Civil Engineering Reference
In-Depth Information
(3)
Check section adequacy:
Assume the distance to stirrups centroid = 1.75 in. (considering 1.5 in. cover and 0.5 in. stirrups diameter)
p
h
= 2(12-2
1.75)+2(20-2
1.75) = 50 in.
A
oh
= (12-2
1.75)(20-2
1.75) = 140.3 in
2
.
V
c
b
w
d
=
24000
=
126.4 psi
2
2
2
2
⎛
⎜
V
u
b
w
d
⎞
⎟
⎛
⎜
T
u
p
h
1.7A
oh
⎞
⎟
=
⎛
⎜
27.4(1000)
12(17.5)
⎞
⎟
⎛
⎜
20.7(12,000)(50)
1.7(140.3)
2
⎞
⎟
+
+
=
393.4
V
bd
+
⎛
⎜
⎞
φ
c
8
f
c
ʹ
⎟
=
0.75(126.4
+
8 4000 )
=
474.3
>
393.4
w
Section is adequate
(4)
Required stirrups for torsion
A
o
= 0.85A
oh
= 0.8(140.3) = 119.2
A
t
T
u
20.7(12,000)
2(0.75)(119.2)(60,000)
=
s
=
A
o
f
y
=
0.023in.
2
/in.(oneleg)
2
φ
(5)
Required stirrups for shear
V
u
V
s
=
φ
−
V
c
=
27.4 / 0.75
−
126.4(12)(17.5) / 1000
=
10 kips
A
v
V
s
s
=
f
y
d
=
10 / 60 / 17.5
=
0.0095 in (two legs)
(6)
Combined stirrups
A
t
A
v
s
+
2s
=
0.023
+
0.0095 / 2
=
0.028in.
2
/in.
maximum stirrups spacing = p
h
/8 or 12 in. = 50/8 = 6.25 in.
For No. 4 closed stirrups required s = 0.2/0.0325 = 6.15 in.
For No. 3 closed stirrups required s = 0.11/.0325 = 3.4 in.
Use No. 4 closed stirrups @ 6 in.
Area for 2 legs = 0.4 in.
2
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