sm11 54

________________________________________________________________________
PROBLEM (11.54) Rework Prob. 11.53, assuming that the load acts at the middle of side
AC (that is, e = 100 mm).
SOLUTION
24
e
LLm
=
=
200 100 160 60A
=
×−×
=× 32
10.4 10 mm
33
1(100 200 60 160 )
12
I=×−×
=×
64
46.19 10 mm
ec100mm
=
=
66.6rI A mm==
Therefore,
22
100 100 2.25 30.03
(66.6) 2e
L
ec
rr
×
== =
Apply Eq.(11.20b) with e
LL
=
:
33
max 39
3
×
5.01
15 10 150 10
1 2.25sec 30.03
10.4 10 70 10 (10.4 10 )
σ
−−
⎡⎤
⎛⎞
××
⎢⎥
=+
⎜⎟
⎜⎟
××
⎢⎥
⎝⎠
⎣⎦
M
Pa=
A
P
200 mm
D
100 mm
C
P
20 mm
B
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.