Civil Engineering Reference
In-Depth Information
or
3, 000
75, 000
87
87
= ×××
+
0.75
0.85
0.85
=
0.01164
75
d
d
=−
0.8
0.9
f
75
3
Q
=
0.01164
× ×=
0.9
0.262 controls (upper bound)
1
75
3
−
3
Q
=×××=
2.67
10
0.8
0.0534
1
40
8
Q
=
0.0568
× ×=
0.9
0.2556
1
40
8
Q
−
3
=×××=
5.0
10
0.8
0.0268 controls (lower bound)
1
0.0268
<< =
Q
0.262
B
1.946
−
1.994 closeto2
1
B
Q
where
=−
=
20.208
1
Assuming
B
2
a
d
=− −
112.77
Q
for
Ψ =
0.85
(5.19)
2
f
f
a
d
=− −
112.35
Q
for
Ψ =
1.0
(5.20)
2
f
f
Example 5.1: Design
Solve Example 15.3 of ACI 440.2R-08 using the direct approach (Figure 5.3).
The problem statement from the design guide goes as follows: A simply sup-
ported reinforced concrete beam is located in an unoccupied warehouse. It is
required to be strengthened by a 50% increase in its live-load carrying capacity.
W
DL
+
W
LL
21.5"
24"
24 ft
3#9
f
c
= 5 ksi
f
y
= 60 ksi
12"
FIGURE 5.3
Example 15.3 of ACI 440.2R-08 showing the beam profile and the cross-
section details.
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