Torsion in RC Beam PDF

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Faculty of Civil Engineering Tel:(+603) 5543 6414 Fax: (+603) 5543 5275Question No. 2The L-beam show in Figure below is 6 m long. The ends of the beam are simply supportedfor vertical load and restrained against torsion.Imposed load on shaded area = 1 kN/m2Dead load on shaded area = 2 kN/m 2ƒcu = 30...

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FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6414 Fax: (+603) 5543 5275

Question No. 2 The L-beam show in Figure below is 6.0 m long. The ends of the beam are simply supported for vertical load and restrained against torsion.

150 mm 600 mm

300 mm

700 mm

Imposed load on shaded area = 1.5 kN/m2 Dead load on shaded area = 2 kN/m2

ƒcu = 30 N/mm2 ƒy = 460 N/mm2 ƒyv = 250 N/mm2 Cover = 30 mm Link  = 10 mm Bar  = 20mm Calculate the reinforcement required to resist bending, shear and torsion.

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FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6414 Fax: (+603) 5543 5275

Reference:

Calculation

Remarks

ANSWER 1. Calculate the total dead load DL= 24 x (0.3x0.6 + 0.15x0.7) = 6.84 kN/m For Shaded area, DL = 2 x 0.7 =1.4 kN/m Total DL = 6.84 + 1.4 = 8.24 kN/m

DL = 8.24 kN/m

2. Calculate the total live load (on shaded area only) LL = 1.5 x 0.7 = 1.05 kN/m

LL = 1.05 kN/m

Ultimate load,  = 1.4x8.24 + 1.6x1.05  = 13.216kN/m

= 13.216kN/m

Therefore, BS8110: Part1: 1997: cl 3.4.1.5

Mmax =

wl 2 13.216 x6 2 = = 59.472kNm 8 8

Mmax = 59.472 kNm

Vmax =

wl 13.216x6 = = 39.648kN 2 2

Vmax = 39.648 kN

3. Calculate the breadth of the flange, bf l 6000 b f = bw + z = 300 + = 900mm 10 10 Since actual flange is bigger, therefore take bf = 900mm (take lesser value)

bf = 900mm

4. Calculate the effective depth, d bar d = h − cov er − − link 2 d = 600-30-10-10 d = 550mm

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d = 550mm

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FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6414 Fax: (+603) 5543 5275

Reference:

Calculation

Remarks

BS8110: Part1: 5. Calculate the value of k ≤ 0.156 for singly reinforced 1997: cl 3.4.4.4 M k=  0 .156 f cu b f d 2

k=

59.472 x10 6 = 0.00730.95d, so take z=0.95d z = 523mm BS8110: Part1: 1997: cl 3.4.4.4

7. Determine the location of N.A. hf If d − z  , 2 or M r = 0.45 f cubh f (d −

hf ) > Mmax 2

then N.A is in the flange

therefore; d - z = 550 – 523 = 27mm

h f 150 = = 75 mm 2 2 Since d − z 

hf , the N.A lies in the flange, 2

therefore design as a rectangular section.

- Torsion in Beam

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FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6414 Fax: (+603) 5543 5275

Reference: BS8110: Part1: 1997: cl 3.4.4.4

BS8110: Part1: 1997: cl 3.12.5.3 Table 3.25

cl 3.12.5.3

BS8110: Part1:

Calculation

Remarks

8. Calculate the value As

As =

M 0.95f y z

As =

59.472x10 6 (0.95)(460)(523)

As = 260 mm2

As, min =

As, max =

( 0.13 )( bw )( h ) ( 0.13 )( 300 )( 600 ) = 100 100

( 4 )( bw )( h ) ( 4 )( 300 )( 600 ) = 100 100

9. Calculate the value Asv

As, req = 260 mm2 As, min = 234 mm2

As, max = 7200mm2 As, prov = 402 mm2 2T16

1997: cl 3.4.5.2

v=

Vmax bv d

v=

39.648 x10 3 (300)(550)

v = 0.240 N/mm2

0 .79 (

v = 0.240 N/mm2

100 As 1 / 3 400 1 / 4 ) ( ) bd d m

BS8110: Part1:

vc =

1997: Table 3.8

1 (100 )( 402 ) 1/ 3 400 1/ 4 ) ( ) 0. 79( 550 ( 300 )( 550 ) vc = 1.25 vc = 0.365 N/mm2

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FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6414 Fax: (+603) 5543 5275

Reference:

Calculation

Remarks 2

Since fcu is greater than 25N/mm , the value of vc must be multiplied by (

fcu 1/ 3 ) , then, 25

vc = ( 0.365 )( 30 )1/ 3 25

vc = 0.388 N/mm2

BS8110: Part1: 1997: Table 3.7

vc = 0.388 N/mm2

0.5vc < v < (vc + 0.4) 0.5(0.388)< 0.205 < 0.788  Provide minimum links for whole length

0. 4bv Asv (0.4)(300) = = S v 0.95f yv (0.95)(250)

Asv = 0.505 Sv 10. Calculate the centre of gravity

A1X 1 + A 2 X 2 A (300x 600)(150)+ (150x 700)(650) = (300x 600) + (150x 700) = 334.2mm =

11. Calculate Torsional moment maximum For dead load (self-weight of beam and slab), 1.4x 6.84x 6x (0.3342 − 0.15) T= = 5.292kNm 2 For dead load on shaded area (partition, wall and superimposed dead load), 1.4 x1.4 x6 x(0.35 + 0.15) T= = 2.94 kNm 2 For live load on shaded area, 1.6x1.05x6x (0.35 + 0.15) T= = 2.52 kNm 2

T = 10.752 kNm

Total T = 5.292+2.94+2.52 = 10.752 kNm

- Torsion in Beam

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FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6414 Fax: (+603) 5543 5275

Reference:

Calculation 12. Calculate the (  h3 min hmax )

maximum

Remarks function

of

1000mm

150mm

(ii) (i)

600mm

300mm

( h

3 min

hmax ) = (3003x600) + (1503x700) = 1.856 x 1010 mm4 1000mm

(ii) 600mm

150mm

(i)

300mm

(  h3min hmax ) = (3003x450) + (1503x1000) = 1.5525 x 1010 mm4 Choose the first diagram since (  h3 min hmax ) is the maximum, 1.856 x 1010 mm4

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FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6414 Fax: (+603) 5543 5275

Reference:

Calculation

BS8110: Part2:

13. Calculate the torsional moment

1985: cl 2.4.4.2

T i = T max(

Remarks

h3minhmax ) ( h3 minhmax )

Section #1 BS8110: Part2: 1985: cl 2.4.4.1

Tw = 10.752x106 (

(300)3 (600) ) 1.856 x 10 10

Tw = 9.385x106 Nm

Tw = 9.385x10 6 Nm

Section #2 6 Tf = 10.752x10 (

(150) 3 (700) ) 1.856 x 1010

Tf = 1.367x106 Nm

T f = 1.367 x10 6 Nm

14. Calculate the torsional shear stress BS8110:Part2:19 85:Table 2.3

2Ti

vt =

h2 min( hmax −

hmin ) 3

Section #1 BS8110:Part2:19

vt, w =

85:cl 2.4.6

2(9.385 x10 6 ) 300 (300 2 )(600 − ) 3

vt , w = 0.417 N / mm2

vt , w = 0.417 N / mm2

Section #2 BS8110:Part2:19 85:cl 2.4.5

vt, f

2(1.367 x10 6) = 150 (1502 )(700 − ) 3

vt , f = 0.187 N / mm2

vt , f = 0.187 N / mm2 vt ,min = 0.37 N / mm2 vtu = 4.38 N / mm2

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FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6414 Fax: (+603) 5543 5275

Reference: BS8110: Part2: 1985: cl 2.4.7

Calculation 15. Check for vt  vt ,min

Remarks

(Section #1 and

Section#2) Section#1

vt  vt ,min Which is 0.417 > 0.37. Therefore Torsional reinforcement is required.

Section#2

vt  vt ,min Which is 0.187 < 0.37. Therefore Torsional reinforcement is not required.

BS8110:Part2:1 985:cl 2.4.8

BS8110:Part2:1

16. Check for (v + vt )  vtu and v t 

( vtu)( y1 ) 550

(v + vt ) = 0.204 + 0.417 = 0.621N / mm2 (v tu )( y 1 ) (4.38)(530) = = 4.221N / mm2 550 550

985:cl 2.4.7 Since (v + vt )  vtu and vt 

( vtu)( y1 ) 550

therefore

shear stress is OK

17. Calculate the torsional reinforcement Additional reinforcement for links

Asvt Ti  S v 0.8 x1 y1( 0.95 fyv )

Asvt 9.385 x10 6 = Sv (0.8)(230)(530)(0.95)(250) Asvt = 0.642 Sv

- Torsion in Beam

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FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6414 Fax: (+603) 5543 5275

Reference:

Calculation

Remarks

Total links required,

Asv Asvt + = 0.505 + 0.642 = 1.147 Sv Sv Provide R10@125mm c/c,

Asv = 1.257 Sv

Additional reinforcement for longitudinal bar

Asl =

Asvt f yv ( x1 + y1 )

Asl = (

Sv f y A svt f yv( x1 + y1 ) )( ) Sv fy

Asl = (0.642)(

250(230 + 530) ) 460

Asl = 265.2mm2 There are six (6) numbers of reinforcement should be distributed evenly round the inside perimeter of the links. Therefore the area for every reinforcement;

A=

265.2 = 44.2 mm2 6

then provide 6T10, where Asl = 471 mm2

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