PHILADELPHIA UNIVERSITY
FACULTY OF ENGINEERING
DEPT. OF CIVIL ENGINEERING
REINFORCED CONCRETE (2)
SECOND SEMESTER 2015-2016
TORSION
LECTURER Dr. SA'AD A. AL-TA'AN
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TORSION
In general, torsion occurs when the resultant of the external forces acting on the
structure does not coincide with the vertical centroidal axis of the member.
Torsion is moment acting about the longitudinal axis of the member
(perpendicular to the bending axis). It is measured in units of (Force ×Distance),
i.e., kN.m.
Types of torsion:
1- Primary (Equilibrium) or statically determinate torsion. The external
load must be supported by torsion and its value can be determined from
the equations of equilibrium.
Shear and bending
moment
Shear, bending
moment, and torsion
X
P
T=P×X
T/2
T/2
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X
T=P×X
2- Secondary (Compatibility)
Secondary or Compatibility or (statically indeterminate torsion), arises from
the requirements of continuity, i.e., compatibility of deformation of adjacent
parts of the structures, and its value can't be determined from the equations of
equilibrium alone but with the conditions of compatibility of deformations.
Tu = wus.Ln2/24
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S. A. Al-Ta'an
Tu =wub.L2n/16
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S. A. Al-Ta'an
Stresses due to torsion
Stresses due to shear and torsion
Hollow Sections
Solid Sections
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S. A. Al-Ta'an
ACI–CODE APPROACH FOR DESIGNING FOR TORSION
The design theory is based on the thin-walled tube / plastic space truss model.
Both solid and hollow members are considered as tubes. Test data for solid and
hollow beams showed that, once torsional cracking has occurred, the concrete in
the center of the member has little effect on the torsional strength of the cross
section and hence can be ignored. This, in effect, produces an equivalent tubular
member.
The ACI Code approach for design for torsion is also based on that the required
strength should be less than or equal to the design strength:
U ≤ Φ.Sn
Tu ≤ Φ.Tn
Tu= factored or ultimate torque acting on the cross-section, Tn = nominal
torsional strength of the cross-section, and Φ is a strength reduction factor =
0.75.
Threshold Torsion
When Tu ≤ 1/4 cracking torsional strength ( T cr
2
. f c' Acp
), the torsion is

3 Pcp
neglected in design:
Tu 
2
 . f c' Acp
12
Pcp
(neglect torsion).
Acp = cross-sectional area and for hollow section use Ag (Net area) instead of
Acp. For beams with slab on one or two sides, part of the slab as shown below
should be taken as an integral part of the section, and Pcp = outer perimeter of
the cross-section.
hw
≤ 4hf
hw
} Whichever
is smaller
Torsion will resisted by closed stirrups and longitudinal bars (so that they form a
space truss), and λ = factor that depends on the type of concrete, =1.0 for normal
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S. A. Al-Ta'an
weight concrete, 0.85 for sand lightweight concrete, and 0.75 for all lightweight
concrete. Linear interpolation may be done between the above values depending
upon the volumetric ratio of the aggregates.
Torsional Reinforcement
Checking the cross-sectional dimensions
a. For solid sections subjected to shear and torsion;
(
Vu 2
T .P
V
2
)  ( u h )2  ( c 
2
bw .d
bw .d 3
1.7 Aoh
f c' )
b. For hollow sections
Vu
T .P
V
2
 u h  ( c 
bw .d 1.7 A 2
bw .d 3
oh
f c' )
Ph = stirrup perimeter, Aoh = area enclosed by the stirrups centre line, and for
rectangular sections equal to:
Aoh =x1.y1
x1 =least dimension of the stirrup dimension = b-2×clear concrete cover – ds , ds
= stirrup diameter, y1= larger dimension of the stirrup dimension = h-2×clear
concrete cover – ds.
If the above equation is not satisfied, use a bigger section.
For flanged section, Aoh and Ph are calculated as below:
x1
x2
y1
y2
Aoh =x1.y1+2[(y2 –x1)x2]
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S. A. Al-Ta'an
Ph = 2.y2+2 (y2-x1)+4.x2+2(y1-2.x2)
Design of closed stirrups
Tn 
2 Ao At f yt
cot 
s
Ao = 0.85Aoh ,
At = area of one leg of the stirrup, fyt = yield strength of the stirrup, s = spacing
of the stirrups, and θ = angle of inclination of the cracks spiraling the crosssection and range from 30 to 60◦ and as an average can be taken as 45◦.
Minimum area of closed stirrups for torsion
If torsion is acting alone without shear:
f c' bw .s
Min. At 
32 f yt
Min . At 
1 bw .s
6 f yt
If shear and torsion are acting together:
Min.( Av  2 At ) 
f c' bw .s
16 f yt
Min .( Av  2 At ) 
1 bw .s
3 f yt
Spacing of Stirrups
s  Ph / 8 or 300 mm
for rectangular stirrup Ph =2(x1 + y1), for L or T shape stirrups Ph = external
perimeter.
Design of longitudinal bars
The area of longitudinal steel should not be less than the stirrups per unit length
of the span:

A
Al  t Ph 

s

Min. Al 
f yt  2
 cot 
f yl 
5 f c' Acp
12 f y
A  f yt
 t Ph 
 fy
s





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S. A. Al-Ta'an
In the above equation, At/s should not be less the minimum At/s =b/(6.fy), or
f c' bw
.
Min. At / s 
32 f yt
Arrangement and Spacing of Longitudinal Bars
i. Sps. ≤ 300 mm in both directions (horizontal and vertical).
ii. One bar in each corner of the cross-section and stirrup.
iii. db ≥ s /24.
Example 1.
A rectangular beam 300 mm wide, 600 mm deep, effective depth d = 540 mm.
The section is subjected to a factored torque Tu of 15 kN.m. fc'=20MPa and fy
=400 MPa. Design the necessary torsional reinforcement.
Solution
Compare the value of Tu with the threshold torsion:
Acp  (0.3 0.6)  0.18 m 2 , Pcp  2 (0.3  0.6) 1.8 m.
2
f c  Acp  0.75  1.0 20  0.18 2 


  5.03 kN.m  Tu 15 kN.m
 P 
12
12
1
.
8
cp





 .
Therefore the torsion must be taken into account in the design.
Check the cross-sectional dimensions to see whether the cross-section can carry
the applied torque or not ?:
Assume a stirrup diameter = 10 mm,
x1  300  2  40 10  210mm
y1  600  2  40  10  510 mm
Aoh  x1  y1  0.21  0.51  0.1071m 2
Ph  2(0.21  0.51)  1.44 m
2
2 
 Vu 
T P 

   u h  
 1.7 A 2 
 bw . d 
oh 

 Vc
2

 bw . d 3
 
2
2
0

  0.015 1.44 
  1.11MPa

  
 0.3  0.54   1.7  0.10712 
 20 2


f c   0.75

20   2.80 MPa  1.11MPa
3

 6

Therefore the section can resist the applied torque.
The nominal torsional strength:
Tn  15 / 0.75  20kN .m
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S. A. Al-Ta'an
Ao  0.85 Aoh  0.85  0.1071  0.091m 2
Tn 
2 Ao At f yt
cot 
s
At
Tn
0.02


 2.747  10 4 m 2 / m
s 2 Ao f yt cot  2  0.091  400  1
Min. At / s 
Min
f c' bw
20 0.3

 1.048  10 6 m 2 / m
32 f yt
32 400
or
At
 1.25  10  4 m 2 / m
s
Therefore min. At/s=1.25×10-6m2/m
Min
At
 1.25  10  4 m 2 / m  Re q. At / s  2.747  10  4 m 2 / m
s
Area of one leg of the stirrup (db =10 mm), Ab = 79 mm2.
2.7474 10  4 
79  10 6
79  10 6
0.79
, s

 288 mm
s
2.747  10  4 2.747
p
Max. s  h  1.44 / 8  0.18m  180 or 300 mm
8
 Max. s = 180 mm
Use # 10 @ 175 mm c/c.
 f yv 
A
 cot2   2.747  10  4  1.44( 400 )1.0
Al  t Ph 
 f yl 
s
400


 3.956  10  4 m 2  396 mm2
To calculate the minimum Al , compare the required (At/s) = 2.747  10 4 m 2 / m
with the minimum = 1.25  10 4 m 2 / m (O.K.).
5 20
 0.18  2.747  10  4  1.44  4.43 10  4 m 2  443 mm 2
12  400
2
Therefore Al =443 mm
Bar diameter ≥ 10 mm, or
≥ S /24 = 175 / 24 = 7.3 mm
No. of longitudinal bars = 443/79 = 5.6
Use 6 bars, #10 mm,
Dist. between bars in the short (horizontal)
direction:
~ 10 ~
600mm
Min. Al 
300mm
S. A. Al-Ta'an
85 mm
130
85 mm
mm
mm
8585
Example 2:
Resolve Example 1 if the section is hollow with a wall
thickness = 85 mm all around.
Solution:
Use Ag instead of Acp:
430
430 mm
mm
300-2×40-2×10-2×20 =
160mm < 300 mm (O.K.)
Dist. between bars in the long (vertical ) direction= (600-2×40-2×10-10) / 2 =
245mm < 300mm (O.K.).
mm
Ag  0.3  0.6  0.13  0.43  0.1241m 2
 f c' Ag2
12
Pcp

85 mm
Pcp  2(0.3  0.6)  1.8m
0.75 20 0.12412

 0.00239 MN .m 
12
1.8
2.39kN.m  Tu  15kN.m
Therefore the torsion must be taken into account in the design.
Check the cross-sectional dimensions to see whether the cross-section can carry
the applied torque or not?:
x1  300  2  40 10  210mm
y1  600  2  40  10  510 mm
Aoh  x1  y1  0.21  0.51  0.1071m 2
Ph  2(0.21  0.51)  1.44 m
Compare (Aoh/Ph=0.1071/1.44)=0.074 m = 74 mm < t =85 mm (O.K.).
2
Otherwise substitute 1.7 Aoh .t instead of 1.7 Aoh
in the second term of the
following equation:
430mm
600mm
 Vu   Tu .Ph  
0
  0.015  1.44 

 
  1.11MPa < 2.8 MPa


2   0.3  0.54  
2

b
.
d
 1.7  0.1071 
 w   1.7 Aoh 
(O.K.) the section can resist the applied torque.
Use the same stirrup size and spacing of example 1
and the same longitudinal reinforcement.
The distance from the centre of the stirrups to
inside section (void) ≥ 0.5 Aoh / Ph
 0.5(0.1071 / 1.44)  37mm .
130mm
Actual distance = 85 – 40 -10/2 = 40 mm > 37 mm (O.K.).
300mm
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S. A. Al-Ta'an
Example 3:
The beam shown in the figure below is in a horizontal plane. The loads shown
are working live loads. f c'  25MPa and f y  345MPa . Choose a suitable cross
section and design for moment, shear, and torsion.
Solution:
To calculate the dead load, the cross-sectional dimensions must be known. Table
(9.5a) from the ACI Code states that h ≥ L/8 = 4000/8 =500 mm. Since the
section is subjected to bending moment, shear and torsion, 500 mm may not be
enough.
Assume b= 300, h = 600 mm and d =540 mm.
Beam weight = 0.3×0.6×1×24=4.32 kN/m.
Design for moment:
Maximum bending moment is at point A
M uA  1.2  4.32(0.85  4  4.15 2 / 2)  1.6(20  4  10  2)  222.27 kN .m
M nA  222 .27 / 0.9  246 .97 kN .m
M uB  1.2  4.32(0.85  2  2.15 2 / 2)  1.6(20  2)  84.79kN .m
M uc  1.2  4.32(1.0 2 / 2)  1.6(20  1)  34.59kN .m
Max.M n  Max.k n .b.d 2  5.69  0.3  0.54 2  497.8kN .m
Therefore the section is a single reinforced one.
PL 10 kN
A
PL 20 kN
B
D
2.0 m
C
2.0 m
1.0 m
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S. A. Al-Ta'an
Design for moment:
Maximum bending moment is at point A
M uA  1.2  4.32(0.85  4  4.15 2 / 2)  1.6(20  4  10  2)  222.27 kN .m
M nA  222 .27 / 0.9  246 .97 kN .m
M uB  1.2  4.32(0.85  2  2.15 2 / 2)  1.6(20  2)  84.79kN .m
M uc  1.2  4.32(1.0 2 / 2)  1.6(20  1)  34.59kN .m
Max.M n  Max.k n .b.d 2  5.69  0.3  0.54 2  497.8kN .m (for ϵt = 0.005, and ɸ
= 0.9)
Max.M n  Max.k n .b.d 2  6.33  0.3  0.54 2  616.03kN.m (for ϵt = 0.004, and ɸ = 0.817)
Therefore the section is single reinforced one.
k n  0.24697 /( 0.3  0.54 2 )  2.82MPa
m  f y /( 0.85 f c' )  345 /( 0.85  25)  16.24

2.k n .m 
1 
1 
2  2.82  16.24 
1  1 
  0.0088
1 1

m
f y  16.24 
345



As  0.0088  300  540  1426 mm2
This area is required at section A, similarly at sections B and C, the area of steel
= 591 and 208 mm2 respectively. Compare these values with the minimum area:
min . As 
1.4
25
300  540  657 mm 2 or min . As 
300  540  587mm 2
345
4  345
Therefore use As  657mm 2 at sections B and C.
Design for shear:
Calculate the factored shear forces
Vu A  1.2  4.32(5)  1.6(10  20)  73.92kN
VnA  73.92 / 0.75  98.56 kN
Vu d  1.2  4.32(5  0.54)  1.6(10  20)  71.12kN
Vnd  71.12 / 0.75  94.83kN
Vu BL  1.2  4.32(5.0  2.0)  1.6(10  20)  63.55kN
VnBL  63.55 / 0.75  84.73kN
Vu BR  1.2  4.32(5.0  2.0)  1.6(20)  47.55kN
VnBR  47.55 / 0.75  63.4kN
Vu C  1.2  4.32(5.0  4.0)  1.6(20)  37.18kN
VnC  37.18 / 0.75  49.58kN
Vu D  1.6(20)  32.0kN
VnD  32 / 0.75  42.67 kN
~ 13 ~
S. A. Al-Ta'an
f c'
25
Vc 
bw .d 
 0.3  0.54  135kN  Vnd  94.83kN
6
6
Vc / 2  135 / 2  67.5kN  Vnd  94.83kN
Vc / 2  135 / 2  67.5kN  VnBR  63.4kN
The beam requires minimum shear reinforcement between points A and B and
no shear reinforcement is required between sections B and C .
Design for torsion:
The torsion is constant between points A and C ,
Tu  1.2  4.32  0.85(0.85 / 2  0.15)  1.6  20  1.0  34.53kN.m
Calculate the threshold torsion:
Acp  0.3  0.6  0.18 m 2
Pcp  2(0.3  0.6)  1.8 m
2
 f c Acp 0.75 25  0.18 2 

 5.63 kN .m  34.53 kN .m
12
Pcp
12
 1.8 


Therefore torsion must be considered in design. Check the cross-sectional
dimensions to see whether it can resist the applied shear and torsion, assume
x1  300  2  40  12  208mm, y1  600  2  40  12  508 mm
Aoh  x1  y1  0.208  0.508  0.1057 m 2
Ph  2(0.208  0.508)  1.432 m
2
T P
 Vu 

   u 2h
 b. d 
 1.7 Aoh
2
2

 0.03453  1.432 
0.07112 

  

  
2 

 0.3  0.54 
 1.7 (0.1057) 

2
 2.64 MPa  0.75  5 25 / 6  3.13 MPa (O.K )
Ao  0.85 Aoh  0.0898 m 2
Tn  34.53 / 0.75  46.04kN .m
At
Tn
0.04604


 7.43  10  4 m 2 / m
s
2 Ao f yt cot  2  0.0898  345  1
Calculate the minimum area for shear and torsion:
0.3
 A  2 At  1 bw
Min.  v

 290  10  6 m 2 / m

s

 3 f y 3  345
f c' bw
25 0.3
 Av  2 At 
Min. 


 272  10  6 m 2 / m

s
16
f
16
345


y
Take the bigger value = 290  10 6 m 2 / m
~ 14 ~
S. A. Al-Ta'an
The table below shows the required strength for shear and torsion and the
combined shear and torsion reinforcement.
Use 30 # 12 closed stirrups
1 @ 50 mm + 15 @ 125 mm c/c + 14 @ 150 mm c/c (4025 mm < 4150 mm)
Design of the longitudinal reinforcement
 f yv 
A
 cot2 
Al  t Ph 


s
 f yl 
 345 
2
Al  743  10  6  1.432 
  1  1064 mm
 345 
Min. Al 
f c
f yv
A
55
Acp  t Ph

 0.18  743  10  6  1.432
12 f y
s
f yl 12  345
5
 23  10  6 m 2  23 mm 2
 Al  1064 mm 2
Point
d
BL
BR
C
D
Vn (kN)
94.83
support)
84.73
63.4
49.58
42.67
Vc (kN)
135
135
135
135
Vc/2 (kN)
135
from
67.5
face
67.5
67.5
67.5
67.5
Vs (kN)
Min.
Min.
--------
--------
--------
(Av/s)103
0.290
0.290
--------
--------
--------
Tn (kN.m)
46.04
46.04
46.04
46.04
--------
(At/s)10³
0.743
0.743
0.743
0.743
--------
(Av+2At)10³/s
1.776
1.776
1.486
1.486
--------
Min.(Av+2At)10³/s
0.29
0.29
0.29
0.29
--------
Req. s (mm)
127
127
152
152
--------
Max. s (shear)
270
270
---------
--------
--------
Max. s (torsion)
179
179
179
179
--------
Max. s (shear and torsion)
179
179
179
179
--------
Provided s
125
125
150
150
--------
This area is distributed to three levels:
 Al / 3  1064 / 3  355 mm 2
When there is torsion and moment acting at the same section, the torsional steel
can be reduced by an amount equal to:
~ 15 ~
S. A. Al-Ta'an
Mu
0.22227

 0.001326 m 2  1326 mm 2  Al / 3
0.9 d f yl 0.9  0.54  345
Therefore do not reduce AƖ .
A
As  1426 mm2 , l  1064 / 3  355mm2 , As  Al / 3  1781 mm2
3
600mm
Use 6#20 bars =6×314 = 1884 mm2 at point A at the top, at the mid-depth of the
cross-section use 2#16 ==402 mm2.
From point B to C, the beam requires:
657+355 = 1012 mm2 , use 4# 20 =1256 mm2.
The torsion reinforcement has to be extended a distance equal to (b+d =
300+540=840 mm beyond point C) therefore extend it to point D.
300mm
Sec. A
Sec. C-D
Sec. B
Example 4:
The slab shown is supported a series of beams. The slab carries 1.0 kPa finishing
DL and 4.0 kPa LL. Calculate the shear and torsion at the critical section of the
edge beam and if the beam require shear or torsion reinforcement or both of
them. f c'  30 MPa and f y  276MPa .
Solution:
Calculate the DL and LL on the slab.
a- On the slab
wds=0.15×1×1×24 =3.6 kPa.
The factored load on the slab wus=1.2(3.6+1)+1.6×4= 11.92 kPa
This load is transferred to the beam in one direction since the slab is oneway-slab.
b- Loads on the beam,
Factored load from the slab to the beam =
wus (3 / 2  0.3)  11.92(1.8)  21.46 kN / m
One meter weight of the beam web = wd .beam  0.3  0.4  1  24  2.88kN / m
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S. A. Al-Ta'an
Total ultimate load on the beam = wu.beam  1.2  2.88  21.46  24.91kN / m
c- Calculation of the factored shear forces, assume d = 485 mm
d- Vud  (7 / 2  0.485)24.55  75.1kN, Vnd  75.1/ 0.75  100.1kN
30
Vc 
 0.3  0.485  132 .8kN  Vnd
6
Vc
 66.4 kN  Vnd
2
(Therefore use min. shear reinforcement)
a
300mm
4hf=600mm
a
hw=400mm
300mm
550mm
3.0m
or hw=400mm
3.0m
300mm
Sec. a-a
300mm
7.0m
300mm
e- Therefore use minimum shear reinforcement.
f- Calculation of the factored torsional moment
According to the ACI recommendation, the negative moment from the slab
will act as a uniformly distributed torsion on the beam:
tu  mus  wus ln2 / 24  11.92  32 / 24  4.47 kN .m / m
The factored torsion at the critical section of the beam (d from face of support),
Tud  tu (ln / 2  d )  4.47 (7 / 2  0.485)  13.48kN .m / m
A cp  0.3  0.55  0.15  0.4  0.225 m 2
Pcp  0.3  0.55  0.7  0.15  0.4  2  2.5 m
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2
f c Acp 0.75 30  0.225  2


  6.93 kN .m  13.4 8kN .m
12 Pcp
12
 2.5 
Therefore torsion must be considered in design.
Ph  0.21  0.46  0.61  0.06  0.4  2  2.14 m

A oh  0.21  0.46  0.06  0.4  0.1206 m 2
A o  0.85 A oh  0.1025 m 2
Check for X – Section Dimensions
2
2 
0.01348  2.14 
 0.0751 
 1.28 MPa

  
2 
 0.3  0.485 
1
.
7
(
0
.
1206
)


5

f c  3.42 MPa  1.28 MPa (O.K )
6
Tn  13.48 0.75  17.97 kN .m
At
Tn
0.01797


 3.18  10  4
s
2 Ao f yv cot  2  0.1025  276  1
2 At
 6.35  10  4 m 2 / m
s
0.3
 A  2 At  1 bw
Min.  v

 3.62  10  4

s

 3 f y 3  276
 A  2 At
Min. v
s

f c bw

 3.72  10  4

 16 f y
Av  2 At
158  10  6
 6.35  3.72 10  4  10.07  10  4 
s
s
s
158  10  6
10.07  10  4

1.58
 157 mm
10.07
Maximum spacing for torsion,
Max. s = 2.14/8 = 268 mm or 300 mm
Maximum spacing for shear,
Max. s = d/2 = 485/2 = 243 mm or 600 mm (Shear)
Maximum spacing for shear and torsion
 Max. s = 243 mm > 157mm
Therefore max. sps. For shear and torsion = 150 mm.
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Design for longitudinal bars
 f yv 
A
 cot 2  ,  3.14  10  4  2.14  400   1  672 mm 2
Al  t Ph 
 f yl 
s
 400 


f yv
5 f c
A
5 30
Min. Al 
Acp  t Ph

 0.225  3.18  10  4  2.14
12 f y
s
f yl 12  400
 603 10  6 m 2  603 mm2 , Al  672 mm2
Distribute the area into three levels, Al/3=672/3=224 mm2
S. A. Al-Ta'an
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S. A. Al-Ta'an