sm7 113

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PROBLEM (7.113) Redo Prob. 7.111 for a C4 x 5.4 channel (refer to Table B.10) with
M = 10 kip⋅in.
SOLUTION
Table B.10: 4.54
×
C
4
.85.3 inIz=
4
.319.0 inIy=
.66.830cos10 inkipM o
z⋅==
.530sin10 inkipM o
y⋅−=−=
(a) 3.85
tan tan( 30 ), 81.8
0.319 oo
φφ =− =−
(b) Point A is farthest from N.A. Thus,
max yA
zA
Azy
M
z
My
I
I
σσ ==− + 33
8.66 10 ( 2) 5 10 ( 1.13)
3.85 0.319
×−−×−
=− +
ksi21.2271.175.4 =+=
z=0.458 in.
y
1.584 in.
.A.
N
0.184 in.
A
C
4 in.
M
z
φ
M
y
z
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