Civil Engineering Reference
In-Depth Information
No. 4 stirrups at 4.5" o.c. for a distance of 20.8"
φ=
×××
0.75
0.450
19.75
V
s
=
65.83 k
4.5
φ+φ= >
VV
88.31
V
ud
No need for external shear strengthening in this
c
s
1
region.
No. 4 stirrups at 6" up to 9.9'
φ=
×××
0.75
0.450
19.75
V
s
=
49.375 k
6
φ+φ=
VV
c
71.86 k
s
2
(
)
(
)
Determine
x
distance from supportto
V
=φ +
VV
:
cs
2
u
c
s
2
98
−
71.86
7
(
)
φ+=−
VV Vwx
x
=
=
3.73 ft
c
s
u
u
cs
2
cs
2
20.8"
12
V
u
=−×
98
7
=
85.87 k
20.8"
φΨ =
ff
85.87
−
71.86
=
14.01 k
14.01
0.75
V
=
=
21.98 k
f
×
0.85
Afd
S
fv fe
fv
=
f
20.0065
×
× ××ε×
wEk
15.75
f
f
v u
V
=
=
21.98 k
f
d
w
+
f
4
550
33,000
*
ε=ε= ×
C
0.95
=
0.0158
fu
E u
k
=
1
1
2, 500
L
e
=
=
2.022 in.
(
)
0.58
10.0065
×
×
33,000
×
10
3
15.75 2.022
15.75
−
k
=
=
0.872
2
10.872
×
×
2.022
k
v
=
=
0.238
<
0.75
468
×
0.0158
19.75
4
21.98
w
+
=
25.41
w
f
f
108.53
3.43
w
f
=
=
31.64 in.
Too big
(
+=
21.98
w
2
25.41
w
f
f
43.96
3.43
w
f
=
=
12.82 in.
se 13" @ 15" c/c(one layer)
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