Unit 10 ( TORSION )

REINFORCED CONCRETE STRUCTURAL DESIGN

UNIT 10 TORSION

O BJE C T I VE VE S

GENERAL OBJECTIVE To understand the principles of torsion reinforcement design. SPECIFIC OBJECTIVES

At the end of this unit you will be able to: 1.

explain torsion failure failur e in beams. bea ms.

2.

explain explain the effects of torsion reinforcements.

3.

design torsion reinforcements for beams.

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INPUT 1

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.

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This is shown diagrammatically in F igure 10.1 below;

T

T

10.3

Figure 10.1: Torsion in a plain concrete beam

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.

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¨ k L  L 2 ¸ ¹¹   1 k 2   L ª º

f ! 0.8 f cu ©©

L!

! c ,1

k !

1.4

0.0022

c ,1

f cu

0.8f cu cu

0

.001

.002

.003

.0035

Figure 10.2 Stress strain curve for rigorous analysis of non non -critical sections

T = ultimate torsion moment of resistance As = total area of longitudinal reinforcement Asv = area of the two legs of eac h link f y = yield strength of t he longitudinal reinforcement f yv yv = 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

)0

!

3) 0 f cu


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Considering a length s v of the beam, we have, Tensile force in links, F=

A sv

v

0.87 f y

s v

Moment of force F about the centre line, =

x1

F v

= F

for vertical legs

2

y1

2

for horizontal legs.

The torsion moment, T=

x y ¨ ¸ ¨ ¸ © F v 1 v 2 ¹ + © F v 1 v 2 ¹ 2 2 ª º ª º

Torsion resistance, T=

A sv

2

¨ y1 ¸¨ x1 ¸ A sv ¹¹© ¹ v 2 0.87 f yv ¨©© x1 ¸¹¹¨© y1 v 2 ¸¹ 2 ª s v ºª 2 º ª s v ºª 2 º

= 0.87 f yv ©©

Therefore, T=

A sv s v

y1 0.87 f yv v 0.8

v x 1

0.8 is the coefficient factor t o be taken into account as inaccuracy may may occur. The closed link should be provided such that, A sv s v

and As

*Note

u

u

T

0.8 x1 y1 (0.87 f yv )

A sv f yv ( x1

 y1 )

s v f y

2 that f y and f yv yv should not be taken as greater than 460 N/mm

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ACTI ITY 10a

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 longitudinal axis. a xis.

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 calcu late 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, x 1 is the ____________ dimension and y1 the _______________ dimension between the corner bars.


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FEED ACK 10a

Did you manage to get all the answers correct? Her e 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



u

A sv f yv  x1 s v f y

10.10 smaller, greater

 y1 


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INPUT 2

10.4

Torsional Shear Stress

Torsion usually exists in combination with shear stress and bending. It is very rare that torsion acts a lone. lone. Therefore T herefore 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, s v must not exceed the least of x1,

y1

2

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:

Fig 10.3 Closed Links (Shape Code 74)


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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 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 v tu. vtu is the maximum combined shear stress (shear plus torsion) and is calculated as follows; Vtu

= 0.8 f cu

2

or 5 N/mm

To avoid chipping of the corner of small section, where y 1 < 550 mm, vt must not exceed vtu

v

¨ y1 ¸ as stated in Clause 2.4.5 of BS 8110. Table 2.4 of BS © ¹ ª 550 º

8110 gives guidelines of providing reinforcement for a combination of shear and torsion as below:

v v

c

v > vc

v v t,min Nominal shear reinforcement, no torsion reinforcement.

v > vt,min Designed torsion reinforcement only.

Designed shear reinforcement, no torsion reinforcement.

Designed shear and torsion reinforcement.

Table 2.4: Providing Reinforcement For A Combination Of Shear And Torsion


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ACTIVITY 10b

Now, let¶s do some calculations based on the questions given below. You can refer to BS 8110 for some technical ter ms. Good luck!

10.11 The beam section given below is subjected t o a torsional moment, T = 150 kNm. A cover of 30 mm is provided.

300mm

700mm

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.


a)

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Calculate vt,min

b)

Calculate vtu

c)

Complete the following table;

Form of reinforceme reinf orcement nt Nominal shear reinforcement. Designed torsion reinforcement only. Designed shear reinforcement, no torsion reinforcement. Designed shear and torsion reinforcement.

FEED ACK 10b

Conditions to be met


Check your answers: 10.11 a)

hmin = 300 mm

b)

hmax = 700 mm

c)

vt =

2T 2

¨ © ª

hmin hmax

=

¸ ¹ 3 º

hmin 

2 v 150 v 10 6

¨ ª

(300) 2 © 700

300 ¸ ¡

= 5.56 N/mm2

d)

y1 = 700 ± 2(30) ± 8 = 632 mm

e)

x1= 300 ± 2(30) ± 8 = 232 mm

10.12 a)

vtmin = 0.8 f cu = 0.8 30 = 0.37 N/mm2

3

¹ º

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b)

vtu = 0.8 f cu = 0.8 30 2

= 4.38 N/mm

c) Form of reinforceme reinf orcement nt

Conditions Conditi ons to be met

Nominal shear reinforcement, no torsion reinforcement Designed torsion reinforcement only Designed shear reinforcement, no torsion reinforcement Designed shear and torsion reinforcement

vt v t ,min v v c vt > vt,min v v c vt v t,min v > vc vt > vt,min v > vc


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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 2

f cu cu = 30 N/mm

2

and 460 N/mm

respectively, calculate the torsion

reinforcement required. Note: As required for bending moment was found to 2

be 1100 mm and

A sv sv

!

0.79 from earlier calculations.

Asv/sv = 0.79 500mm

As = 1100 mm

2

300mm

Figure 10.3: Cross Section Of Rectangular Beam

Solution


Step 1:

Step 2:

A sv s v

! 0.79

2T

vt =

¨ ª

hmin © hmax 2

hmin ¸

¹ º

¢

3

2 v 10 6

=

¨ ª

300 2 © 500

300 ¸ £

3

¹ º

2

= 0.56 N/mm

Step 3:

0.56 > 0.37 (From Table Tabl e 2.4, BS 8110) Therefore torsion reinforcement is required. required.

Step 4:

v=

=

V bd

160 v 10 3 300 v 450

= 1.19 N/mm2

Therefore,

Vt <

v t u y1

550

v t u y1

550

!

4.38 v 440 550

as required.

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Step 5: Additional

A sv s v

!

T

0.8 x1 y1 ( 0.87 f yv )

=

10 v 10 6 0.8( 240 v 440) v 0.87 v 250

= 0.55

Full

A sv s v

!

0.79  0.55

= 1.34

¨ A sv

Provide R10 at 100mm centre rectangular closed link ©©

ª s v

Step 6: Longitudinal steel;

As =

¨ A sv ¸¨© f yv ¸¹ ©© ¹¹  x1  y1  ª sv º©ª f y º¹

= 0.55 v

250 460

240  440

= 203 mm2

Full steel area = 1100 mm 2 + 203 mm2

¸ ! 1.57 ¹¹ º

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The reinforcement details are a re shown below;

T12

T12

R10 at 100mm centre

T12

T12

T25

Step 7:

T20

T25

The torsional reinforcement is is to be extended at least a distance equal to 500 mm beyond the point where it ceases to be required.


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SUMMARY

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 torsi onal shear stress by using equation 2 of Part 2, BS8110. The equation is reproduced below; 2T

vt =

¨ ª

hmin © hmax 2

¤

hmin ¸

3

¹ º

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. A sv s v

6.

T !

0.8 x1 y1 (0.87 f yv )

Calculate additional longitudinal longitudinal reinforcement by using the equation,

As =

7.

¨ A sv ¸¨© f yv ¸¹ ©© ¹¹  x1  y1  © ¹ s f ª v ºª y º

Fulfill detailed requirements for; a)

spacing of links

b)

form of links

c)

distance to be extended

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SELF-ASSESSMENT

Given the following information: hmin = 350 mm hmax = 800 mm T

= 105 kNm

As

= 762 mm

A sv s v

2

= 0.35

A. Calculate;

1.

vt

2.

vtu

3.

vtmin

4.

Decide whether torsion reinforcement is r equired.

5.

x1

6.

y1

7.

the additional

8.

the additional As

9.

the total

A sv s v

A sv s v

(torsion) and

A sv s v

(shear)


10.

the proposed size and spacing of closed links required. State the A sv s v

11.

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proposed.

the proposed longitudinal torsion bars required (from item 7) . State As provided.

B. Sketch the reinforcement details.


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FEED FEED ACK ACK ON SEL SELF F-ASS -ASSES ESSM SMEN ENT T

A. 1.

2T

vt = 2

¨ © ª

hmin hmax

¸ ¹ 3 º

hmin ¥

2 v 105 v 10 6

=

¨ ª

350 2 © 800

350 ¸ ¦

3

¹ º

= 2.51 N/mm2

2.

vtu = 5 N/mm2 (From Table 2.3 of BS8110)

3.

vtmin = 0.4 N/mm (From Table 2.3 of BS8110)

4.

Since vt > vtmin , torsion reinforcement is r equired.

5.

x1 = 350 ± (2)(30) ± 10

2

= 280 mm

6.

y1 = 800 ± 2(30) ± 10 = 730 mm

7.

A sv s v

!

105 v 10 6 ( 0.8)(280)(730 )(0.87 )(460)

= 1.60 mm

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¨ 460 ¸ ¹280  730 ª 460 º

As = (1.60) ©

8.

= 1616 mm2

9.

Total

A sv s v

= 1.60 + 0.35

= 1.95 mm 10.

R12 at 100mm centre (

11.

T16 (As = 1809 mm2 )

A sv s v

= 2.26 mm)

B. Reinforcement details;

3T16

2T16

2T16

R12 at 100mm centres (closed link)

2T25 + 2T16

END OF UNIT 10

Unit 10 ( TORSION )

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