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( )

1

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2

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& C

,

3

. A ,

. , , B

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&

5

ULS in Flexure!

ULS in Shear!

Cracked Section

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.

6

, .

1 , 1.

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= V Ay/Ib

7

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, 1

45 . ,

, .. 45

, (=/A) 1.

8

, , . () .

, .

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( )

F A

, , .

=

.A/ C, = /2 + (/4 + )

9

= fcp

= fs

1111 = ft

=

B 8110

= 0.24

F

/A = 0.67

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( )

= (/A)( + )

Vco = 0.67bvh(ft2 + 0.8fcpft)0.5

10

Vco

bv breadth of the member or for T-, I- and L-beams, widthof the web

if grouted duct is present in the web,

bv = bw 0.67dd (dd diameter of duct)

ft = 0.24fcu0.5

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( )

Mo moment which produces zero stress at extreme tension fibre;

Mo = 0.8fptI/ywhere fpt is the level of prestress in concrete at the tensile face

11

As sum of area of prestressing steel and

non-prestressing steel

d depth from compression face to centroid of

total steel

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,

According to Clause 4.3.8.3 BS 8110,

Vc

= Vco

at uncracked section (M < Mo)

Vc = is the smaller of Vco and Vcr at cracked section (M >= Mo)

For deflected tendon, the vertical component of the prestressingforce will help to resist the shear force.

12

The total shear resistance then becomes:Vc + Pe sin where is the angle of inclination of theprestressing tendon

For Parabolic Profile, e(x) = (- 4/L2)x2 + (4/L)x

(x) = (-8d/L2)x + (4/L) in radian

where = abs(es ems)

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(x) = - (8/L2) x + (4/L)

13

( ( ( (

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1.

2. ( ...2)

( /) 0. /

. .

. .

(M >= Mo) (M < Mo)

14

. co cr .

.

Vc = Vcr or Vco + Pesin

Vc = Vco + Pesin

. 0. Vc, ( ...)

7. If 0.5 Vc < V

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6. 1.8Vc, the maximum spacing should be reduced to 0.5dt.11. The lateral spacing of the individual legs of the links provided at a

cross-section should not exceed dt

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, 85/ 15

2000. D . :

= 40/; A = 2.9 10 ; = 3.54 10

= =

16

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1. D BD FD

()=0.5() & ()=(0.5 ) = 85 /

17

2. Check maximum allowable shear stress (Clause 4.3.8.2)

V = 637.5 kN, v = 637.5*10/(500*500) = 2.55 N/mm

vmax = lesser of (0.8*40 = 5.06N/mm2 and 5N/mm) ok

3. BD. = 0.8 / & =

/A +

/

e(x) = (- 4/L2)x2 + (4/L)x where = 425mm

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18

M > Mo ; Section Cracked in Flexure

M < Mo ; Section Uncracked in Flexure

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20

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Nominal Shear Reinforcement

Use R8 ; Asv = 2*0.25**82 = 101 mm

Sv = 101*0.87*250/ 0.4*150 = 366 mm

Sv mak = 0.75d = 0.75*908 = 681 mm

Use R8 350 mm

21

Design Shear ReinforcementUse R8 ; Asv = 2*0.25**82 = 101 mm2

V - Vc = 107.12 kN

Sv = 101*0.87*250*772/ (107.12*1000) = 158 mm

Sv mak = 0.75d = 0.75*772 = 579 mm

Use R8 150 mm

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R8 - 150 R8 - 350

22

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A 28, , , 4 / 10 /. 14 15.7 7 (A = 150 ) 7 2

23

.

1044 . :

= 50 /; A = 5.08 10 ; = 134 10 ;

= 912 ; = 1770 / ; = 250 /

D .

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24

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1.

, = 0.508 24 = 12.19 /

, = 1.4(12.19 + 4) + 1.6 10 = 38.67 /2. 2

= 38.67(0.5 28 2) = 464

25

= . x . x - = m

3. Calculate Mo e = 814 mm; d = 1500 98 = 1402 mm

fpt = (1044x103/508000)+(1044x103 x 814 x 912 / 134 x 109)

= 7.84 /2

= 0.8 7.84 134 109 / 912 = 922 < = 1005

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4. C

Vco = 0.67bvh(ft2 + 0.8fcpft)0.5

= 1044 103

/50800 = 2.06 N/mm2

= 0.24 500.5 = 1.70 N/mm2 ; = 1500 ; v= 175

Vco = 0.67x175x1500(1.702 + 0.8x2.06x1.70)0.5 /103 = 420

26

. a cu at on o cr

= 1044 10 / 7 150 = 994 N/mm2

100As/bvd= 100x7x150/175x1402 = 0.43 40 N/mm2; Use fcu = 40 N/mm2

vc = 0.79 x 0.431/3 x 1 x (40/25)1/3/1.25 = 0.556 N/mm2

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5. Calculation of Vcr contd

Vcr = (1 0.55x994/1770)x0.556x175x1402x10-3 +

992x464/1005 = 520 kN Vcr 0.1x175x1402x501/2x10-3 = 174 kN ok

6. Shear resistance provided by the concrete, Vc

27

Vc = 420 kN

7. Design of shear reinforcement

= 464 > 0.5 = 210 < + 0.4 = 518

A/ = 0.4 175 / 0.87 250 = 0.322 /

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7. Design of shear reinforcement

= 464 > 0.5 = 210 < + 0.4 = 518

A/ = 0.4 175 / 0.87 250 = 0.322 /

10 /, A = 157 2

28

= . = < . = . . =

Use R10 450 mm c/c

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29

Arrangement of Shear Links

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Shear in

30

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31

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F ,

,

32

F ,

, ,

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33

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34

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35

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,

F ,

36

. A ,

.

, ,

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37

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= C Interface in the compression zone

V = C

38

The average horizontal shear stress is given as: (vh)av = Vh/ bv l

Where bv is the width of contact surface and l is the distance

between point of maximum moment and the point of zeromoment

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, vh

(vh

)av

l shear force diagram

For a uniformly distributed load, this stress distribution is linear as

39

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.

The extent of shear resistance depend on the strength class of in situ

concrete and the texture of precast beam at the interface

40

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,

41

= .

4 600

, :

Ah = 1000 bv vh/ 0.87fyv