sm11 62

________________________________________________________________________
PROBLEM (11.62) Rework Prob. 11.59 for a steel pipe column of outer diameter 160 mm
and inner diameter 140 mm.
SOLUTION
22
(160 140 ) 4712.4
4
Am
2
m
π
=−=
From Example A.4:
22 2 2
11
160 140 53.2
44
rDd m=+= +=m
From solution of Prob. 11.59: 128.8
c
C
=
(a) 2
e
Lm=
37.59
ec
Lr C=<
Equation (11.24):
3
5 3 37.59 1 37.59
( ) 1.77
3 8 128.8 8 128.8
s
n=+ − =
62
250 10 1 37.59
[1 ( ) ] 135.2
1.77 2 128.8
all
M
Pa σ
×
=−=
Thus,
66
135.2 10 (4712.4 10 ) 637
all
Pk
−
=× ×=N
(b) 5
e
Lm=
3
510 93.98
53.2
ec
Lr C
×
==<
Use Eq. (11.24):
3
5 3 93.98 1 98.98
( ) 1.89
3 8 128.8 8 128.8
s
n=+ − =
62
250 10 1 93.98
[1 ( ) ] 97.1
1.89 2 128.8
all
M
Pa σ
×
=−=
Hence,
66
97.1 10 (4712.4 10 ) 457
all
Pk
−
=× × = N
Continued on next slide
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(c) 10
e
Lm=
3
10 10 187.97
53.2
ec
Lr C
×
==>
Use Eq. (11.25):
29
2
(210 10 ) 30.55
1.92(187.97)
all
M
Pa
π
σ
×
==
So,
66
30.55 10 (4712.4 10 ) 144
all
Pk
−
=× ×=N
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.