Civil Engineering Reference
In-Depth Information
5.9.8 Examples—Shear wall strength
and shear friction
Example 5.5
Calculate the design shear strength of the shear wall detailed as follows:
Concrete:
f
´
c
= 4.0 ksi
E
c
(ksi) = 57√
f
´
c
(psi) = 3600 ksi
Vertical reinforcement:
f
ffu
=
f
fd
= 60 ksi
E
f
= 6000 ksi
A
f
1
= #5@12 in. = 0.3 in.
2
/ft
Layer 1
A
f
2
= #5@8 in. = 0.45 in.
2
/ft Layer 2
Horizontal reinforcement:
f
ffu
= 60 ksi
E
fv
= 6000 ksi
A
fv
= #3@12 in. Two layers
Size:
b
= 12.0 in. Wall thickness
l
w
= 16 ft Wall length
Solution:
Concrete contribution,
V
c
:
ρ
f
= (
A
f
1
+
A
f
2
)/
b
= 0.0625 in./
ft = 0.00521
n
f
=
E
f
/
E
c
= (6000 ksi)/(3600 ksi)
= 1.67
n
f
ρ
f
= 0.00868
k
= 0.0852 Equation (5.64)
c
=
kl
w
= 16.4 in.
V
c
= 62.2 kip
Shear reinforcement contribution,
V
f
:
A
fv
= 2#3@12 in. = 0.22 in.
2
s
= 12.0 in.
f
fv
= 0.004
E
fv
≤
f
ffb
f
ffb
=
f
ffu
= 60 ksi
No bends
f
fv
= 24.0 ksi
d
= 0.8
l
w
= 153.6 in.
V
f
=
A
fv
f
fv
d
/
s
= 67.6 kip
The nominal shear strength:
V
n
=
V
c
+
V
s
= 129.8 kip
And the design shear strength:
ϕ
V
n
= 97.4 kip
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