sm7 115 117

________________________________________________________________________
PROBLEMS (7.115 through 7.117) A beam of cross section shown (Figs. P7.115 through
P7.117) is acted upon by a moment M with its vector forming an angle
α
with the horizontal
axis. Determine:
(a) The orientation of the neutral axis.
(b) The maximum bending stress,
Given: M = 2 kN⋅m,
α
= 15°
For Z-section: 64 64 64
3.11 10 , 12.43 10 , 4.56 10
zy yz
I
mm I mm I mm=× = × =× .
SOLUTION (7.115)
mmr 110
=
446.7
3r
y
mm
π
==
4644 101.16)110(11.011.0 mmrIz×===
446
(110) 57.5 10
88
yr4
I
mm
ππ
== =×
mNM o
y⋅=×= 51815sin102 3
mNM o
z⋅=×= 193215cos102 3
(a) 16.1
tan tan tan15
57.5 o
z
y
I
I
φα == ,
o
3.4= φ
(b) Point A is farthest from N.A. From geometry:
mmry o
A637.463.4cos −=+−=
mmrz o
A25.83.4sin ==
Thus,
yA
zA
Azy
M
z
My
I
I
σ =− +
66
1932( 0.063) 518(0.00825) 7.56 0.074
16.1 10 57.5 10
−−
−
=− + = +
××
= MPa63.7
SOLUTION (7.116)
Continued on next slide
r
y
M
φ
y
A
N
.A.
z
y
z
MC
φ
120 m
m
40 m
m
A
N
.A.
z
y
y
M
z
M
b=100 mm
C
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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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33
64
100(120) 4.8 10
36 36
ybh
I
mm== =×
33
64
120(100) 2.5 10
48 48
zhb
I
mm== =×
mNM o
y⋅=×= 51815sin102 3, mNM o
z⋅−=×−= 932.115cos102 3
(a) 2.5
tan tan tan( 15 )
4.8 o
z
y
I
I
φα ==−
,
o
9.7−= φ
(b) Point A is farthest from N.A. Thus,
6
1932(0.05) 518(0.04)
2.5 10 4.8 10
y
z
Azy
Mz
My
II
σ
−
=− + = +
××
6−
MPa4332.464.38 =+=
SOLUTION (7.117)
Dimensions are
In millimeters.
y
'
φ
M
z’
M
y’
z’
This part is already given (not required by student)
33
26
4
15(60) 150(15)
2[ 15 60(37.5) ] 3.11 10
12 12
z
I
mm=+×+=×
33
26
60(15) 15(150)
2[ 15 60(67.5) ] 12.43 10
12 12
y4
I
mm=+×+=×
)5.67)(5.37)(15(600)5.67)(5.37)(15(600
−
−
+
+
+=
yz
I
=
46
1056.4 mm×
Continued on next slide
z
y
’
'
A
y
p
θ
C
cos
Ap
z
θ
N
.A. sin
Ap
y
60
θ
15
15
15
A
120
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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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22 (4.56) (4.66)r=+
52.6
=
4.56
tan2 ' 4.66
p
θ =
' 22.2o
p
θ =
O
6
(10 )
yz
I
C
2'
(12.43, 4.56)
64
'1
7.77 6.52 14.29 10
y
I
Im== + = × m
46
2' 1025.152.677.7 mmIIz×=−+==
mkNM oo
y⋅−=+−= 21.1)2.2215sin(2
'
mkNM oo
z⋅=+= 59.1)2.2215cos(2
'
(a) '
'
tan ' tan( 15 22.2 )
oo
z
y
I
I
φ =−−
,
o
8.3−= φ
(b) Point A farthest from N.A.
'cos sin
AApA
zz yp
θ
θ =− −
=− 60cos22.2 67.5sin22.2 81.1
oo
mm− =−
p
'sin cos
AA pA
yz y
θ
θ =−
=− 60sin 22.2 67.5cos22.2 39.8
oo
mm=−
Thus,
'
'
''
'
'yA
zA
Azy
M
z
My
II
σ =− +
33
66
1.59 10 ( 0.0398) 1.2 10 ( 0.0811)
1.25 10 14.29 10
−−
×− −×−
=− +
××
=98.763.50 +58.61
M
Pa
=
p
θ
6
,(10)
yz
II
R
'
I
=7.77 4.66
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
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