unit 10 ( torsion )
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REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
UNIT 10
TORSION
GENERAL OBJECTIVE
To understand the principles of torsion reinforcement design.
At the end of this unit you will be able to:
1. explain torsion failure in beams.
2. explain the effects of torsion reinforcements.
3. design torsion reinforcements for beams.
1
OBJECTIVES
SPECIFIC OBJECTIVES
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
10.1 Introduction
For this unit, reference should be made to Section 2.4, Part 2 of BS8110
regarding the design of torsion reinforcement. This unit is concerned with the
design calculations for torsional reinforcement when torsion is of a particular
importance. In normal slab-and-beam or framed construction, torsional
reinforcement is normally not provided because torsional cracking is
adequately controlled by shear reinforcement. This is stated in Clause 2.4.1 of
the code.
10.2 Torsion in plain concrete beams
When a plain concrete beam is subjected to pure torsion, the torsional
moment, T induces shear stresses, which produce tensile stresses at 45◦ to the
longitudinal axis. When the maximum tensile stress reaches the tensile
strength of the concrete, diagonal cracks form, which tend to spiral round the
beam.
2
INPUT 1
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
This is shown diagrammatically in Figure 10.1 below;
10.3 Effects of torsional Reinforcement
A plain concrete beam fails as soon as diagonal cracking occurs. Torsion
reinforcement in the form of longitudinal bars and closed links will carry the
force resulting from the torsional moments. The longitudinal bars are
distributed evenly round the inside perimeter of the links. The truss analogy is
used to calculate the shear resistance of the beam. In this analogy, longitudinal
bars act as stringers, the legs of the links acting as posts and the concrete
between the cracks as the compression diagonals. Refer to Figure 10.2 on the
next page.
3
T
T Figure 10.1: Torsion in a plain concrete beam
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
T = ultimate torsion moment of resistance
As = total area of longitudinal reinforcement
Asv = area of the two legs of each link
fy = yield strength of the longitudinal reinforcement
fyv = yield strength of the link
sv = longitudinal spacing of the link
x1 = the smaller dimension between the corner bars
y1 = the larger dimension between the corner bars
4
0 .001 .002 .003 .0035
0.8fcu
cucu
c
c
cu
ffk
k
kff
001,
1,
2
34.1
0022.0
218.0
Figure 10.2 Stress strain curve for rigorous analysis of non-critical sections
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
Considering a length sv of the beam, we have,
Tensile force in links,
F =
Moment of force F about the centre line,
= for vertical legs
= for horizontal legs.
The torsion moment,
T = +
Torsion resistance,
T =
=
Therefore,
T =
0.8 is the coefficient factor to be taken into account as inaccuracy may occur.
The closed link should be provided such that,
and As
*Note that fy and fyv should not be taken as greater than 460 N/mm2
5
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
Fill in the blanks.
10.1 Torsional reinforcement is designed according to Clause
______________ Part 2 of BS 8110.
10.2 Torsional crack is normally adequately controlled by ______________
reinforcement. Thus torsion need not be designed in a framed
construction.
10.3 Torsional shear stresses induce tensile stress at __________________
to the longitudinal axis.
10.4 Diagonal cracks occur when the maximum torsional tensile stress
reaches the tensile strength of the _______________________.
10.5 Diagonal cracks form due to torsional failure _______ round the beam.
10.6 _______________ analogy is used to calculate torsional shear stresses.
10.7 To determine torsional shear resistance of a beam, the ____________
analogy is used.
10.8 Torsion reinforcement is provided consisting of _____and _____ links.
10.9 The area of links required is calculated using this equation:
___________________.
10.10 The equation in Question 9 above, x1 is the ____________ dimension
and y1 the _______________ dimension between the corner bars.
6
ACTIVITY 10a
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
Did you manage to get all the answers correct? Here are the answers.
10.1 2.4.1
10.2 shear
10.3 45◦
10.4 concrete
10.5 spiral
10.6 sand-heap
10.7 truss
10.8 longitudinal bars, closed
10.9 Asv
10.10 smaller, greater
7
FEEDBACK 10a
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
10.4 Torsional Shear Stress
Torsion usually exists in combination with shear stress and bending. It is very
rare that torsion acts alone. Therefore when bending reinforcement is required,
the longitudinal torsion reinforcement area can be increased. This is done
either by providing additional bars or by increasing the bar size. Because of
the combination of bending, shear and torsional forces, a greater amount of
reinforcement is needed.
10.5 Detailing Requirements
As required by Clause 2.4.8 of the code, spacing of links, sv must not exceed
the least of x1, or 200 mm. The links are of the closed type complying with
the shape code 74 of BS 4466 as shown in Figure 10.3 below:
8
Fig 10.3 Closed Links (Shape Code 74)
INPUT 2
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
The longitudinal reinforcement is to be distributed evenly round the inside
perimeter of the links. So, the clear distance between these bars is not to
exceed 300 mm and there are at least four bars. One on each corner of the
links is to be used. All longitudinal torsion bars should extend a distance at
least equal to the largest dimension of the section beyond where it ceases to be
required. For more information please refer to Clause 2.4.8, 2.4.9.and 2.4.10
of the code.
10.6 Reinforcement for Torsion
Torsional reinforcement is required when the torsional shear stress, vt exceed
the minimum torsional shear stress, vt,min . Values of vt,min are given in Table
2.3 of the code. In order to ensure that the crushing of concrete will not occur,
(v + vt) must not be greater than vtu. vtu is the maximum combined shear stress
(shear plus torsion) and is calculated as follows;
Vtu = 0.8 or 5 N/mm2
To avoid chipping of the corner of small section, where y1 < 550 mm, vt must
not exceed vtu as stated in Clause 2.4.5 of BS 8110. Table 2.4 of BS
8110 gives guidelines of providing reinforcement for a combination of shear
and torsion as below:
v ≤ vt,min v > vt,min
v ≤ vc Nominal shear reinforcement, no torsion reinforcement.
Designed torsion reinforcement only.
v > vc Designed shear reinforcement, no torsion reinforcement.
Designed shear and torsion reinforcement.
9
Table 2.4: Providing Reinforcement For A Combination Of Shear And Torsion
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
Now, let’s do some calculations based on the questions given below. You can
refer to BS 8110 for some technical terms. Good luck!
10.11 The beam section given below is subjected to a torsional moment, T =
150 kNm. A cover of 30 mm is provided.
Calculate;
a) hmin
b) hmax
c) vt
d) y1 (Use R8 link)
e) x1 (Use R8 link)
10.12 For questions 6 and 7, a concrete of grade 30 is used.
10
700mm
300mm
ACTIVITY 10b
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
a) Calculate vt,min
b) Calculate vtu
c) Complete the following table;
Form of reinforcement Conditions to be met
Nominal shear reinforcement.
Designed torsion reinforcement only.
Designed shear reinforcement, no torsion
reinforcement.Designed shear and torsion
reinforcement.
11
FEEDBACK 10b
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
Check your answers:
10.11
a) hmin = 300 mm
b) hmax = 700 mm
c) vt =
=
= 5.56 N/mm 2
d) y1 = 700 – 2(30) – 8
= 632 mm
e) x1= 300 – 2(30) – 8
= 232 mm
10.12
a) vtmin = 0.8
= 0.8
= 0.37 N/mm2
b) vtu = 0.8
12
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
= 0.8
= 4.38 N/mm 2
c)
Form of reinforcement Conditions to be met
Nominal shear reinforcement, no torsion reinforcement
vt ≤ vt,min
v ≤ vc
Designed torsion reinforcement only
vt > vt,min
v ≤ vc
Designed shear reinforcement, no torsion reinforcement
vt ≤ vt,min
v > vc
Designed shear and torsion reinforcement
vt > vt,min
v > vc
13
INPUT 3
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
10.7 Design Example
A rectangular beam section is shown in Figure 10.3. It is subjected to a
bending moment of 170 kNm, shear force of 160 kN and torsional moment of
10 kNm. If the characteristic strength of concrete and steel reinforcement are
fcu = 30 N/mm2 and 460 N/mm2 respectively, calculate the torsion
reinforcement required. Note: As required for bending moment was found to
be 1100 mm2 and from earlier calculations.
Solution
Step 1:
14
500mm
300mm
As = 1100 mm2
Asv/sv = 0.79
Figure 10.3: Cross Section Of Rectangular Beam
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
Step 2: vt =
=
= 0.56 N/mm 2
Step 3: 0.56 > 0.37 (From Table 2.4, BS 8110)
Therefore torsion reinforcement is required.
Step 4: v =
=
= 1.19 N/mm 2
Therefore,
Vt < as required.
Step 5: Additional
=
15
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
= 0.55
Full
= 1.34
Provide R10 at 100mm centre rectangular closed link
Step 6: Longitudinal steel;
As =
=
= 203 mm 2
Full steel area = 1100 mm2 + 203 mm2
The reinforcement details are shown below;
16
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
Step 7: The torsional reinforcement is to be extended at least a
distance equal to 500 mm beyond the point where it ceases
to be required.
17
R10 at 100mm centre
T12T12
T25T20 T25
T12T12
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
1. Determine the area of reinforcement, As and Asv to carry bending moment
and shear using the methods discussed in Unit 3, 4, 5, 6 and 9.
2. Calculate torsional shear stress by using equation 2 of Part 2, BS8110.
The equation is reproduced below;
vt =
3. When vt > vtmin torsional reinforcement is required.
4. Check that v + vt is not greater than vtu for section having
y1 < 550 mm.
5. Calculate additional shear reinforcement by using the equation.
6. Calculate additional longitudinal reinforcement by using the equation,
As =
7. Fulfill detailed requirements for;
a) spacing of links
b) form of links
c) distance to be extended
18
SUMMARY
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
Given the following information:
hmin = 350 mm
hmax = 800 mm
T = 105 kNm
As = 762 mm2
= 0.35
A. Calculate;
1. vt
2. vtu
3. vtmin
4. Decide whether torsion reinforcement is required.
5. x1
6. y1
7. the additional
8. the additional As
9. the total (torsion) and (shear)
19
SELF-ASSESSMENT
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
10. the proposed size and spacing of closed links required. State the
proposed.
11. the proposed longitudinal torsion bars required (from item 7) .
State As provided.
B. Sketch the reinforcement details.
20
FEEDBACK ON SELF-ASSESSMENT
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
A.
1. vt =
=
= 2.51 N/mm 2
2. vtu = 5 N/mm2 (From Table 2.3 of BS8110)
3. vtmin = 0.4 N/mm2 (From Table 2.3 of BS8110)
4. Since vt > vtmin , torsion reinforcement is required.
5. x1 = 350 – (2)(30) – 10
= 280 mm
6. y1 = 800 – 2(30) – 10
= 730 mm
7.
= 1.60 mm
21
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT10/
8. As = (1.60)
= 1616 mm 2
9. Total = 1.60 + 0.35
= 1.95 mm
10. R12 at 100mm centre ( = 2.26 mm)
11. T16 (As = 1809 mm2 )
B. Reinforcement details;
22
R12 at 100mm centres (closed link)
3T16
2T16
2T16
2T25 + 2T16
END OF UNIT 10
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