Civil Engineering Reference
In-Depth Information
Example 4.8
Design a beam for the nominal flexural strength of
M
n
= 5400 kip-in.
= 450 ft-kip. Other properties are as follows:
Concrete:
f
′
c
= 4.0 ksi
β
1
= 0.85
Reinforcement:
f
fu
= 60 ksi
E
f
= 6000 ksi
ε
fu
= 0.010
Size:
b
= 16.0 in.
h
= 25.0 in.
c
= 3.0 in.
d = h
−
c
= 22.0 in.
Solution:
m
= 0.2051
ω′
f
= 0.1972
e
= 3.333
f
= 0.799 < 1: Failure by concrete crushing
ω
f
= 0.2468
A
f
= 5.79 in.
2
: Use 5#10
Example 4.9
Design a beam for the nominal flexural strength of
M
n
= 3600 kip-in.
= 300 ft-kip. Other properties are similar to those in Example 4.8.
Solution:
m
= 0.1367
ω′
f
= 0.1255
e
= 3.333
f
= 1.427 > 1: Failure by FRP rupture
ω
fb
= 0.1667
ω
f
=
0.1289
A
f
= 3.02 in.
2
: Use 3#9
(
b2
) Applied moment and the stress level in reinforcement are
known; dimensions and the reinforcement area are unknown:
In this case, if a predetermined value is assigned to the stress level
(
f
<
1), then:
0.85
1
β
1
ω=
f
+
ef
The reinforcement area and the dimensions can be calculated from
ω
f
ω=
f
f
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