sm7 121 123

________________________________________________________________________
PROBLEMS (7.121 through *7.123) 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
= 80 MPa
and
α
= 30o .
SOLUTION (7.121)
Refer to Solution of Prob.7.115:
46
105.577.46110 mImmymmr y− ×===
46
101.16 mIz− ×=
We now have
MMMMMM o
z
o
y866.030cos5.030sin ====
16.1
tan tan30 tan30 , 9.2
57.5
oo
z
y
I
I
φφ == =
o
Point A is farthest from N.A.
mmyry o
A9.612.9cos −=+−=
mmrz o
A6.172.9sin ==
Therefore,
yA
zA
Azy
M
z
My
I
I
σ =− +
666
0.866 ( 0.0619) 0.5 (0.0176)
800 10 16.1 10 57.5 10
MM
−−
−
×=− +
××
or 80 MM 153.03.3103+=×
from which
mkNM ⋅= 97.22
SOLUTION (7.122)
See solution of Prob.7.116:
4646 105.2108.4 mImI zy −− ×=×=
We now have
MMMMMM o
z
o
y866.030cos5.030sin −=−===
2.5
tan tan( 30 ) tan( 30 )
4.8
oo
z
y
I
I
φ =−= −
,
o
7.16−= φ
Point A is farthest from N.A. Thus,
yA
zA
Azy
M
z
My
I
I
σ =− +
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Continued on next slide

666
0.866 (0.05) 0.5 (0.04)
80 10 2.5 10 4.8 10
MM
−−
×= +
××
or
MM 17.432.171080 3+=×
from which
mkNM ⋅= 72.3
SOLUTION (*7.123)
See solution of Prob.7.117:
64
'1
14.29 10 ' 22.2o
yp
II m θ
−
== × =
64
'2
1.25 10 ' 39.8 ' 81.1
zAA
I
Imymmz
−
== × =− =− mm
We now have
'
'
1.25
tan ' tan( 30 ) tan( 30 )
14.29
oo
z
y
I
I
φ =−= − o
9.2' −= φ ,
MMM oo
y79.0)302.22sin(
'−=+−=
MMM oo
z613.0)302.22cos( =+=
Point A is farthest from N.A. Thus,
'
'
''
'
'yA
zA
Azy
M
z
My
II
σ =− +
666
0.613 ( 0.0398) 0.79 ( 0.0811)
80 10 1.25 10 14.29 10
MM
−−
−
−−
×=− +
××
or
3
80 10 19.518 4.423
M
M×= +
Solving,
mkNM ⋅= 34.3
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.