Background Knowledge - Strengthening Design of Reinforced Concrete with FRP - page 27

Vw l

28
2

n

= = ×= 5.5 77 kips at support u u

2

19.75
12

V =−× 77 5.5 = 67.95 kips at critical section

ud

V 2 f bd 0.7524,000

×× =

12 19.75 22,483.8lb = 22.48 kips

φ=φ× = × c c w

φ= <= V 22.48 kipsV 67.95 kips c ud

Therefore, shear reinforcement is required.

φ=−φ = VV V 67.95 − 22.48 = 45.47 kips < φ× 8 f bd = 89.94 kips

s u c c w

Af d
VV

φ 0.75 × 0.22 × 50 × 19.75

vyt

S = = = 3.58 in. reqd '

−φ 45.47

u c

Try#4stirrups

0.75 ××× 0.450 19.75

S = = 6.52 in. Use 6" spacing reqd '

45.47

Check S :

max

89.94
2

φ 4 fbd = = 44.97 kips < 45.47 kips

cw

d

S == < 4.94 in. 12" (theformercontrols)

max

4

Use#4stirrups at 4.5" o.c.onlybetween supportand adistanceof 20.8"

Minimum shear reinforcement:

Af
fb

0.450,000
0.75 4,000

× Af

b

0.450,000
50

×

vyt vyt

S = =

× =≤ =

35"

0.75 12 50 × 12 cw w

d

=>= 33" 9.9in. (controls)

2

( )

Determine x distance from supportto V =φ V

c u c

VV
w

−φ 77 − 22.48

5.5

u c

φ=− VVwx x = = = 9.9ft

c u uc c

u

V

c

Determine x distance from supportto V =φ m u

2

V 22.48

2

c

V

−φ 77 −

V

u

2

c

φ=− Vwx x = = = 11.96 ft u u mm

2 w 5.5

u