sm11 13

________________________________________________________________________
PROBLEM (11.13) A 60-mm outer-diameter and 50-mm inner-diameter steel pipe (E = 20
GPa) 2 m long acts as a spreader in the assembly of Fig. P11.13. Determine for a factor of safety
ns = 2, the value of F that will cause buckling of the pipe.
SOLUTION
44 4 4
()(3025)
44
Icb
π
π
=−= − 44 32.94 10 mm=×
22 9 6
81.28 kN
22
(200 10 )(0.3294 10 )
2(2)
cr
ss
PEI
nnL
ππ
−
××
==
=
2(81.28) 162.6Fk== N
2
2
F
1
Pcr
2 1
1 Pcr
2
1
cr
F
P
=
2 F
Justification of the formula used:
22 2 2
( ) (30 25 ) 863.94Acb mm ππ =−= −=
329,400 19.5
863.94
I
rm m
A
== =
and 2000 102.6
19.5
L
r== O.K.
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.