sm11 59

________________________________________________________________________
PROBLEM (11.59) Determine the allowable axial load P for a W 250 x 80 steel column
(see Table B.8) for each of three effective lengths:
(a) Le = 2 m.
(b) Le = 5 m.
(c) Le = 10 m.
Given: E = 210 GPa, σy = 250 MPa.
SOLUTION
WF250x80:
32 min
10.2 10 65
y
Ammrrmm=× ==
229
3
2 2 (210 10 ) 128.8
250 10
cyp
E
C
ππ
σ
×
== =
×
(a) 2
e
Lm=
2000 65 30.77
ey
Lr==
Since
yc
<Lr C, use Eq. (11.24):
3
5 3 30.77 1 30.77
( ) 1.75
3 8 128.8 8 128.8
s
n=+ − =
62
2
250 10 (30.77)
[1 ] 138.8
1.75 2(128.8)
all
M
Pa σ
×
=−=
Thus,
63
138.8 10 (10.2 10 )
all
P− =× × 1416 kN
=
(b) 5
e
Lm=
5000 65 76.92
ey
Lr==
Since
yc
C<Lr , use Eq. (11.24):
3
5 3 76.96 1 76.96
( ) 1.87
3 8 128.8 8 128.8
s
n=+ − =
62
2
250 10 (76.96)
[1 ] 109.8
1.87 2(128.8)
all
M
Pa σ
×
=−=
Therefore,
63
109.8 10 (10.2 10 ) 1120
all
Pk
−
=× ×= N
Continued on next slide
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(c) 10
e
Lm=
3
10 10 65 153.85
ey c
Lr C=× = >
Use Eq. (11.25):
29
2
(250 10 ) 45.61
1.92(153.85)
all
M
Pa
π
σ
×
==
Thus,
63
45.61 10 (10.2 10 )
all
P− =× × 465.2 kN
=
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.