________________________________________________________________________
PROBLEM (*11.27) The truss ABC shown in Fig. P11.27 supports a load P = 2 kN.
Determine the factor of safety ns with respect to buckling of the members.
Assumption: Each member is a circular steel pipe having E = 210 GPa.
*SOLUTION
110, 1.414
2
xAB AB
FF F=−= =
∑kN
1(1.414) 1.732 0
2
yBC
FF
=
+− =
∑
0.732
BC
FkN
=
Member AB
44 34
(20) 7.854 10
64 64
AB d
I
mm
ππ
== = ×
2.828
AB
Lm
=
229 9
22
(210 10 )(7.854 10 )
( ) 2.035
(2.828)
AB
AB cr AB
EI
FkN
L
ππ
−
××
== =
() 2.035 1.44
1.414
AB cr
sAB
F
n
F
===
Member BC
44 34
(15) 2.485 10
64 64
BC d
I
mm
ππ
== = × 2
BC
Lm
=
229 9
22
(210 10 )(2.485 10 )
( ) 1.288
(2)
BC
BC cr BC
EI
Fk
L
ππ
−
××
== =
N
() 1.288 1.76
0.732
BC cr
sBC
F
n
F
===
The smallest governs: 1.44
s
n
=
FBC
FAB
1 kN
1
732 kN
B
1.
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