sm11 79

________________________________________________________________________
PROBLEM (*11.79) An aluminum alloy 6061-T6 compression member AB is fixed at its
base and free at its top as shown in Fig. P11.79. Using the interaction formula with (σall )b = 140
MPa, Calculate the largest permissible eccentric load P.
*SOLUTION
We have (from Fig. 11.3a). 2
e
L=L
2
50(100) 5000Abh mm== =
The smallest moment of inertia is
33
11
(100)(50) 1.042 10
12 12 64
I
bh mm== =×
3
1.042 10 14.44
5
I
rA
×
== =
The largest slenderness ratio is thus
2(1800) 249.3
14.44
e
L
r==
Assume Eq.(11.29) applies (249.3 > 66).
39
6
23
350 10 350(10 )
( ) (10 ) 5.63
( ) (249.3)
all c e
M
Pa
Lr
σ
×
===
Also
3200
510
PP P
A−
==
×
33
(0.025)(0.05) 300
12 (0.05)(0.1) 12
Mc Pec P P
Ibh
== =
Equation (11.34) becomes
66
200 300
1; 1
( ) ( ) 5.63 10 140 10
c
all c all b
MI
PA P P
σσ
+≤ +
××
≤
N
or
26.5Pk=
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.