sm7 118 120

________________________________________________________________________
PROBLEMS (7.118 through 7.120) For the beam cross section shown in Figs. P7.115
through P7.117, determine the largest permissible value of the moment M if all
σ
= 100 MPa
and
α
= 0o.
SOLUTION (7.118)
mmr 110
=
0
=
α
0
=
φ
mmy 7.46
=
y
From solution of Prob. 7.115:
4646 105.57101.16 mmImmI yz ×=×=
We have
zA
all A z
M
y
I
σσ ==
from which
66
100 10 (16.1 10 )
(0.11 0.0467)
all
zA
My
σ
−
××
== −
mkN ⋅= 4.25
SOLUTION (7.119)
Refer to solution of Prob.7.116: 4646 105.2108.4 mImI zy −− ×=×=
We now have and 0=
y
MMMz
−
=
. Thus,
A
all A
M
y
I
σσ
−
==− or all z
zA
I
M
y
σ
=
66
100 10 (2.5 10 ) 5
0.05
z
M
kN m
−
××
==⋅
SOLUTION (7.120)
Refer to solution of Prob. 7.117:
46
2'
46
1' 1025.11029.14 mIImII zy −− ×==×==
' 22.2 ' 39.8 ' 81.1
o
PA A
y
mm z mm θ ==− =−
We now have Continued on next slide
r
A
N
.A.
z
y
M
C
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MMM o
y378.02.22sin
'−=−=
MMM o
z926.02.22cos
'=−=
'
'
1.25
tan tan( 22.2 ) tan( 22.2 )
14.29
oo
z
y
I
I
φ =−= −
,
o
2−= φ
Point A is farthest from N.A. Thus,
66
1932(0.05) 518(0.04)
2.5 10 4.8 10
y
z
Azy
Mz
My
II
σ
−−
=− + = +
××
666
0.926 ( 0.0398) 0.378 ( 0.0811)
100 10 1.25 10 14.29 10
z
MM
−−
−
−
−
×=− +
××
or 3
100 10 29.48 2.15 z
M
M×= +
from which 3.16
z
M
kN m=⋅
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.