Unit 10 ( TORSION )

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


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


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


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


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


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


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


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


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


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


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


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


= 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


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


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


= 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


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


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


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


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


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


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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Unit 10 ( TORSION )