sm7 114

________________________________________________________________________
PROBLEM (7.114) Redo Prob. 7.112 for an L 5 x 3 x ¼ angle (refer to Table B 12) with
M = 15 kip-in.
SOLUTION
Table B.12: 4135 ××L
.657.0 inz
=
.66.1 iny
=
4
.44.1 inIy=
4
.11.5 inIz=
3.0tan 71
=
p
θ
o
p4.20=
θ
2
.94.1 inA =
.663.0
min inr
=
o
p4.40204.20 =+=+
αθ
,
.72.94.40sin15
'inkipM o
y⋅== .42.114.40cos15
'inkipM o
z⋅==
,
422
min' .853.0)663.0(94.1 inArIy=== 4 '' .697.5 inIIII yyzz =−+=
(a) '
'
5.697
tan ' tan(40.4 ) tan(40.4 )
0.853
oo
z
y
I
I
φ == ,
o
80= φ
(b) Point A is farthest from N.A. From geometry:
pApAA zyy
θ
θ
sincos
'+=
pApAA zyz
θ
θ
cossin
'+−=
Substitute the numerical values into Eqs.(1):
oo
A
y4.20sin657.04.20cos)66.15(
'+−= .36.3 in
=
oo
A
z4.20cos657.04.20sin)66.15(
'+−−= .548.0 in
−
=
Thus,
''
''
''
yA
zA
Azy
M
z
My
I
I
σ =− +
11.42(3.36) 9.72( 0.548) 6.74 6.25
5.697 0.853
−
=− + =− − ksi13
−
=
y
'
φ
cos
A
p
y
θ
M
z’
M
y’
y
z
’
z
y’
z’A
z
p
θ
C
N
.A.
A
cos
Ap
z
θ
siy
sin
Ap
z
θ
n
Ap
θ
(1)
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