By T. Viswanathan
1
Elements not requiring shear reinforcement.
Estimate the shear capacity of RCC slab in
which no shear reinforcement is provided.
The particulars of slab are as follows:
Thickness of slab
750mm, cover 50mm
concrete grade M30.
Reinforcement 25dia 125mm.
Area of reinforcement = 39.27 cm2/m
25
Effective depth = 750 – 50 – = 687.5mm.
2
2
Effective width = 1000mm
Asl = 39.27 cm2
𝐴𝑠𝑙
39.27
ρl =
=
= .0057 ≤ .02
𝑏𝑤 𝑑
100 𝑥 68.75
k = 1+
200
= 1.53
687.5
Axial Load = 0
 𝜎𝑐𝑝 = 0.
VRDC
=[0.12 x 1.53 80 x .0057 x 30)0.33 𝑥 1000 x 687.5 x 10−3
= 302 kN/m
VRDCmin = .031 x 1.533 2 x 301 2 x 1000 x 687.5 x 10−3 =
221.0kN/m
The slab can with stand a shear of 302 kN/m without
providing any shear reinforcement.
3
Elements Requiring shear reinforcement:
Determine the shear reinforcement in RCC Beam at Various
sections. The particulars of the beams are as follows:
RCC girder
M35 Grade concrete
Span = 21.0M
Girder cross section as shown.
Fig : Cross section of girder
4
Summary of Ultimate Shear Force (Value in kN)
Section
1M
2M
5M
6.71M
10.0M
1234
1072
856
759
544
650
250
250
250
250
2.42
2.15
1.54
Distance
from
Support
Shear Force
in kN
Width of
Web in MM
Now Proceed to Flow Chart
Shear stress
in N/mm2 =
𝑉𝑤
0.9 𝑧 𝑥 𝑏
1.34
3.03
z = 1570mm
5
Max Allowable shear stress 0.225 𝑥 0.6 1 −
35
310
𝑥 35 = 4.19 N/mm2
As the shear stress at the various section are less than 4.19 N/mm2
no redesign is required
Allowable shear stress corresponding  = 21.80 0.155 𝑥 0.6 1 −
𝐴𝑠𝑤 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑥 𝑏𝑤
=
𝑠
𝑓𝑦𝑤𝑑 cot 𝜃
fywd = 0.87 fyk
1M
Using fyk 415
0.97mm2/m
𝐴𝑠𝑤 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑥 𝑏𝑤
=
𝑠
0.87 𝑥 415 𝑥 2.5
Providing reinforcement 12 @ 200
Reinforcement mm2/m 1.13mm2/m
2M
5M
6.71M
10M
0.97mm2/m
0.60mm3/m 0.43mm3/m
12 @ 250
0.90mm2/m
10 @ 220
10 @ 300
0.71mm2/m 0.52mm2/m
Not
applicable
6
The minimum reinforcement ratio as per codal equation 16.9
.072 𝑓𝑐𝑘
𝐴𝑠𝑤
=
𝑏𝑤 𝑠
𝑓𝑦𝑘
.072 𝑓𝑐𝑘
𝐴𝑠𝑤
=
𝑏𝑤 𝑠
𝑓𝑦𝑘

𝐴𝑠𝑤
= 1.026 𝑥 10−3 𝑥 𝑏𝑤
𝑠
𝐴𝑠𝑤
= 1.026 𝑥 10−3 𝑥 650 = 0.667𝑚𝑚2 /𝑚
𝑠
𝐴𝑠
For bw 250mm 𝑤 = 1.026 𝑥 10−3 𝑥 250 = 0.250𝑚𝑚2 /𝑚
𝑠
For bw 650mm
Design reinforcement is much higher.
Suppose at section 2M the shear force is made 1.4 times of
the shear force worked out there shear stress works out to
4.03 N/mm2 > 2.89 N/m2 but less than 4.19 N/mm2
Hence no redesign of section is required:
As the shear stress is more then 2.89N/mm2 the  value has
to be worked out.
𝜃 = 0.5 𝑠𝑖𝑛−1
4.03
4.19
= 37.060
Cross checking from codal equation 10.8
7
 4.03 =
1 𝑥 0.532 𝑥 0.45 𝑥 35
1𝑥0.532 𝑥0.45𝑥35
=
= 4.03
cot 37.06+tan 37.06
2.08
LHS = RHS
Hence the formula shown for estimating the angle can be
𝐴𝑠
4.03 𝑥 250
used.  𝑤 =
= 2.1076𝑚𝑚2 /𝑚
𝑠
0.87 𝑥 415 𝑥 cot 37.06
For 3.03N/𝑚𝑚2 the angle is 23.150 and the steel
requirement is 0.897𝑚𝑚2 /𝑚
The addition longitudinal tensile steel required at various
sections for the original calculation.
Using the codal expression
∆ 𝐹𝑡𝑑 = 0.5 𝑉𝐸𝐷 = (cot 𝜃 − cot 𝛼) 𝜃 = 21.8 𝛼 = 90
∆ 𝐹𝑡𝑑 = 0.5 𝑉𝐸𝐷 2.5 − 0 = 1.25 𝑉𝐸𝐷 𝐴𝑠𝑡 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 =
1.25 𝑉𝐸𝐷
=
415/1.15
3.46 𝑥 10−3 𝑉𝐸𝐷
Section
1M
2M
5M
6.71M
10M
Addition
steel mm2
4.269
1.17
𝑥 3.709
1.25
= 3.467
2.961
2.626
1.882
8
Additional longitudinal steel over and above that required to
resist the moment has to be provided. It is to be noted that θ
becomes shallower, the longitudinal steel will increase. But
the shear steel will reduce.
If θ increases the additional longitudinal steel will reduce
and shear steel will increase. Alternatively following shift
rule the additional longitudinal steel requirement can be
provided.
9
Example of Prestressed Box Design for Shear:
Span of box 31.00M, Grade of Concrete M40, Loading class
70R or Two lane of class A. Prestressing 10 cable of 19 T 13
and 2 cable of 12 T 13 cross section of box shown below.
Cross Section of the Box Girder
10
The following are the Design Parameters 𝑓𝑐𝑡𝑑 =
Allowable tensile stress 𝑓𝑐𝑡𝑑 =
𝑓𝑐𝑡𝑘.05
2.1
=
= 1.4𝑀𝑃𝑎
𝟏.𝟓
1.5
𝑓𝑐𝑡𝑑.05
2.1
=
= 1.4𝑀𝑃𝑎
𝟏.𝟓
1.5
1–1
Support
2.2
θ9D
3.3
L/8
4.4
L/4
5.5
3L/8
6.6
L/2
Distance in mm from
support
0
1.8
3.88
7.75
11.63
15.5
Cross sectional area of box
m2
6.90
6.90
6.90
5.38
5.14
5.14
CG of Section form bottom
in m
1.09
1.09
1.09
1.2
1.22
1.22
Cable force after all losses
in kN
20784
21040
21400
21838
22136
22192
Vertical Component
of prestressing force in kN
1825
1472
898
365
0
0
Cable eccentricity from CG
of section in m
0.32
0.5
0.62
0.83
0.9
0.91
Section
11
Average compressive stress
P/A in kn/m2
3012
3049
3101
4059
4307
4318
Zt m3
4.16
4.16
4.16
3.99
3.96
3.96
Zb m3
3.5
3.5
3.5
2.68
2.52
2.52
𝑃
𝑃𝑒
+
𝐴
𝑍𝑡
1.41
0.52
-0.09
-0.48
-0.72
-0.78
4.91
6.05
6.89
10.82
12.21
12.33
I = Moment of Inertia in m4
3.80
3.80
3.80
3.21
3.08
3.08
Ultimate shear force in kNm
5640
5000
4100
3150
1750
620
vertical component of
prestress force in kN
1825
1472
898
365
0
0
Top fiber stress due to
prestress in MPa
𝑃
𝐴
. +
𝑃𝑒
𝑍𝑏
Bottom fiber
stress due to prestress in MPa
12
Net shear force in kN
3815
3528
3202
2785
1750
620
Ultimate moment in kN
0
9400
18606
31720
38700
41147
Stress at bottom due to
moment kN/m2
0
2685
5316
11836
15357
16328
Resultant at Stress due to
prestress effect and applied
moment in MPa
4.91
3.35
1.59
-0.98
-3.09
-3.97
uncracked Uncracked Uncracked Uncracked Cracked
Cracked
Comparing with allowable
tensile stress of 1.4 MPa
S = 𝐴𝑥 = moment of the area
above CG about CG in m3
Breadth of web bw in mm
𝐼 𝑏𝑤
𝑆
= 𝑚2
2.415
2.415
2.415
1.95
1.88
1.88
2(600-0.5
x 90) =
1110
1110
1110
2(338-0.5 x
90) 586
600
600
1.746
1.746
1.746
0.964
0.983
0.983
13
2
𝑓𝑐𝑡𝑑
+ 𝜎𝑐𝑝 𝑓𝑐𝑡𝑑 in kN/m2 .
2485
2495
2590
2764
4338
4356
4382
2664
Note: σcp axial stress due to
prestress at CG
Shear capacity
𝐼𝑏𝑤
𝑆
2
𝑓𝑐𝑡𝑑
𝑥 𝜎𝑐𝑝 𝑓𝑐𝑡𝑑
in kN
Shear reinforcement
Not
Not
Not
required required required
Min to be Min to be Min to be
Provided Provided Provided
Required
Formula not
applicable section is
cracked.
In case of section not cracked but shear capacity is less than
the applied shear adequate shear reinforcement to be provided.
Which is to be based on codal equation 10.28.
If the section is cracked also shear reinforcement need to be
provided as the capacity of section to resist the shear without
shear reinforcement will be virtually negligible due to absence
of sizable amount of tensile reinforcement.
14
To calculate the shear reinforcement the most important parameter
required is lever arm Z which is obtained form bending analysis.
Analyzing Section 4 – 4
Assuming the stress in the cable corresponding to the yield strain
(Assumption steel yields)
Force in
10 cables of 19T13 = 0.87 x 3492 x 10 = 30380 kN
02 cables of 12T13 = 0.87 x 2205 x 2 = 3837 kN
Total Tensile force = 34217 kN
To balance this tensile force, the NA axis will occur at 0.18M for top
Lever arm = 0.80 + 0.83 - .09 = 1.54 M. Note: all cables are provided
in one row.
Moment this force can resist = 34217 x 1.54
52694 kNm > 31720M.
For the actual moment the Lever arm will be slightly low. But
ignoring this take Z = 1.6 M.
Mean compressive stress at section 4 – 4
allowable compression
0.67 𝑥 4000
2 2
=
17866𝑘𝑁/𝑚
𝑚 𝛼𝑐𝑤 =
1.5
θ = 45
4059 kN/m2. Max
4059
1 + 17866 = 1.227 𝛼𝑐𝑤 = 1.23
15
At
Capacity at θ = 450 = 1.23 x 0.135 x 40000 x 2 x 0.293 x 1.6 =
6227 kN
θ = 21.8 = 4290 kN. Actual shear force at section 4-4 is 2785 kN
2785
= 2970 𝑘𝑁/𝑚2 : Taking cot θ = 21.8 min
2 𝑥 0.293 𝑥 1.6
𝐴𝑠𝑤
2.97𝑥293
=
= 1.048𝑚𝑚2 /𝑚𝑚 10 @ 150 will give =
𝑠
0.8 𝑥 415𝑥2.5
Shear stress =
1.048mm2/mm
For section 5.5 mean compressive stress = 4307 kN/m2
0.67
𝑥 40000)
1.5
Allowable compressive stress = fcd = 17866 kN/m2 (
𝜎𝑐𝑝 < 0.25 𝑓𝑐𝑑 , 𝛼𝑐𝑚 = 1 + 0.24 ≈ 1.25
Max allowable shear force = 1.25 x 0.135 x 40000 x 2 x 0.30 x 1.6
= 6480 kN if θ = 450
It θ = 21.8
= 1.25 x 0.093 x 40000 x 2 x 0.3 x 1.6 =
4464 kN
Shear force at the section = 1750 kN
16
Shear stress =
1750
= 1823𝑘𝑁/𝑚2
2 𝑥 0.3 𝑥 1.6
Area of shear reinforcement
=
𝐴𝑠𝑤
1.823 𝑥 30
=
= .0.66𝑚𝑚2 /𝑚𝑚
𝑠
0.87 𝑥 415 𝑥 2.5
Adopting 2L10 @ 200 mm 2L reinforcement provided is =
0.7854 𝑥 2 𝑥 100
= 0.785𝑚𝑚2 /𝑚𝑚
200
Hence adopt 10 @ 200: in each web and provide same
reinforcement in the next section 6 – 6 also.
other sections provide minimum reinforcement
Minimum reinforcement =
.072 𝑓𝑐𝑘
𝐴𝑠𝑤
300 𝑥 .072 40
𝑏𝑤 =
𝑏𝑤 =
=
𝑠
𝑓𝑦𝑘
415
0.329𝑚𝑚2 /𝑚𝑚
By examining the capacity at a point 350mm below deck slab.
350 form top (below cantilever)
Prestress effect and Moment effect are shown below.
17
Section
1–1
2.2
3.3
4.4
5.5
6.6
Effect of prestress in
N/mm2 at the above
point
2.02
1.48
1.13
1.79
1.54
1.51
Moment effect at the
above point
0
+1.38
+2.74
+4.44
5.40
5.75
Total effect at the
above point
2.02
2.86
2.87
5.93
6.94
7.86
1st Moment of are a
of Deck and Hench
about CG
1.91
1.91
1.91
1.73
1.70
1.70
𝐼𝑏𝑤
𝑠
2.20
2.20
2.20
1.086
1.087
1.087
14002 + 1400𝑥2020
= 2188
2442
2716
3203
3417
3274
4813 kN
5372
5975
3478
Not applicable
2
𝑓𝑐𝑡𝑑
+ 𝜎𝑐𝑝 𝑓𝑐𝑡𝑑
𝐼𝑏𝑤
2
𝑓𝑐𝑡𝑑
+ 𝜎𝑝 𝑓𝑐𝑡𝑑
𝑠
18
The section has more shear resisting capacity than at CG of
section. Hence the designer has to check at other location on the
cross section if required in case if there is a doubt that capacity
may work out less than the capacity at the CG of section.
19
Punching shear worked Examples:
Following are the details
1.
Open Foundation
:
2.
Column
:
3.
Load on Column
:
4.
Moment on Column
:
5.
Material Properties
:
6.
Footing thickness
Size
Size
:
2000
Base pressure on foundation 4 𝑥 4 ±
4m x 4m
1m x 1m
2000 Kn
1000 kNm
fck = 35 MPa
fyk = 500 MPa
0.7 M
6 𝑥 1000
43
= 125 ± 93.75 = 219𝑘𝑁/𝑚2 31𝑘𝑁/𝑚
To Resist the bending moment reinforcement provided is : 25 MM
@ 200 c/c in both direction.
Reinforcement provided is 24.5cm2/m in each direction:
Effective depth dy = 700 – 50 – 25/2 = 637.5 mm
Effective depth for other direct dz = 637.5 – 25 = 612.5 mm
20
The effective depth for punching shear calculation
d=
637.5+612.5
= 625𝑚𝑚
2
Basic Control Perimeter u at 2d
= 2 (C1 +C2) + 4 x П x
2𝑑
𝑑
= 2 (C1 +C2) + 4 П d
= 2 (1.0 + 1.0) + 4 x П x 0.625
= 11.85m
Area within perimeter = (4 𝑥 𝛱 𝑥 0.6252 +
4 𝑥 0.625 𝑥 2 𝑥 1.00 𝑥 1.00 𝑥 1.0 = 10.908𝑚2
21
Base Pressure
Average pressure
187.70+42.7
2
=
115.2
𝑘𝑁/𝑚
2
Total upward force = 10.90 x 115.2 = 1255 kN
Net shear force = VED.red = VED – Δ VED
VED.red = 2000 – 1255 = .745kN
2450
% Steel in longitudinal direction = 637.5 𝑥 1000 = 0.00384
2450
% Steel in Transverse direction = 612.5 𝑥 1000 = 0.004
𝜌𝐿 = 0.00384 𝑥 0.004 = 0.00392
𝑘 =1+
200
= 1.562
625
22
0.18
𝑥1.565 (80𝑥0.00392𝑥35)1/3 = 0.417
0.5
𝑉𝑚𝑖𝑛 = 0.031𝑥1.5653/2 𝑥351/2 = 0.359 𝑀𝑃𝑎
𝑉𝑅𝑑𝑐
Governing VRdc = 0.417 MPa
For an Internal columns subject to moment
(𝐶1 )2
𝑊1 =
+ 𝐶1 𝐶2 + 4 𝐶2 𝑑 + 16𝑑 2 + 2𝛱 𝑑 𝐶1
2
1
= + 1 + 4 𝑥 1 𝑥 0.625 + 16 𝑥 0.6252 + 2𝛱 𝑥 0.625 𝑥 1.0
2
= 0.5 + 1 + 2.5 + 6.25 + 3.92 = 13.17 𝑚2
𝑉𝐸𝐷.𝑟𝑒𝑑
𝑀𝐸𝐷 𝑢
745 𝑥 103
𝑉𝐸𝐷 =
1+𝑘
=
𝑢1 𝑑
𝑉𝐸𝐷.𝑟𝑒𝑑 𝑊
11.85 𝑥 1000 𝑥 625
0.6 𝑥 1000 𝑥 106 𝑥 11.85 𝑥 1000
1+
745 𝑥 103 𝑥 13.17
= 0.100 1 + 0.724 = 0.172 𝑀𝑃𝑎 < 𝑉𝑅𝐷𝐶
23
As per clause 10.4.2 (4) and 10.4.5 (1) the punching shear
should also be verified at a distance less than 2d. Checking at a
section a d distance away from the column face.
Perimeter length u = 2 П x 0.625 + 2 (1.0 +1.0) = 7.925m
Area with in this perimeter ( П x 0.6252 + 4 x 0.625 x 1.0 + 1.02 )
= 4.7m2
Relieving Average Pressure: 125 kN/m2
Total upward face = 125 x 4.7 = 587 kN
Net shear force= 2000 – 587 = 1413 kN
Shear resistance of concrete = 0.417 x
2𝑑
= 0.417𝑥2 = 0.834 𝑀𝑃𝑎
𝑎
The value of W what was estimated earlier can not hold good.
as the plane has come closer: the W applicable for this plane is
1.02
0.625
0.625 2
0.625
𝑊=
+ 1 𝑥 1 + 4 𝑥 1.0 𝑥
+ 16 (
) + 2𝛱 𝑥 1.0 𝑥
2
2
2
2
= 0.5 + 1 + 1.25 + 3.125 + 1.963 = 6.838𝑚2
= 𝑉𝐸𝐷 =
𝑉𝐸𝐷.𝑟𝑒𝑑
𝑢𝑑
1+𝑘𝑉
𝑀𝐸𝑑
𝑢
𝐸𝐷.𝑟𝑒𝑑 𝑊
24
1413 𝑥 103
=
7.925 𝑥 103 𝑥 625
0.6 𝑥 1000 𝑥 106 𝑥 7.925 𝑥 103
1+
1413 𝑥 103 𝑥 6.838 𝑥 106
0.285 [1+0.492] = 0.425 MPa < 0.834 MPa
Checking the punching shear stress at face of column as per
clause 10.4.6.(2)
Wo = 4 x 1.0 = 4.0m
1.02
Wo =
+ 1.0 𝑥 1.0 = 1.5𝑚2
2
𝑉𝐸𝐷
𝑀𝐸𝑑 𝑢𝑜
2000 𝑥 103
𝑣𝐸𝐷 =
1+𝑘
=
𝑢𝑜 𝑑
𝑉𝐸𝐷.𝑟𝑒𝑑 𝑊𝑜
4000 𝑥 625
0.6 𝑥 1000 𝑥 106 𝑥 4000
1+
2000 𝑥 103 𝑥 1.5 𝑥 106
= 0.8[1+0.8] = 1.44 MPa
Allowable shear stress 0.3 𝑥 1 −
35
310
𝑥
0.67 𝑥 35
= 4.16 𝑀𝑃𝑎 > 𝑉𝐸𝐷
1.5
Hence the section is safe.
25
The box girder shown in example 10.3.3 is subjected to a
torsion of 5000 kNm at support. Design the shear and
longitudinal reinforcement.
Ak = (2 – 0.25) x (5.5 – 0.6) = 8.57 m2
Codal Equation 10.46
τti =
𝑇𝐸𝐷
2 𝐴𝑘 𝑡𝑒𝑓𝑖
VEdi =τti x tefi x Zi
Substituting for τti in the above equation
VEdi =
𝑇𝐸𝐷
𝑇
𝑥 𝑡𝑒𝑓𝑖 𝑥 𝑍𝑖 = 𝐸𝐷 𝑍𝑖
𝐴𝑘 𝑡𝑒𝑓𝑖
2 𝐴𝑘
VEdi is the design shear force due to torsion
𝑇𝐸𝐷 𝑍𝑖
𝐴
= 𝑠𝑡 𝑍𝑖 𝑓𝑦𝑑 cot 𝜃
2 𝐴𝑘
𝑠𝑡
𝐴
𝑇𝐸𝐷
 𝑠𝑡 =
𝑠𝑡
2 𝐴𝑘 𝑓𝑦𝑑 cot 𝜃
But
θ should be same as what has been assumed in shear
analysis.
26
θ = 21.80
fyk = 0.8 x 415 = 332 N/mm2
𝐴𝑠𝑡
5000 𝑥 106
2 /𝑚
=
=
0.35
𝑚𝑚
𝑠𝑡
2 𝑥 8.57 𝑥 106 𝑥 332 𝑥 2.5
Adopting 2L – 10 MM @400 c/c will give a reinforcement of
78.5 𝑥 2
= 0.39𝑚𝑚2 /𝑚.
400
Longitudinal Reinforcement
𝐴
𝑠𝑡
 𝑠𝑡 =
𝑇𝐸𝐷 cot 𝜃
2 𝐴𝑘 𝑓𝑦𝑑
𝐴𝑠𝑡
5000 𝑥 106
=
𝑥 2.5 = 2.19 𝑚𝑚2 /𝑚𝑚
6
𝑠𝑡
2 𝑥 8.57 𝑥 10 𝑥 332
2.19
on each face
= 1.1 mm2/mm 12 @ 100 mm c/c will give
2
1.13 mm2 /mm
checking for TRDmax for web capacity
2 𝑥 𝑣 𝑥 𝜎𝑐𝑤 𝑥 𝑓𝑐𝑑 𝑥 𝐴𝑘 𝑥 𝑡𝑒𝑓𝑖 sin 𝜃 cos 𝜃
2 0.6
1−
27
𝑇𝐸𝐷
5000
=
= 0.150
𝑇𝑅𝐷𝐶
33077
15% of allowable torsion the section is carrying. Hence
𝑉𝐸𝐷
𝑉𝑅𝐷𝐶
will workout to 0.85. Hence, the section can be subjected
85% allowable shear.
Similar exercise has to be done for other sections.
28
I beam is subjected To Torsion of 50 kNm. Design the section
for torsion. Concrete is M40 grade. The Reinforcement fyk is
415 N/mm2
I idealized section is show below.
Cross Section
29
As a first step torsional inertia of each rectangular is to be
evaluated. The torsional constant can be obtain from any
standard reference books.
Torsional Inertia of Deck Slab.
b = 1500mm
= 0.305
𝐼𝑥𝑥 =
t = 200mm
𝑏
= 7.5
𝑡
k (torsional constant)
1
x 0.305 x 1500 x 2003 = 18.3 x 108 mm4
2
As the deck slab is acting in two directions the torsional
inertia has been halved.
Torsional Inertia of Top flange:
b = 400mm
t = 250mm
𝑏
= 1.6
𝑡
k = 0.203
𝐼𝑥𝑥 = 0.203 x 400 x 2503 = 12.68 x 108 mm4
Torsional Inertia of web
b = 900mm
t = 300mm
𝑏
= 3.0
𝑡
k = 0.263
𝐼𝑥𝑥 = 0.263 x 900 x 3003 = 63.9 x 108mm4
Torsional Inertia of Bottom flange
30
b = 600mm
t = 350mm
𝑏
= 1.714
𝑡
k = 0.207
𝐼𝑥𝑥 = 0.207 x 600 x 3503 = 53.25 x 108mm4
Total Torsional Inertia of Section = 18.3 x 108 + 12.68
x 108 + 63.9 x 108 +53.25 x 108 = 148.13 x108
The torsion will be shared in proportion to their
torsional stiffness.
Deck slab TED
Top flange TED
Web TED
Bottom Flange TED
18.3
x 50 = 6.2 kNm
148.13
12.68
148.13
63.9
148.13
53.25
148.13
x 50 = 4.28 kNm
x 50 = 21.57 kNm
x 50 = 17.97 kNm
Design of Reinforcements for Torsion:
Deck slab torsion should be combined with Deck slab
design.
31
Effective thickness of the members can be found as follows.
For top flange
𝑡𝑒𝑓𝑖 =
𝐴
400 𝑥 250
=
= 76.9𝑚𝑚.
𝑊
2(400+250)
But this thickness should not be taken less than 2 twice the
distance of longitudinal bar from the surface [effective
cover].
Taking cover as 40mm dia for longitudinal and transverse bars
of 10mm.
Effective cover 40 + 10 + 5 = 55m. Twice the 110mm.
As 110 > 76.9mm consider 110mm.
Ak = (400 – 110) (250 – 110) = 40600mm2
The value of θ will be same as what has been worked out in
shear analysis. However for simplicity assume θ = 450
𝐴𝑠𝑡
𝑇𝐸𝐷
4.28 𝑥 106
=
=
= 0.1587𝑚𝑚2 /𝑚𝑚
𝑠𝑡
2 𝐴𝑘 𝑓𝑦𝑑 cot 𝜃
2 𝑥 40600 𝑥 332 𝑥 1
Adopt 8M @ 300 two legged stirrup = Reinforcement provided
32
= 50/300 = 0.166mm2/mm
Longitudinal reinforcement =
𝐴𝑠𝐿
𝑇
= 𝐸𝐷 cot 𝜃. Since θ = 450
𝑠𝐿
𝐴𝑘 𝑓𝑦𝑑
reinforcement will be same. Hence provide same
reinforcement.
The can be provided at corner of links instead of distributing
it throughout the section.
Checking the crushing resistance =𝑇𝑅𝑑𝑚𝑎𝑥 =
2𝑣 𝛼𝑐𝑤 𝑓𝑐𝑑 𝐴𝑘 𝑡𝑒𝑓𝑖 sin 𝜃 cos 𝜃.
𝑇𝑅𝐷𝑚𝑎𝑥 = 2 𝑥 0.6 1 −
33
Design of Web:
Thickness 𝑡𝑒𝑓𝑖 =
𝐴
900 𝑥 300
−
= 112.5𝑚𝑚
𝑤
2(900+300)
Twice the distance of longitudinal bar = 110mm
𝐴𝑘 = 300 − 112.5 900 − 112.5 = 147656 𝑚𝑚2
𝐴𝑠𝑡
𝑇𝐸𝐷
21.57 𝑥 106
=
=
= 0.22𝑚𝑚2 /
𝑠𝑡
𝐴𝑘 𝑓𝑦𝑑 cot 𝜃
2 𝑥 147656 𝑥 352 𝑥 1
𝑚𝑚 𝜃 𝑖𝑠 𝑡𝑎𝑘𝑒𝑛 𝑎𝑠 450
The reinforcement in each vertical leg is to be combined
with shear reinforcement and provided. The same amount of
longitudinal steel to be provided as θ has been taken as 450
Torsion Resisting Capacity
𝑇𝑅𝐷𝑚𝑎𝑥 = 2𝑣 𝜎𝑐 𝑓𝑐𝑑 𝐴𝑘 𝑡𝑒𝑓𝑖 sin 𝜃 cos 𝜃
=2𝑥0.6 1 −
40
310
𝑥
0.67 𝑥 40
147656 𝑥 112.5
1
1
𝑥
𝑥
𝑥
= 155 𝑘𝑁𝑚
1.5
106
2
2
= 155 𝑘𝑁𝑚 > 21.57 𝑘𝑁𝑚
Similarly bottom flange can be analyzed.
34
Thank
You
35