sm11 40

________________________________________________________________________
PROBLEM (*11.40) Determine the critical load Pcr
a
for an aluminum alloy 6061-T6 pipe
with pinned ends. The column has outside diameter d0 = 75 mm, inside diameter di = 72 mm,
and length L = 800 mm.
*SOLUTION
70 260
y
EGPa MP
σ
== (from Table B.4)
22 2 2
( ) (75 72 ) 346.361
44
oi
Add m
2
m
π
π
=−= −=
44 4 4 3
( ) (75 72 ) 233.99(10 )
64 64
oi 2
I
dd mm
π
π
=−= −=
3
233.99(10 ) 26
346.36
I
rm
A
== = m
So,
800 30.8
26
L
r==
Equation (11.10):
3
70 10 51.5
260
y
LE
r
ππ
σ
×
== =
Since
30.8 51.5<, Johnson formula should be used.
2
2
()
[1 ]
4
y
cr y
Lr
PA E
σ
σπ
=−
2
23
260(30.8)
(346.361)(260)[1 ] 82
4 (70 10 ) kN
π
=− =
×
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