PLANNING, ANALYSIS AND DESIGN OF A
SHOPPING MALL
A PROJECT REPORT
Submitted by
R. ANEES FATHIMA
110119103002
S.A.C. WASIF MUKTHAR
110119103026
N. SAIFULLAH
110119103304
in partial fulfillment for the award of the degree
of
BACHELOR OF ENGINEERING
IN
CIVIL ENGINEERING
AALIM MUHAMMED SALEGH COLLEGE OF
ENGINEERING
ANNA UNIVERSITY : CHENNAI 600025
DECEMBER - 2022
ANNA UNIVERSITY : CHENNAI 600025
BONAFIDE CERTIFICATE
Certified that this project report “PLANNING, ANALYSIS & DESIGN OF A
SHOPPING MALL”
is the bonafide work of “R. ANEES FATHIMA (110119103002) ; S.A.C. WASIF
MUKTHAR (110119103026) ; N. SAIFULLAH (110119103304)” who carried out the
project work under my supervision.
SIGNATURE
SIGNATURE
Prof. Dr. M. AFZAL ALI BAIG, M.E.,
Ph.D., M.I.S.T.E.,
Er. Mr. D. Prakash, M.E.,
Assistant Professor,
HEAD OF THE DEPARTMENT
SUPERVISOR
DEPARTMENT OF CIVIL
ENGINEERING
DEPARTMENT OF CIVIL
ENGINEERING
AALIM MUHAMMED SALEGH
COLLEGE OF ENGINEERING
MUTHAPUDUPET, AVADI IAF
CHENNAI – 600055.
AALIM MUHAMMED SALEGH
COLLEGE OF ENGINEERING
MUTHAPUDUPET, AVADI IAF
CHENNAI – 600055.
TABLE OF CONTENTS
CHAPTER NO.
TITLE
PAGE
NO.
1
ABSTRACT
01
2
LIST OF SYMBOLS
02
3
INTRODUCTION
03
4
ANALYSIS
05
5
DESIGN
FLAT SLAB (TERRACE)
09
FLAT SLAB (FLOORS)
11
III.
TWO WAY SLAB
15
IV.
EDGE BEAM
17
AXIALLY LOADED COLUMN
19
UNI-AXIALLY LOADED COLUMN
21
BI-AXIALLY LOADED COLUMN
22
CIRCULAR COLUMN
23
DOG LEGGED STAIRCASE
25
RETAINING WALL
27
ISOLATED COLUMN FOOTING
30
I.
II.
V.
VI.
VII.
VIII.
IX.
X.
XI.
6
CONCLUSION
33
7
REFERENCE
34
ABSTRACT
Report to accompany the Planning, Analysis & Design of a
“SHOPPING MALL”
The main of this project is to design a shopping mall, which is designed in
achievement of an acceptable probability that the structure being designed within appropriate
factor of safety and perform satisfactorily during their intended life.
This design is done according to the limit state method and conforming to the Indian
standard code IS 456-2000 & SP 16. The beam and column are designed by considering the
maximum loading condition and maximum bending moment. Isolated Rectangular footing is
provided for the Foundation. Doglegged staircase is provided for emergency exit.
The plot area is 21,920 sq.m the building consists of (G+2) floors with a Basement
for Car Parking. For simplicity of the design, the loads of doors and windows are neglected.
The site is located in Seismic Zone III and the soil is of type III : medium soil. This Project
comprised of Planning, Structural Analysis and Design of “SHOPPING COMPLEX” (G+2) in
Chennai. Good ventilation, Lighting is given special importance to satisfy the various function
requirements. A Good front elevation was also giving to enhance appearance of building.
The proposed building was planned to make comfortable aesthetic, economical,
good orientation and to meet all requirements. The project is analysed by using moment
distribution method. The structure is designed by using limit state method as per IS 456:2000
& SP 16. It is considered as live load and dead load referred as per IS 875:1987- Part - 1 &
Part – 2.
The detail drawing is drawn by using AutoCAD software. At the end of the project,
we are able to analyse and design any type of RC structures in professional practice.
1
LIST OF SYMBOLS
A
Ac
Ag
Ast
B
BV
D
DL
D
fck
fy
LL
lx
ly
Mf
Mu
Mulim
Pu
Vu
Sv


Mx
I
rv
rc
M
P
R
T
V
W
W
wd
X

K
K
L
lef
lex
ley
Area
Net area of concrete
Gross area of section
Area of steel on section
Breadth of beam
Basic value
Overall depth of beam or slab
Dead Load
Effective depth of beam or slab
Characteristic compressive strength of concrete
Factored Load
Live Load
Length of shorter side of slab
Length of longer side of slab
Modification Factor
Factored moment
Moment of resistance of limiting section
Axial Load
Factored shear force
Spacing of stirrups
Bending moment of co-efficient for shorter span
Bending moment of co-efficient for longer span
Moment per unit width along shorter span
Unsupported length of column
Nominal shear stress
Shear stress in concrete
Bending moment
Axial load on a compression member
Radius
Torsional moment
Shear force
Total load
Distributed load per unit area
Distributed Dead load per unit area
Depth of neutral axis
Diameter of bar
Stiffness of member
Constant or co-efficient or factor
Length of a column or beam
Effective span of beam or slab or column
Effective length about x-x axis
Effective length about y-y axis
2
INTRODUCTION
Construction of shopping complex are useful in saving the land as many types of
shops can be included in a single building. Moreover, we can get all the required things in a
place.
In our project we have provided different shops for clothing, household
appliances, electrical items, furniture, toys & stationery items, shoes, ice cream shop, saloon,
sports store, service centre pets shops pharmacies, restaurants, theatres, etc.
The planning of this project is done keeping in view all the requirements for a
better assembling of people.
The analysis for all elements of the building is done as per IS code of practice for
Plain and Reinforced cement concrete IS456-2000 and SP-16 design aids for designing the
elements.
3
ANALYSIS
4
ANALYSIS OF A COMPONENT ( BEAM )
Required Data’s :
Size of a beam - - - B = 300 mm ; D = 650 mm
Factored load on beam = 29.34 kn/m
E = 5000 x √fck = 29580.40 mm2
I = ( b d3 ) / 12 = 68.66 x 108 mm4
[ 300 x 650 mm ]
STEP – 1 : ( Calculate F.E.M )
SPAN NUMBER
SPAN LENGTH
(m)
UDL ( kn/m )
L1
8
29.34
L2
8
29.34
L3
8
29.34
L4
8
29.34
Fixed End Moments ( kn.m )
Mf AB
-156.48
Mf BA
156.48
Mf BC
-156.48
Mf CB
156.48
Mf CD
-156.48
Mf DC
156.48
Mf DE
-156.48
Mf ED
156.48
STEP – 2: ( Calculate D.F )
Joint
Member
Stiffness ( K )
BA
2.54 x 1013
BC
2.54 x 1013
CB
2.54 x 1013
CD
2.54 x 1013
DC
2.54 x 1013
DE
2.54 x 1013
B
C
D
Total Stiffness ( ∑ K )
AB
I
68.66 x 108
BA
68.66 x
108
0.5
5.08 x 1013
0.5
0.5
5.08 x 1013
0.5
0.5
5.08 x 1013
STEP – 3 : ( Calculate Distribution of moment )
JOINT
A
B
MEMBER
Distribution Factor
( D.F = K / ∑K )
BC
68.66 x
108
CB
68.66 x
108
5
0.5
C
D
CD
68.66 x
108
DC
68.66 x
108
E
DE
68.66 x
108
ED
68.66 x
108
L
8
8
8
8
8
8
8
8
K
2.54 x 1013
2.54 x
1013
2.54 x
1013
2.54 x
1013
2.54 x
1013
2.54 x
1013
2.54 x
1013
2.54 x
1013
D.F
0
0.5
0.5
0.5
0.5
0.5
0.5
0.5
F.E.M
-156.48
156.48
-156.48
156.48
-156.48
156.48
-156.48
156.48
BALANCE
-
0
0
0
0
0
0
-
STEP – 4 : ( Analysis Diagram )
CONTINUOUS BEAM --- REACTIONS & END MOMENT
6
SHEAR FORCE DIAGRAM
BENDING MOMENT DIAGRAM
SLOPE DIAGRAM
7
DESIGN
8
FLAT SLAB DESIGN [ TERRACE ]
REQUIRED DATAS :
Panel Size = 8 m x 8 m
Column Size = 0.6 m x 0.6 m
Height = 4.5 m
Live Load = 2.5 kn/m2
Super Imposed Dead Load = 2.8 kn/m2
fck = 25 n/mm2
&
fy = 500 n/mm2
A] INTERIOR PANEL –
STEP – 1 : ( Thickness of Slab )
Span / effective depth = 40 x 0.8
8000/d = 32
d = 8000/32 = 250 mm > 125 mm [ minimum thickness of slab ]
assuming 12 mm and clear cover = 20 mm
Overall Depth, D = d + /2 + clear cover = 250 + (12/2) + 20 = 276 mm
Adopt a Overall Depth of D = 280 mm
Effective depth, d = 254 mm
STEP – 2 : ( Size of Drop Panel )
Drop Panel size should not be less than = 8000/3 = 2666.67 mm
Minimum depth of Drop Panel = 1/4th of D = ( 1/4 ) x 280 = 70 mm
Depth of Slab including Drop Panel = 280 + 70 = 350 mm
Width of Column strip = Width of middle strip = 8000/2 = 4000 mm
Drop Panel Size = 4000 mm x 4000 mm
STEP – 3 : ( Load Calculations )
Live Load = 2.5 kn/m2
Super Imposed Dead Load = 2.8 kn/m2 [ finishes ]
Dead Load of Slab = 0.28 x 25 = 7 kn/m2
Dead Load of Drop Panel = 0.07 x 25 = 1.75 kn/m2
Total Characteristics Load, WC = 2.5 + 2.8 + 7 + 1.75 = 14.05 kn/m2
Factored Load, Wf = 14.05 x 1.5 = 21.08 kn/m2
L1 & L2 = c/c span = 8 m ;
Ln = 8 – ( 0.6 / 2 ) – ( 0.6 / 2 ) = 7.4 m
Design Load, Wn = Wf x Ln x L2 = 21.08 x 7.4 x 8 = 1247.94 kn
STEP – 4 : ( Bending Moment Calculation )
9
Mo = ( Wn . Ln ) / 8 = (1247.94 x 7.4) / 8 = 1154.34 kn.m  1155 kn.m
As per IS 456 : 2000, Clause – 31.4.3.2
-ve design moment = 0.65 x Mo = 0.65 x 1155 = 750.75 kn.m
+ve design moment = 0.35 x Mo = 0.35 x 1155 = 404.25 kn.m
As per IS 456 : 2000, Clause – 31.5.5.1 & 31.5.5.3
TYPE
COLUMN STRIP ( kn.m )
MIDDLE STRIP ( kn.m )
- ve moment
750.75 x 0.75 = 563.06
750.75 – 563.06 = 187.69
+ ve moment
404.25 x 0.60 = 242.55
404.25 – 242.55 = 161.70
STEP – 5 : ( Check for effective depth, d )
dreq. (slab) = √ (Mu / 0.138 x fck x b ) = √ [ ( 691.28 x 106 ) / ( 0.138 x 25 x 4000 ) ]
= 223.81 mm < 254 mm
dprov. (drop panel) = 254 + 70 = 324 mm < 350 mm
Hence, it’s safe..
dreq. < dprov.
STEP – 6 : ( Main Reinforcement calculation )
BM = 0.87 x fy x Ast x d x [ 1 – ( fy x Ast ) / ( fck x b x d ) ]
1. COLUMN STRIP : [ -ve reinforcement ]
563.06 x 106 = 0.87 x 500 x Ast x 324 x [ 1 – ( 500 x Ast ) / ( 25 x 4000 x 324 ) ]
Ast = 3917.5 mm2
Assuming 12 mm, ast = 113.10 mm2
Spacing, Sv = ( 4000 x ast ) / Ast = ( 4000 x 113.10 ) / ( 3917.5 ) = 110 mm c/c
[ +ve reinforcement ] :
Ast = 1631.11 mm2
Sv = 260 mm c/c
2. MIDDLE STRIP : [ -ve reinforcement ]
187.69 x 106 = 0.87 x 500 x Ast x 254 x [ 1 – ( 500 x Ast ) / ( 25 x 4000 x 254 ) ]
Ast = 1759.66 mm2
Sv = 250 mm c/c
10
[ +ve reinforcement ] :
Ast = 1508.26 mm2
Sv = 280 mm c/c
STEP – 7 : ( Check for punching shear )
d/2 distance from column
600/2 = 300 mm
Perimeter of critical section = bo = 4 ( a + d ) = 4 ( 600 + 300 ) = 3600 mm
Shear Force on this plane, Vu = W [ L x L – ( a + d ) ( a + d ) ] = 14.05 [ 82 – 0.81 ] = 887.82 kn
Nominal Shear stress = ( Vu / b . d ) = (887.82 x 103 ) / ( 4000 x 350 ) = 0.63 n/mm2
ks = ( 0.5 + ßc ) = 0.5 + 1 = 1.5 but not greater than 1.
ßc = 0.6/0.6 = 1 [ Ratio of short side to long side of the column ]
τc = 0.25 √fck = 0.25 x √25 = 1.25 n/mm2
ks . τc = 1 x 1.25 = 1.25 n/mm2
... The actual shear stress of 0.63 n/mm2 is within safe permissible limits.
STEP – 8 : ( Integrity Reinforcement )
As = ( 0.5 . wf . L1 . L2 ) / ( 0.87 . fy )
= [ ( 0.5 . 21.08 . 8 . 8 ) / ( 0.87 . 500 ) x 103 ] = 1550.71 mm2
Provide two 25 mm diameter bars, with a length of 2.Ld = ( 2 x 1071.43 ) = 2142.86 mm2
passing through the column cage each way.
FLAT SLAB DESIGN [ FLOORS ]
REQUIRED DATAS :
Panel Size = 8 m x 8 m
Column Size = 0.6 m x 0.6 m
Height = 4.5 m
Live Load = 6 kn/m2
Super Imposed Dead Load = 2.5 kn/m2
fck = 25 n/mm2
&
fy = 500 n/mm2
Note : Live Loads taken from IS 875 : 1987 ( PART – II )
As per IS 456 : 2000
A] INTERIOR PANEL –
STEP – 1 : ( Thickness of Slab )
Span / effective depth = 40 x 0.8
8000/d = 32
d = 8000/32 = 250 mm > 125 mm [ minimum thickness of slab ]
11
assuming 12 mm and clear cover = 20 mm
Overall Depth, D = d + /2 + clear cover = 250 + (12/2) + 20 = 276 mm
Adopt a Overall Depth of D = 280 mm
Effective depth, d = 254 mm
STEP – 2 : ( Size of Drop Panel )
Drop Panel size should not be less than = 8000/3 = 2666.67 mm
Minimum depth of Drop Panel = 1/4th of D = ( 1/4 ) x 280 = 70 mm
Depth of Slab including Drop Panel = 280 + 70 = 350 mm
Width of Column strip = Width of middle strip = 8000/2 = 4000 mm
Drop Panel Size = 4000 mm x 4000 mm
STEP – 3 : ( Load Calculations )
Live Load = 6 kn/m2
Super Imposed Dead Load = 2.5 kn/m2 [ finishes ]
Dead Load of Slab = 0.28 x 25 = 7 kn/m2
Dead Load of Drop Panel = 0.07 x 25 = 1.75 kn/m2
Total Characteristics Load, WC = 6 + 2.5 + 7 + 1.75 = 17.25 kn/m2
Factored Load, Wf = 17.25 x 1.5 = 25.88 kn/m2
L1 & L2 = c/c span = 8 m ;
Ln = 8 – (0.6/2) – (0.6/2) = 7.4 m
Design Load, Wn = Wf x Ln x L2 = 25.88 x 7.4 x 8 = 1532.10 kn
STEP – 4 : ( Bending Moment Calculation )
Mo = ( Wn . Ln ) / 8 = (1532.10 x 7.4) / 8 = 1417.19 kn.m  1418 kn.m
As per IS 456 : 2000, Clause – 31.4.3.2
-ve design moment = 0.65 x Mo = 0.65 x 1418 = 921.70 kn.m
+ve design moment = 0.35 x Mo = 0.35 x 1418 = 496.30 kn.m
As per IS 456 : 2000, Clause – 31.5.5.1 & 31.5.5.3
TYPE
COLUMN STRIP ( kn.m )
MIDDLE STRIP ( kn.m )
- ve moment
921.70 x 0.75 = 691.28
921.70 – 691.28 = 230.42
+ ve moment
496.30 x 0.60 = 297.78
12
496.30 – 297.78 = 198.52
STEP – 5 : ( Check for effective depth, d )
dreq. (slab) = √ (Mu / 0.138 x fck x b ) = √ [ ( 691.28 x 106 ) / ( 0.138 x 25 x 4000 ) ]
= 223.81 mm < 254 mm
dprov. (drop panel) = 254 + 70 = 324 mm < 350 mm
Hence, it’s safe..
dreq. < dprov.
STEP – 6 : ( Main Reinforcement calculation )
BM = 0.87 x fy x Ast x d x [ 1 – ( fy x Ast ) / ( fck x b x d ) ]
1. COLUMN STRIP : [ -ve reinforcement ]
691.28 x 106 = 0.87 x 500 x Ast x 350 x [ 1 – ( 500 x Ast ) / ( 25 x 4000 x 350 ) ]
Ast = 4880.74 mm2
Assuming 12 mm, ast = 113.10 mm2
Spacing, Sv = ( 4000 x ast ) / Ast = ( 4000 x 113.10 ) / ( 4880.74 ) = 90 mm c/c
[ +ve reinforcement ] :
Ast = 2013.80 mm2
Sv = 220 mm c/c
2. MIDDLE STRIP : [ -ve reinforcement ]
230.42 x 106 = 0.87 x 500 x Ast x 324 x [ 1 – ( 500 x Ast ) / ( 25 x 4000 x 324 ) ]
Ast = 2178.89 mm2
Sv = 200 mm c/c
[ +ve reinforcement ] :
Ast = 1440.57 mm2
Sv = 300 mm c/c
STEP – 7 : ( Check for punching shear )
d/2 distance from column
600/2 = 300 mm
Perimeter of critical section = bo = 4 ( a + d ) = 4 ( 600 + 300 ) = 3600 mm
Shear Force on this plane, Vu = W [ L x L – ( a + d ) ( a + d ) ] = 15.5 [ 82 – 0.81 ] = 979.45 kn
Nominal Shear stress = ( Vu / b . d ) = (979.45 x 103 ) / ( 4000 x 350 ) = 0.70 n/mm2
ks = ( 0.5 + ßc ) = 0.5 + 1 = 1.5 but not greater than 1.
ßc = 0.6/0.6 = 1 [ Ratio of short side to long side of the column ]
τc = 0.25 √fck = 0.25 x √25 = 1.25 n/mm2
ks . τc = 1 x 1.25 = 1.25 n/mm2
... The actual shear stress of 0.70 n/mm2 is within safe permissible limits.
13
STEP – 8 : ( Integrity Reinforcement )
As = ( 0.5 . wf . L1 . L2 ) / ( 0.87 . fy )
= [ ( 0.5 . 25.88 . 8 . 8 ) / ( 0.87 . 500 ) x 103 ] = 1903.82 mm2
Provide two 25 mm diameter bars, with a length of 2.Ld = ( 2 x 1071.43 ) = 2142.86 mm2
passing through the column cage each way.
COLUMN STRIP
= DROP WIDTH
COLUMN STRIP
= DROP WIDTH
MIDDLE STRIP
4000
COLUMN STRIP
= DROP WIDTH
# 12 - 110 c/c
4000
MIDDLE STRIP
# 12 - 260 c/c
# 12 - 250 c/c
4000
COLUMN STRIP
= DROP WIDTH
# 12 - 280 c/c
8000
BOTTOM BARS
TOP BARS
REINFORCEMENT DETAILS OF FLAT SLAB
260
# 12 - 110 c/c
# 12 - 260 c/c
800
REINFORCEMENT DETAILS OF COLUMN STRIP
350
# 12 - 110 c/c
# 12 - 260 c/c
800
REINFORCEMENT DETAILS OF MIDDLE STRIP
14
TWO WAY SLAB [ AHU Room, Electric room, others, etc., ]
REQUIRED DATAS :
Lx = 5 m
Ly = 6 m
fck = 25 n/mm2
fy = 500 n/mm2
( ly / lx ) = ( 6 / 5 ) = 1.2 --- Two way slab simply supported on all sides with provision for
torsion at corners.
STEP – 1 : ( Depth of Slab )
As the span is more than 3.5 m, adopt a span/depth ratio of 25.
Effective depth, d = Span / 25 = 5000 / 25 = 200 mm
assuming 12 mm and clear cover = 20 mm
Overall Depth, D = d + /2 + clear cover = 200 + (12/2) + 20 = 226 mm
Adopt a Overall Depth of D = 230 mm
Effective depth, d = 204 mm
STEP – 2 : ( Effective Span )
Effective span = Clear span + effective depth = 5 + 0.204 = 5.204 m
STEP – 3 : ( Loads )
Dead Load = 0.2 x 25 = 5 kn/m2
Live Load = 2.5 kn/m2
Super Imposed Dead Load = 1 kn/m2 [ finishes ]
Considering 1 m width, Total Characteristic Loads = 8.5 kn/m
Factored Load = 8.5 x 1.5 = 12.75 kn/m
STEP – 4 : ( Ultimate Design moments and shear forces )
From IS 456 : 2000, Table 27 – co-efficients
αx = 0.084
;
αy = 0.059
Mx = αx . w. lx2
= 0.084 x 12.75 x 5.2042 = 29 kn.m
My = αy . w. lx2
= 0.059 x 12.75 x 5.2042 = 20.37 kn.m
Vux = 0.5 . w . lx = 0.5 x 12.75 x 5.204 = 33.18 kn
STEP – 5 : ( Check for effective depth, d )
Mmax = 0.138 fck b d2
d = √ [ ( 29 x 106 ) / ( 0.138 x 25 x 1000 ) ] = 91.68 mm < 204 mm
15
Hence, the effective depth selected is sufficient to resist the design ultimate moment.
STEP – 6 : ( Reinforcements along Short & Long span direction )
Astmin = (0.12/100) x b x D = (0.12/100) x 1000 x 230 = 276 mm2
BM = 0.87 x fy x Ast x d x [ 1 – ( fy x Ast ) / ( fck x b x d ) ]
29 x 106 = 0.87 x 500 x Ast x 204 x [ 1 – ( 500 x Ast ) / ( 25 x 1000 x 204 ) ]
Ast = 338 mm2
Spacing, Sv = ( 1000 x ast ) / Ast = ( 1000 x 113.10 ) / ( 338 ) = 300 mm c/c
Adopt 12 mm diameter bars at 300 mmc/c in shorter span direction.
Effective depth, d = 204 - 12 = 192 mm
20.37 x 106 = 0.87 x 500 x Ast x 192 x [ 1 – ( 500 x Ast ) / ( 25 x 1000 x 192 ) ]
Ast = 250.43 mm2
Spacing, Sv = ( 1000 x ast ) / Ast = ( 1000 x 113.10 ) / ( 250.43 ) = 300 mm c/c
Hence, provide 12 mm diameter bars at 300 mm c/c in the long span direction.
STEP – 7 : ( Check for Shear stress )
Considering the short span lx and unit width of slab, the shear stress is given by.,
τv = Vu / b.d = 33.18 x 103 / ( 1000 x 204 ) = 0.16 n/mm2
pt = ( 100 x Ast ) / ( b . d ) = ( 100 x 338 ) / ( 1000 x 204 ) = 0.17 %
k . τc = 1.12 x 0.304 = 0.34 n/mm2 > 0.16 n/mm2 ( τv )
Hence, the slab is safe against shear.
STEP – 8 : ( Check for Deflection )
( L / d ) basic = 20 x Modification Factor
FS = 0.58 . fy . ( Ast req. / Ast prov. )
FS = 0.58 . 500 . ( 338 / 452.39 ) = 216.67 n/mm2
M F = 1.8
( L / d )max = 20 x 1.8 = 36
( L / d )provided = 5204 / 204 = 25.51 < 36
Hence, deflection control is satisfied.
STEP – 9 : ( Torsion Reinforcements at corners )
Area of reinforcements in each of the four layers = 0.75 x 338 = 253.5 mm2
Distance over which torsion reinforcement is provided = ( L / 5 ) x short span
= ( L / 5 ) x 5000 = 1000 mm
Provide 8 mm diameter bars at 150 mm c/c for a length of 1000 mm at all four corners in 4
layers.
16
MIDDLE STRIP
4500
EDGE STRIP
EDGE STRIP
600
600
# 8 - 150 c/c
EDGE STRIP
625
0.10 Ly = 600
X
EDGE STRIP
625
Ly = 6 m
MIDDLE STRIP
3750
X
TORSION REINFORCEMENT
Lx = 5 m
# 12 - 300 c/c
230
# 12 - 300 c/c
5000
SECTION X X
REINFORCEMENT DETAILS IN TWO WAY SLAB
( with provision for torsion at corners )
EDGE BEAM
REQUIRED DATAS :
fck = 25 n/mm2
fy = 500 n/mm2
Clear Span = 8 m
Bearing wall thickness = 300 mm
STEP – 1 : ( Cross sectional Dimensions )
Assuming effective depth = ( Span / 15 ) = [ ( 8 x 103 ) / 15 ] = 533.33 mm
Adopt d = 600 mm, D = 650 mm, b = 300 mm
Cover to tension steel = 50 mm
STEP – 2 : ( Loads )
Live Load, q = 6 kn/m
Self weight of beam = 0.65 x 0.3 x 25 = 4.88 kn/m
Total Dead Load, g = 3.8 + 4.88 = 8.68 kn/m
17
Total characteristics load = 19.56 kn/m
STEP – 3 : ( Bending moments and Shear forces )
Negative B.M. at interior support is computed as,
Mu (-ve) = 1.5 [ gL2/10 + qL2/9 ] = 1.5 [ ( 8.68 x 82 ) / 10 + ( 6 x 82 ) / 9 ] = 147.33 kn.m
Positive B.M. at centre of span is computed as,
Mu (+ve) = 1.5 [ gL2/12 + qL2/10 ] = 1.5 [ ( 8.68 x 82 ) / 12 + ( 6 x 82 ) / 10 ] = 127.04 kn.m
Maximum shear force at the support section is given by,
Vu = 0.6 L ( g + q ) 1.5 = ( 0.6 x 8 ) ( 6 + 8.68 ) 1.5 = 105.70 kn.m
STEP – 4 : ( Limiting moment of resistance )
Mu,lim = 0.138 fck b d2 = [ 0.138 x 25 x 300 x 6002 ] x 10-6 = 372.6 kn.m
Since Mu < Mu,lim, the section is under reinforced.
STEP – 5 : ( Main Reinforcements )
Mu (-ve) = 0.87 x fy x Ast x d x [ 1 – ( fy x Ast ) / ( fck x b x d ) ]
147.33 x 106 = 0.87 x 500 x Ast x 600 x [ 1 – ( 500 x Ast ) / ( 25 x 300 x 600 ) ]
Ast = 605.18 mm2
Using 2 bars of 20 mm diameter on the tension side (Ast = 628.32 mm2)at supports.
Mu (+ve) = 0.87 x fy x Ast x d x [ 1 – ( fy x Ast ) / ( fck x b x d ) ]
127.04 x 106 = 0.87 x 500 x Ast x 600 x [ 1 – ( 500 x Ast ) / ( 25 x 300 x 600 ) ]
Ast = 516.37 mm2
Using 2 bars of 20 mm diameter on the tension face (Ast = 628.32 mm2) at midspan.
STEP – 6 : ( Shear Reinforcements )
τv = Vu / b.d = 105.70 x 103 / ( 300 x 600 ) = 0.59 n/mm2
pt = ( 100 x Ast ) / ( b . d )
= ( 100 x 628.32 ) / ( 300 x 600 ) = 0.35 %
τc = 0.412 n/mm2 < 0.59 n/mm2 ( τv )
Since τv > τc , shear reinforcements are to be designed to resist the balance shear computed as
Vus = [ 105.70 – ( 0.412 x 300 x 600 ) 10-3 ] = 31.65 kn
Sv = [ 0.87 fy Asv d / Vus ]
= [ 0.87 x 500 x 2 x 50.27 x 600 / 31.65 x 103 ] = 829.10 mm
Using 8 mm diameter 2 legged stirrups at 300 mm centres throughout the beam.
18
Y
X
Y
X
# 8 - 300 c/c
650
2 - # 25
2 - # 20
6000
6000
650
2 - # 25
2 - # 20
300
SECTION - XX
SECTION - YY
REINFORCEMENT DETAILS IN CONTINUOUS BEAM
AXIALLY LOADED COLUMN
REQUIRED DATAS :
Number of story = Basement + G + 2 ( 3 STORY )
Height of floor = 4.675 m
Size of column = 600 mm x 600 mm
External and Internal wall thickness = 300 mm both
fck = 25 n/mm2
&
fy = 500 n/mm2
SBC = 200 kn/m2
I.
INTERIOR COLUMN :
Load Calculations -1. Factored Load on Roof Slab = 21.08 kn/m2
2. Factored Load on Floor Slab = 15.05 kn/m2
3. Factored Load of Wall
= 34.88 kn/m
Floor Area surrounded by column = 8 x 8 = 64 m2
Roof to 2nd Floor –
Roof Load = 21.08 x 8 x 8 = 1349.12 kn
Walls Load = 34.88 x 4.075 x 0.15 = 21.32 kn
Total Load = 1370.44 kn
2nd Floor to 1st Floor –
Floor Load = 15.05 x 8 x 8 = 963.20 kn
Walls Load = 34.88 x 4.075 x 0.15 = 21.32 kn
Load from Top Floor = 1370.44 kn
Cummulative Load = 2354.96 kn
1st Floor to Plinth –
19
Floor Load = 15.05 x 8 x 8 = 963.20 kn
Walls Load = 34.88 x 4.075 x 0.15 = 21.32 kn
Load from Top Floor = 2354.96 kn
Cummulative Load = 3339.48 kn
II Floor –
STEP – 1 : ( Slenderness Ratio )
L / D = 4675 / 600 = 7.79 < 12
Hence, the column is designed as a Short Column.
nd
STEP – 2 : ( Minimum Eccentricity )
emin = [ L / 500 + D / 30 ] = [ ( 4675 / 500 ) + ( 600 / 30 ) ] = 29.35 > 20 mm
0.05 D = 0.05 x 600 = 30 mm
STEP – 3 : ( Longitudinal Reinforcements )
Pu = [ 0.4 . fck . Ag + ( 0.67 . fy + 0.4 . fck ) Asc ]
1370.44 x 103 = ( 0.4 . 25 . 600 . 600 ) + ( 0.67 . 500 + 0.4 . 25 ) Asc
Asc = 5568.87 mm2
Assume 25 mm & 20 mm diameter bars
Provide 7 nos. of 25 mm dia bars, Asc = 3436.12 mm2
Provide 7 nos. of 20 mm dia bars, Asc = 2199.11 mm2
Total Asc = 5635.23 mm2 > 5568.87 mm2
Minimum Asc = 0.8 % of B . D = 0.8/100 x 600 x 600 = 2880 mm2
Maximum Asc = 6 % of B . D = 6/100 x 600 x 600 = 21600 mm2
STEP – 4 : ( Lateral Ties )
I.
( 1 / 4 ) x 25 = 6.25 mm
II.
6 mm
STEP – 5 : ( Spacing )
I.
Least Lateral Dimensions = 600 mm
II.
16 x D =16 x 25 = 400 mm
III.
300 mm
Provide 8 mm diameter ties at 300 mm c/c.
20
520
600
4 - # 25 mm
4 - # 20 mm
TIES - # 8 - 300 c/c
REINFORCEMENT DETAILS
UNIAXIAL ECCENTRIC LOADED COLUMN
REQUIRED DATAS :
Size of column = 600 mm x 650 mm
fck = 25 n/mm2
&
fy = 500 n/mm2
Ultimate Load, Pu = 1370.44 kn
Ultimate moment, Mu = 200 kn.m
STEP – 1 : ( Non-Dimensional Parameters )
( Pu / fck b D ) = 0.14
( Mu / fck b D2 ) = 0.03
STEP – 2 : ( Longitudinal Reinforcements )
Adopting an effective cover of 50 mm = d /
( d / / D ) = 0.10
Refer chart 33 of SP : 16 and read out the ratio ( p /fck ) = 0.02
P = 25 x 0.02 = 0.50
Asc = ( p b D / 100 ) = 1950 mm2
Adopt 5 bars of 25 mm dia.
Area provided = 2454.37 mm2
Provide 5 bars on each of the short faces at an effective cover of 50 mm.
STEP – : 3 ( Lateral ties )
I.
( 1 / 4 ) x 25 = 6.25 mm
II.
6 mm
STEP – 4 : ( Spacing )
I.
Least Lateral Dimensions = 600 mm
II.
16 x D =16 x 25 = 400 mm
III.
300 mm
21
Provide 8 mm diameter ties at 300 mm c/c.
650
600
8 - # 25 mm
# 8 mm TIES
@ 300 mm c/c
REINFORCEMENTS IN COLUMNS
WITH UNIAXIAL BENDING
BIAXIAL ECCENTRIC LOADED COLUMN
REQUIRED DATAS :
Size of column = 600 mm x 600 mm
fck = 25 n/mm2
&
fy = 500 n/mm2
Ultimate Load, Pu = 1370.44 kn
Ultimate moment, Mu = 120 kn.m @ Major Axis
Ultimate moment, Mu = 90 kn.m @ Minor Axis
STEP – 1 : ( Equivalent Moment )
The reinforcement in section is designed for the axial compressive load Pu and the equivalent
moment given by the relation,
Mu = 1.15 √ ( M2ux + M2uy ) = 172.5 kn.m
STEP – 2 : ( Non-Dimensional Parameters )
( Pu / fck b D ) = 0.15
( Mu / fck b D2 ) = 0.03
STEP – 3 : ( Longitudinal Reinforcements )
Adopting an effective cover of 50 mm = d /
( d / / D ) = 0.10
Refer chart 33 of SP : 16 and read out the ratio ( p / fck ) = 0.06
P = 25 x 0.04 = 1
Asc = ( p b D / 100 ) = 3600 mm2
22
Adopt 8 bars of 25 mm dia.
Area provided = 3926.99 mm2
Provide 8 bars of 25 mm dia with 3 bars in each face.
P = ( 100 x 3926.99 ) / ( 600 x 600 ) = 1.09 and the ratio ( p / fck ) = 0.04
( Mux1 / fck b D2 ) = 0.04
Mux1 = 0.04 x 25 x 600 x 6002 = 235.44 kn.m
Due to symmetry, Mux1 = Muy1 = 235.44 kn.m
Pu = [ 0.45 fck Ac + 0.75 fy As ] = 5478.44 kn
Pu / Puz = 0.25
STEP – : 4 ( Check for safety under Biaxial Loading )
( Mux / Mux1 )αn + ( Muy / Muy1 ) αn < 1
0.51 + 0.38 = 0.89 < 1
Hence, the section is safe under specified loading.
STEP – : 5 ( Lateral ties )
III.
( 1 / 4 ) x 25 = 6.25 mm
IV. 6 mm
STEP – 6 : ( Spacing )
IV. Least Lateral Dimensions = 600 mm
V.
16 x D =16 x 25 = 400 mm
VI. 300 mm
Provide 8 mm diameter ties at 300 mm c/c.
600
600
8 - # 25 mm
# 8 mm TIES
@ 300 mm c/c
REINFORCEMENTS IN COLUMNS
WITH BIAXIAL BENDING
CIRCULAR COLUMN
REQUIRED DATAS :
Diameter of column, D = 600 mm
Unsupported Length, L = 4675 mm
fck = 25 n/mm2 ; fy = 500 n/mm2
23
STEP – 1 : ( Slenderness Ratio )
L / D = 4675 / 600 = 7.79 < 12
Hence, the column is designed as a Short Column.
STEP – 2 : ( Minimum Eccentricity )
emin = [ L / 500 + D / 30 ] = [ ( 4675 / 500 ) + ( 600 / 30 ) ] = 29.35 > 20 mm
0.05 D = 0.05 x 600 = 30 mm
STEP – 3 : ( Longitudinal Reinforcements )
Pu = [ 0.4 . fck . Ag + ( 0.67 . fy + 0.4 . fck ) Asc ]
1370.44 x 103 = ( 0.4 . 25 . 600 . 600 ) + ( 0.67 . 500 + 0.4 . 25 ) Asc
Asc = 5568.87 mm2
Minimum Asc = 0.8 % of B . D = 0.8/100 x 600 x 600 = 2880 mm2
Maximum Asc = 6 % of B . D = 6/100 x 600 x 600 = 21600 mm2
STEP – 4 : ( Helical Reinforcement spirals )
Assuming clear cover of 40 mm over spirals
Core diameter = [ 600 – ( 2 x 40 ) ] = 520 mm
Area of core = Ac = 206736.43 mm2
Volume of core/m = Vc = 206736.43 x 103 mm3
Gross area of section = Ag = 282743.34 mm2
Using 8 mm diameter helical spirals at a pitch ‘p’ mm, the volume of helical spiral per metre
length is given by
Vns = (80859 x 103) / p mm3 / m
( Vns / Vc ) < 0.36 [ Ag /Ac ) – 1 ] ( fck / fy )
Solving pitch, p = 58.93 mm
p < 75 mm or ( core diameter / 6 ) = 86.67 mm
p > 25 mm or ( 3 x diameter of helix ) = 24 mm
Hence, provide 10 mm diameter spirals at a pitch of 86 mm.
24
600
86
# 10 - spiral @ pitch 36 mm
12 - # 25
CLEAR COVER
40 mm
# 10 - spiral @ pitch 36 mm
REINFORCEMENTS IN HELICALLY
REINFORCED COLUMN
DOG LEGGED STAIRCASE [ Emergency Exit ]
REQUIRED DATAS :
No. of steps = 10
Tread = 250 mm ; Rise 150 mm
UDL = 4 kn/m2
Weight of finishes = 1 kn/m2
STEP – 1 : ( Effective Span )
The landing slabs span parallel with the risers ( perpendicular to the flight direction )
Effective span of flight slab = 2.5 + 1 = 3.5 m
STEP – 2 : ( Loads )
Consider 1 m of width of flight slab.
Live Load = 4 kn/m
Self weight of steps = 1.5 kn/m
Assume the total thickness of waist slab as 3.5 x 47 =164.5  165 mm
Self weight of slab per m2 of inclined area = 1 x 0.165 x 25 = 4.13 kn/m
Self weight of slab per m of horizontal length = 4.82 kn/m
Weight of finishes = 1 kn/m
Total Characteristics load = 15.45 kn/m
Design Load = 23.18 kn/m
The same Loading is assumed in the landing portions also.
25
STEP – 3 : ( Bending moments and shearing force )
Design B.M @ midspan = 35.49 kn.m
Design S.F @ support = 40.57 kn
STEP – 4 : ( Depth required )
For balanced sections of M25 grade concrete and Fe500 grade steel,
Moment of resistance = 2.66 b d2
Effective depth required = 115.51
Providing 10 mm dia. Bars with a nominal cover of 20 mm,
Effective depth provided = 165 – 20 – 5 = 140 mm > 115.51 mm
STEP – 5 : ( Reinforcements )
BM = 0.87 x fy x Ast x d x [ 1 – ( fy x Ast ) / ( fck x b x d ) ]
23.18 x 106 = 0.87 x 500 x Ast x 140 x [ 1 – ( 500 x Ast ) / ( 25 x 1000 x 140 ) ]
Ast = 403.93 mm2
Anchorage Length required is, Ld = ( 0.87 x 500 x 8 ) / ( 4 x 1.4 x 1.6 ) = 388.39 mm
In addition to Ld, we have to provide sufficient balancing moment to prevent overturning.
Distribution reinforcement
= 0.12% of b.D = 198 mm2
Spacing, Sv = ( 1000 x ast ) / Ast = ( 1000 x 50.27 ) / ( 198 ) = 253.89 mm  240 mm
Provide 8 mm dia. bars at 240 mm c/c
STEP – 6 : ( Check for shear & stiffness )
Nominal shear stress = 0.25 n/mm2
Minimum value of permissible shear stress = 0.28 n/mm2
Hence safe.
For stiffness, basic value of L / d ratio = 20
% of tension steel = 0.47 % < 0.72 %
Stress in steel, fs = ( 0.58 x 500 x 8 ) / 10 = 232 n/mm2
Modification factor = 1.5
Effective depth required for stiffness = 93.33 mm
Effective depth provided = 140 mm > 93.33 mm
Since the effective depth provided is more than both the requirements, a lesser value of overall
depth may be assumed ( D = 140 mm ) and checked.
26
165
T = 300
400
R = 160
WAIST SLAB
# 12 - 240 c/c ( main )
LANDING BEAM
4000
REINFORCEMENT IN STAIRCASE FLIGHT
RETAINING WALL / GUIDE WALL
REQUIRED DATAS :
Height of Embankment above ground level = 4 m
Safe Bearing capacity of soil = 200 kn/m2
Density of Earth = 18 kn/m3
Angle of Repose = 30°
Co-efficient of friction = 0.5
M 25 & Fe 500
STEP – 1 : ( Dimensions of retaining wall )
Minimum depth of Foundation = ( p / ɣe ) [ (1-sin) / (1-sin) ]2 = 1.23 m
∴ Provide depth of foundation = 1.2 m
Overall Depth of wall = 4 + 1.2 = 5.2 m
Thickness of base slab = (h/12) = 5200/12 = 433.33 mm
Adopt thickness of base slab, tb = 450 mm
∴ Height of stem, hs = 5.2 – 0.45 = 4.75 m
Co-efficient of Active earth pressure is computed as,
Ca = [ (1-sin) / (1-sin) ] = 0.333
Width of base slab = 0.4 H to 0.7 H = 1.9 m to 3.33 m
Adopt B = 3 m
Width of heel slab = 0.5 B = 0.5 x 3 = 1.5 m
Width of toe slab = 3 – ( 1.5 + 0.45 ) = 1.05 m
STEP – 2 : ( Stability computations )
27
Vertical
Load
Parts
Force (kn)
Distance from heel (m)
Moment (kn.m)
W1
Soil on Heel
128.25
0.75
96.19
W2
Stem ( Triangle )
38.59
1.63
62.71
W3
Stem ( Rectangle )
23.75
1.85
43.94
W4
Base Slab
33.75
∑W = 224.34
1.50
50.63
∑M = 253.46
Total =
STEP – 3 : ( Design of Stem )
Mu = ( 1.5 Ca ɣe h3 ) / 6 = 159.15 kn.m
Consider cover = 50 mm
Mu = 0.87 x fy x Ast x d x [ 1 – ( fy x Ast ) / ( fck x b x d ) ]
253.46 x 106 = 0.87 x 500 x Ast x 400 x [ 1 – ( 500 x Ast ) / ( 25 x 1000 x 400 ) ]
Ast = 1581.77 mm2
Provide 20 mm dia. bars @ 150 mm c/c.
STEP – 4 : ( Calculation of Earth pressure )
Distance of point of application of resultant from edge of heel, Z = ∑M / ∑W = 1.13
Eccentricity, e = [ Z – ( 0.5 x B ) ] = - 0.37 < 0.5
B / 6 = 0.5
Pmax & Pmin = ( W / B ) [ 1 + 6 e / B ]
Pmax = 130.12 kn/m2
Pmin = 19.44 kn/m2
STEP – 5 : ( Calculation of Moment )
By interpolating the Pmax & Pmin values with required distance, we get the pressure distribution
values
below for the heel and toe design.
P1
-118.87 kn/m2
P2
-80.13 kn/m2
P3
21.97 kn/m2
P4
77.31 kn/m2
Moment for the design of Heel = 1.5 x (P3 x (1.52 / 2) + [ (P4 - P3) x (1.52) / 3 ]
= 37.07 + 41.51 = 78.58 kn.m
28
Moment for the design of Toe = 1.5 x (P2 x (1.052 / 2) + [ (P1 – P2) x (1.052) / 3 ]
= 66.26 + 14.24 = 80.50 kn.m
STEP – 6 : ( Design of Heel )
78.58 x 106 = 0.87 x 500 x Ast x 400 x [ 1 – ( 500 x Ast ) / ( 25 x 1000 x 400 ) ]
Ast = 462.30 mm2
Provide 20 mm dia. bars @ 300 mm c/c.
STEP – 7 : ( Design of Toe )
80.50 x 106 = 0.87 x 500 x Ast x 400 x [ 1 – ( 500 x Ast ) / ( 25 x 1000 x 400 ) ]
Ast = 473.87 mm2
Provide 20 mm dia. bars @ 300 mm c/c.
STEP – 8 : ( Distribution reinforcements )
Ast = 0.12% of b .D
0.12 / 100 x b x D = 432 mm2
Assuming 12 mm dia. bars
Spacing, Sv = ( 1000 x ast ) / Ast = 261.8  250 mm
Provide 12 mm dia. bars @ 250 mm c/c as distribution reinforcements for Stem, Heel slab &
Toe slab.
STEP – 9 : ( Check for overturning )
Mo = ( Ca ɣe hs3 ) / 6 = 106.10 kn.m
Ms = ∑W ( B – Z ) = 419.52 kn.m
F.0.S against overturning = ( 0.9 x Ms ) / Mo = 3.56 > 1.4
Hence it’s safe against overturning.
STEP – 10 : ( Check for overturning )
P = 0.5 Ca ɣe h2 = 80.31 kn.m
∑W = 224.34 kn
F.0.S against sliding = ( 0.9 x ∑W ) / P = 2.51 > 1.4
Hence it’s safe against sliding.
29
0.2
# 20 - 150 c/c
1.75
4.75
3
# 12 - 250 c/c
0.45
# 20 - 300 c/c
# 20 - 300 c/c
1.05
0.45
1.5
REINFORCEMENT DETAILS IN
CANTILEVER RETAINING WALL
ISOLATED COLUMN FOOTING
REQUIRED DATAS :
Factored Load = 3600 kn
Size of Column = 600 x 600 mm
S.B.C of Soil = 200 kn/m2
M 25 & Fe 500 HYSD bars
STEP – 1 : ( Size of Footing )
Load on Column = 3600 kn
Weight of footing and backfill at 10% = 360 kn
Total Load = 3960 kn
Area of footing = ( 3960 / 200 ) = 19.8 m2
Size of Footing = L = B = √19.8 = 4.45 m
Adopt 4.5 m x 4.5 m square footing.
Net soil pressure at ultimate loads with a load factor of 1.5 is given by
qu = ( 3600 x 1.5 ) / ( 4.5 x 4.5 ) = 266.67 kn/m2 = 0.27 n/mm2
STEP – 2 : ( One way Shear )
The critical section is at a distance ‘d’ from the column face.
Factored shear force, Vu1 = ( 0.27 x 4500 ) ( 1950 – d ) = 1215 ( 1950 – d )
Assuming percentage of reinforcement in footing, pt = 0.25 %
τc = 0.36 n/mm2
30
One way shear resistance = Vct = ( 0.36 x 4500 x d ) = ( 1620 d ) N
Vu1 < Vct
1215 ( 1950 – d ) < 1620 d
d > 835.71 mm
STEP – 3 : ( Two way Shear )
Assuming the effective depth of slab, d = 835.71 mm and computing the 2 way shear resistance
at a critical section (d/2) from the face of the column, we have the relation,
Vu2 = 0.26 [ 45002 – ( 600 + d2 ] = 0.26 [ 45002 – ( 600 + 835.712 ] = 5083257.09 N
Two way shear resistance Vc2 is computed as,
Vc2 = ks τc [ 4 ( 600 + d ) d ] where ks = 1 & τc = 0.25 = 1.118 n/mm2
Vc2 = 2683.2 d + 4.47 d2
Vu2 < Vct
5083257.09 N < ( 2683.2 d + 4.47 d2 )
Solving, d = 807.69 mm
Hence, one way shear is more critical.
Adopt effective depth, d = 1712.5 mm and Overall depth, D = 1800 mm
STEP – 4 : ( Design of Reinforcements )
Ultimate moment at column face, Mu = 3960 x 1.950 x 0.5 = 3861 kn.m / m
( Mu / bd2 ) = 1.32
Refer table – 2 of SP : 16 and interpolate the percentage reinforcement as, pt = 0.32 % assumed
for 1 way-shear.
Ast = ( pt x b x d ) / 100 = 5480 mm2 / m
Using 25 mm dia bars,
Spacing of bars, S = 100 mm c/c
Adopt 25 mm dia bars @ 100 mm c/c in both directions.
31
600
COLUMN REINFORCEMENT
1800
# 25 - 100 mm c/c
( BOTH WAYS )
4500
SECTION X - X
SECTION FOR MOMENT
SECTION FOR 1 WAY SHEAR
X
600
4500
X
SECTION FOR 2 WAY SHEAR
600
d/2
d
4500
PLAN
DESIGN OF COLUMN FOOTING
32
CONCLUSION
➢ The Shopping Complex is designed for about 21.920 sq. m.
➢ For the safety of the Shopping Complex, we checked out the Bending
Moment, Shear force and Deflection control by using Limit State
Method.
➢ Thus the Shopping Complex is designed with safe and economical.
33
REFERENCES
➢ MODEL BUILDING BYE – LAWS, 2016
➢ IS code of practice for Plain and RCC - IS 456-2000
➢ IS code of practice for design loads - IS 875 (Part 1 – Super imposed
Loads) – 1987 ; (Part 2 – Imposed Loads) – 1987
➢ Design aids for RCC structures (S.P. 16), Bureau of Indian Standards,
New Delhi.
➢ Reinforced Concrete Design – AUTHOR – N. Krishna Raju & R.N.
Pranesh : New Age International (P) Limited, Publishers
34