Quantcast Design shear reinforcement

Share on Google+Share on FacebookShare on LinkedInShare on TwitterShare on DiggShare on Stumble Upon
Custom Search
 
  
 


a = A -sfdy/(0.49f -dcb)
= 0.341 X'66,000/(3062.5 X 3.75)
= 1.96 in
O.K.
Use two No. 4 in each stem.
A -s = 0.4 in2
Check maximum rebound reinforcement
A -s < /=
0.49fdc K1
(87,000-0.36nf'dc) bd -
fdy
(87,0000-0.36nf'dc + fdy)
=
3062.5 x 0.7375
66,000
(87.000-0.36 x 6.76 x 6250)
(87,000-0.36 x 6.76 x 6250 + 66,000)
x (3.75 x 23.62)
=
1.58 in2 > 0.40 in2
O.K.
(16) Design shear reinforcement.
Calculate shear at distance, d, from support
vu = ru(L/2-d)/bwd
= 124(20 x 12 - 22.0)/(2 x 4.75 x 22.0)
= 129 psi < 10[phi](fc')1/2
= 10 x 0.85 x (5000)1/2
= 601 psi
O.K.
vc = [phi][1.9(fc')1/2 + 2500p] < /=
2.28[phi](fc')1/2
= 0.85[1.9(5000)1/2 + 2500 x .612/(2 x 4.75 x
22.0)]
= 120 psi < /= 2.28(0.85)(5000)1/2
< /= 137 psi
O.K.
Av
= [(vu - vc)bss]/[phi]fy
vu - vc >/= vc
vu - vc = 129-120
= 9 psi < vc so use vc.
Assume No. 3 stirrups,
Av = 2 x 0.11
= 0.22 in2
s
= (0.85 x 60,000 x 0.22)/(120 x 4.75)
= 19.7 in
But s < /= d/2 = 11.0 in
2.08-264





 


Privacy Statement - Copyright Information. - Contact Us

Integrated Publishing, Inc.
6230 Stone Rd, Unit Q Port Richey, FL 34668

Phone For Parts Inquiries: (727) 493-0744
Google +