Civil Engineering Reference
In-Depth Information
f
fv
= 0.004
E
fv
= 24 ksi ≤
f
ffb
f
ffb
= (0.05
r
b
/
d
b
+ 0.3)
f
fu
≤
f
fu
f
ffb
= 0.45
f
fu
= 27.0 ksi
f
fv
= 24.0 ksi
V
f
=
A
fv
f
fv
d
/
s
= 14.2 kip
And the nominal shear strength:
V
n
=
V
c
+
V
s
= 40.2 kip
ϕ
V
n
= 30.2 kip
Example 5.3
Calculate the concrete shear strength
V
c
of the square column in
Example 5.2 for various values of
P
u
, M
u
when accounting for the
effect of the axial load. In Example 5.2, the shear strength of the col-
umn is calculated disregarding the axial load. To account for the effect
of the axial load, Equation (5.59) is used to evaluate
c
/
d
that replaces
k
= 0.159, calculated in Example 5.2. Therefore:
0 ≤
c
/
d
= 0.159 + 0.01197 (
P
u
d
/
M
u
) ≤ 0.4
The contribution of concrete to the shear strength is calculated for
the loading cases that follow:
1. No axial force
c
/
d
= 0.159
c
= 3.42 in.
V
c
= 26.0 kip
2.
M
u
= 0, tensile force
c
/
d
= 0
c
= 0
V
c
= 0
3.
M
u
= 0, compressive
force
c
/
d
= 0.4
c
= 8.6 in.
V
c
= 65.3 kip
4.
P
u
= 500 kip,
M
u
= 100 ft-kip
c
/
d
= 0.266
c
= 5.72 in.
V
c
= 43.4 kip
5.
P
u
= 500 kip,
M
u
= 200 ft-kip
c
/
d
= 0.213
c
= 4.58 in.
V
c
= 34.8 kip
6.
P
u
= 500 kip,
M
u
= 300 ft-kip
c
/
d
= 0.195
c
= 4.19 in.
V
c
= 31.8 kip
5.9.5 Circular sections
For circular sections, the same equations introduced in rectangular sections
to compute shear contributions of concrete and transverse reinforcement
can be used, if an equivalent rectangular section is defined as
b
w
=
D; d
= 0.8
D;
A
1
=
A
2
=
A
f
/ 4
(5.63)
The equivalent dimensions replicate the provisions of ACI 318-11 for
shear strength of circular sections. Similarly, the strength-reduction factor
of rectangular sections equal to 0.75 applies to circular ones under the
same limitations for
V
c
and
V
f
.
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