Design for Torsion in reinforced concrete

1. Chapter 5
Torsion in RC Beams
1
May 2019

2. Torsional Stresses in Un-cracked Members
 Members subjected to a torsional moment, commonly
known as a torque ( ), develop shear stresses ( ).
 In general, these tend to increase in magnitude from the
longitudinal axis of the member to its surface. If the shear
stresses are sufficiently large, cracks will propagate
through the member and, if torsion reinforcement is not
provided, the member will collapse suddenly.
 The elastic behavior of un-cracked concrete members
with torsion, particularly non-circular members, is
difficult to model precisely. In a circular member
subjected to a torque T, such as the member in Fig. 1(a),
the circumferential shear stress at a given cross-section
varies linearly from the longitudinal axis of the member
to a maximum value, τmax , at the periphery of the
section.
2

3. 3
Torsional Stresses in Circular members

4. The stress at any distance r from the longitudinal axis of
a circular member is given by:
= … … … … … … … … … . (1)
Where is the polar second moment of area of the
section and is equal to =
∅
, where ∅ is again the
diameter of the member. The maximum shear stress
is thus found by setting =
∅
in equation (1).
= =
∅
2
∅
32
=
16
∅
4
Torsional Stresses in Circular members

5. For a non-circular member, the distribution of shear
stress is not so straightforward. The stress distribution is
not linear as we can see from Figure 2 below.
Unlike in the circular member, the stress distribution in a
rectangular member is non-linear.
5
Torsional Stresses in Rectangular members

6. 6
The maximum shear stress occurs at the midpoint of
the longer side and the shear stress at the corners of
the section is zero indicating that the corners of the
section are not distorted under torsion.
Analytical studies have shown that the maximum
shear stress, in rectangular section is given by:
= … … … … … … … … … … (2)
Where x and y are the lengths of the shorter and longer
sides, respectively. The value of the parameter
depends on the relative values of x and y. For a square
section, = 0.208, while for a section with x/y = 0.1, =
0.312

7. • The stress distribution in thin-walled hollow members is
much easier to determine than for solid non-circular
members, even for a member with complex shape and
varying thickness such as illustrated in Fig. 3(a).
• The shear stress in the walls is reasonably constant and is
given by:
=
2
… … … … … … … … … … . (3)
Where t is the thickness of the wall of the member and
is the area within a perimeter bounded by the center line of
the wall (Fig. 3(b)). On a given section, the shear stress is
maximum where the thickness of the wall is minimum.
7
Torsional Stresses in thin-walled hollow
members

8. 8

9. Failure of concrete members with torsion
9
 Consider the rectangular member of Fig. 4 subjected to
a torque T. Since there are no other external forces (and
ignoring self-weight) the member is considered to be in
pure torsion. Then torque causes the member to twist
and to develop shear stresses.

10.  Consider small elements on each face of the member,
as illustrated in the figure 5. As for members with
applied shear, shear stresses act on the sides of each
element in the directions shown in fig.5(a).
 The equivalent principal stresses, inclined at an angle of
45° to the horizontal, are illustrated in Fig. 5(b). In the
same way as for shear, the principal tensile stresses
cause the development of cracks inclined at a angle of
45°.
 However, in the case of torsion, they form a spiral all
around the member, as illustrated in Fig. 5(c). Since the
shear stresses in members with torsion are greatest at
the surface, these cracks develop inwards from the
surface of the member. 10

11. 11

12. • The member illustrated in Fig. 6 is subjected to a
vertical force, V, in addition to the applied torque. This
results in a combination of bending, shear and torsion
and alters the orientation of the inclined cracks. As for
members with shear, the introduction of priestess has
the effect of delaying the onset of torsional cracking
and altering the orientation of the inclined cracks.
12

13. • The torsional strength of a concrete member can be
significantly increased by providing suitable torsion
reinforcement across the cracks. This is usually
provided in the form of ‘closed’ four-sided stirrups, as
illustrated in Fig. 7, in combination with longitudinal
bars distributed around the periphery of the section.
This reinforcement controls the propagation of cracks
and ensures that when failure occurs due to yielding of
the reinforcement, it is not sudden.
13

14. • To quantify the behavior of members with such
torsional reinforcement, an equivalent space truss
model, similar to the plane truss model for shear, can
be used. This theory, developed by Lampert and
Collins(1972), assumes that solid members can be
designed as equivalent hollow members. Extensive tests
proven that the presence of the concrete at the center
of the member does not have a very significant effect
on its torsional resistance. Thus, members are designed
as equivalent thin-walled members. The thickness of
the wall, t, is commonly taken as:
= … … … … … … … … … (4)
Where is the gross cross-sectional area of the
member and is the length of the perimeter. 14

15. • The space truss model proposed by Lampert and Collins is
illustrated in Fig.8 for the member of Fig. 7. Each leg of the closed
stirrups acts as a tension member, the longitudinal steel bars act
as continuous top and bottom chords and the concrete in
compression between the cracks acts as compression struts. The
concrete struts are inclined at an angle ϴ which, like shear, varies
in the range 220< ϴ < 680.
• The truss dimensions, xo and yo, are measured from center to
center of the notional thin walls.
15

16. • Referring to Fig. 9(a), the number of stirrups crossing a
side face of height yo(cotϴ)/s where s is the stirrup
spacing. Hence, assuming the reinforcement has yielded,
the total vertical shear force transmitted across the cracks
by the stirrups on one side face is:
=
(cot )
… … … … … (5)
Where is the area of one leg of the stirrup.
Similarly the shear force transferred across the top or
bottom face is:
=
(cot )
… … … … … (6)
16

17. The shear force on the side wall of a thin walled member
is the product of the average stress due to applied load
and the surface area. Hence the shear on a side wall is
(see Fig. 9(b)):
=
Where: t is the wall thickness. 17

18. Substituting for from equation (3) gives:
=
2
=
2
… … … … … . … … (7)
= 2 … … … … … … … … … … . … . … (8)
Similarly it can be shown that:
= 2 … … … … … … … … … … … . . … (9)
Substituting from equation (5) into equation (8) (or from
equation (6) into equation (9) gives torsions at which the
stirrups yield:
=
2 (cot )
… … … (10)
Thus the stirrup reinforcement required to resist a torsion
of T is: 18

19. = … … … … … (11)
In addition to the stirrup reinforcement, longitudinal
reinforcement is required to resist torsion. As can be seen in
the truss model of Fig. 10(a), diagonal compression struts join
the vertical members of the truss. Equilibrium at a joint of the
truss where these members meet is considered in Figs 10(b)
and (c).
19

20. The compressive force in the diagonal members, C, must
equal V1/sinϴ. There must also be a force in the
longitudinal members of N = Ccosϴ = V1cosϴ. Substitution
from equation (5) gives:
= (cot ) … … … (12)
The total force in the longitudinal members from all four
joints at a given cross section is:
2 cot + 2 cot = 2( + ) cot
If the total area of longitudinal reinforcement is Along,
then:
= 2( + ) cot
20

21. Substituting from equation (5.8) and (5.9) gives:
= 2
2
+
2
cot =
+ cot
=
+
( / )
… … … … … … … (13)
This is the total area (all bars) of longitudinal
reinforcement required to resist an applied torsion of T. It
is additional to whatever longitudinal reinforcement is
required to resist bending moment.
Alternatively, equation (13) can be rearranged to give the
maximum torsion possible without leading to yielding of
the longitudinal reinforcement:
21

22. =
( / )
+
… … … . . (14)
Regardless of how much stirrup and longitudinal
reinforcement is provided, the torsion must not be of such
magnitude as to cause crushing of the concrete in the
diagonal struts.
As mentioned above, equilibrium at the joints of the truss
requires a compressive force in the struts of C =V1/sinϴ.
This force is resisted by stresses in the concrete between
the diagonal cracks. The surface area of concrete to which
this is applied is, from Fig. 11, yo(cos ϴ) t, where t is the
thickness of the notional wall. Hence the stress in the struts
is:
( / sin )
(cos )
=
(sin )( cos )
22

23. 23

24. This must not exceed the compressive strength of
concrete, / , where is the effectiveness factor
for torsion, that is:
(sin )( cos )
≤ … … … … … … … … . (15)
Substitution for V1 from equation (7) gives:
/2
(sin )( cos )
≤
From which the torsion which could cause crushing of
the concrete struts is Tw, where:
=
2 (sin )(cos )
… … … … … … … . (16)
24

25. Design of members for torsion in accordance with
EBCS2
The torsional resistance of members is calculated on the
basis of an equivalent thin-walled section (i.e. the truss
analogy) as described above. As for shear, the strut
inclination angle, , can have any value in the range
22° ≤ ≤ 68°. The thickness of the equivalent wall is
given by equation (4) but must not be less than twice the
cover to the longitudinal steel. In the case of hollow
members, the equivalent wall thickness should not
exceed the actual wall thickness. For sections of complex
(solid) shape, such as T-sections, the torsional resistance
can be calculated by dividing the section into individual
elements of simple (say, rectangular) shape. The torsional
resistance of the section is equal to the sum of the
capacities of the individual elements, each modeled as an
equivalent thin-walled section.
25

26. Members with Pure Torsion:
For members with pure equilibrium torsion, EBCS2
requires that:
(a) The applied ultimate torque, T, does not exceed the
torsional capacity, as dictated by the quantities of stirrup
and longitudinal reinforcement present; and
(b) The applied ultimate toque, T, does not exceed the
level that would cause crushing of the compressive
struts, Tw.
The longitudinal reinforcement limits the capacity for
torsion to that given by equation (14) while the stirrup
reinforcement limits the capacity to the value given by
equation (10).
26

27. Alternatively, equations (13) and (11) can be used to
determine the areas of longitudinal and stirrup
reinforcement required to resist a torque T. The torque
that would cause crushing of the compression struts, Tw,
is calculated from equation (16). However, for torsion,
the effectiveness factor, , is restricted to 70 per cent of
the level allowed for shear, that is:
≥
0.7 0.7 −
200
0.35
… … … … … . . (17)
In order to prevent diagonal compression failure in the
concrete, the torsional resistance:
= 0.80 ℎ … … … … … … (18)
= area enclosed within the center line of the thin-wall
cross-section including inner hollow areas =
27

28. • The torsional resistance of concrete for the torque To
shall be taken as (see figure 12)
= . … … … … … … (19)
Minimum torsional reinforcement in the form of stirrups
and longitudinal reinforcement shall be provided. The
minimum web reinforcement shall be, , = . /
and the spacing shall not exceed / .
The longitudinal bars required for torsion shall be
distributed uniformly (at least one in each corner) around
the perimeter of the closed stirrups at a spacing not
exceeding 350mm.
28

29. 29

30. Members with Combined Actions:
For members subjected to combined moment and
torsion, EBCS2 recommends that the requirements for
each action be determined separately and that the
following rules are then applied:
(a) In the flexural tension zone, the longitudinal
reinforcement required for torsion should be provided in
addition to the amount required for moment.
(b) In the flexural compression zone, if the tensile stress
in the concrete due to torsion is less than the
compressive stress due to moment, then n longitudinal
torsion reinforcement need be provided.
30

31. For members with combined torsion and shear, the
ultimate torque, T, and the ultimate shear force, V,
should satisfy the condition:
+ ≤ 1 … … … … … (20)
Where and are the torque and shear force
respectively that would, acting alone, cause crushing of
the concrete struts.
The calculations for the design of stirrups may be made
separately for torsion and shear. However, the angle, ϴ,
for the concrete struts must be the same in both cases.
The requirements for shear and torsion are, of course,
additive.
31

32. The limiting values of Torsional and Shear resistance shall
be taken as the basic values and respectively,
multiplied by the reduction factors and .
= 0.80 ℎ ∗
= 0.25 ∗
ℎ ; =
1
1 +
/
/
=
1
1 +
/
/
32

33. The torsional and shear resistance of the concrete shall be
taken as the basic values and respectively,
multiplied by the following reduction factors and .
= 1.2 ℎ
= 0.25
ℎ ; =
1
1 +
/
/
=
1
1 +
/
/
33

34. Example 1: Member with pure torsion
Determine the maximum torque which can be applied to
the member of Fig. E1 given that = 30 N/mm2, the
yield strength for the longitudinal reinforcement is =
460 N/mm2 and the yield strength for the stirrup
reinforcement is = 250 N/mm2.
34

35. Step 1: Material Properties
−∗∗ ∶
=
∗∗
1.25
= 30 (given)
=
0.85 ∗
1.5
=
0.85 ∗ 30
1.5
= 17
=
. ∗
.
=
. ∗
.
= 1.352
− 460 (Longitudinal):
= 460 & =
1.15
=
460
1.15
= 400
− 250 (Stirrups):
= 250 & =
1.15
=
250
1.15
= 217.39
35

36. Step 2: Determine the Torsional Resistance (T):
The total area of longitudinal reinforcement available to
resist torsion is:
= 4
∗ 16
4
= 804
From Fig. Example 1, the dimensions of the analogous thin
walled section are:
= =
350 ∗ 600
2 350 + 600
= 110
= 350 − = 350 − 110 = 240
= 600 − = 600 − 110 = 490
36

37. Hence assuming a compression strut angle of 45°,
equation (6.14)
=
( / )
+ cot
=
(840)(240)(490)(460/1.15)
240 + 490 1
= 52
Similarly, equation (6.10) gives the torsional capacity as
dictated by the area of stirrup reinforcement. The label
‘R10’ indicates a 10 mm diameter mild steel stirrup with a
characteristic yield strength of fy = 250 N/mm2
The area of one leg is:
=
∗ 10
4
= 78.5
37

38. : =
2 (cot )
=
2 240 490 1
150
78.5
250
1.15
= 26.758
The effectiveness factor for torsion is, from equation (17):
≥
0.7 0.7 −
200
= 0.7 0.7 −
24
200
= 0.385
0.35
Hence the torque that would cause crushing of the
compression struts is, from equation (16)
=
2 (sin )(cos )
38

39. =
2(240)(490)(110)(1/ 2)(1/ 2)(0.385)(30)
1.5
= 99607200
= 100
Thus, with a compression strut inclined at an angle of 45°,
the torsion capacity is governed by the area of stirrup
reinforcement. However, the strut inclination angles can
have any value in the range 22° ≤ ≤ 68° . Hence, cot
can vary in the range 2.5 ≥ ≥ 0.4 . By trial and error (or
by equating the two equations for T), an optimum value for
cot can be found. Taking cot =1.4 ( =36°), the
longitudinal reinforcement dictates a torsional capacity of 37
kNm and the stirrup reinforcement dictates a capacity of 38
kNm. The corresponding value for Tw is 94 kNm. It can
therefore be concluded that this beam has the capacity to
resist torsion of 37 kNm. 39

40. Example 2: Torsion –RC Beams
Design a rectangular section of 300mm*600mm overall dimension
for torsion. The design torsion to be resisted is 26kNm. Materials to
be used are C-25 and S-300 steel.
Step 1: Material Properties
− 25 ∶
= 25/1.25 = 20
=
0.85 ∗
1.5
=
0.85 ∗ 20
1.5
= 11.33
=
0.21 ∗
1.5
=
0.21 ∗ 20
1.5
= 1.032
− 300 :
= 300 & =
1.15
=
300
1.15
= 260.87
40

41. Step 2: Determine the torsion to be resisted by reinforcement
= 26
ℎ =
300 ∗ 600
2(300 + 600)
= 100
= 300 − 100 ∗ 600 − 100 = 100000
= 200 ∗ 2 + 500 ∗ 2 = 1400
= 1.2 ℎ = 1.2 1.032 100000 100 ∗ 10
= 2.06
Torsion to be resisted by reinforcement ( ) will then be:
= − = 26 − 2.06 = 23.94 ≈ 24
=
2 ∗ ∗
41

42. Spacing of stirrups:
=
2 100000 260.87 8
4 ∗ 24 ∗ 10
= 109.27
Provide a spacing of 100mmC/C.
Maximum permitted spacing;
=
8
=
1400
8
= 175 > 100 . . . !
=
2
=
24 1400
2(100000)(260.87)
= 643.99
12 ∅ , =
643.99
∗ 12 /4
= 5.69
Provide 6 numbers of 12 mm diameter rods as torsional
longitudinal bar apart from longitudinal flexural
reinforcement.
42

43. Example 3: Torsion –RC Beams
A Rectangular section of 250mm breadth and 400mm effective
depth is reinforced with 3 number of 22mm diameter bar for
flexure. It has to resist a shear force of 160kN and torsional
moment of 10kNm. Materials C-30 Concrete, S-460 for
Longitudinal reinforcement and S-250 for stirrups.
Step 1: Material Properties
− 30 ∶
=
30
1.25
= 24
=
0.85 ∗
1.5
=
0.85 ∗ 24
1.5
= 13.6
=
. ∗
.
=
. ∗
.
= 1.165
43

44. − 460 (Longitudinal):
= 460 & =
1.15
=
460
1.15
= 400
− 250 (Stirrups):
= 250 & =
1.15
=
250
1.15
= 217.39
Step 2: Design the section for shear and Torsion
= 160 = 10
ℎ = 400 + 11 + 8 + 25 = 445
ℎ = =
250 ∗ 445
2(250 + 445)
= 80
= 250 − 80 ∗ 445 − 80 = 62050
= (250 − 80) ∗ 2 + (445 − 80) ∗ 2 = 1070 44

45. 45
Actual Values of and :
= 1.2 ℎ = 1.2 1.165 62050 80 ∗ 10
= 6.9
= 0.25 ℎ = 1 + 50 ≤ 2.0
= =
3 ∗
22
4
250 ∗ 400
= 0.0114
= 1 + 50 = 1 + 50[0.0114] = 1.57 ≤ 2.0 Ok!
H = .
= 1.6 − = 1.6 − 0.4 = 1.2 ≥ 1.0 Ok!
H = .

46. Therefore, = 0.25 ∗ ∗ ∗ ∗ ∗
= 0.25 ∗ 1.165 ∗ 1.57 ∗ 1.2 ∗ 250 ∗ 400 ∗ 10 = .
Reduction factor for torsion ( ) will then be;
=
1
1 +
/
/
=
1
1 +
160/54.87
10/6.9
= 0.445
= 0.445 ∗ 6.9 = 3.07
ℎ :
= 10 − 3.07 = 7
=
2 ∗ ∗
, =
∗ 8
4
= 50.3
46

47. , =
2 ∗ ∗
=
2 62050 (217.39)(50.3)
7 ∗ 10
= 193.85
Maximum Spacing of reinforcements permitted, ;
=
8
=
1070
8
= 133.75 < 193.85 … !
Hence provide a spacing of 130mm
=
2
=
7 1070
2(62050)(400)
= 150.88
12 ∅ , =
150.88
∗ 12 /4
= 1.334
Provide 2 numbers of 12 mm diameter rods as torsional
longitudinal bar apart from longitudinal flexural reinforcement.
47

48. Example 4: Torsion –RC Beams
A RCC beam of rectangular cross section having b =
350mm and d = 460mm is subjected to a design torsional
moment of 44kNm. The material used are C-20 concrete;
S-460 steel for longitudinal bars; S-400 steel for stirrups.
There are 5 number of 14mm diameter longitudinal
flexural reinforcement provided using 8mm stirrups;
design the member for torsion.
Step 1: Material Properties
− 20 ∶
=
20
1.25
= 16
=
0.85 ∗
1.5
=
0.85 ∗ 16
1.5
= 9.07
48

49. =
0.21 ∗
1.5
=
0.21 ∗ 16
1.5
= 0.8889
− 460 (Longitudinal):
= 460 & =
1.15
=
460
1.15
= 400
− 400 (Stirrups):
= 400 & =
1.15
=
250
1.15
= 347.83
Step 2:Determine the
= 41 = 44
Overall height of the section
ℎ = 460 + 25 + 7 + 8 = 500
49

50. ℎ = =
350 ∗ 500
2(350 + 500)
= 102.94 = 103
= 350 − 103 ∗ 500 − 103 = 98059
= (350 − 103) ∗ 2 + (500 − 103) ∗ 2 = 1288
Actual Values of and :
= 1.2 ℎ = 1.2 0.8889 98059 103 ∗ 10
= 10.77
= 0.25 ℎ = 1 + 50 ≤ 2.0
= =
5 ∗
14
4
350 ∗ 460
= 0.00478
= 1 + 50 = 1 + 50[0.00478] = 1.24 ≤ 2.0 Ok!
H = .
50

51. = 1.6 − = 1.6 − 0.46 = 1.14 ≥ 1.0 Ok!
H = .
= 0.25 ∗ 0.8889 ∗ 1.24 ∗ 1.14 ∗ 350 ∗ 460 ∗ 10
= .
Reduction factor for torsion ( ) will then be;
=
1
1 +
/
/
=
1
1 +
41/50.58
44/10.77
= 0.980
= 0.980 ∗ 10.77 = 10.6
ℎ :
= 44 − 10.6 = 733.4
51

52. =
2 ∗ ∗
, =
2 ∗ ∗
=
2 ∗ ∗
=
2 98059 (347.83)(50.24)
33.4 ∗ 10
= 102.68
Maximum Spacing of reinforcements permitted, ;
=
8
=
1288
8
= 161 > 102.68 … !
Hence provide 8mm stirrups @c/c spacing of 100mm
=
2
=
33.4 1288 ∗ 10
2(98059)(400)
= 548.33
14 ∅ , =
548.89
∗ 14 /4
= 3.56
Provide 4 numbers of 14mm diameter rods at corner of the beam as
torsional reinforcement apart from the longitudinal flexural
reinforcements. 52

53. Exercises: Torsion – RC Beams
1. Do Examples 1 and 2 using EBCS-2, 1995 design
procedures and Example 3 using equivalent truss analogy.
2. Given a cantilever beam with l=2.5m ,P=70kN applied
at an eccentricity of 400mm from the beam center line
and materials C-25 concrete, S-460(longitudinal bars) and
S-300(stirrups), Design the beam for flexure, shear and
torsion. (Assume θ=450 & b/D =300/500.
3. Design a rectangular RC beam to sustain a design torque
of 56kN-m; where this beam is made from C-25 concrete
and S-300 reinforcing steel (Both Long. & lat.). Assume the
concrete work is Class-II Work.
53

54. Thank You!
54