x!. l-169
:z O-748 SE l-180
0706 O-760
O-717
O-728 :z
0.739 O-795
O-750 0807
O-761
K&
.
O-804
0.815 O-878
.
X’E .
t’E
Om8 O-913
O-859 O-925
1% DElION AIDS FOR RRINKRkCRD CDNCRR’I’B

130
1 WORKING STRESS METHOD ( 140
190
TABLE 70 FLEXURE-MOMENT OF RESISTANCE FACTOR, hf/bd¶, N/mm* FOR
230
275
%bc ’
a, N/mm’
pt
’ 130 140 190 230 275-
pt
’ 130 140 190 230 275 .
85
O-244 0.262 O-431 O-515 0.96 1.096 1.180
0.288 x::;: O-473 O-565 0.97 1.192
8::;: O-313 O-425 @514 0.615 0.98 x: 1.203
O-314 O-459 O-556 O-664 O-99 1.128 1.215
O-337 Ei O-493 0.597 O-714 1.00 1.139 1.227
O-30 0.361 O-388 O-527 O-638 O-763 1.238
O-32 O-394 O-413 O-561 0.679 O-812 ::E 1.250
0.34 O-407 O-595 O-720 O-861 1.171
O-36 O-430 tee O-628 0.761 1.182
O-38 O-453 @488 O-662 O-801 E 1.193
O-476 O-512 Ow2 1.203

O-498
O-521 x:::: ::tiz ::z
O-544 O-586 O-962 1.236
O-567 0.610 l-002 1.246
O-50 O-589 O-634 1.042 1.11

0.52 O-612 O-659 l-082 1.12
O-54 O-634 0%83 1.13
056 0.707 1.14
O-58 x:::3 0.731 1.15
0.701 l-025 1.16 1.310 l-411

0.723 l-057 1.17 l-321 l-423
0.746 l-090 1.18 1.332 1.434
O-768 1.19
O+Ml ::i:: 1.20 ::E: .
:‘%
0.70 0.812 O-875 l-187 1.21 1.364 1468
O-72 O-834 O-898 l-219 1.22 1.374
O-74 0.856 0.922 1.23
O-76 0.878 0946 :::;:
0.78 0900 0.969 E 1406
O-80 0993 1.26 1.417
O-82 XE 1.427
O-83 O-955 .
:% ::f; 1.438
O-84 O-966 i-29 1448
0.85 0.977 :zE 1.30 1.459
0.987 1.31 1469
8:: 0998 1,480
O-88 ltM9 :::z 1.491
O-89 l-020 1a501
090 l-031 .
:*:: I.512
l-042 1.122 1.36 1.522
l-134 1.37 1.533
E:. 1.145
l-074 l-157
l-085 l-169
WORKING STRESSDESIGN 197

%t
130
140
190
230 I TABLE 71 FLEXURE - MOMENT OF RESISTANCE FACTOR, M/bd2, N/mm2 FOR
275 SINGLY REINFORCED SECTIONS
WC = 10.0N/mm’
%bc USI,N/mm*
Pl
.
100 230 ’ 130 140 190 230 27;
O-433 I.10 1.257
0.475 1.12 l-279 %i
O-517 1.14 1.301 1.401
I.16 1.322 1.424
X:E 1.18 l-344 1’447
OTi42 1.365 1.470
O-683 1.387 1.494
O-724 1.408
l-430 E.
ALE l-451 l-563
0.847 1’30 l-473 1.586
1’31 l-483 l-597
.
x’% l-32 l-494
O-969 l-33 l-505 :z
1*009 1’34 l-515 1,632
1.049 l-35 1.526 1.643
l-537 l-655
;:g :::4 1.547
1.38 l-558 t%i
.
l-210 l-39 1.569 l-689
1.250 1’701
l-289 ;:g
. t ::i:
1.43
l-44
1.45 l-632
1.46 1.643
1.47 1.653
l-48 l-664
1.49 1.675
1.50 1.685
1.51
1.52 EE
.
1.53 l-717
1.54 l-727
135 1.738
1.56 l-749
1.759
:z. 1.770
1.59 l-780
l-60 l-791
DESIGNAIDS FOR RBINFORCEDCON-

1 WORKING STRESS METHOD 1
TABLE 73 FLEXURE - REINFORCEMENT PERCENTAGES FOR DOUBLY

REINFORCED SECTIONS
6dc = 7.0 N/mm2
qt = 140 N/mm2
d’Jd is 0.05 d’/d pi O-10 d’ld = 0.15 d’/d = 0.20

MJbd a
I 3 r- -A-------\
N/mm2
pt PO . pt PO- pt PC p, PC-
1.22 1a05 oaK lTM6 1.006 0.013
1.25 l-028 O-033
O-078
l-031
l-073
X:E
O-123
1.033
l-077
oa69
O-163
:::i KG O-124 1*:15 O-193 l-122 O-257
140 l-140 O-169 l-157 0.264 1.167 0.351
1.45 1.178 1.188 0.264 I.199 0.335 1.211 0.445

150 1.216 1228 O-319 1.241 1.255 0539
1.55 1.253 l-267 O-375 1.283 :% l-301 0.633
l-60 1.291 0.431 l-325 0.547 1.345 0.727
l-65 l-328 ::z:: O-486 1.367 O-618 1.390 0.821
1.70 1.386 1409 0.689 1.435 0.915

::z ::z 1.426 :::ii 1.451 O-760 1.479 1009
::ii 1441 O-531 1.466 0,653 1.493 0.830 1.524 l-103
l-85 1.479 0.576 1.505 0.709 1.535 0901 l-568 1.197
1.90 l-516 0.621 1.545 0.765 l-577 0.972 l-613 l-291
1.95 O-666 1.585 O-821 1.619 1a43 1.658 1.385

2.00 0.712 1.624 0.876 1.661 1.113 1.702 1.479
2.05 O-757 O-932 l-703 1.184 1.747 1.573
2.10 @802 Kz O-988 l-745 1.255 1.792 1.667
2.15 O-847 I.743 l-043 1.787 I.326 1.836 1.761
l-783 1.099 1.829 I.396 1.881 1.855

z : ::;3 Ef 1.823 1.155 1.871 1.467 1.926 1.949
2.30 0.983 l-862 1.913 l-538 1.970 2.043
2.35 ::i:: l-028 1902 ::ti: l-955 l-609 2.015 2.137
240 1.892 l-073 l-942 1.322 l-997 1.680 2.060 2.231
245 1.119 1.981 1.378 2.039 1.750 2.104 2.325

2.50 t:;zt 1.164 1.433 2.081 1.821 2.149 2.419
255 2005 l-209 Et 1.489 2123 l-892 2.193 2.513
2.043 2101 1.545 2165 l-963 ’ 2.238 2.607
E 2080 1::;; 2140 l%OO 2-207 2.033 2283 2701
2-70 2-118 l-345 2180 1.656 2.249 2.104 ‘2.327 2.795

2’75 2.155 1.390 1.712 2.291 2.175 2.372 2.888
280 2193 l-435 3% 2.333 2.246 2.982
285 2231 l-480 2.299 :34: 2375 2.316 ::t :: 3.076
290 2268 1.526 2339 1.879 2-417 2387 2506 3.170
295 2306 l-571 2378 1.934 2.459 2.458 2.551 3.264

2343 l-616 2501 2-529 2-595 3.358
::z 2381 1.661 3% ::Ez 2.543 2.599 3.452
3.10 2.419 1a707 2497 2.102 2585 2.670 ;zg 3.546
3.15 2.456 1.752 2-537 2.157 2.627 2741 3-540
2.494 1.797 2.577 2.213 2.669 2.812 2.774 3.734

::z 2531 I.842 2.616 2.269 2.711 2883 2818 3.828
2569 2.656 2324 2.754 2.953 2.863 3.922
:::: f:E 2696 2.380 2796 3.024 2908 4.016
f:fg 1.978 2-735 2.436 2.838 3.095 2-952 4.110
340

1 WORKING !3T@ESS METHOD 1
TABLE 80 SHEAR - PERMISSIBLE SHEAR STRESS IN CONCRETE, sc, N/mm*
M30 M35 la40
o-21 O-21 O-21

O-25 O-25 0.25
E LE iz
033 O-34
O-36 :3$
O-38 XG O-39
0.39 040 O-41
o-41 O-42 O-42
0.43 O-43 O-44
O-45 O-45
E 0.47
E O-48
gz O-49 O-49
0.49 o-50 851
O-51 O-52
8:: 0.52 0.53
O-52 0.53 O-54
O-53 O-54 0.55
O-54 O-55 O-56
0.54 O-56 @57
0.55 O-57 O-58
@56 0.59
o-57 iE: O-60
O-59
x::: 8::
O-59
O-59 :::i
TABLE 81 SHEAR - VERTICAL. STIRRUPS
Valuesof 9 for hvo leggedstirrups,kN/cm
Gv 3: MN/mm* UN - 230N/mma
STIRRUP Dmmnm, mm I)l-. mm
sPAuNo,
col
:
7
x
16
::
13
l-979
1.863
I -759
1.667
l-583
1.267 2081
l-056 1a734
0905
f’ii!
:z 1:156
WOltKlNO STRBSSDSlCiN 207

1 WORKING SfRESS METHOD 1
TABLE 82 SHEAR - BENT UP BARS
Valuewf V, for single bar, kN
%V = 140 N/mm* up to 20 mm diameter

= 130 N/mm’ over 20 mm diameter %v=230 N/mm’
, b r-
0=4S0 a=60° a=450
778 9.52 1277

1120 13.71 iii*%3
lP90 2438 3270
25-19 3086 41.39
31.10 38.09 51.09
61.82
7Pa3
E . 6P32
PO-54
100-14
130-80
93.57 11460 16554
NOTE- a is the an6 between the bent up bar and the axis of the member.
TABLE 83 DEVELOPMENT LENGTH FOR PLAIN BARS
us,= 140 N/mm* for bars up to 20 mm diameter

= 130 N/mm* for bars over 20 mm diameter
0IE = 130 N/mm’ for all diameter
TENSIONBARS tikXlPRE%SlON BARS

Gnrse OF CONCRETE GRAIX OF CONCRETEi
D&R. , 3
mm Ml5 M20 T ‘M1S MU) M2S M30
6 35-o 263 23.3 21.0 z:(: lP-5 17.3 15.6
8 35.0 31.1 28-O 23-l 208
f8 z:
70-O :s:: 46.7
38-P 42.0
35.0 43.3
520 E3
3PO z:; 26-o
31.2
:; 105.0
93-3 78-a
70.0 622
70.0 63.0
56-O 78-o
6P3, ;:3 46.2
520 41.6
46.8
z 1167
llP2 875
w-4 77.8
7P4 71.5
70-O 993
867 E 63.6
57-a .
:27p
25
28 151.7
135-4 101.6
113-a 1Z 91-o
81.3 108.3
121.3 - 81.3
91-o za z::
:: 195.0
173.3 146.3
130-O 1300
115.6 117-o
104-o 1560
138.7 1170
104-o 1E 93.6
83.2
208 DESIGN AIDS FOR REINFORCED CONCRBIg

WORKING STRESS METHOD
I
TABLE 84 DEVELOPMENT LENGTH FOR DEFORMED BARS

%t = 230 N/mm*
SC = 190 N/mm’
TENSIONBARS COMPREWON BARS

DIAMETER, PA_
mm xi----- M20 M25 M30 ’ TiiY M20 M25 M3d
41.1 30.8 27-4 24.6 27.1 20.4 18.1 16.3

: 54.8 41.1 36.5 329 36.2 27.1 24-l 21-7
10 68-5 51.3 45.6 41.1 45.2 33.9 302 27.1
12 82-l 61.6 548 49.3 ,54*3 40.7 36.2 32-6
16 109.5 82-l 73.0 65.7 724 54.3 48.3 43.4
18 123.2 92.4 82.1 73.9 81.4 61.1 54-3 48-9
136-9 1027 91.3 82.1 67-9 EC: 54-3
150.6 1129 100.4 90.4 z*: 74.6 597
171’1 128.3 114.1 1027 113.1 84.8 75.4 67-9
191.7 143.8 127.8 115*0 126.7 95.0 84.4
219.0 164.3 146.0 131.4 144.8 108.6 96.5 L!&!
2464 184.8 164.3 147.9 1629 122.1 108.6 97.7
TABLE 85 DEVELOPMENT LENGTH FOR DEFORMED BARS

Qll = 275 N/mm*

0,=19ON/mm
TENSIONBARS
_.___.___ COMPRESSION
BARS
DLAMEI-ER, ,_-__h_ > \
mm Ml5 M20 M25 M30 M20 M25 M30
49-l 36.8 327 29.5 27.1 20.4 18.1 16.3

:::; 491
61.4 43,7
54.6 49.1
39.3 36.2 27.1 241 21.7
45.2 33.9 30.2 27-l
98.2 73.7 65.5 58.9 54-3 40.7 36.2 326
16 131-o 98.2 87.3 78,6 72.4 z: 48-3 43.4

18 147.3 ::I; 98.2 88.4 81.4
163.7 109*1 98.2 67.9 z:: E
;: 180.1 135.0 120.6 108.0 E: 74-6 66.3 597
% f 171.9
153.5 152.8
136.4 1228 113.1 848 75.4 67.9
126.7 95.0 84.4 76.0
261.9 196.4 174.6 z: 144.8 108.6 96.5 86.9
2946 221.0 1964 176.8 162.9 1221 108.6 97.7
WORKXNQ STRBSS DESIGN 209

7. DEFLECTION CALCULATION
7.1 EFFECTIVE MOMENT OF The. chart takes into account the condition
INERTIA
4 > 1. After finding the value of Zd it has
I
A method of calculating the deflections is
given in Appendix E of the Code. This to be compared with Z* and the lower of
method requires the use of an effective the two values should be used for calcula-
moment of inertia I& given by the following ting the deflection.
equation For continuous beams, a weighted average
Z* value of Z~lr should be used, as given in
I&r - B-2.1 of the Code.
1.2-s; 1-2 +_
( )
but, Ir < Za < b 7.2 SHRINKAGE AND CREEP
Whrn DEFLECTIONS
Ir is the moment of inertia of the cracked
section ; Deflections due to shrinkage and creep can
is the cracking moment, equal to -fcllll and B-4 of the Code. This is illustrated in
be calculated in accordance with clauses B-3
Mr
Yt
where Example 12.
fa is the modulus of rupture of con-
crete, Zmis the moment of inertia of
the gross section neglecting the re- Example 12 Check for deflection
inforcement and yt is the distance Calculate the deflection of a cantilever
from the centroidal axis of the beam of the section designed in Example 3,
gross section to the extreme fibre in with further data as given below:
tension ; Span of cantilever 4.0 m
M is the maximum moment under service Re\didimoment at service 210 kN.m
loads;
z is the lever arm; Sixty percent of the above moment is due to
d is the effective depth; rmanent loads, the loading being distri-
x is the depth of neutral axis; %uted uniformly on the span.
b, is the breadth of the web; and
b is the breadth of the compression face. BP 300 x @O)* _ 5.4 x 10’ mm’
ZE =-i-T= 12
The values of x and z are those obtained
by elastic theory. Hence z = d - x/3 for From clause 5.2.2 of the Code,
rectangular sections; also b = b, for rec-
Flexural tensile strength,
tangular sections. For flanged sections where
the flange is in compression, b will be equal fcr= 0.74 z N/mm9
to the flange width br. The value of z for
flanged beams will depend on the tlange fcrP O-7 t/E = 2.71 N/mm’
dimensions, but in order to simplify the
calculations it is conservatively assumed the Yt -D/2=~=3OOmm
value of z for ganged beam is also d - x/3. 2.71 x 5.4 x 10’
With this assumption, the expression e&c-
tive moment of inertia may be written as
follows : - 488 x 10’ N.mm
- O-067
a’/d II 005 will be used in referring to Tables.
From 5.2.3.Z of the Code,
but, F
> 1 EC = 5700 q/fck N/mm*
r
and Zen< Zm I 5 700 d/13= 22-l x 10’ N/mm*
Chorr 89 can be used for finding the value of A?& P 200 kN/mm* = 2 x 10s N/mm%
F in accordance with the above equation.
I
DEFLECTION CALCULATION 213

From Example 3, Deflection due to creep,
p, = 1.117.p, =0.418 a,, (pcrm)= a,,, (p,r,nj- a, ,,,cmr,
p,(m - I)/@, m) = (0.418 X 8.05)/ In the absence of data, the age at loading
(1.117 X 9.05) = 0.333 is assumed to be 28 days and the value of
PJ?? = 1.117 x 9.05 = 10.11 creep coefficient, 8 is taken as 1.6 from
Referring to Table 87, 5.2.5.1 of the Code.
I,/(bd’/ 12) = 0.720
.. I, = 0.720 X 300 X (562.5)‘/ 12 EC, = E,
= 3.204 X IO9 mm4 1 +e
Referring to Table 91, = 22.1 x IO3
1 + 1.6 = 8.5 X 10’ N/mm2
J = 0.338
E, 2x IO5
= 23.53
Moment at service load, M = 210 kN.m m = z = 8.5 X lo3
= 21.0 X 10’ N.mm p, = 1.117, pL = 0.418
4.88 X 10’ pc (m - l)/(p,m) = 0.418(23.53 - I)/
Mr/ M =
21.0 x lo’= o.232 (1.117 X 23.53)
Referring to Chart 89. = 0.358
I,,,/ I, = 1.0
. Ierr = I, = 3.204 X IO9 mm’ Referring to Table 87,
For a cantilever with uniformly distributed t,/(bd’/ 12) = I,.497
load, I, = 1.497 X 300 (562.5)3/ 12
2 = 6.66 X lo9 mm”
Elastic deflection = f .g I, < Lrr d I$q
cll
6.66)X 10” d I,,, < 5.4 x IO9
z -__---2 1.o X 10’ x (4000)? .*. Ierr = 5.4 X 10’ mm4
4 x 22.1 X lo3 x 3.204 X 10”
. ..( 1) alcc (,,rr,,r,= Initial plus creep deflection due to
= I I.86 mm
permanent loads obtained using the
Deflection due to shrinkage (see clause B-3 Above modulus of elasticity
of the Code): 1 Ml2
---
IILo= k+ Vv, I‘ = 4 E&r
ki = 0.5 for cantilevers (0.6 X 21 X 10’) (4 000)2
= $X
p, = l.l17,p, = 0.418 8.5 X IO3 X 5.4 X IO9
pi--p<= 1.117-0.418=0.699< 1.0 = 10.98 mm
aI (pwn\ = Short term deflection due to
... ~4=0.72Xy& ’
permanent load obtained using EC
Pt
1 (0.6 X 21 X 10’) (4 000)’
= 0,72 x (1.1.17 - 0.418) =i-x- 22.1 X 10’ X 3.204 X IO9
fii-iT = 7.12 mm
==0.476 ... a‘r(pc.,m)= 10.98 - 7.12 = 3.86 . ..(3)
In the absence of data, the value of the . . Total deflection (long term) due to initial
ultimate shrinkage strain &, is taken as load, shrinkage and creep
0.000 3 as given in 5.2.4.1 of the Code.
= 1 1.86 + 1.90 + 3.86 = 17.62 mm.
L)=6OOmm
According to 22.2(a) of the Code the final
.’. Shrinkagecurvature q\Ir,,= k4 g deflection should not exceed span/2SO.
. ‘&!I?!!?
= 0.476 X 0.000 3 = 2 38 x 1o-7 Permlsslble deflection = 250 = 16 mm.
600 The calculated deflection is only slightly

a,, = 0.5 X 2.38 X 10e7 X (4 000)2 greater than the permissible value and hence
= 1.90 mm . ..(2) the section may not be revised.

i
w
t
ii
L
1.0
10
RATIO bf/bw

Chart 89 EFFECTIVE MOMENT OF INERTIA FOR
CALCULATING DEFLECTION
f
i-
b-0
3i
1.0
1.1 L_;______;-__
_;-c_$ IY
I I I I I I hi--
- 1
I I I i I I I I I I
I I Al I I I I I%++-II“8
l-4
l-5
1.6
1.7
1.8
l-9
2~0"""""""'~""""""~
Chart 90 PERCENTAGE, AREA AND SPACING ‘OF BARS
IN SLABS
29
24
23
22
21
20
19
10
0 1 2 J b s 9 1 D 9 10 11 12 13 14 19
AREA OF REINFORCEMENT cm’ PER METER WIOTH
+USf ECfCCtlVL OCfTM OR OVERALL WHICHEVER IS USLO POR CALCUlATING p
DBFLECTION CALCULATION 217

.
Chart 91 EFFECTIVE LEhlGlH OF COLUMNS-

Frame Restrained Against Sway
FUED 0
0.6 0*7 0.8 0.9 l*O
P
E Pa 8
ii z
r
BXand Paare the valuesof 19at the top and bottom of the column when, p-
sKc the summation being
done for the members framin8 into r joint; KC and Kb arc the fkxural r;K, + tKb ’
stiffacosesof column and &m mpiwfy.
218
DESk3N AIDS FOR REINFORCED CONCRHI
chrrt 92 EFFECTIVE LENGTH OF COLUMNS -
Frame Wiiut Restraint to Sway
0*9
04
0.6
P1
FIXED 0
o\ 0.1 0.2 0.3 04 03 0*6 0.7 0.8 0.9 1-O
wp
x
P2
ii
hand @,are the values of b at the top and bottom of the column. where b’~Kc~sKb,
tbe swnmation
being
done for tbo mombem framing into a joint; KC and Kb are the llex~rd dfhessu of &Umn and beamrespeCtivclY.
DFMACMON CALCULATION 219

TABLE 86 MOMENT OF INERTIA- VALUES OF M/12 000
b, cm
d. cm
15 20 25 30 35 40 45 50
10 1.2 4-2
l-7 ;; ::: ::g 5.5
5-8 6-5
$; ;:; ;‘2
46 ;:; 1:: 11.4
11.3 14-l
g:‘: g 1rt :::: 17-l
10-2 14-3 z 18.4
12-l 146 17-O 19-4 21.9 z
143 17-l 20-O m9 25.7 28.6
10-O 23.3 33.3
27-o 38-6
:::4 31.1
15.2 35.5 iii::
17.3 40.3 57-6
19.5 326 45.6 52-l 58.6 65-l
220 36-6 51.3 65.9 73-2
24-6 41-o :!:f 82-O
27-4 45-7 z:t 73-2 i::!! 91.5
30-5 50-8 71-l 81.3 91.5 101.6
33.8 45.0 67-5 78.8 90.0 101.3 1125
41.0 546 z:: 81.9 95.6 109.2 122.9
49.1 65.5 81.9 98.3 114.6 131.0 147-4 %5 .
58.3 77’8 97.2 136-l 175.0 194-4
68.6 91-5 1143 fK4 160.0 :z . 205.8 228.6
53-3 106-7 186.7 213.3 2400 266.7
61.7 it: 123.5 ::::: 216-l 277.8
1420 177-5 248.5 E8 3194 :::I:
E :z3 1622 202-8 283.9 324-5 365.0 405-6
922 138-2 :843 23@4 322.6 368.6 4147 460-8
lW2 208-3 260-4 3125 364.6 416-7 5208
117-2 :f:; 234.3 2929 351.5 410-l E:
131.2 196.8 393.7 459-3 iii:; 590-5 i%;
146-3 219-5 i;::‘: ~%I: 439-o 5122 585-4 658.6 731.7
1626 243-9 325.2 406.5 487.8 569-l 650-4 731.7 813-O
180-O 270-O 360.0 450.0 5400 630.0 8AO.O

343.3 5721 686.6 801.0 ES . 10298 1E
%:X 428.7 ::::7’ 7146 857.5 1286-2
351.6 527-3 703.1 878.9 10547 fzz tz:: 15820 :%-ii
4267 640.0 853.3 1066-7 1280-O 1493’3 1706-7 192@0 2133:3
511.8 767.7 1023.5 1279-4 1791.2 2047-l

607-s 911’3 1518.8 z:: 2126-3 %T .
7145 1071.7 %X 1786-2 2143.4 ii% . 3572’4
833.3 1250.0 1666-7 2083.3 2500.0 K:: 3333’3 4166-7
220 DEMON AIDS FOR REINFORCED CONCRETE

TABLE 87 MOMENT OF INERTIA OF CRACKED SECTION -
VALUES OF &/(R”)
.
0-O O-3 0.8 1.6
O-100 O-100
11::;1;: Ki
O-226 0229
O-264 0269
O-298 0.310
O-348
:::z O-386
0398 O-424
O-430 0460
O-472 O-490 0.4%
iz O-525 0332
FiTi 0.559 O-567
.
x’:;; O-596 zi o”:zz
O-601 O-625 O-670

O-628 O-654 .
8’:;; O-704
@653 O-682 O-723 O-738
O-678 O-710 O-755 0.771
O-703 0,738 O-787 0.804
0765 O-818 O-837
O-792 0.850
O-818 0.880 x:ii;
O-844 O-911 0.934
O-870 0942 0.966
0.839 0998
O-860 l-030
O-881 l-061
1.093
:zi l-124
1.121 1.155
l-179 l-217
I.278
.
:-zz l-340
l-351 1400
l-123 1461
l-156 ;:g l-521
l-188 l-581
1.220
l-250
l-575
l-630 ::z
l-685 l-758
l-739 l-817
l-876
::z 1.934
1902 1993
DEFLECTIONCALCULATION 221
VALUES OF h,(z)
0.0 @l O-3 04 O-6 O-8 1.0

O-100 &lOO 0.100 O-100
E O-144 0.144 O-144 O-144
O-185 O-185 0.185 O-186 Q186
O-224 O-225 0225 O-226
O-262 O-263 O-263 0264 ::z
0.298 O-300 O-300 0.302 O-303
E! O-333 0.335 @336 0338 O-340 F$$
O-366 O-367 0.370 O-371 O-373 O-376
O-398 0403 0405 O-411 0.413
O-430 Et! O-436 O-438 zz O-445 O-448
O-463 0468 0.470 O-479 0.483
O-493 O-499 O-502 O-512 o-517
0523 O-530 0.550
O-551 O-560 8:::: 8:::: 0.583
O-580 O-589 O-593 0.609 O-616
O-607 O-618 0.622 O-632 O-648
O-634 o-651
::zE 0680 z: ::tE!
;:z O-701 O-708 O-720 0.743
o-71 1 O-727 0735 0.749 O-774
0.727 O-792 0.805

8:::; O-836
X:E O-815 z:: O-866
O-795 O-841
O-818 0867 Ez. f:E
O-839 O-854 0956

O-860 O-876 I:E :z: O-986
0881 0.898 0.993 l-015
8E l-021
Ez :E l-022 lft49 fZ
o-942 0962 0999 l-077 1.103

0.980 l-045 l-131 1.160
1.018 EZ l-185 l-217
l-082 :d~ l-239 1.274
1:E 1*120 l-176 1.292 I.330
l-123 1.157 1.218 l-247 l-298 l-344 1.386

l-156 1.193 I.260 1.291 1.347 1.3% 1441
I*188 1301
f:E
::E!
l-296 ::z:
:::z
I.419
::z
l-489
::g
I.551
f:Zf
I.606
l-329 1461 l-535 lTjO1

12$ 1582 1.652
1.337 :::: ::E
1.425 l-583 .
t-E :E
:::9”: l-455 l-623 l-718 1*801
222 DESIGN AIDS FOR RRINFORCED CONCRETE

VALUES OF It/ s
t )
r
o-0 0.1 O-2 0.3 O-4 o-6 O-8 J-0
O-100 0.100 O-100 @JO0 o-100 O-100 O-100 O*lOO

0.143 O-143 O-143 O-143 0.143 @I43 0.143 0.143
O-185 0.185 O-185 0.185 0.185 0.185
O-224 O-224 O-224 O-224 OQ24 xi!: .
8‘:::
x:z 0.262 O-262 O-262 0262 0.262 O-263 O-263
O-298 O-298 O-298 O-298 0.299 0.300

O-333 O-333 0334 O-334 x:E
EZ 0.367 O-367 O-368 O-368 O-369 FL:::
O-398 O-399 0401 O-402 0403 O-405
0.430 0.43 1 z! O-433 0,434 0.436 O-438
O-462 0.463 O-465 0468 0.471 0473

x:z @494 O-495 zz! o-503 O-505
0519 X:iE O-523 O-525 0.527 Ei O-534 O-537
0547 0.550 O-552 O-555 O-557 O-561 O-565 O-569
O-575 O-578 O-581 O-583 O-586 O-591 0.596 O-600
0.601 O-605 O-608 0.611 O-614 0.620 O-626 O-631

O-628 O-632 O-635 0639 O-643 O-649 O-655 0.661
O-653 O-658 0.662 0.670 0.678 O-685 O-691
O-678 O-683 O-688 8:E 0.697 O-706 O-713 O-721
o-703 o-708 0.714 O-719 0724 O-733 O-742 O-750
O-727 O-733 O-739 0,745 O-750 0,761 O-770 O-779

0750 0.757 O-764 @770 0.776 O-788 0.798 P808
0773 0.781 O-788 O-795 O-802 0814 O-826 O-836
0.795 O-804 O-812 0820 0827 O-841 O-853
0.818 O-827 @836 O-844 O-852 0.867 O-880 ::8896:
0.839 0.849 O-859 O-868 O-876 O-893 0.92 I

O-860 0871 O-882 0.891 0901 O-918 X:K O-948
O-881 0.893 O-915 0.925 0943 0.960 O-976
0902 0914 ;z O-938 0,949 O-969 0.987 l-003
O-922 O-935 O-960 0.972 0.993 l-013 1 a30
O-942 0956 0.970 0.983 J-039 l-057

O-980 0997 I-012 1.027 E :::t l-090 1~111
l-018 l-036 I .054 l-070 1.086 l-115 I.141 1.164
l-054 1.075 I.094 1.112 l-130 l-162 l-191 1.216
1 a89 J-112 I.134 J-154 1.173 1 a208 l-240 l-268
1.123 1.148 I.172 J-194 I.216 I.320

l-156 1.184 l-210 1234 I.257 1.371
I.188 l-219 1.247 I.274 I.299 I -422
1.220 I *252 I.283 1.312 l-339 I.473
l-250 J-286 I.319 1.350 1.379 I.523
1.280 1.318 1.354 1.387 1.419 I.476 I:627 1.573

1.308 J-350 1.388 J-424 l-574
I.337 l-381 I.422 J-461 : ::z: zzi ‘I:x:3
l-364 I-411 1455 1.4% l-535 l-605 1E I.721
1.391 1441 1488 1.532 1.573 1 a7 I.712 1 a770
DEFLMXION CALCULATION 223

VALUES OF Jr/ $
( )
d'/d-O-20
Am--l)/(m)
0-l O-2 0.3 O-4 O-6 O-8 19’
O-298 O-298 a298 0.298

;; O-333 O-333 O-333 O-333 Ei
@367 O-367 O-368
O-399 EZ O-401
O-430 !E! O-431 X% .
%i O-434
O-462
O-492 8% yg :%!
0.523 0528
:::: 0.552 O-554 O-558
O-579 @580 O-583 @588
O-603 O-605 O-607 @614 @617
O-632 0.634 0o:g O-643
zz O-658 O-663 O-671 ic
O-681 iT% O-689 O-699
O-706 Ei 0.712 O-715 O-726 O-731
O-730 O-734 O-741 0.753

0754 O-758 0.766 O-779 .
x’:::
0.782 O-791 O-806 O-812
Ei O-815 O-832 0839
Of23 EE O-839 O-857 O-865
0.863 0.874 0.883
O-887 O-898
0910 O-922 Ei
@958
z:: ii% O-983
o-961 0969 O-978 O-993 l-007

Ei l-012 l-022 1so39 1955 ST%
1931 :izz la65 l-085 1.103 1.119
l-082 :z!z l-129 l-168
:%i l-120 l-135 xi. l-173 :z 1:216
1141 l-174 1.189 l-217 l-241

l-176 l-212 l-259 l-287 Et
:zz l-331 1358
.
:-iii .
:% 1.307 1:E 1a5
l-276 l-324 1.345 l-384 .
:*:z 1.451
1.308 l-334 l-383 1.425 1463 1.497

l-339 1368 1,420 l-506 l-542
::zz l-587
.
:%i ::zi :1E l-545 :::;; l-632
l-429 1464 1.528 l-584 l-633 1.677

TABLE 91 DEPTH OF NEUTRAL AXES - VALUES OF x/d
BY ELASTIC THEORY
d’/d=@OS
O-132 O-131 o-131 0.130 O-128

O-159 O-158 O-157 O-155 O-153
O-181 O-180 0.178 O-176 O-173
O-198 0.197 O-194 @190
EY O-215 O-213 O-209 0205
0227 0225 O-223 0.218 O-214

O-238 0235
E!i O-249 tz! x:z
O-263 i!Lg O-251 O-245
O-274 z: O-260 O-253
0.287 0.276 O-268 0.261

@297 EJ O-284 0.276 0.269
O-292 0.276
Ei 0.305 ;ij 0.282
O-323 O-312 Ez 0.289
O-336 0.330 @325 0.319 O-314 0.304 O-294

O-338 O-326 0:321 O-310 O-300
8E 0345 E O-333 O-327 O-316 O-305
O-358 0.352 O-345 O-339 O-333 O-321
O-365 O-358 @351 O-345 O-339 C+326 :::::
@358 0.351
O-363 :::t ::E
0.369 EZ 0.341 0328
0374 O-367 0.345 0332
O-380 O-372 O-350 0.336
MO2 O-393 O-385
Oar7 O-398 O-390 EZ
O-413 O-403 O-394 O-378
O-418 O-399 O-382
0,423 z! O-404 @386
O-418 O-399
O-427 :iti
0.425 :zi
E O-432 O-421
O-451 O-439 0428
O-383
x- @388
O-459 0392
0465 O-396
0.471 0400
O-507 O-491
O-513 0497 Ez ::g
0519 0503 0488 O-411
O-525 O-493 0.414
O-531 8E 0.498 O-417
DEFLECTlON CALCULAflON 225

TABLE 92 DEPTH OF NEUTRAL AXIS - VALUES OF x/d
BY ELASTIC THEORY
01 0.2 0.4 0.6 0.8 1.6

0.132 0.131 0.131 0.130 0.130 0.130
@158 0.158 0.157 0.156 0.155 0.154
@MO 0.179 0.178 ml76 P175 0.174
0.199 0.198 0.196 0.194 0.192 0.190
0.215 0.214 0211 0209 0.206 0.204
0230 0228 0.225 0.219
0242 0.238 XE 0.231
odz 0254 0249 0245 0241
0.268 0.265 0.260 0.255 0.251
0.278 0275 0.270 0.265 0.260
0285 0.279 0.273 0263
0288 0.282 0.270
ii 0289
. 812Z 0296 x:18’:
0.319 0.311 0303 0289
0331 0.327 0.318 0.310 0.295
0.339 0.334 0.325 0316 X:E 0.300
0.341 0.331 0322 0.313 0.305
%: 0347 0.337 @318 0310
0.359 0.354 0.343 “0:::: 0.323 0.314
0.349 0.338
8% 0.354 0.343 8% x::::
0.371 0.359 0.348
0.377 0.364 0.352 ::::: “0::::
0.382 0.369 0.357 0.345 0.335
0.394 0.387 0.361 0.349

0.392 0365 0.353
x:z 0.397 0.357
0.410 8% 0.360
0.414 ;:z 0377 0.363
0.419 0.411 0.395 0.380 0.367 0354
0.428 0403 O-387 0.373 0.360
0.437 00::;; @394 O-365
0445 0.435 X% 0400 8% 0.370
0.453 0442 0.423 0406 0.389 0.375
0449 0.411
fii$ 0456 0.416
0.421
0481 E; 0.426
0487 0.475 0.431
0.493
Ei
8%
226 DESIGN AIDS FOR REINFORCBD CONCRBTB

BY ELASTIC THEORY
’ o-0 0.1 O-2 0.3 o-4 0.6 0.8 1-d

0.132 O-132 0.132 0.132 o-133 0.133 0.133
O-159 O-159 O-159 @159 0.158 0.158 O-lS8
O-181 O-180 @180 O-179 0.178
0.199 .
8’:: O-198 O-198 :::;Y O-196 O-195
O-216 O-215 O-214 O-214 0.212 O-211 O-209
O-231 0229 O-228 0224 O-222

z-z 0.244 O-242 OatI 0.236 0.234
0:258 0.257 0.254 0.252 0.247 0.244
O-270 O-268 O-265 0.263 0.257 O-254
O-281 0.279 O-275 O-273 0266 0.262
O-292 O-289 O-287 0.285 0.282 0.278 O-274 0.270

0301 O-299 O-296 0294 O-291 0.287 O-282 O-278
O-311 @308 0.305 @302 O-299 0.294 O-290 O-285
0.316 O-313 O-310 o-307 @297
ii::: O-324 O-321 0.318 o-315 x:z 0.303 8:ZE
O-336 0.332 O-329 O-325 0322 O-315 0309 O-304

0.344 O-340 O-336 0.332 O-329 O-322 O-315
o-351 O-347 0.343 fF339 O-335 O-328 O-321 8Z
O-358 0.354 O-349 O-345 O-341 O-326 0.320
0.365 O-360 O-356 0.351 o-347 x:::: O-332 O-324
0372 0.367 0.362 0.353 O-344 0.336 0.329

O-378 O-373 O-368 8:::: O-358 0.349 O-341 O-333
O-384 O-379 O-374 O-369 0.364 O-346 O-338
O-390 O-385 0.379 0.374 O-369 x::::: 0.350 O-342
O-396 O-390 0.384 O-379 0.374 0.364 O-354 O-345
0402 0.396 0.390 0384 0.378 O-368 O-358 0.349

O-407 O-401 O-395 0.389 0.383 O-372 0.362 O-353
O-413 0406 O-393 O-387 0.376 O-366 0.356
O-418 0.411 :‘iz 0.398 0.392 O-380 O-369
O-423 O-416 O-409 0402 O-396 0.384 O-373 “o::zi
O-428 O-420 0.413 O-407 OaO O-388 O-376 O-366

O-437 0.429 O-422 O-415 O-395 O-383 O-372
0.438 O-430 O-422 8::; O-401 0.377
Et 0.438 O-430 0422 0.408 “0::;; O-382
0.463 iEi O-445 O-437 0.429 O-414 O-un O-387
0462 O-452 O-444 0.435 0419 0405

O-469 0.459 O-450 O-441 0.425 O-410 ZE
O-476 O-466 0.456 O-447 0.430 0.415 O-401
O-482 O-462 O-435 O-419 0405
O-489 8:::; O-468 ::t:: O-440 O-423 O-408
0495 @484 0.473 O-463 0.427 0.412

O-501 0.489 0.478 0468 ::I$ . 0431 O-415
O-506 0.494 0.453 O-435 0.419
0.512 X:E ::t:‘7 0.457 0438 O-422
O-517 x:z 0493 0.481 O-460 O-442 0.425

BY ELASTIC THEORY
d'jd=0.20
PC@+-lMptm)
ptm
. 0.0 0.1 0.2 0.3 0.4 0.6 0.8 1.6
1.0 O-132 0.133 0.133 0134 O-135 0.135 0136

1.5 O-159 0.160 O-160 O-160 0.161 O-161 0162
O-181 O-181 O-182 O-182 0.182 0.182 O-183
;:; 0.200 0.200 0.200 0.200
3-o O-217 0.216 O-216 8:% 8% ::z 0215
3.5 0.232 O-231 O-231 0.231 O-230 0230 0229 0228

0246 0.244 O-243 0.242 O-241 0.240
9:; 0.258 8:;;; x:z 0.256 @255 O-254 0.252 0.251
5.0 O-270 0.269 O-268 0.267 0.266 O-264 0.262 O-261
5.5 0.281 O-280 O-279 O-277 0.276 0.274 O-272 0.270
0.292 0290 0.287 0.286 ,0.283 0280 O-278

:I’: 0.301 O-300 x:4:: 0*29., 0.295 0.291 0.286
@311 O-309 O-307 0305 0.300 8:;;: 0.293
3:: 0.319 0.315 0.313 ii:::: 0.307 0303
8-O O-328 x:::: 0.323 O-321 0319 0.314 0.310 ~:~
0.336 O-333 O-331 O-328 0.326 0.321 0317 0.313

O-344 0.341 0.338 0.335 0.333 O-328 0.323 0.318
O-351 O-348 O-345 O-342 0339 O-334 0329 0.324
O-358 0.355 O-352 O-348 0.345 0.340 0.334 0.329
0365 0362 0.358 O-355 0.351 0.345 0.340 O-334
11-o O-368 o-364 0.357 0.351 0.339

11’5 O-374 O-370 0.363 0356 O-343
0.380 0.376 O-368 0361 0.348
:I:; 0.386 O-382 0.366 0352
13.0 0.391 0387 :%I O-370 0.356
13.5 0.397 O-392 0388 O-383 0375 0.360

140 8% O-402 0.397 0.392 0.379 8% 0364
14-5 O-413 0407 0402 0.397 8% 0.375
15.0 O-418 O-412 0407 O-402 0397 k% 0.379 ::z:
15-5 0.423 O-417 0.411 O-406 0401 0391 O-382 O-374
16-o o-428 0.416 O-410 0405 O-395 O-386 0377

17.0 0.437 8% O-425 O-419 0.413 0402 0.393 0384
18-o 0446 0.439 0.433 0.427 f-b.421 O-399 0.389
19.0 O-448 O-441 O-434 0.428 %z 0405 0.395
20.0 t%: O-456 O-448 0441 0434 O-422 0410 0400
21-O 0471 O-455 0448 0441 O-428 O-416 O-405

0.479 0462 O-454 0447 a433 0.421
3:*8 O-486 0.469 0461 O-453 0.439 O-426 8:Z
24-o @493 O-475 0.459 O-418
25-O O-500 0.481 :z: O-464 :zi x:1:‘: O-422
@478 0469 0.453 0.439 0426

0.483 @458 O-443 O-429
O-488 X% @462 O-447 @433
O-484 0.466 O-450
tz;ii 0488 O-470 0.454 x:1:9”
228 DESlGN AIDS FOR REINFORCEDCONCRET’E

TABLE 95 AREAS OF GIVEN NUMBERS OF BARS IN Cm)
BAR DIANKIER. mm
NUMBER ,
OPBAW -6 10 12 14 16 18 20 22 25 28 32 36
::i; 2.26
1.13 3.07
I -54 4.02
2o1 z 3.14 3.80
. 491 6.16 8 01 IO.18
1231 1608 2035
235 339 461 6.03 763 i% 1% l:% 18.47 24.12 30.53
:z: 452
5.65 769
6.15 10.05
8.04 10.17
12.72 1256
15.70 15.20
1900 2454
1963 2463 32.17 40.71
30.78 40.21 50.89
6.78 923 12% 15.26 l&85 22.80 29.45 36.94 48.25 61.07
7.91 10.77 14.07 17.81 2199 43.10 56.29 71.25
9.04 1231 l&O8 20.35 25.13 zz ii::; 49.26 6434 81.43
10.17 13.85 18.09 2290 34.21 44.17 55.41 7238 91 a
11.31 15.39 20.10 2544 iK 38.01 49.08 61.57 80.42 101.78
8.63 1244 1693 211 2799 34.55 41.81 5399 67.73 88.46 111.96
9.42 13.57 18.47 24.12 30.53 37.69 45.61 5890 73.89 96.5 I 122.14
10.21 1470 20.01 26.13 33.08 4941 63.81 80.04 101.55 132.32
1099 15.83 21.55 28.14 35.62 z% :;g . g*;; 86.20 142.50
11.78 16.96 23.09 30.15 38.17 47.12 92.36 : g:;;
152.68
18.09 24.63
-. __ 3217 40.71 50% 98.52 128.68 162.86
1::::
1413
i9*i2
20.35
2617
27.70
3i*iS
36.19
43.26
45.80
53.40
56.54
z%
68.42
ix
88.35
104.67
1 IO.83
136.72
144.76 f ;;:;;
1492 21.48 2924 38.20 48.34 5969 ;;g ;;g 116.99 152.80 193.39
5.65 10.05 15.70 22.62 30.78 40.21 50.89 6283 123.15 160.85 203.57
DEFLECTION CALCULATlON 229

TABLE 96 AREAS OF BARS AT GIVEN SPACINGS
Values in cm2 per Meter Width
BAR DIAMEIER.mm
1
cm
6 8 10 12 14 16 18 20 22 25 28 32 .
5.65 10.05 15.71 22.62 30-79 40-21 50.89 6283 76.03 98.17 123’15 16085
471 8.38 13.09 18.85 25.66 33.51 4241 52.36 63.36 81.81 102%8
7-18 11-22 16.16 2199 36.35 44.88 54.30 70.12 87.96 :4qs
%
3.14
6.28
5.58
9.82
8.73
14-14
12-57
19-24
17.10
f E
22.34
31.81
28.27
39.27
3491
47.52
4224
61.36
54-54
76-9
68’42
10053
89-36
10 2.83 5.03 7.85 11.31 20-11 25.45 31.42 38.01 49-09 61.57 80-42
257 4.57 7.14 10’28 ::-sz
* 18.28 23.13 28.56 34.56 44.62 55.98 73.11
t:. 4.19 6.54 9.42 1283 1675 21.21 26.18 31.68 40-91 51-31 6702
z7” a.70 11.84 15.47 19-57 24.17 29.24 37.76 47-37 61.86
:: 2.02 ZJ 4:: S-08 11-00 1436 18.18 22.44 27.15 35.06 43.98 57.45
15 1.88 3.35 5.24 10.26 1340 1696 20.94 25.34 41.05 53.62
16 l-77 3.14 4.91 % 9-62 1257 15?w 19.63 23.76 :z 38.48 50-27
17 l-66 2.96 4-62 9.05 11-83 14-97 18.48 22.36 28.87 36.22 47.3 1
18 1.57 2.79 4.36 f:E 8-55 11.17 1444 17.45 21.12 27.27 2421 44.68
19 1.49 2.65 4.13 5.95 8.10 IO-58 13.39 16.53 20-01 25.84 32-41 42.33
1.41 2.51 3.93 5.65 7.70 10.05 12.72 15.71 19.01 24.54 30-79 40.21
1.35 239 3-74 5.39 7.33 9-57 12.12 14.96 18.10 23.37 29.32 3830
l-28 2-28 3.57 5-14 9.14 11.57 4.28 17.28 22.31 2799 36-56
1.23 218 3’41 4-92 ::z! 8.74 11.06 13’66 16-53 21.34 26.77 34.97
1.18 2-09 3.27 4’71 6.41 8.38 10-60 13.09 15.84 20.54 25.66 33.51
l-13 2.01 3.14 4.52 6.16 8.04 12.57 15.20 19.63 24.63
1.09 1.93 4.35 5.92 7-73 1208 14.62 18.88 g:;;
l-05
1 *Ol
1.86
1.79
;:Ff
2.80
419
4.04
5-70
5.50
7.45
7.18
11.64
1 l-22
1408
13.58
18.18
17.53
f E
21.99
29.79
28.76
0.97 1.73 271 3-90 5-31 6-93 10.83 13.11 16-93 21.23 27.73
30 O-94 1.68 2.62 3.77 5.13 6-70 8.48 10-47 12.67 16.36 20.52 26.81
32 0.88 1.57 245 3-53 4.81 6-28 7.95 9-82 11.88 15.34 19.24 25.13
34 0.83 1.48 2.31 3.33 4.53 5.91 7.48 5.24 11-18 14.44 18.11 23.65
36 0.78 2.18 3.14 4,28 5.58 7.07 8.73 10-56 13.63 17.10 22.34
0.74 :z 2.07 2.98 4.05 5.29 6.70 8.27 12.92 16.20 21.16
:: 071 1.26 1.96 2.83 3.85 5.03 6.36 7.85 ‘!?z. 12.27 15.39 20.11
230 DESICSNAIDS FOR REINFORCED CONCRETE

Table 97 FIXED END MOMENTS FOR PRISMATIC BEAMS
LOAD TYPE M rr Mm
Pab’ Pdb
(’ 1’
I-
PI
%
w, [’ 12 ad+ s2 (I-3b)l
12 1’
-7 2 I (31~e4+3s’) ‘,f+-3S)
121’
w I’ w I2
+12 -12
5Wl’
A -96
WI’ w I2
+20 -30
t-----‘----l
231
IXSf’LECT’ION CALCULATION
Table 98 DEFLECTION FORMULAE FOR PRISMATIC BEAMS
&it I
I+‘/2 48EI
Pl’ P 1’
192
L&! 23PI’
6=
5 PI’
646 EI
L ( w
, rrrrcrrl H!t
,,
,’
6E1
,P 1 --I
P 1’ 2
L,MI
3 16 EI
Note:- W is total distributed load
DESIGN AIDS FOR REINEORCED CONCRETE

SP16-Design Aid For RC To IS456-1978

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DESIGN AIDS

BUREAU OF INDIAN STANDARDS

0 BUREAU OF INDIAN STANDARDS

UDC 624.0 12.45.04 (026)

PRICE Rs. 500.00

a) Development programme on Code implementation for building and civil engineering

A Special Committee for Implementation of Science and Technology Projects (SCIP)

Functional Requirements of Buildings

Construction Safety Practices

One of the explanatory handbooks identified is on IS : 456-1978 Code of practice for

The design aids cover the following:

a) Material Strength and Stress-Strain Relationships;

The format of these design aids is as follows:

a) Assumptions regarding material strength;

LIST OF TABLES M THE EXPLANATORY -TEXT ... ... x

1. MATERIAL STRENGTH AND STRESS-STRAIN RELATIONSHIPS 3

1.1 Grades of Concrete .. . .. 3

2. FLEXURAL MEMBERS ... .. . 9

3. COMPRESSION MEMBERS ... . .. 99

3.1 Axially Loaded Compression Members . .. . . . 99

4. SHEAR AND TORSION . .. . .. 175

5. DEVELOPMENT LENGTH AND ANCHORAGE . .. .. . 183

5.1 Development Length of Bars . .. . .. 183

6. WORKING STRESS DESIGN .. . ... 189

6.1 Flexural Members . .. . .. 189

7. DEFLECTION CALCULATION .. . . .. 213

7.1 Effective Moment of Inertia . .. . .. 213

LIST OF TABLES IN THE EXPLANATORY TEXT

B Values of F for Different Grades of Steel . .. . .. 9

C Limiting Moment of Resistance and Reinforcement Index for Singly

1 CL= 15 N/mm’, fy= 250 N/mm’ d- 5 to 30 cm ... 17

FLEXURE - Doubly Reinforced Section

19 fr I 250 N/mm’, d-d’ - 20 to 50 cm ... .. . 41

21 fr - 250 N/mm’ ... ... 43

24 h - 250 N/mm’ . .. .“. 109

COMPRESSION WITH BENDING - Rectangular Section -

COMPRESSION WlTH BENDlNG - Circular Section

86 Axial Compiession (Working Stress Design) 0, - 130 N/mm* . . . 193

1 fft -15 N/mm’ ...... 47

FLEXURE - Moment of Resistance of Slabs, kN.m Per Metre Width

5 fd - 15 N/mm’ /r- 250 N/mm* Thickness - 10.0 cm ... 51

35 f ck - 20 N/mm2 h - 415 N/mm2 Thickness - 100 cm . .. 73

FLEXURE - Reinforcement Percentages for Doubly

FLEXURE - Limiting Moment of Resistance Factor, Mo,u,&,# /a, for

WORKING STRESS METHOD - FLEXURE - Moment of

72 uck- 5.0 N/mm1 aa - 140 N/mm8 . .. . ..

80 Permiklble Shear Stress in Concrete rc, N/mm* .. . .. .

WORKING STRESS METHOD - DEVELOPMENT LENGTH

MOMENT OF INERTIA OF CRACKED SECTION-Values of Ir/

87 &Id - 095 ... . .. 221

DEPTH OF NEUTRAL &US - Values of n/d by Elastic Theory

91 d’/d I 005 1.. ... 225

95 Areas of Given Numbers of Bars in cm* ... .. . 229

AC = Area of concrete E Flexural tensik strength