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b.
Computing ns :
d , EI , and P c are same as before
0
C m 0.6 0.4
0.6
14
0.6
1 48 4.16
0.75 731
ns
0.66
Use 1.0
c.
Computing s :
d 0 since there is no dead load involved in this case
EI (0.4)(3605)(1210)
1 0
1.74 10 6 k-in. 2
2 )(1.74 10 6 )
(1.95 12 11.25) 2 247.84 k
(
P c
1
s
1.35
(2)(48) 4.16 4.16
0.75
1
2
247.84
d.
Compute the magnified moment:
M c (1.0)(126.2) (1.35)(76.8) 229.9 ft- k
5. Consider the loading case 0.9D. Analysis results are shown in Figure 11.12.
a.
Are column moments ACI minimum?
e min 0.6 (0.03)(12) 0.96 in.
M 2,min (13.5)(0.96) 12.96 in.- k 1.08 ft- k
OK
Figure 11.12
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