sm7 110 112

________________________________________________________________________
PROBLEM (*7.110 through 7.112) As shown in cross section in Figs. P7.110 through P7.112, a
beam is subjected to a moment M with its vector forming an angle
α
with the horizontal axis. Calculate:
(a) The orientation of the neutral axis.
(b) The maximum bending stress.
SOLUTION (*7.110)
33
22
6(1) 1(6)
6 1(1.75) 1 6(1.75)
12 12
z
I=+× ++×
4
.25.55 in=
33 4
1(6) 6(1) 18.5 .
12 12
y
I
in=+=
.6.9615cos100 inkipM o
z⋅−=−=
.9.2515sin100 inkipM o
y⋅−=−=
(a) 55.25
tan tan tan15
18.5 o
z
y
I
I
φα == ,
o
7.38= φ
(b) Point A is farthest from N.A.
Thus,
yA
zA
Azy
M
z
My
I
I
σ =− +
33
96.6 10 ( 4.75) 25.9 10 (0.5) 8.305 0.7
55.25 18.5
−×− −×
=− + =− −
ksi01.9−=
SOLUTION (7.111)
Table B.10: 75.188
×
C
44 .98.1,.44 inIinI yz ==
o
z
M30cos30=.98.25 inkip⋅
=
o
y
M30sin30−= .15 inkip⋅
−
=
(a) 44
tan tan tan( 30 )
1.98 o
z
y
I
I
φα ==−
,
o
5.85−= φ
y
2.527 in.
z=0.565 in.
z
A
0.39 in.
C
N
.A.
8 in.
φ
M
z
M
y
y
z
φ
2.25 in.
n.
6 i
M
z
M
y
1 in
.
N
.A
1 in.
A
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Continued on next slide

(b) Point A is farthest from N.A. Thus,
max yA
zA
Azy
M
z
My
I
I
σσ ==− +
33
25.98 10 ( 4) 15 10 ( 1.962)
44 1.98
×−−×−
=− +
ksi22.1786.1436.2 =+=
SOLUTION (7.112)
Table B.11: 4388 ××L
y
A
2
.4.11 inA =
.58.1
min inr
=
4
.7.69 inII zy ==
We have
22
'min
14.4(1.58) 28.46 .
y2
I
Ar in== =
4
'1' .94.110 inIIIII yzyz =−+==
(a) '
'
110.94
tan ' tan(45 20 ) tan65
28.46
oo
z
y
I
I
φ =+= o
,
o
2.83'= φ
(b) From geometry:
' 0 ' 2 3.224 .
AB
yzy===in
' sin45 2 cos45 2.433 .
oo
A
za y i=− + − n
' cos45 5.657 .
o
A
y
ai==n
.
52.84)2045sin(200
'inkipM oo
z⋅=−=
.26.181)2045cos(200
'inkipM oo
y⋅=−=
Apply Eq.(7.45):
'
'
''
'
'yA
zA
Azy
M
z
My
II
σ =− +
84.52(5.657) 181.26( 2.433) 4.31 15.5
110.94 28.46
−
=− + =− −
= ksi81.19−
45o
'
φ
C
N
.A.
a =8 in.
α
M
z’
M
y’
z’
y’
A
y’
A
y
z==2.28 in.
z’
z
B
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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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Continued on next slide

Similarly, 181.26(3.224)
0 20.53
28.46
Bksi
σ =+ =
Thus, ksi53.20
max =
σ
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
Copyright Act without the permission of the copyright owner is unlawful.