sm11 39

________________________________________________________________________
PROBLEM (11.39) A structural steel column is fixed at its base and pinned at its top. The
column has a rectangular cross section of width b = 40 mm and depth h = 12 mm. Calculate the
buckling
load Pcr for the following lengths:
(a) L =200 mm.
(b) L = 600 mm.
SOLUTION
From Table B.4: 200 250
y
EGPa MPa
σ
=
=
The properties of area are
,
2
(40)(10) 400Abh mm== = 33
11
(40)(12) 5760
12 12 4
I
bh mm== =
5760 3.795
400
I
rm
,
m
A
== = 3
200 10
( ) 88.9
250
ecy
LE
r
ππ
σ
×
== =
mm== =
(a) From Fig. 11.3c, LL . Hence 0.7 0.7(200) 140
e
140 36.9
3.795
e
L
r==
Since
36.9 88.9<, John Formula should be used. Thus:
2
2
()
[1 ]
4
ye
cr y
Lr
PA E
σ
σπ
=−
2
23
250(36.9)
(400)(250)[1 ] 95.71
420010 kN
π
=− =
××
(b) Now we have
0.7(600) 110.7 88.9
3.795
e
L
r==>
Euler formula should used. So
22 9 12
64.5 kN
22
2
(200 10 )(5760 10 )
(20)
cr EI
PL
ππ
−
××
==
=
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