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PROBLEM (12.50) A truss ABC of three steel members supports a vertical load P at C as
shown in Fig. P12.50. The cross-sectional areas of members are AAB , ABC , and AAC . What is the
vertical displacement δ v of joint C? Apply Castigliano’s theorem.
Given:
AAB = 0.002 m,
ABC = 0.003 m,
AAC = 0.0026 m,
P = 25 kN,
E = 210 GPa
SOLUTION
F ∂F
F ∂F
F ∂F
∂U 1
= ∑ ( AB AB + BC BC + AC AC )
∂P E
AAB ∂P
ABC ∂P
AAC ∂P
Joint C
δv =
P
5.4
2.25
FAC
(1)
We have
C
LAC = 5.42 + 2.252 = 5.85 m
2.25
LBC = 32 + 2.252 = 3.75 m
LAB = 2.4 m
3
FBC
5.4
3
FAC +
FBC = 0
5.85
3.75
2.25
2.25
∑ Fy = 0 : − 5.85 FAC + 3.75 FBC = 0
Solving,
FAC = 3.25 P
FAC = 3.75 P
∑F
x
= 0: −
Joint B
FBC
2.25
B
3
FAB
∑F
x
3
FBC = 0
3.75
= 3P
= 0 : FAB +
FAB
RB
Equation (1) is therefore
P
3
3.75
3.25
P
[( )(3) + (
)(3.75) + (
)(3.25)] = 13.25(103 )
δv =
−3
3
2.6
E (10 ) 2
E
The value of displacement at C is
25(103 )
δ v = 13.25(103 )
= 1.577 mm ↓
210(109 )
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