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Reinforced Concrete Design Worked examples to BS 8110
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Reinforced Concrete Design
Worked examples to BS 8110
Mohamed Elhassan B.Sc. M.Sc.
University of Khartoum, Sudan
© 2018 Mohamed Elhassan
All rights reserved. No part of this publication may be reproduced in any other form
without the written consent of the copyright holder.
Disclaimer
The author bears no responsibility for any consequence that may occur as a result of
the usage of this work. This book should be treated more like a design note book than
as a standard design textbook. Kindly refer to more standard books on structural
design for more knowledge.
i
Introduction
This work is intended to provide students, and practising engineers, with a guide to
meeting the requirements of BS 8110: Structural Use of Concrete part 1 and contains
worked examples which have been prepared to give a detailed indication of the
process of designing reinforced concrete members to this code.
ii
Contents
Page
Introduction
ii
Symbols
iv
Example 1
Beam sizing
1
Example 2
Simply supported singly reinforced rectangular beam
2
Example 3
Simply supported doubly reinforced rectangular beam
4
Example 4
Continuous beam
6
Example 5
Transfer beam
9
Example 6
Continuous solid slab spanning in one direction
11
Example 7
Solid slab spanning in two direction
13
Example 8
Flat slab
15
Example 9
Ribbed slab
18
Example 10
Waffle slab
20
Example 11
Stair slab
22
Example 12
Short column subjected to axial load
24
Example 13
Short column subjected to axial load and moment in one direction
25
Example 14
Short column subjected to axial load and biaxial moment
26
Example 15
Braced slender column
27
Example 16
Shear wall
29
Example 17
Axially loaded footing
30
Example 18
Isolated footing subjected to axial load and moment
32
Example 19
Combined footing
34
Appendix
Selected tables from BS 8110
37
iii
Symbols
Symbol
Definition
𝑨𝑪
Area of concrete
𝑨𝑺
Area of tension reinforcement
𝑨𝑺,𝒎𝒊𝒏
Minimum area of reinforcement
𝑨𝑺,𝒑𝒓𝒐𝒗
Provided area of reinforcement
𝑨𝑺,𝒓𝒆𝒒
Required area of reinforcement
𝑨𝑺𝑪
Area of vertical reinforcement in a column
𝑨𝑺𝑽
Total cross-sectional area of shear reinforcement
𝒃
Width or effective width of the section or flange in the compression zone
𝒃𝒆
Breadth of the effective moment transfer strip
𝒃𝑽
Breadth of section (for a flanged beam width, below the flange)
𝑪
Cover
𝒅
Effective depth of the tension reinforcement
𝒆
Eccentricity of lateral load and shear centre
𝑭
Total design ultimate load (1.4 𝐺𝐾 + 1.6 𝑄𝐾 )
𝑭𝑪
Ultimate load capacity of wall
𝒇𝑪𝑼
Characteristic strength of concrete
𝒇𝒔
Estimated design service stress in the tension reinforcement
𝒇𝒚
Characteristic strength of reinforcement
𝒇𝒚𝒗
Characteristic strength of shear reinforcement
𝑮𝑲
Characteristic dead load
𝒉
Overall slab depth
𝒉
Depth of column section
𝒉𝒇
Thickness of the flange
𝑰
Second moment of area of shear walls
𝑲
A measure of the relative compressive stress in a member in flexure
𝑲′
Value of K above which compression reinforcement is required
𝑲𝑺
Horizontal coefficient of earth pressure
𝑳
Length of wall
𝑳𝒕
Length of wall in tension
𝒍
Effective span
iv
𝒍𝒆𝒙 , 𝒍𝒆𝒚
Effective height of a column about x and y axis
𝒍𝒙
Length of shorter side of two-way spanning slab
𝒍𝒚
Length of longer side of two-way spanning slab
𝑴
Design ultimate moment at the section considered
𝑴𝑹
Moment of resistance
𝑴𝒕
Design moment transferred between slab and column
𝒎𝒔𝒙
Maximum design ultimate moments of unit width and span lx
𝒎𝒔𝒚
Maximum design ultimate moments of unit width and span ly
𝑵
Design ultimate axial load on a column or wall
𝑸𝑲
Characteristic imposed load
𝑺
Distance between points of inflection
𝑺𝑽
Spacing of links along the member
𝒕
Thickness of wall
𝒖
Effective length of the outer perimeter of the zone
𝒖°
Effective length of the perimeter that touches the loaded area
𝑽
Design shear force due to ultimate loads
𝒗
Design shear stress at a cross section
𝒗𝒄
Design concrete shear stress
𝑽𝒆𝒇𝒇
Design effective shear including allowance for moment transfer
𝒗𝒔𝒙 , 𝒗𝒔𝒚
Shear force per unit width in x and y directions
𝑾
Total design ultimate load
𝒛
Lever arm
𝜶𝒔𝒙 , 𝜶𝒔𝒚
Moment coefficients for simply supported two-way spanning slabs
𝜷
Effective height factor for columns
𝜷𝒃
Redistribution ratio
𝜷𝒔𝒙 , 𝜷𝒔𝒚
Moment coefficients for restrained two-way spanning slabs
𝜷𝒗𝒙 , 𝜷𝒗𝒚
Shear force coefficients for restrained two-way spanning slabs
𝝈𝒕
Tensile stress in shear wall
∅𝒃
Bar diameter
∅𝒍
Link diameter
v
Job Title
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Subject:
Worked Example 1 - Beam Sizing
Client
UofK
Made by
ME
Date
2018
Checked by
Unless otherwise A concrete beam carries 𝐺 = 150 KN and 𝑄 = 50 KN Has an
𝐾
𝐾
stated, all
effective span of 6 m supports a 250 mm brick wall. Determine
references are
from: BS8110-1
suitable dimension for the beam if grade 35 for concrete is to be used.
It is recommended that the beam breadth match the wall thickness, b = 250 mm
so that b = 250 mm
Section 3.2.1.2.2
Loading
Assume 15 KN for the self-weight of the beam
Ultimate Load
F = 1.4 x (150 + 15) + 1.6 x 50 = 311 KN
F = 311 KN
Therefore maximum shear = 155.5 KN
Distributed load along the beam length = 311/6 = 51.83 KN/m
F = 51.83 KN/m
Bending
Maximum moment =
𝑊 × 𝐿2 51.83 × 62
=
= 233.25 KN. m
8
8
The case with no compression steel will be considered, So
Section 3.4.4.4
Table 3.3, 3.4
𝑀
< .156,
𝑏𝑑 2 𝑓𝑐𝑢
233.25 × 106
< .156,
250 × 𝑑 2 × 35
M = 233.25
KN.m
∴ d > 414 mm
For mild conditions of exposure, the cover = 20 mm. so for 8 mm
links and 16 mm bars Overall depth h = d + 20 + 8 + 16/2 = 450
Cover = 20 mm
mm. take h = 500 mm, then d = 464 mm
Shear
Section 3.4.5.2
shear stress v =
𝑉
155.5 × 1000
=
= 1.34 N/𝑚𝑚2
𝑏×𝑑
250 × 464
For C35 concrete, maximum v allowed = 0.8√35 = 4.73 N/mm2.
Therefore
v < 4.73
Shear OK
OK
Deflection
Table 3.9
basic span − effective depth ratio =
6000
= 12.93 < 20 OK
464
∴ A beam size of 250 mm by 500 mm deep would be suitable.
Deflection OK
1
Job Title
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Subject:
Worked Example 2 - Simply Supported singly reinforced Rec. Beam
Client
UofK
Made by
ME
Date
2018
Checked by
Unless otherwise A simply supported reinforced concrete beam carries 𝐺 = 30 KN/m
𝐾
stated, all
including self-weight and 𝑄𝐾 = 10 KN/m. Has a span of 5 m. Design
references are
from: BS8110-1
the beam for bending and shear if grade 30 and 460 are to be used
for concrete and all steel respectively. Take d = 550mm and b =
250mm
Section 3.2.1.2.2
Loading
Ultimate Load
F = 1.4 x 30 + 1.6 x 10 = 58 KN/m
Bending
𝑊 × 𝐿2 58 × 52
=
= 181.25 KN. m
8
8
Maximum moment =
F = 58 KN/m
M = 181.25 KN.m
6
Section 3.4.4.4
𝐾=
𝑀
181.25 × 10
=
= 0.079 < .156
2
𝑏𝑑 𝑓𝑐𝑢 250 × 5502 × 30
∴ No compression reinforcement is required
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐾
)}
0.9
z = 0.903d = 496.5 mm
𝐴𝑆 =
Z = 496.5 mm
𝑀
181.25 × 106
=
= 836 𝑚𝑚2
0.95𝑓𝑦 𝑧 0.95 × 460 × 496.5
Provide 3T20, area provided 942.6 mm2
Provide 3T20
Shear
𝑊×𝐿
= 58 × 5/2 = 145 KN
2
𝑉
145 × 1000
shear stress 𝑣 =
=
= 1.05 N/mm2 < 0.8√30
𝑏×𝑑
250 × 550
maximum design shear load =
Section 3.4.5.2
𝑣 = 1.05 𝑁/𝑚𝑚
shear links
at distance d from face of support V = 145-58x0.55 = 113.1 KN
shear stress 𝑣 =
𝑉
131.1 × 1000
=
= 0.954 N/mm2
𝑏×𝑑
250 × 550
1/3
Table 3.8
𝑣𝑐 =
0.79 100𝐴𝑆
(
)
𝛾𝑚
𝑏𝑑
400 1/4 𝑓𝑐𝑢 1/3
(
) ( )
𝑑
25
Only two 20 mm bars extent a distance d from the support face
2
Mohamed Elhassan-Civil Engineering Department-UofK
1
1
100𝐴𝑆 3
100 × 628.4 3
∴(
) =(
) = 0.77 < 3 ,
𝑏𝑑
250 × 550
1
1
400 4
400 4
(
) =(
) = 0.923 (𝑠ℎ𝑜𝑢𝑙𝑑 𝑛𝑜𝑡 𝑏𝑒 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑡 𝑜𝑛𝑒)
𝑑
550
(
𝑓𝑐𝑢 1/3
) = 1.05
25
vc = 0.51N/mm2
∴ 𝑣𝑐 = 0.51 𝑁/𝑚𝑚2
Table 3.7
𝐴𝑆𝑉 𝑏(𝑣 − 𝑣𝑐 ) 250 × (0.954 − 0.51)
=
=
= 0.254
𝑆𝑉
0.95𝑓𝑦𝑣
0.95 × 460
Provide T8 at 200 mm c/c
𝐴𝑆𝑉
2 × 50.3
=
= 0.5
𝑂𝐾
𝑆𝑉
200
Deflection
Table 3.9
basic span-effective depth ratio =
5000
550
= 9.1
basic ratio for simply supported rectangular beam = 20
Table 3.10
modification factor for tension reinforcement
𝑓𝑠 =
2𝑓𝑦 𝐴𝑆,𝑟𝑒𝑞
3𝐴𝑆,𝑝𝑟𝑜𝑣
∗
1
2 × 460 × 836
=
= 407.98
𝛽𝑏
3 × 942.6
𝑀
181.25 × 106
=
= 2.4
𝑏𝑑 2
250 × 5502
modification factor = 0.55 +
(477 − 𝑓𝑠 )
≤ 2.0 = 0.73
𝑀
120 (0.9 + 2 )
𝑏𝑑
basic ratio = 20 x 0.725 = 14.5
actual ratio = 9.1 < basic ratio = 14.5 OK
Deflection OK
3
Job Title
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Subject:
Worked Example 3 - Simply Supported doubly reinforced Rec. Beam
Client
UofK
Made by
ME
Date
2018
Checked by
Unless otherwise A rectangular beam is 250 mm wide by 600 mm depth. The beam
stated, all
is simply supported and spans 8 m. carries dead load 𝐺𝐾 = 20 KN/m
references are
from: BS8110-1
including self-weight and imposed load 𝑄𝐾 = 15 KN/m. The concrete
and steel grades are 30 and 460 respectively. Design the beam.
Loading
Section 3.2.1.2.2
Ultimate Load
F = 1.4 x 20 + 1.6 x 15 = 52 KN/m
F = 52 KN/m
Bending
Maximum moment =
Section 3.4.4.4
Table 3.3, 3.4
𝐾=
𝑊 × 𝐿2 52 × 82
=
= 416 KN. m
8
8
M = 416 KN.m
𝑀
𝑏𝑑 2 𝑓𝑐𝑢
For mild conditions of exposure the cover = 20 mm. so for 8 mm
links and 16 mm bars Overall depth h = d + 20 + 8 + 16/2 = 600
mm. d = 564 mm
416 × 106
= 0.1744 > 0.156
250 × 5642 × 30
∴ Compression reinforcement is required
Section 3.4.4.4
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐾′
)}
0.9
z = 0.777d = 438.2 mm
𝐴𝑆 ′ =
Z = 438.2 mm
(𝐾 − 𝐾 ′ )𝑏𝑑 2 𝑓𝑐𝑢 (0.1744 − 0.156) × 250 × 5642 × 30
=
= 190.25 𝑚𝑚2
0.95𝑓𝑦 (𝑑 − 𝑑 ′ )
0.95 × 460 × (564 − 36)
Provide 2T16, area provided 402.2 mm2
𝐴𝑆 =
𝐾′𝑏𝑑 2 𝑓𝑐𝑢
0.156 × 250 × 5642 × 30
+ 𝐴𝑆 ′ =
+ 190.25 = 2133.77 𝑚𝑚2
0.95𝑓𝑦 𝑧
0.95 × 460 × 438.2
Provide 5T25, area provided 2454.5 mm2
Section 3.4.5.2
Shear
𝑊×𝐿
maximum design shear load =
= 52 × 8/2 = 208 KN
2
𝑉
208 × 1000
shear stress 𝑣 =
=
= 1.475 N/mm2 < 0.8√30 OK
𝑏×𝑑
250 × 564
𝑣 = 1.5 𝑁/𝑚𝑚2
4
Mohamed Elhassan-Civil Engineering Department-UofK
shear links
at d from face of support V = 208- 52 x 0.565 = 178.67 KN
shear stress𝑣 =
Table 3.8
𝑉
178.67 × 1000
=
= 1.267 N/mm2
𝑏×𝑑
250 × 564
1/3
𝑣𝑐 =
0.79 100𝐴𝑆
(
)
𝛾𝑚
𝑏𝑑
400 1/4 𝑓𝑐𝑢 1/3
(
) ( )
𝑑
25
Only three 25 mm bars extent a distance d from the support face
1
1
100𝐴𝑆 3
100 × 1472.7 3
∴(
) =(
) = 1.023 < 3 ,
𝑏𝑑
250 × 550
1
1
400 4
400 4
(
) =(
) = 0.923 (𝑠ℎ𝑜𝑢𝑙𝑑 𝑛𝑜𝑡 𝑏𝑒 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑡 𝑜𝑛𝑒)
𝑑
550
(
𝑓𝑐𝑢 1/3
) = 1.0626
25
vc = 0.687N/mm2
∴ 𝑣𝑐 = 0.687 𝑁/𝑚𝑚2
Table 3.7
𝐴𝑆𝑉 𝑏(𝑣 − 𝑣𝑐 ) 250 × (1.267 − 0.687)
=
=
= 0.33
𝑆𝑉
0.95𝑓𝑦𝑣
0.95 × 460
Provide T8 at 200 mm c/c
𝐴𝑆𝑉
2 × 50.3
=
= 0.5
𝑂𝐾
𝑆𝑉
200
Deflection
Table 3.9
basic span-effective depth ratio =
8000
564
= 14.19
basic ratio for simply supported rectangular beam = 20
Table 3.10
modification factor for tension reinforcement
𝑓𝑠 =
2𝑓𝑦 𝐴𝑆,𝑟𝑒𝑞
3𝐴𝑆,𝑝𝑟𝑜𝑣
∗
1
2 × 460 × 836
=
= 266.6 𝑁/𝑚𝑚2
𝛽𝑏
3 × 942.6
𝑀
416 × 106
=
= 5.23
𝑏𝑑 2 250 × 5642
modification factor for tension reinforcement:
= 0.55 +
(477 − 𝑓𝑠 )
= 0.836 ≤ 2
𝑀
120 (0.9 + 2 )
𝑏𝑑
Modification factor for compression reinforcement:
Table 3.11
100𝐴𝑆,𝑝𝑟𝑜𝑣
𝑏𝑑
′
= 0.285
Mod. Factor = 1.086
∴ basic ratio = 20 x 0.836 x 1.086 = 18.15
Deflection OK
actual ratio = 14.2 < basic ratio = 18.15 OK
5
Job Title
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Subject:
Worked Example 4 - Continuous Beam
UofK
Client
ME
Made by
Date
2018
Checked by
Unless otherwise A reinforced concrete continuous beam carries 𝐺 = 80 KN/m
𝐾
stated, all
including self-weight and 𝑄𝐾 = 60 KN/m. the beam is 250 mm wide
references are
from: BS8110-1
by 800 mm depth and has three equal spans of 6m. The beam also
carries RC slab 200 mm thick spans 4 m centres in the transverse
direction. Design the beam for bending and shear if grade 30 and
460 are to be used for concrete and all steel respectively.
Loading
Section 3.2.1.2.2
Ultimate Load
F = 1.4 x 80 + 1.6 x 60 = 208 KN/m
Total ultimate load on a span = 208 x 6 = 1248 KN
As the loading is uniformly distributed, 𝐺𝐾 > 𝑄𝐾 and the spans are
F = 1248 KN
equal, the coefficients in the table below may be used to calculate the
design ultimate bending moments and shear forces.
Table 1 Design ultimate bending moments and shear forces
Table 3.5
At outer
Near
At first
At middle
At interior
support
middle of
interior
of interior
support
end span
support
span
Moment
0
0.09Fl
- 0.11Fl
0.07Fl
-0.08Fl
Shear
0.45F
__
0.6F
__
0.55F
NOTE
l is the effective span;
F is the total design ultimate load (1.4Gk + 1.6Qk).
No redistribution of the moments calculated from this table should be made
Bending
a) Mid span of 1st and 3rd spans - design as T section
Table 3.5
Maximum moment = 0.09 𝐹𝑙 = 0.09 x 1248 x 6 = 673.92 KN.m
Section 3.4.1.5
Effective width of flange = 𝑏𝑤 +
M = 673.92 KN.m
0.7 × 𝐿
6000
= 250 + 0.7 ×
= 1090 𝑚𝑚
5
5
For mild conditions of exposure, the cover = 20 mm. so for 10 mm
Table 3.3, 3.4
links and 20 mm bars Overall depth
h = d + 20 + 10 + 20/2 = 800 mm.
d = 760 mm
d = 760 mm
The moment of resistance of the concrete only
𝑀𝑅𝐶 = 0.45 × 𝑏 × ℎ𝑓 × 𝑓𝑐𝑢 (𝑑 − 0.5 × ℎ𝑓 )
= 0.45 × 1090 × 200 × 30 × (760 − 100) = 1942.4 KN. m
6
Mohamed Elhassan-Civil Engineering Department-UofK
The neutral axis lies in the flange.
Section 3.4.4.4
𝐾=
𝑀
673.92 × 106
=
= 0.155 < 0.156
𝑏𝑑 2 𝑓𝑐𝑢 250 × 7602 × 30
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐾
)}
0.9
z = 0.779d = 591.95 mm
Z = 591.95 mm
6
𝐴𝑆 =
𝑀
673.92 × 10
=
= 2605.2 𝑚𝑚2
0.95𝑓𝑦 𝑧 0.95 × 460 × 591.95
Provide 6T25, area provided 2945 mm2 (bottom steel for 1st and 3rd spans)
b) Mid span of 2nd span - design as T section
Table 3.5
Section 3.4.4.4
Maximum moment = 0.07 𝐹𝑙 = 0.07 x 1248 x 6 = 524.16 KN.m
𝐾=
M = 524.16 KN.m
𝑀
524.16 × 106
=
= 0.121 < 0.156
𝑏𝑑 2 𝑓𝑐𝑢 250 × 7602 × 30
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐾
)}
0.9
z = 0.84d = 638.4 mm
𝐴𝑆 =
Z = 638.4 mm
𝑀
524.16 × 106
=
= 1879 𝑚𝑚2
0.95𝑓𝑦 𝑧 0.95 × 460 × 638.4
Provide 4T25, area provided 1963.6 mm2 (bottom steel for 2nd span)
c) Interior support - design as rectangular section
Maximum moment = 0.11 𝐹𝑙 = 0.11 x 1248 x 6 = 823.68 KN.m
M = 823.68 KN.m
6
Table 3.5
𝐾=
𝑀
823.68 × 10
=
= 0.19 > 0.156
2
𝑏𝑑 𝑓𝑐𝑢 250 × 7602 × 30
Thus, compression steel is required.
𝐴𝑆 ′ =
(𝐾 − 𝐾 ′ )𝑏𝑑 2 𝑓𝑐𝑢 (0.19 − 0.156) × 250 × 7602 × 30
=
= 468.11 𝑚𝑚2
0.95𝑓𝑦 (𝑑 − 𝑑 ′ )
0.95 × 460 × (760 − 40)
This area will be provided by extending the span reinforcement
beyond the supports.
𝐴𝑆 =
𝐾′𝑏𝑑 2 𝑓𝑐𝑢
0.156 × 250 × 7602 × 30
+ 𝐴𝑆 ′ =
+ 468.11 = 3093.6 𝑚𝑚2
0.95𝑓𝑦 𝑧
0.95 × 460 × 0.775 × 760
Provide 4T32, area provided 3217 mm2 (Top steel for interior support)
Shear
Table 3.5
maximum design shear load = 0.6 𝐹 = 1248 x 0.6 = 748.8 KN
Section 3.4.5.2
shear stress 𝑣 =
𝑉
748.8 × 1000
=
= 3.95 N/mm2 < 0.8√30 OK
𝑏×𝑑
250 × 760
7
Mohamed Elhassan-Civil Engineering Department-UofK
shear links
shear at d from face of support= 748.8- 208 x 0.76 = 590.72 KN
Table 3.8
shear stress 𝑣 =
𝑉
590.72 × 1000
=
= 3.1 N/mm2
𝑏 ∗ 𝑑
250 × 760
1/3
𝑣𝑐 =
0.79 100𝐴𝑆
(
)
𝛾𝑚
𝑏𝑑
𝑣 = 3.1 𝑁/𝑚𝑚2
400 1/4 𝑓𝑐𝑢 1/3
(
) ( )
𝑑
25
Only four 25 mm bars extent a distance d from the support face
1
1
100𝐴𝑆 3
100 × 1963.6 3
∴(
) =(
) = 1.011 < 3 ,
𝑏𝑑
250 × 760
1
1
400 4
400 4
(
) =(
) = 0.85 (𝑠ℎ𝑜𝑢𝑙𝑑 𝑛𝑜𝑡 𝑏𝑒 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑡 𝑜𝑛𝑒)
𝑑
760
(
𝑓𝑐𝑢 1/3
) = 1.0626
25
∴ 𝑣𝑐 = 0.679 𝑁/𝑚𝑚2
Table 3.7
vc = 0.68 N/mm2
𝐴𝑆𝑉 𝑏(𝑣 − 𝑣𝑐 ) 250 × (3.1 − 0.68)
=
=
= 1.38
𝑆𝑉
0.95𝑓𝑦𝑣
0.95 × 460
T10 at 100
Provide T10 at 100 mm c/c
𝐴𝑆𝑉
2 × 78.54
=
= 1.57
𝑂𝐾
𝑆𝑉
100
Deflection
Table 3.9
Table 3.10
basic span-effective depth ratio =
6000
760
= 7.9
basic ratio for continuous flanged beam = 20.8
modification factor for tension reinforcement
𝑓𝑠 =
2𝑓𝑦 𝐴𝑆,𝑟𝑒𝑞
3𝐴𝑆,𝑝𝑟𝑜𝑣
∗
1
2 × 460 × 2605.2
=
= 271.3 𝑁/𝑚𝑚2
𝛽𝑏
3 × 2945
𝑀
673.92 × 106
=
= 4.667
2
𝑏𝑑
250 × 7602
modification factor for tension reinforcement =
= 0.55 +
(477−𝑓𝑠 )
𝑀
)
𝑏𝑑2
120(0.9+
= 0.858 ≤ 2
∴ basic ratio = 20.8 x 0.858 = 17.85
actual ratio = 7.9 < basic ratio = 17.85 OK
Deflection OK
8
Job Title
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Subject:
Worked Example 5 - Transfer Beam
Client
UofK
Made by
ME
Date
2018
Checked by
Unless otherwise A reinforced concrete transfer beam carries 𝐺 = 1400 KN/m
𝐾
stated, all
including self-weight and 𝑄𝐾 = 1000 KN as shown in the figure below.
references are
from: BS8110-1
Design the beam for bending and shear if grade 30 and 460 are to
be used for concrete and all steel respectively.
Loading
Section 3.2.1.2.2
Ultimate Load F = 1.4 x 1400 + 1.6 x 10000 = 3560 KN
Shear stress not to exceed 4 N/mm2 (avoid reinforcement congestion).
F = 3560 KN
Take b = 600 mm
∴d=
𝑉
3560 × 103
=
= 1483.33 𝑚𝑚
𝑣𝑏
4 × 600
Take overall depth = 1600 mm (d = 1500)
Bending
Maximum moment = 3560 x 1.5 = 5340 KN.m
Section 3.4.4.4
𝐾=
M = 5340 KN.m
𝑀
5340 × 106
=
= 0.132 < .156
2
𝑏𝑑 𝑓𝑐𝑢 600 × 15002 × 30
∴ No compression reinforcement is required
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐾
)}
0.9
z = 0.821d = 1232 mm
𝑀
5340 × 106
𝐴𝑆 =
=
= 9918.6 𝑚𝑚2
0.95𝑓𝑦 𝑧 0.95 × 460 × 1232
Provide 8T40, area provided 10053 mm2
Z = 1232 mm
Provide 8T40
Shear
at distance d from face of support V = 3560 KN
shear stress 𝑣 = 4 N/mm2
1/3
Table 3.8
𝑣𝑐 =
0.79 100𝐴𝑆
(
)
𝛾𝑚
𝑏𝑑
400 1/4 𝑓𝑐𝑢 1/3
(
) ( )
𝑑
25
9
Mohamed Elhassan-Civil Engineering Department-UofK
1
1
100𝐴𝑆 3
100 × 10053 3
∴(
) =(
) = 1.038 < 3,
𝑏𝑑
600 × 1500
1
1
400 4
400 4
(
) =(
) = 0.719 (𝑠ℎ𝑜𝑢𝑙𝑑 𝑛𝑜𝑡 𝑏𝑒 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑡 𝑜𝑛𝑒)
𝑑
1500
(
𝑓𝑐𝑢 1/3
) = 1.05
25
∴ 𝑣𝑐 = 0.69 𝑁/𝑚𝑚2
Table 3.7
vc = 0.69 N/mm2
𝐴𝑆𝑉 𝑏(𝑣 − 𝑣𝑐 ) 600 × (4 − 0.69)
=
=
= 4.5446
𝑆𝑉
0.95𝑓𝑦𝑣
0.95 × 460
Provide 4 Legs T12 at 90 mm c/c
𝐴𝑆𝑉
4 × 113.1
=
= 5.03
𝑂𝐾
𝑆𝑉
90
Deflection
Table 3.9
basic span-effective depth ratio =
1500
1500
= 1.0
basic ratio for cantilever beam = 5.6
Table 3.10
modification factor for tension reinforcement
𝑓𝑠 =
2𝑓𝑦 𝐴𝑆,𝑟𝑒𝑞
3𝐴𝑆,𝑝𝑟𝑜𝑣
×
1
2 × 460 × 9919
=
= 302.58
𝛽𝑏
3 × 10053
𝑀
5340 × 106
=
= 3.95
𝑏𝑑 2 600 × 15002
modification factor = 0.55 +
(477−𝑓𝑠 )
120(0.9+
𝑀
)
𝑏𝑑2
= 0.58 ≤ 2.0
basic ratio = 5.6 x 0.58 = 3.248
actual ratio = 1 < 3.25 OK
Deflection OK
10
Job Title
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Subject:
Worked Example 6 - Continuous solid slab spanning in one direction
UofK
Client
ME
Made by
Date
2018
Checked by
Unless otherwise A four-span reinforced concrete slab spanning in one direction carries
stated, all
𝐺𝐾 = 1.5 KN/m2 and 𝑄𝐾 = 2.5 KN/m2. The effective span is 6m and
references are
from: BS8110-1
Grade 30 and 460 are to be used for concrete and all steel
respectively. Design the slab.
6000
6000
6000
6000
Basic span-effective depth ratio = 26
6000
= 230.76 𝑚𝑚,
26
𝑇𝑟𝑦 𝑠𝑙𝑎𝑏 𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑑𝑒𝑝𝑡ℎ = 250 𝑚𝑚
Loading
Section 3.2.1.2.2
Ultimate Load F = 1.4x(1.5+0.25x24) +1.6x2.5 = 14.5 KN/m/m
F = 14.5 KN/m/m
Total ultimate load on a span = 14.5 x 6 = 87 KN
As the area of each bay exceeds 30 m2,
𝑄𝐾
𝐺𝐾
≤ 1.25 and The
characteristic imposed load does not exceed 5 KN/m2, the coefficients
in the table below may be used to calculate the design ultimate
bending moments and shear forces.
Table 1 Design ultimate bending moments and shear forces
Table 3.12
End support/slab connection
At first
middle
interior
Simple
interior
interior
support
support
spans
Continuous
At outer
Near
At outer
Near
support
middle
support
middle
of end
of end
span
Moment
Shear
NOTE
span
0
0.086Fl
- 0.04Fl
0.075Fl
- 0.086Fl
0.063Fl
- 0.063Fl
0.45F
__
0.46F
__
0.6F
__
0.5F
l is the effective span;
F is the total design ultimate load (1.4Gk + 1.6Qk).
Bending
Table 3.12
Maximum moment = 0.086 𝐹𝑙 = 0.086 x 87 x 6 = 44.9 KN.m
Table 3.3, 3.4
For mild conditions of exposure the cover = 20 mm. so for 10 mm
M = 45 KN.m
links and 20 mm bars Overall depth
h = d + 20 + 16/2 = 250 mm.
d = 222 mm
Section 3.4.4.4
𝐾=
d = 222 mm
𝑀
45 × 106
=
= 0.03
𝑏𝑑 2 𝑓𝑐𝑢 1000 × 2222 × 30
11
Mohamed Elhassan-Civil Engineering Department-UofK
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐴𝑆 =
𝐾
0.9
)}, z = 0.95d = 210.9 mm
Z = 210.9 mm
𝑀
45 × 106
=
= 488.3 𝑚𝑚2 /𝑚
0.95𝑓𝑦 𝑧 0.95 × 460 × 210.9
Provide T10 at 150 mm centres, area provided 523 mm2/m
Transverse reinforcement
Provide minimum area of steel =
0.13𝑏ℎ
= 0.13 × 1000 × 250/100
100
𝐴𝑆 = 325 𝑚𝑚2 /𝑚
provide T10 at 200 mm centres, area provided 393 mm2/m
Shear
Table 3.12
maximum design shear load = 0.6 𝐹 = 87 × 0.6 = 52.2 KN
Section 3.4.5.2
shear stress 𝑣 =
𝑉
52.2 × 1000
=
= 0.235 N/mm2 < 0.8√30 OK
𝑏×𝑑
1000 × 222
shear at d from face of support= 52.2- 14.5 x 0.22 = 49 KN
𝑉
49 ∗ 1000
=
= 0.22 N/mm2
𝑏 ∗ 𝑑
1000 ∗ 222
shear stress 𝑣 =
1/3
Table 3.8
𝑣𝑐 =
0.79 100𝐴𝑆
(
)
𝛾𝑚
𝑏𝑑
400 1/4 𝑓𝑐𝑢 1/3
(
) ( )
𝑑
25
1
1
𝑣 = 0.22𝑁/𝑚𝑚2
100𝐴𝑆 3
100 × 523 3
∴(
) =(
) = 0.618 < 3 ,
𝑏𝑑
1000 × 222
1
1
400 4
400 4
(
) =(
) = 1.16 (𝑠ℎ𝑜𝑢𝑙𝑑 𝑛𝑜𝑡 𝑏𝑒 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑡 𝑜𝑛𝑒)
𝑑
222
(
𝑓𝑐𝑢 1/3
) = 1.0626
25
∴ 𝑣𝑐 = 0.48 𝑁/𝑚𝑚2
Since v < vc, shear reinforcement is not required
Deflection
basic span-effective depth ratio =
Table 3.9
6000
222
= 27, basic ratio = 26
modification factor for tension reinforcement
𝑓𝑠 =
2𝑓𝑦 𝐴𝑆,𝑟𝑒𝑞
3𝐴𝑆,𝑝𝑟𝑜𝑣
×
1
2 × 460 × 488
=
= 286 𝑁/𝑚𝑚2
𝛽𝑏
3 × 523
𝑀
45 × 106
=
= 0.913
𝑏𝑑 2 1000 × 2222
modification factor for tension reinforcement =
= 0.55 +
(477 − 𝑓𝑠 )
= 1.43 ≤ 2
𝑀
120 (0.9 + 2 )
𝑏𝑑
∴ basic ratio = 26 x 1.43 = 37.2
actual ratio = 27 < basic ratio = 37.2
Deflection OK
12
Job Title
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Subject:
Worked Example 7 - Two-way Slab
Client
UofK
Made by
ME
Date
2018
Checked by
Unless otherwise A reinforced concrete slab spanning in two directions, as shown in
stated, all
the figure below, carries 𝐺𝐾 = 7.5 KN/m2 including self-weight and
references are
from: BS8110-1
𝑄𝐾 = 5 KN/m2. Grade 30 and 460 are to be used for concrete and
all steel respectively. Design the slab if the cover is equal to 25 mm.
𝑇𝑟𝑦 𝑠𝑙𝑎𝑏 𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑑𝑒𝑝𝑡ℎ = 250 𝑚𝑚
Loading
Section 3.2.1.2.2
Ultimate Load F = 1.4 x 7.5 + 1.6 x 5 = 18.5 KN/m2
F = 18.5 KN/m2
Bending
𝒍𝒚
=𝟏
𝒍𝒙
Bending strength and deflection are usually the governing criteria in
the design of slabs. The corner panel should be checked because the
moments are larger in this panel.
Table 3.14
𝛽𝑠𝑥 = −0.047 𝑎𝑛𝑑 𝛽𝑠𝑥 = 0.036
Maximum support moment:
Section 3.4.4.4
𝑚𝑠𝑥 = 𝛽𝑠𝑥 𝑛𝑙𝑥 2 = −0.047 × 18.5 × 72 = −42.6 𝐾𝑁. 𝑚
M = -42.6 KN.m
b = 1000 mm, d = 220 mm and 𝑓𝑐𝑢 = 30 𝑁/𝑚𝑚2 .
d = 220 mm
𝐾=
𝑀
42.6 × 106
=
= 0.03
𝑏𝑑 2 𝑓𝑐𝑢 1000 × 2202 × 30
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐴𝑆 =
𝐾
0.9
)}, z = 0.95d = 209 mm
Z = 209 mm
𝑀
42.6 × 106
=
= 466.5 𝑚𝑚2 /𝑚
0.95𝑓𝑦 𝑧 0.95 × 460 × 209
Provide T10 at 150 mm centres, area provided 524 mm2/m
13
Mohamed Elhassan-Civil Engineering Department-UofK
Maximum sagging moment:
𝑚𝑠𝑥 = 𝛽𝑠𝑥 𝑛𝑙𝑥 2 = 0.036 × 18.5 × 72 = 32.6 𝐾𝑁. 𝑚
𝐾=
Section 3.4.4.4
M = 32.6 KN.m
𝑀
32.6 × 106
=
= 0.0225
2
𝑏𝑑 𝑓𝑐𝑢 1000 × 2202 × 30
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐴𝑆 =
𝐾
0.9
)}, z = 0.95d = 209 mm
Z = 209 mm
𝑀
32.6 × 106
=
= 357 𝑚𝑚2 /𝑚
0.95𝑓𝑦 𝑧 0.95 × 460 × 209
Provide T10 at 200 mm centres, area provided 393 mm2/m
Shear
Table 3.15
𝛽𝑣𝑥 = 0.40
maximum shear load = 𝛽𝑣𝑥 𝑛𝑙𝑥 = 0.4 x 18.5 x 7 = 51.8KN/m width
𝑉
51.8 × 1000
=
= 0.235 N/mm2
𝑏×𝑑
1000 × 220
100𝐴𝑆
100 × 524
=
= 0.238
𝑏𝑑
1000 × 220
shear stress 𝑣 =
Table 3.8
∴ 𝑣𝑐 = 0.45 𝑁/𝑚𝑚2
Since v < vc, shear reinforcement is not required
𝑣 = 0.45𝑁/𝑚𝑚2
Deflection
Table 3.9
basic span-effective depth ratio =
basic ratio = 26
Table 3.10
7000
220
= 31.8
modification factor for tension reinforcement
𝑓𝑠 =
2𝑓𝑦 𝐴𝑆,𝑟𝑒𝑞
3𝐴𝑆,𝑝𝑟𝑜𝑣
∗
1
2 × 460 × 357
=
= 278.6 𝑁/𝑚𝑚2
𝛽𝑏
3 × 393
𝑀
32.6 × 106
=
= 0.674
2
𝑏𝑑
1000 × 2202
modification factor for tension reinforcement =
= 0.55 +
(477−𝑓𝑠 )
𝑀
)
𝑏𝑑2
120(0.9+
= 1.6 ≤ 2
∴ basic ratio = 26 x 1.6 = 41.6
actual ratio = 31.8 < basic ratio = 41.6
Deflection OK
14
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Job Title
Subject:
Worked Example 8 - Flat Slab
Client
UofK
ME
Made by
Date
2018
Checked by
Unless otherwise A reinforced concrete Flat slab spanning 7 m in both directions, as
stated, all
shown in the figure below, carries superimposed dead load 1.5 KN/m2
references are
from: BS8110
and imposed load 𝑄𝐾 = 3 KN/m2. Grade 30 and 460 are to be used
for concrete and steel respectively. Design the slab if the cover is equal
to 25 mm.
7000
= 269 𝑚𝑚,
26
𝑠𝑎𝑦 300 𝑚𝑚
Loading
Section 3.2.1.2.2
Ultimate Load F = 1.4 x (1.5 + 24 x 0.3) + 1.6 x 3 = 17 KN/m2
F = 17 KN/m2
Ultimate load per panel = 17 x 7 x 7 = 833 KN
Bending
Section 3.5.2.3
Section 3.7.2.7
As the area of each bay exceeds 30 m2,
𝑄𝐾
𝐺𝐾
≤ 1.25 and The
characteristic imposed load does not exceed 5 KN/m2, the coefficients
in the table 3.12 BS8110 may be used to calculate the design
ultimate bending moments and shear forces.
1. First interior support:
Table 3.12
𝑀 = −0.086𝐹𝐿 = −0.086 × 833 × 7 = −501.5 𝐾𝑁. 𝑚
Section 3.7.2.7
This moment may be reduced by 0.15Fh c
Assume column 500 x 500 mm2
0.15 x 833 x 0.5 = 62.5 KN.m
Then the net negative moment = - 501.5 + 62.5 = - 439 KN.m
Middle strip moment = 0.25 x - 439 = - 110 KN.m
Column strip moment = 0.75 x - 439 = - 330 KN.m
15
Mohamed Elhassan-Civil Engineering Department-UofK
For Middle strip:
b = 3500 mm, d = 260 mm and 𝑓𝑐𝑢 = 30 𝑁/𝑚𝑚2 .
Section 3.4.4.4
𝐾=
𝑀
110 × 106
=
= 0.015
𝑏𝑑 2 𝑓𝑐𝑢 3500 × 2602 × 30
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐴𝑆 =
𝐾
0.9
)}, z = 0.95d = 247 mm
𝑀
110 × 106
=
= 1020 𝑚𝑚2
0.95𝑓𝑦 𝑧 0.95 × 460 × 247
Provide T12 at 300 mm centres, area provided 1319 mm2/strip
For column strip:
b = 3500 mm, d = 260 mm and 𝑓𝑐𝑢 = 30 𝑁/𝑚𝑚2 .
Section 3.4.4.4
𝐾=
𝑀
330 × 106
=
= 0.05
𝑏𝑑 2 𝑓𝑐𝑢 3500 × 2602 × 30
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐴𝑆 =
𝐾
0.9
)}, z = 0.94d = 245 mm
𝑀
330 × 106
=
= 3082.3 𝑚𝑚2
0.95𝑓𝑦 𝑧 0.95 × 460 × 245
Provide T12 at 100 mm centres, area provided 3958 mm2/strip
2. Near middle of end span:
Table 3.12
𝑀 = 0.075𝐹𝐿 = 0.075 × 833 × 7 = 438 𝐾𝑁. 𝑚
Middle strip moment = 0.45 × 438 = 198 KN.m
Column strip moment = 0.55 × 438 = 241 KN.m
For Middle strip:
b = 3500 mm, d = 260 mm and 𝑓𝑐𝑢 = 30 𝑁/𝑚𝑚2 .
Section 3.4.4.4
𝐾=
𝑀
198 × 106
=
= 0.028
𝑏𝑑 2 𝑓𝑐𝑢 3500 × 2602 × 30
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐴𝑆 =
𝐾
0.9
)}, z = 0.95d = 247 mm
𝑀
198 × 106
=
= 1835 𝑚𝑚2
0.95𝑓𝑦 𝑧 0.95 × 460 × 247
Provide T12 at 150 mm centres, area provided 2639 mm2/strip
For column strip:
b = 3500 mm, d = 260 mm and 𝑓𝑐𝑢 = 30 𝑁/𝑚𝑚2 .
Section 3.4.4.4
𝐾=
𝑀
241 × 106
=
= 0.034
2
𝑏𝑑 𝑓𝑐𝑢 3500 × 2602 × 30
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐾
0.9
)}, z = 0.95d = 247 mm
16
Mohamed Elhassan-Civil Engineering Department-UofK
𝐴𝑆 =
𝑀
241 × 106
=
= 2232.74 𝑚𝑚2
0.95𝑓𝑦 𝑧 0.95 × 460 × 247
Provide T12 at 125 mm centres, area provided 3167.5 mm2/strip
Shear
Table 3.12
𝑉𝑡 = 833 𝐾𝑁
𝑉𝑒𝑓𝑓 = 1.15 𝑉𝑡 = 833 × 1.15 = 957.95 𝐾𝑁
Assume column 500 × 500 mm2
At column face:
𝑣=
Table 3.8
𝑉𝑒𝑓𝑓
958 × 1000
=
= 1.84 < 0.8√𝑓𝑐𝑢
𝑢° ∗ 𝑑 4 × 500 × 260
100𝐴𝑆 100 × 3958
=
= 0.435
𝑏𝑑
3500 × 260
∴ 𝑣𝑐 = 0.55 𝑁/𝑚𝑚2
Punching shear perimeter u = (500 + 3×260) × 4 = 5120 mm
𝑉𝑒𝑓𝑓
958 × 1000
𝑁
=
= 0.72
< 1.6𝑣𝑐
𝑢 ∗ 𝑑 5120 × 260
𝑚𝑚2
(𝑣 − 𝑣𝑐 )𝑢𝑑 (0.72 − 0.55) × 5120 × 260
∴ ∑ 𝐴𝑠𝑣 sin 𝛼 ≥
=
= 517.86 𝑚𝑚2
0.95𝑓𝑦𝑣
0.95 × 460
𝑣=
Equation 29a
Provide 8 two legs T12 shear links.
Area provided 1810 mm2 >
0.4𝑢𝑑
= 1218 𝑚𝑚2
0.95𝑓𝑦𝑣
Deflection
at the centre of the span:
Table 3.9
Table 3.10
basic span-effective depth ratio =
7000
basic ratio = 26
260
= 26.93
modification factor for tension reinforcement
𝑓𝑠 =
2𝑓𝑦 𝐴𝑆,𝑟𝑒𝑞
3𝐴𝑆,𝑝𝑟𝑜𝑣
∗
1
2 × 460 × 2233
=
= 216.2 𝑁/𝑚𝑚2
𝛽𝑏
3 × 3167.5
𝑀
241 × 106
=
= 1.019
𝑏𝑑 2 3500 × 2602
modification factor for tension reinforcement =
= 0.55 +
(477 − 𝑓𝑠 )
= 1.6825 ≤ 2
𝑀
120 (0.9 + 2 )
𝑏𝑑
∴ basic ratio = 26 × 1.6825 = 43.745
actual ratio = 26.93 < basic ratio = 43.745
17
Job Title
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Subject:
Worked Example 9 - Ribbed slab
Client
UofK
Made by
ME
Date
2018
Checked by
Unless otherwise A four-span reinforced concrete ribbed slab spanning in one direction.
stated, all
The characteristic dead load including self-weight is 𝐺𝐾 = 4.5 KN/m2
references are
from: BS 8110-1 and the characteristic live load is 𝑄𝐾 = 2.5 KN/m2. The effective span
is 6m and Grade 30 and 460 are to be used for concrete and all
steel respectively. Design the slab considering a 0.5 m width of floor
is supported by each rib.
Loading
Section 3.2.1.2.2
Ultimate Load = 1.4x4.5+1.6x2.5 = 10.3KN/m2
Total UL on the span = 10.3x6x0.5 = 31 KN
F = 10.3 KN/m2
Bending
1. At mid-span:
The maximum positive moment occurs near middle of end span, So
Table 3.12
the max. moment = 0.075 𝐹𝑙 = 0.075 x 31 x 6 = 13.95 KN.m/rib M = 14 KN.m
Assume the cover = 20 mm. so for 8 mm links and 16 mm bars
Overall depth h = d + 20 + 8 + 16/2 = 280 mm. ∴ d = 244 mm
𝑀𝑅 = 0.45𝑓𝑐𝑢 𝑏𝑓 ℎ𝑓 (𝑑 −
d = 244 mm
ℎ𝑓
) = 0.45 × 30 × 500 × 80 × (244 − 40) = 110 𝐾𝑁. 𝑚
2
∴ 𝑁. 𝐴. 𝑖𝑛 𝑓𝑙𝑎𝑛𝑔𝑒, 𝑑𝑒𝑠𝑖𝑔𝑛 𝑎𝑠 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑠𝑒𝑐𝑖𝑜𝑛
Section 3.4.4.4
𝐾=
𝑀
14 × 106
=
= 0.0156
𝑏𝑑 2 𝑓𝑐𝑢 500 × 2442 × 30
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐴𝑆 =
𝐾
0.9
)}, z = 0.95d = 231.8 mm
Z = 231.8 mm
𝑀
14 × 106
=
= 138 𝑚𝑚2
0.95𝑓𝑦 𝑧 0.95 × 460 × 231.8
Provide 2T10, area provided 157 mm2
1. At a support: design as rectangular section
Table 3.12
Maximum moment = −0.085 𝐹𝑙 = −0.085 × 31 × 6 = − 15.8 KN. m/rib
M = - 15.8 KN.m
6
Section 3.4.4.4
𝐾=
𝑀
15.8 × 10
=
= 0.0176
𝑏𝑑 2 𝑓𝑐𝑢 500 × 2442 × 30
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐴𝑆 =
𝐾
)} , z = 0.95d = 231.8 mm
0.9
𝑀
15.8 × 106
=
= 156 𝑚𝑚2
0.95𝑓𝑦 𝑧 0.95 × 460 × 231.8
18
Mohamed Elhassan-Civil Engineering Department-UofK
Provide 2T10, area provided 157 mm2
Shear
Table 3.12
maximum design shear load = 0.6 𝐹 = 0.6 x 31 = 18.6 KN
shear stress 𝑣 =
𝑉
18.6 × 1000
=
= 0.5 N/mm2 < 0.8√30 OK
𝑏×𝑑
150 × 244
shear at d from face of support = 18.6- 5.15 x 0.244 = 17.3 KN
shear stress 𝑣 =
𝑉
17.3 × 1000
=
= 0.47 N/mm2
𝑏×𝑑
150 × 244
1/3
0.79 100𝐴𝑆
𝑣𝑐 =
(
)
𝛾𝑚
𝑏𝑑
400 1/4 𝑓𝑐𝑢 1/3
(
) ( )
𝑑
25
Only four 25 mm bars extent a distance d from the support face
1
Table 3.8
1
100𝐴𝑆 3
100 × 157 3
∴(
) =(
) = 0.754 < 3 ,
𝑏𝑑
150 × 244
1
1
400 4
400 4
(
) =(
) = 1.13 (𝑠ℎ𝑜𝑢𝑙𝑑 𝑛𝑜𝑡 𝑏𝑒 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑡 𝑜𝑛𝑒)
𝑑
244
(
𝑓𝑐𝑢 1/3
) = 1.0626
25
𝑣𝑐 = 0.57
∴ 𝑣𝑐 = 0.57 𝑁/𝑚𝑚2
Since v < vc, shear reinforcement is not required
𝑁
𝑚𝑚2
Deflection
Table 3.9
basic span − effective depth ratio =
basic ratio = 20.8
6000
= 24.59
244
modification factor for tension reinforcement
𝑓𝑠 =
2𝑓𝑦 𝐴𝑆,𝑟𝑒𝑞
3𝐴𝑆,𝑝𝑟𝑜𝑣
∗
1
2 × 460 × 138
=
= 269.55 𝑁/𝑚𝑚2
𝛽𝑏
3 × 157
𝑀
14 × 106
=
= 0.47
𝑏𝑑 2 500 × 2442
Table 3.10
modification factor for tension reinforcement =
= 0.55 +
(477 − 𝑓𝑠 )
= 1.8 ≤ 2
𝑀
120 (0.9 + 2 )
𝑏𝑑
∴ basic ratio = 20.8 x 1.8 = 37.44
actual ratio = 24.59 < basic ratio = 37.44
Deflection OK
19
Job Title
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Subject:
Worked Example 10 - Waffle Slab
Client
UofK
Made by
ME
Date
2018
Checked by
Unless otherwise A reinforced concrete waffle slab spanning 6 m in each direction. The
stated, all
characteristic dead load including self-weight is 𝐺𝐾 = 5.0 KN/m2 and
references are
from: BS8110-1
the characteristic live load is 𝑄𝐾 = 2.5 KN/m2. Grade 30 and 460
are to be used for concrete and all steel respectively. Design the slab
for an internal panel considering a 0.5 m width of floor is supported
by each rib.
Loading
Section 3.2.1.2.2
Ultimate Load n = 1.4 x 5 + 1.6 x 2.5 = 11 KN/m2
Bending
Table 3.14
n = 11 KN/m2
𝒍𝒚 6
= =𝟏
𝒍𝒙 6
The slab will be designed as two way slab, so
𝛽𝑠𝑥 = −0.031 𝑎𝑛𝑑 𝛽𝑠𝑥 = 0.024
1. At mid-span:
𝑚𝑠𝑥 = 𝛽𝑠𝑥 𝑛𝑙𝑥 2 = 0.024 × 11 × 62 = 9.5 𝐾𝑁. 𝑚/𝑚
𝑚𝑠𝑥 = 9.5 × 0.5 = 4.75 𝐾𝑁. 𝑚/𝑟𝑖𝑏
M = 4.75 KN.m
Assume the cover = 20 mm. so for 8 mm links and 16 mm bars
Overall depth h = d + 20 + 8 + 16/2 = 280 mm. ∴ d = 244 mm
𝑀𝑅 = 0.45𝑓𝑐𝑢 𝑏𝑓 ℎ𝑓 (𝑑 −
d = 244 mm
ℎ𝑓
) = 0.45 × 30 × 500 × 80 × (244 − 40) = 110 𝐾𝑁. 𝑚
2
∴ 𝑁. 𝐴. 𝑖𝑛 𝑓𝑙𝑎𝑛𝑔𝑒, 𝑑𝑒𝑠𝑖𝑔𝑛 𝑎𝑠 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑠𝑒𝑐𝑖𝑜𝑛
b = 500 mm, d = 244 mm and 𝑓𝑐𝑢 = 30 𝑁/𝑚𝑚2 .
Section 3.4.4.4
𝐾=
𝑀
4.75 × 106
=
= 0.005
𝑏𝑑 2 𝑓𝑐𝑢 500 × 2442 × 30
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐴𝑆 =
𝐾
0.9
)}, z = 0.95d = 231.8 mm
Z = 231.8 mm
𝑀
4.75 × 106
=
= 46.89 𝑚𝑚2 /𝑟𝑖𝑏
0.95𝑓𝑦 𝑧 0.95 × 460 × 231.8
Provide 2T8, area provided 100.6 mm2/rib
20
Mohamed Elhassan-Civil Engineering Department-UofK
1. At a support: design as rectangular section
Table 3.14
𝑚𝑠𝑥 = 𝛽𝑠𝑥 𝑛𝑙𝑥 2 = −0.031 × 11 × 62 = −12.3 𝐾𝑁. 𝑚/𝑚
𝑚𝑠𝑥 = 9.5 × 0.5 = −6.15 𝐾𝑁. 𝑚/𝑟𝑖𝑏
Section 3.4.4.4
M = - 6.15 KN.m
𝑀
6.15 × 106
𝐾= 2
=
= 0.0069
𝑏𝑑 𝑓𝑐𝑢 500 × 2442 × 30
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐴𝑆 =
𝐾
0.9
)}, z = 0.95d = 231.8 mm
Z = 231.8 mm
𝑀
6.15 × 106
=
= 60.7 𝑚𝑚2 /𝑟𝑖𝑏
0.95𝑓𝑦 𝑧 0.95 × 460 × 231.8
Provide 2T8, area provided 100.6 mm2/rib
Shear
Table 3.15
𝛽𝑣𝑥 = 0.33
For one rib maximum design shear load = 𝛽𝑣𝑥 𝑛𝑙𝑥
= 0.33 x 11 x 6 x 0.5 = 10.9 KN
𝑉
10.9 × 1000
=
= 0.3 N/mm2
𝑏×𝑑
150 ∗ 244
100𝐴𝑆 100 × 100.6
=
= 0.275
𝑏𝑑
150 × 244
shear stress 𝑣 =
Table 3.8
∴ 𝑣𝑐 = 0.48 𝑁/𝑚𝑚2
Since v < vc, shear reinforcement is not required
.
Table 3.9
Table 3.10
Deflection
basic span-effective depth ratio =
basic ratio = 20.8
6000
244
= 24.6
modification factor for tension reinforcement
𝑓𝑠 =
2𝑓𝑦 𝐴𝑆,𝑟𝑒𝑞
3𝐴𝑆,𝑝𝑟𝑜𝑣
×
1
2 × 460 × 47
=
= 143.3 𝑁/𝑚𝑚2
𝛽𝑏
3 × 100.6
𝑀
4.75 × 106
=
= 0.16
𝑏𝑑 2 500 × 2442
modification factor for tension reinforcement =
= 0.55 +
(477 − 𝑓𝑠 )
= 3.17 ≤ 2
𝑀
120 (0.9 + 2 )
𝑏𝑑
∴ basic ratio = 20.8 x 3.17 = 66
actual ratio = 24.6 < basic ratio = 66
Deflection OK
21
Job Title
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Subject:
Worked Example 11 - Stair Slab
Client
UofK
Made by
ME
Date
2018
Checked by
Unless otherwise A reinforced concrete stair slab spanning longitudinally and
stated, all
supported on a beam at the top and the bottom as shown in the
references are
from: BS8110-1
figure below. the slab carries 𝐺𝐾 = 10 KN/m2 including self-weight
and 𝑄𝐾 = 5 KN/m2. The effective span is 4m and the rise is 1.5 m,
with 30 cm goings and 16 cm risers. Grade 30 and 460 are to be
used for concrete and all steel respectively. Design the stair slab.
Basic span-effective depth ratio = 20
4000
= 200 𝑚𝑚,
20
𝑇𝑟𝑦 𝑎 200 𝑚𝑚 𝑡ℎ𝑖𝑐𝑘 𝑤𝑎𝑖𝑠𝑡
Assume the cover is 25 mm and the bar diameter is 10 mm.
d = 200 – 25 – 10 = 165 mm
Loading
Section 3.2.1.2.2
Ultimate Load F = 1.4 × 10 + 1.6 × 5 = 22 KN/m/m width
Bending
F = 22 KN/m
With no effective end restraint (simply supported).
𝑀𝑚𝑎𝑥 =
Section 3.4.4.4
𝐾=
𝑤𝑙 2 22 × 42
=
= 44 𝐾𝑁. 𝑚
8
8
𝑀
44 × 106
=
= 0.054
2
𝑏𝑑 𝑓𝑐𝑢 1000 × 1652 × 30
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐴𝑆 =
𝐾
0.9
)}, z = 0.936d = 154.5 mm
𝑀
44 × 106
=
= 651.7 𝑚𝑚2 /𝑚
0.95𝑓𝑦 𝑧 0.95 × 460 × 154.5
Provide T10 at 100 mm centres, area provided 785 mm2/m
22
Mohamed Elhassan-Civil Engineering Department-UofK
Transverse distribution reinforcement:
Table 3.25
𝐴𝑠 =
0.13𝑏ℎ
200
= 0.13 × 1000 ×
= 260 𝑚𝑚2 /𝑚
100
100
provide T10 at 200 mm centres, area provided 393 mm2/m
Shear
maximum design shear load = 22 * 4 = 88 KN
𝑉
88 × 1000
𝑁
=
= 0.533
< 0.8√𝑓𝑐𝑢
𝑏 × 𝑑 1000 × 165
𝑚𝑚2
Section 3.4.5.2
𝑣=
Table 3.8
0.79 100𝐴𝑆
𝑣𝑐 =
(
)
𝛾𝑚
𝑏𝑑
1/3
400 1/4 𝑓𝑐𝑢 1/3
(
) ( )
𝑑
25
1
1
100𝐴𝑆 3
100 × 785 3
∴(
) =(
) = 0.78 < 3 ,
𝑏𝑑
1000 × 165
1
1
400 4
400 4
(
) =(
) = 1.2478 (𝑠ℎ𝑜𝑢𝑙𝑑 𝑛𝑜𝑡 𝑏𝑒 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑡 𝑜𝑛𝑒)
𝑑
165
(
𝑓𝑐𝑢 1/3
) = 1.0626
25
∴ 𝑣𝑐 = 0.6536 𝑁/𝑚𝑚2
Since v < vc, shear reinforcement is not required
Deflection
Table 3.9
Table 3.10
basic span-effective depth ratio =
4000
165
= 24.24
basic ratio = 20 (simply supported)
modification factor for tension reinforcement
𝑓𝑠 =
2𝑓𝑦 𝐴𝑆,𝑟𝑒𝑞
3𝐴𝑆,𝑝𝑟𝑜𝑣
×
1
2 × 460 × 651.7
=
= 254.6 𝑁/𝑚𝑚2
𝛽𝑏
3 × 785
𝑀
44 × 106
=
= 1.616
𝑏𝑑 2 1000 × 1652
modification factor for tension reinforcement =
= 0.55 +
(477 − 𝑓𝑠 )
= 1.2866 ≤ 2
𝑀
120 (0.9 + 2 )
𝑏𝑑
Section 3.10.2.2 ∴ basic ratio = 20 × 1.2866 × 1.15 = 29.5895
actual ratio = 24.24 < basic ratio = 29.5895
Deflection OK
23
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Job Title
Subject:
Worked Example 12 - Short column subjected to axial load
Client
UofK
Made by
ME
Date
2018
Checked by
Unless otherwise A reinforced concrete short column is 400 mm * 400 mm and
stated, all
support 2500 KN at the ultimate limit state. Find the steel area
references are
from: BS8110-1
required if grade 30 concrete and grade 460 reinforcement are used.
Section 3.8.4.4
𝑁 = 0.35𝑓𝑐𝑢 𝐴𝑐 + 𝐴𝑠𝑐 (0.7𝑓𝑦 − 0.35𝑓𝑐𝑢 )
2500000 = 0.35 × 30 × 4002 + 𝐴𝑠𝑐 (0.7 × 460 − 0.35 × 30)
𝐴𝑠𝑐 =
2 500 000 − 1 680 000
= 2632.5 𝑚𝑚2
311.5
Provide 4T32. Area steel provided 3216.8 mm2
24
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Job Title
Subject: Worked Example 13
Short column subjected to axial load and moment about one axis
Client
UofK
Made by
ME
Date
2018
Checked by
Unless otherwise A reinforced concrete short braced column is 400 mm * 400 mm,
stated, all
subjected to 1500 KN and 150 KN.m at the ultimate limit state.
references are
from: BS8110-1
Design the column using design charts if grade 30 concrete and grade
460 reinforcement are used.
Assume 20 mm cover, 8 mm for links and 25 mm for the main
reinforcement.
d = 400 – 20 – 8 - 25/2 = 359.5 mm
d/h = 0.9
by using the chart No.29 form BS 8110-3
𝑁
1500 × 1000
=
= 9.375
𝑏ℎ
400 × 400
𝑀
150 × 106
=
= 2.34
𝑏ℎ2 400 × 4002
100𝐴𝑠𝑐
= 0.75
𝑏ℎ
0.75
𝐴𝑠𝑐 = 400 × 400 ×
= 1200 𝑚𝑚2
100
Provide 4T25. Area steel provided 1963.6 mm2
4T25
25
Job Title
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Subject: Worked Example 14
Short column subjected to axial load and biaxial moment
Client
UofK
Made by
ME
Date
2018
Checked by
Unless otherwise A reinforced concrete short column is 500 mm * 500 mm, subjected
stated, all
to 2000 KN plus Mxx = 100 KN.m and Myy = 80 KN.m at the ultimate
references are
from: BS8110-1
limit state. Design the column if grade 30 concrete and grade 460
reinforcement are used.
Assume 25 mm cover, 8 mm for links and
.
32 mm for the main reinforcement.
d = 500 – 25 – 8 - 32/2 = 451 mm
b’ = h’ = d
d/h = 0.9
Equation 40
𝑀𝑥 100
=
= 0.222
ℎ′
451
𝑀𝑦
80
=
= 0.177
′
𝑏
451
𝑏′
𝑀𝑥 ′ = 𝑀𝑥 + 𝛽 ′ 𝑀𝑦
ℎ
𝑁
2000 × 1000
=
= 0.267
𝑏ℎ𝑓𝑐𝑢 500 × 500 × 30
Table 3.22
∴ by interpolation β = 0.6896
𝑀𝑥 ′ = 100 + 0.6896 ×
451
× 80 = 155.2 𝐾𝑁. 𝑚
451
by using the chart No.29 form BS 8110-3
𝑁
2000 × 1000
=
=8
𝑏ℎ
500 × 500
𝑀
155.2 × 106
=
= 1.242
𝑏ℎ2 500 × 5002
100𝐴𝑠𝑐
∴
= 0.4
𝑏ℎ
0.4
𝐴𝑠𝑐 = 500 × 500 ×
= 1000 𝑚𝑚2
100
Provide 4T25. Area steel provided 1963.6 mm2
26
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Job Title
Subject:
Worked Example 15: Braced slender column
Client
UofK
Made by
ME
Date
2018
Checked by
Unless otherwise A reinforced concrete braced column is 250 mm * 500 mm,
stated, all
subjected to 1500 KN plus end moments of 80 KN.m and 30 KN.m
references are
from: BS8110-1
about X-X axis as shown in the figure below. Design the column if
grade 30 concrete and grade 460 reinforcement are to be used. Take
the effective heights lex = 7 m and ley = 8 m.
.
Slenderness ratios:
lex / h = 7/0.50 = 14 < 15
ley / b = 8/0.25 = 32 > 15
therefore, the column is slender.
Assume 25 mm cover, 8 mm for links and
32 mm for the main reinforcement.
h’ = 500 – 25 - 8- 16 = 451 mm
b’ = 250 – 25 – 8 - 16 = 201 mm
M1 = - 30 KN.m
M2 = 80 KN.m
Equation 36
Mi = 0.4 M1 + 0.6 M2 >= 0.4 M2
Mi = 0.4 × -30 + 0.6 × 80 = 36 KN.m >= 0.4M2
The additional moment induced by deflection of the column:
Equation 35
𝑀𝑎𝑑𝑑 = 𝑁𝑎𝑢
Equation 32
𝑎𝑢 = 𝛽𝑎 𝐾ℎ
Equation 33
𝐾=
Equation 34
𝑁𝑢𝑧 − 𝑁
≤1
𝑁𝑢𝑧 − 𝑁𝑏𝑎𝑙
𝛽𝑎 =
1
𝑙𝑒 2
( ′)
2000 𝑏
Taking an initial value of K = 1.0
𝑀𝑎𝑑𝑑 =
Figure 3.2
1500 × 500 8000 2
(
) = 384 𝐾𝑁. 𝑚
2000
250
The maximum design moment will be the greatest of:
1. M2,
2. Mi + Madd,
3. M1 + Madd/2
4. emin * N
27
Mohamed Elhassan-Civil Engineering Department-UofK
∴ MT = Mi + Madd = 36 + 384 = 420 KN.m
d/h = 0.9
by using the chart No.29 form BS 8110-3-1985
𝑁
1500 × 1000
=
= 12
𝑏ℎ
250 × 500
𝑀
420 × 106
=
= 6.72
𝑏ℎ2 250 × 5002
100𝐴𝑠𝑐
∴
=4
𝑏ℎ
K = 0.8
Calculate the moment again, So
𝑀𝑎𝑑𝑑 =
1500 × 500 8000 2
(
) × 0.8 = 307.2 𝐾𝑁. 𝑚
2000
250
∴ MT = Mi + Madd = 36 + 307.2 = 343.2 KN.m
𝑀
343.2 × 106
=
= 5.5
𝑏ℎ2 250 × 5002
100𝐴𝑠𝑐
∴
= 3.2
𝑏ℎ
K = 0.75
Asc = 250 × 500 × 3.2 /100 = 4000 mm2
As a check on the final value of K interpolated from the chart:
𝐾=
𝑁𝑢𝑧 − 𝑁
≤1
𝑁𝑢𝑧 − 𝑁𝑏𝑎𝑙
Nbal = 0.25fcubd = 0.25 × 30 × 250 × 500 × 10-3 = 937.5 KN
Nuz = 0.45fcuAc + 0.95fyAsc
Nuz = 0.45 × 30 × 250 × 500 + 0.95 × 460 × 3875 = 3380.9 KN
K = (3380.9-1500) / (3380.9-937.5) = 0.77
Which agreed with the final value obtained from the table above.
Provide 6T32. Area steel provided 4825 mm2
28
Job Title
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Subject:
Worked Example 16 - Shear Wall
Client
UofK
Made by
ME
Date
2018
Checked by
Unless otherwise A reinforced concrete shear wall carries 𝐺 = 30 KN/m including self𝐾
stated, all
weight, 𝑄𝐾 = 15 KN/m and wind load 𝑊𝐾 as shown in the figure below.
references are
from: BS8110-1
the wall thickness is 200 mm and the story height is 3m. Design the
wall if grade 30 and 460 are to be used for concrete and all steel
respectively.
Loading
Section 3.2.1.2.2
Fc = (1.4 x 30 + 1.6 x 15) x 5 = 990KN/m length
Equation 42
Table 3.25
𝐹𝑐 ≤ 0.35𝑓𝑐𝑢 𝐴𝑐 + 0.67𝑓𝑦 𝐴𝑠𝑐
Using min. area of steel required (0.004 Ac)
= 0.35 × 30 × 103 × 200 + 0.67 × 460 × 103 × 200 × 0.004
= 2346.6 𝐾𝑁/𝑚
Section 3.2.1.2.2
OK
Assume critical combination is (1.0 Gk + 1.4 wk)
N = 1.0 x 30 x 5 x 3 = 450 KN
M = 1.4 (25 x 15 + 50 x (12+9+6+3)) = 2625 KN.m
Table 3.19
Referring to table 3.19 in the code the end conditions are condition
1 at top and bottom, so
Assume the clear height = 2750 mm
0.75 x 2750 /200 = 10.3125 < 15
∴ The wall is stocky
𝑓𝑡 =
𝑁
𝑀
450 × 103
2625 × 106
−
=
−
= 0.75 − 8.75 = − 8𝑁/𝑚𝑚2
𝑡𝐿2
30002
𝐿𝑡
3000 × 200
( )
(200 ×
)
6
6
Assume that the tension will be resisted by 1 m at the end of the
wall
𝐴𝑠 =
0.5𝑓𝑡 𝐿𝑡 𝑡 0.5 × 8 × 1000 × 200
=
= 1830.6 𝑚𝑚2 𝑜𝑟 915 𝑚𝑚2 /𝑓𝑎𝑐𝑒
0.95 𝑓𝑦𝑣
0.95 × 460
Provide T16@200. Area steel provided 1005 mm2/face
29
Job Title
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Subject:
Worked Example 17 - Axially loaded footing
Client
UofK
Made by
ME
Date
2018
Checked by
Unless otherwise A reinforced concrete column 500 mm x 500 mm carries G = 1000
K
stated, all
KN and Q K = 500 KN. Design a square foundation to resist the load
references are
from: BS8110-1
if the safe bearing capacity is 300 KN/m2. The concrete is grade 35
and the reinforcement is grade 460.
Table 3.2
The condition of exposure is moderate for non-aggressive soil. Take
the cover 50 mm.
Size of base
Assume the base weight = 100 KN
Service load = 1000 + 500 + 100 = 1600 KN
Base area = 1600 / 300 = 5.33 m2
Let the base area = 2.4 m x 2.4 m
Base area provided = 5.76 m2
Bending
Section 3.2.1.2.2
ultimate load = 1.4 × 1000 + 1.6 × 500 = 2200 KN
ultimate pressure = 2200 ÷ 5.76 = 382 KN/m2
The critical section XX at the column face is
shown in the figure above.
Mxx = 382 × 0.95 × 2.4 ×
0.95
= 413.7 KN. m
2
Try overall depth of 550 mm with 16 mm bars. The effective
depth d = 550 – 50 – 16 – 8 = 476 mm
Section 3.4.4.4
M
413.7 × 106
=
= 0.0217
2
bd fcu 2400 × 4762 × 35
z = d {0.5 + √(0.25 −
K
0.9
)},
z = 0.95d = 452.2 mm
AS =
M
413.7 × 106
=
= 2093.5 mm2
0.95fy z 0.95 × 460 × 452.2
Provide T16@200 mm, area steel provided 2412
Vertical shear
The critical section YY at distance d from the column face as shown
in the figure above.
30
Mohamed Elhassan-Civil Engineering Department-UofK
V = 382 × 2.4 × 0.474 = 434.56 KN
Section 3.4.5.2
Table 3.8
Shear stress v =
V
434.56 × 1000
N
=
= 0.38
< 0.8√𝑓𝑐𝑢
b×d
2400 × 476
mm2
1/3
𝑣𝑐 =
0.79 100𝐴𝑆
(
)
𝛾𝑚
𝑏𝑑
400 1/4 𝑓𝑐𝑢 1/3
(
) ( )
𝑑
25
1
1
100𝐴𝑆 3
100 × 2412 3
∴(
) =(
) = 0.595 < 3 ,
𝑏𝑑
2400 × 476
1
1
400 4
400 4
(
) =(
) = 0.957 (𝑠ℎ𝑜𝑢𝑙𝑑 𝑛𝑜𝑡 𝑏𝑒 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑡 𝑜𝑛𝑒)
𝑑
476
(
𝑓𝑐𝑢 1/3
) = 1.08775
25
∴ 𝑣𝑐 = 0.41 𝑁/𝑚𝑚2
Since v < vc, shear reinforcement is not required
Punching shear
Punching shear is checked on a perimeter 1.5d from the column
face.
The critical perimeter = 4 × (1.928) = 7.712 m
Punching shear force = 382 × (2.42 − 1.9282 ) = 780.355 KN
Section 3.4.5.2
Shear stress v =
V
780.355 × 1000
N
=
= 0.2125
< 𝑣𝑐
b×d
7712 × 476
mm2
The base is satisfactory, no shear reinforcement is required.
31
Job Title
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Subject:
Worked Example 18 - Isolated footing subjected to axial load and moment
UofK
Client
ME
Made by
Date
2018
Checked by
Unless otherwise A reinforced concrete column 500 mm x 500 mm carries load and
stated, all
moment as shown in the table below. Design a square footing to resist
references are
from: BS8110-97 the load if the safe bearing capacity is 300 KN/m2. The concrete is
grade 35 and the reinforcement is grade 460.
Table 3.2
Vertical load (KN)
Moment (KN.m)
Dead
1000
44
Imposed
500
24
The condition of exposure is moderate for non-aggressive soil. Take
the cover 50 mm.
Size of base
Assume the base weight = 100 KN
Total axial load = 1000 + 500 + 100 = 1600KN
Total moment = 44 + 24 = 68 KN.m
Assume base area = 6.25 m2
the base dimensions = 2.5 m x 2.5 m
pressure P =
total load
Moment
±
base area modulus Z
Area = bl
Z = bl2 /6
Pmax =
1600
68
KN
+
= 282.12 2 < 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑂𝐾
2
6.25 2.5 × 2.5 /6
m
Bending
Section 3.2.1.2.2
ultimate load = 1.4 × 1000 + 1.6 × 500 = 2200 KN
ultimate moment = 1.4 × 44 + 1.6 × 24 = 100 KN. m
ultimate pressure =
2200
100
+
= 390.4 KN/m2
6.25 2.53 /6
The critical section XX at the column face is
shown in the figure above.
Mxx = 390.4 × 1.0 × 2.5 ×
1.0
= 468.5 KN. m
2
Try overall depth of 650 mm with 16 mm bars. The effective
depth d = 650 – 50 – 16 – 8 = 576 mm
32
Mohamed Elhassan-Civil Engineering Department-UofK
Section 3.4.4.4
M
468.5 × 106
=
= 0.016
bd2 fcu 2500 × 5762 × 35
K
z = d {0.5 + √(0.25 −
0.9
)},
z = 0.95d = 547.2 mm
M
468.5 × 106
=
= 1959.2 mm2
0.95fy z 0.95 × 460 × 547.2
AS =
Provide T16@250 mm, area steel provided 2010
Vertical shear
The critical section YY at distance d from the column face as shown
in the figure above.
V = 390.4 × 2.5 × 0.424 = 413.824 KN
Section 3.4.5.2
Table 3.8
Shear stress v =
V
413.824 × 1000
N
=
= 0.287
< 0.8√𝑓𝑐𝑢
b×d
2500 × 576
mm2
1/3
𝑣𝑐 =
0.79 100𝐴𝑆
(
)
𝛾𝑚
𝑏𝑑
400 1/4 𝑓𝑐𝑢 1/3
(
) ( )
𝑑
25
1
1
100𝐴𝑆 3
100 × 2010 3
∴(
) =(
) = 0.519 < 3 ,
𝑏𝑑
2500 × 576
1
1
400 4
400 4
(
) =(
) = 0.913 (𝑠ℎ𝑜𝑢𝑙𝑑 𝑛𝑜𝑡 𝑏𝑒 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑡 𝑜𝑛𝑒)
𝑑
576
(
𝑓𝑐𝑢 1/3
) = 1.08775
25
∴ 𝑣𝑐 = 0.36 𝑁/𝑚𝑚2
Since v < vc, shear reinforcement is not required
Punching shear
Punching shear is checked on a perimeter 1.5d from the column
face.
The critical perimeter = 4 × (2.228) = 8.912 m
Punching shear force = 390.4 × (2.52 − 2.2282 ) = 502.06 KN
Section 3.4.5.2
Shear stress v =
V
502.06 × 1000
N
=
= 0.097
< 𝑣𝑐
b×d
8912 × 576
mm2
The base is satisfactory, no shear reinforcement is required.
33
University of Khartoum
Faculty of Engineering
Department of Civil Engineering
Reinforced Concrete Design
Job Title
Subject:
Worked Example 19 - Combined footing
UofK
Client
Made by
ME
Date
2018
Checked by
Unless otherwise
A reinforced concrete combined footing supports two columns 400
stated, all
references are
mm square with characteristic dead and imposed loads as shown in
from: BS8110-97
the table below. Design the footing to resist the columns’ load if the
safe bearing capacity is 300 KN/m2. The concrete is grade 35 and
the reinforcement is grade 460. Take the distance between column
equal to 3 m.
Table 3.2
Column one
Column two
Dead (KN)
1000
1000
Imposed (KN)
300
300
The condition of exposure is moderate for non-aggressive soil. Take
the cover 50 mm.
Size of base
Assume the base weight = 150 KN
Total axial load = 2 x 1000 + 2 x 300 + 150 = 2750 KN
Base area = 2750 / 300 = 9.167 m2
Assume base length = 4.6 m
Therefore, the base width = 2 m
Base area provided = 9.2 m2
Bending
Section 3.2.1.2.2
ultimate load = 1.4 × 1000 × 2 + 1.6 × 300 × 2 = 3760 KN
34
Mohamed Elhassan-Civil Engineering Department-UofK
ultimate pressure = 3760/9.2 = 408.7 KN/m2
bending moment diagram:
Try overall depth of 700 mm with 16 mm bars. The effective
depth d = 700 – 50 – 16 – 8 = 626 mm
1. Mid-span between the columns
Section 3.4.4.4
M
658 × 106
=
= 0.024
bd2 fcu 2000 × 6262 × 35
z = d {0.5 + √(0.25 −
K
0.9
)},
z = 0.95d = 594.7 mm
AS =
M
658 × 106
=
= 2531.9 mm2
0.95fy z 0.95 × 460 × 594.7
Provide T16@150 mm at top, area steel provided 2680 mm2
2. At the columns
Section 3.4.4.4
M
261.57 × 106
=
= 0.0095
bd2 fcu 2000 × 6262 × 35
z = d {0.5 + √(0.25 −
K
0.9
)},
z = 0.95d = 594.7 mm
AS =
M
261.57 × 106
=
= 1006.5 mm2
0.95fy z 0.95 × 460 × 594.7
Provide T16@300 mm, area steel provided 1340 mm2
3. Transverse bending
M = 408.7 × 4.6 × 0.8 ×
Section 3.4.4.4
0.8
= 601.6 KN. m
2
M
601.6 × 106
=
= 0.022
bd2 fcu 2000 × 6262 × 35
z = d {0.5 + √(0.25 −
K
0.9
)},
z = 0.95d = 594.7 mm
AS =
M
601.6 × 106
=
= 2314.87 mm2
0.95fy z 0.95 × 460 × 594.7
Provide T16@150 mm, area steel provided 2680 mm2
35
Mohamed Elhassan-Civil Engineering Department-UofK
Vertical shear
The critical section is at the face of the column
Column reaction V = 1880 KN
Shear stress v =
V
1880 × 1000
N
=
= 1.9
< 0.8√𝑓𝑐𝑢
column perimeter × d
1600 × 626
mm2
Shear at d from face of support:
V = 1266.08 − (407.7 × 2 × 0.2) − (408.7 × 2 × 0.626) = 590.8 KN
Table 3.8
v =
V
590.8 × 1000
=
= 0.47 N/mm2
b×d
2000 × 626
𝑣𝑐 =
0.79 100𝐴𝑆
(
)
𝛾𝑚
𝑏𝑑
400 1/4 𝑓𝑐𝑢 1/3
(
) ( )
𝑑
25
1
1
1/3
100𝐴𝑆 3
100 × 1340 3
∴(
) =(
) = 0.475 < 3 ,
𝑏𝑑
2000 × 626
1
1
400 4
400 4
(
) =(
) = 0.894 > 0.67
𝑑
626
(
𝑓𝑐𝑢 1/3
) = 1.12
25
∴ 𝑣𝑐 = 0.3 𝑁/𝑚𝑚2
Since v > vc, shear reinforcement is required
Or the overall depth should be increased to 800 mm.
Punching shear
Can not be checked, because the critical perimeter 1.5d lies outside
the base area.
36
Appendix: Selected tables from BS 8110
Table 1 Form and area of shear reinforcement in beams (table 3.7 of BS 8110)
Value of v (N/mm2)
Form of shear reinforcement
to be provided
Less than 0.5vc
throughout the beam
See Note 1
Area of shear reinforcement
to be provided
__
0.5𝑉𝑐 < 𝑉 < (𝑉𝑐 + 0.4)
Minimum links for whole length
of beam
𝐴𝑠𝑣 ≥ 0.4b𝑠𝑣 /0.95 (see Note 2)
(𝑉𝑐 + 0.4) < 𝑉 < 0.8√𝑓𝑐𝑢
𝑜𝑟 5 𝑁/𝑚𝑚2
Links or links combined with
bent-up bars. Not more than 50%
of the shear resistance provided
by the steel may be in the form of
bent-up bars (see Note 3)
Where links only provided:
Asv ≥ bvsv (v – vc)/0.95fyv. Where
links and bent-up bars
provided: see Cl 3.4.5.6 of BS
8110
Notes
1 While minimum links should be provided in all beams of structural importance, it will be
satisfactory to omit them in members of minor structural importance such as lintels or where the
maximum design shear stress is less than half Vc.
2 Minimum links provide a design shear resistance of 0.4 N/mm2.
3 See Cl.3.4.5.5 of BS 8110 for guidance on spacing of links and bent-up bars.
Table 2 Values of design concrete shear strength, Vc (N/mm2) (table 3.8 of BS 8110)
𝟏𝟎𝟎𝑨𝒔
𝒃𝒗 𝒅
Effective depth (mm)
125
150
175
200
225
250
300
400
≤ 0.15
0.45
0.43
0.41
0.40
0.39
0.38
0.36
0.34
0.25
0.53
0.51
0.49
0.47
0.46
0.45
0.43
0.40
0.5
0.67
0.64
0.62
0.60
0.58
0.56
0.54
0.50
0.75
0.77
0.73
0.71
0.68
0.66
0.65
0.62
0.57
1.00
0.84
0.81
0.78
0.75
0.73
0.71
0.68
0.63
1.50
0.97
0.92
0.89
0.86
0.83
0.81
0.78
0.72
2.00
1.06
1.02
0.98
0.95
0.92
0.89
0.86
0.80
≥ 3.00
1.22
1.16
1.12
1.08
1.05
1.02
0.98
0.91
For characteristic concrete strengths greater than 25 N/mm2, the values in this table may be
multiplied by (fcu/25)1/2. The value of fcu should not be taken as greater than 40.
Table 3 Basic span/effective depth ratio (table 3.9 of BS 8110)
Support conditions
Rectangular section
Flanged beams with 𝒃𝒘 /𝐛 ≤ 𝟎. 𝟑
Cantilever
7
5.6
Simply Supported
20
16.0
Continuous
26
20.8
37
Table 4 Modification factor for tension reinforcement (table 3.10 of BS 8110)
Service stress
𝑴/𝒃𝒅𝟐
0.5
0.75
1.0
1.5
2.0
3.0
4.0
5.0
6.0
100
2.00
2.00
2.00
1.86
1.63
1.36
1.19
1.08
1.01
150
2.00
2.00
1.98
1.69
1.49
1.25
1.11
1.01
0.94
160 (𝑓𝑦 = 250)
2.00
2.00
1.91
1.63
1.44
1.21
1.08
0.99
0.92
200
2.00
1.95
1.76
1.51
1.35
1.14
1.02
0.94
0.88
250
1.90
1.70
1.55
1.34
1.20
1.04
0.94
0.87
0.82
300
1.60
1.44
1.33
1.16
1.06
0.93
0.85
0.80
0.76
333 (𝑓𝑦 = 460)
1.41
1.28
1.18
1.05
0.96
0.86
0.79
0.75
0.72
Note: The design service stress in the tension reinforcement may be estimated from the equation
𝑓𝑠 =
2𝑓𝑦 𝐴𝑠,𝑟𝑒𝑞
3𝐴𝑠,𝑝𝑟𝑜𝑣 𝛽𝑏
Table 5 Modification factor for compression reinforcement (table 3.11 of BS 8110)
𝟏𝟎𝟎𝑨𝒔,𝒑𝒓𝒐𝒗 /𝒃𝒅
Factor
0
0.15
0.25
0.35
0.50
0.75
1.00
1.50
2.00
2.50
≥ 3.00
1.00
1.05
1.08
1.10
1.14
1.20
1.25
1.33
1.40
1.45
1.50
Table 6 Form and area of shear reinforcement in solid slab (table 3.16 of BS 8110)
Value of v (N/mm2)
Form of shear reinforcement
to be provided
Area of shear reinforcement
to be provided
𝑣 < 𝑣𝑐
None required
None
𝑣𝑐 < 𝑣 < (𝑣𝑐 + 0.4)
Minimum links in areas where
𝑣 > 𝑣𝑐
𝐴𝑠𝑣 ≥ 0.4b𝑠𝑣 /0.95𝑓𝑦𝑣
Where links only provided:
𝐴𝑠𝑣 ≥ b𝑠𝑣 (v − 𝑣𝑐 )/0.95𝑓𝑦𝑣
Where bent-up bars only
provided:
𝐴𝑠𝑣 ≥ b𝑠𝑣 (v − 𝑣𝑐 )/0.95𝑓𝑦𝑣
(cos 𝛼 + sin 𝛼 × cos 𝛽)
Note In slabs less than 200 mm deep, it is difficult to bend and fix shear reinforcement so that its
effectiveness can be assured. It is therefore not advisable to use shear reinforcement in such slabs.
(𝑣𝑐 + 0.4) < 𝑣 < 0.8√𝑓𝑐𝑢
or 5 N/mm2
Links and/or bent-up bars in
any combination (but the
spacing between links or bentup bars need not be less than d)
38
Table 7 Minimum percentages of reinforcement (table 3.25 of BS 8110)
Definition of
percentage
Situation
Minimum percentage
fy = 250
fy = 460
N/mm2
N/mm2
Tension Reinforcement
100𝐴𝑠 /𝐴𝑐
0.80
0.45
𝑏𝑤 /b < 0.4
100𝐴𝑠 /𝑏𝑤 ℎ
0.32
0.18
𝑏𝑤 /b ≥ 0.4
100𝐴𝑠 /𝑏𝑤 ℎ
0.24
0.13
T-beam
100𝐴𝑠 /𝑏𝑤 ℎ
0.48
0.26
L-beam
100𝐴𝑠 /𝑏𝑤 ℎ
0.36
0.20
100𝐴𝑠 /𝐴𝑐
0.24
0.13
Sections subjected mainly to pure tension
Sections subjected to flexure:
Flanged beams, web in tension:
Flanged beams, flange in tension:
Rectangular section (in solid slabs this
minimum should be provided in both directions)
Compression Reinforcement (where such reinforcement is required for the ULS)
100𝐴𝑠𝑐 /𝐴𝑐𝑐
0.40
0.40
100𝐴𝑠 /𝐴𝑐
0.40
0.40
Flange in compression
100𝐴𝑠 /𝑏ℎ𝑓
0.40
0.40
Web in compression
100𝐴𝑠 /𝑏𝑤 ℎ
0.20
0.20
100𝐴𝑠𝑐 /𝐴𝑐
0.20
0.20
General rule
Simplified rules for particular cases:
Rectangular column or wall
Flanged beam:
Rectangular beam
Transverse reinforcement in flanges or flanged beams (provided over full effective flange
width near top surface to resist horizontal shear)
100𝐴𝑠𝑡 /ℎ𝑓 𝑙
0.15
0.15
Table 8 Values of B for braced and unbraced columns (table 3.19 and 3.20 of BS 8110)
End condition
at top
Values of 𝜷 for braced columns
Values of 𝜷 for unbraced columns
End condition at bottom
End condition at bottom
1
2
3
1
2
3
1
0.75
0.80
0.90
1.20
1.30
1.60
2
0.80
0.85
0.95
1.30
1.50
1.80
3
0.90
0.95
1.00
1.60
1.80
4
-
-
-
2.20
-
-
39
Table 9 Bending moment coefficients for rectangular panels supported on four sides with provision
for torsion at corners (table 3.14 BS 810)
Type of panel and
moments
considered
Short span coefficients, 𝜷𝒔𝒙 Values of 𝒍𝒚 /𝒍𝒙
Long span
coefficients
, 𝜷𝒔𝒚 Values
of 𝒍𝒚 /𝒍𝒙
1
1.1
1.2
1.3
1.4
1.5
1.75
2
0.031
0.37
0.042
0.046
0.050
0.053
0.059
0.063
0.032
0.024
0.028
0.032
0.035
0.037
0.040
0.044
0.048
0.024
0.039
0.044
0.048
0.052
0.055
0.058
0.063
0.067
0.037
0.029
0.033
0.036
0.039
0.041
0.043
0.047
0.050
0.028
0.039
0.049
0.056
0.062
0.068
0.073
0.082
0.089
0.037
0.030
0.036
0.042
0.047
0.051
0.055
0.062
0.067
0.028
Interior panels
Negative moment at
continuous edge
Positive moment at midspan
One short edge discontinuous
Negative moment at
continuous edge
Positive moment at midspan
One long edge discontinuous
Negative moment at
continuous edge
Positive moment at midspan
Two adjacent edges discontinuous
Negative moment at
continuous edge
Positive moment at midspan
0.047
0.056
0.063
0.069
0.074
0.078
0.087
0.093
0.045
036
0.042
0.047
0.051
0.055
0.059
0.065
0.070
0.034
0.046
0.050
0.054
0.057
0.060
0.062
0.067
0.070
-
0.034
0.038
0.040
0.043
0.045
0.047
0.050
0.053
0.034
-
-
-
-
-
-
-
-
0.045
0.034
0.046
0.056
0.065
0.072
0.078
0.091
0.100
0.034
Two short edges discontinuous
Negative moment at
continuous edge
Positive moment at midspan
Two long edges discontinuous
Negative moment at
continuous edge
Positive moment at midspan
Three edges discontinuous (one long edge continuous)
Negative moment at
continuous edge
Positive moment at midspan
0.057
0.065
0.071
0.076
0.081
0.084
0.092
0.098
-
0.043
0.048
0.053
0.057
0.060
0.063
0.069
0.074
0.044
Three edges discontinuous (one short edge continuous)
Negative moment at
continuous edge
Positive moment at midspan
-
-
-
-
-
-
-
-
0.058
0.042
0.054
0.063
0.071
0.078
0.084
0.096
0.105
0.044
0.065
0.074
0.081
0.087
0.092
0.103
0.111
0.056
Four edges discontinuous
Positive moment at midspan
0.055
40
Table 10 Shear force coefficients for rectangular panels supported on four sides with provision for
tension at corners (table 3.15 of BS 8110)
Type of panel and
location
𝜷𝒗𝒙 Values of 𝒍𝒚 /𝒍𝒙
𝜷𝒗𝒚
1
1.1
1.2
1.3
1.4
1.5
1.75
2
0.33
0.36
0.39
0.41
0.43
0.45
0.48
0.50
0.33
Interior panels
Continuous edge
One short edge discontinuous
Continuous edge
0.36
0.39
0.42
0.44
0.45
0.47
0.50
0.52
0.36
Discontinuous edge
-
-
-
-
-
-
-
-
0.24
One long edge discontinuous
Continuous edge
0.36
0.40
0.44
0.47
0.49
0.51
0.55
0.59
0.36
Discontinuous edge
0.24
0.27
0.29
0.31
0.32
0.34
0.36
0.38
-
Two adjacent edges discontinuous
Continuous edge
0.40
0.44
0.47
0.50
0.52
0.54
0.57
0.60
0.40
Discontinuous edge
0.26
0.29
0.31
0.33
0.34
0.35
0.38
0.40
0.26
Two short edges discontinuous
Continuous edge
0.40
0.43
0.45
0.47
0.48
0.49
0.52
0.54
-
Discontinuous edge
-
-
-
-
-
-
-
-
0.26
Two long edges discontinuous
Continuous edge
-
-
-
-
-
-
-
-
0.40
Discontinuous edge
0.26
0.30
0.33
0.36
0.38
0.40
0.44
0.47
-
Three edges discontinuous (one long edge continuous)
Continuous edge
0.45
0.48
0.51
0.53
.055
0.57
0.60
0.63
-
Discontinuous edge
0.30
0.32
0.34
0.35
0.36
0.37
0.39
0.41
0.29
Three edges discontinuous (one short edge continuous)
Continuous edge
-
-
-
-
-
-
-
-
0.45
Discontinuous edge
0.29
0.33
0.36
0.38
0.40
0.42
0.45
0.48
0.30
0.36
0.39
0.41
0.43
0.45
0.48
0.50
0.33
Four edges discontinuous
Discontinuous edge
0.33
Table 11 Bending moment coefficients for slab spanning in two directions at right angles, simply
supported on four sides (table 3.13 of BS 8110)
𝒍𝒚 /𝒍𝒙
1
1.1
1.2
1.3
1.4
1.5
1.75
2
𝜶𝒔𝒙
0.062
0.074
0.084
0.093
0.099
0.104
0.113
0.118
𝜶𝒔𝒚
0.062
0.061
0.059
0.055
0.051
0.046
0.037
0.029
41
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