________________________________________________________________________
PROBLEM (11.10) Redo Prob. 11.9 for W = 30 kN and α = 40°.
SOLUTION
30
0: 46.67
sin40
yBC o
FF===
BC
FkN
∑
0: 46.67cos40o
xcrAB
FPF===
∑
35.75kN
=
W=30 kN
40o
B
232
64
229
35.75 10 (3)(2.5)
,0
(200 10 )
cr AB s
EI
PF I m
fL
π
π
−
×
== = = ×
×.3410
Therefore,
46
0.34 10 , 25.6
4
c
I
cm
π
−
==× = m
m
and
22
(25.6) 2066Am
π
==
Justification of the formula used:
12.8rI A mm==
2500 12.8 195Lr
=
= O.K.
AB
F
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