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PROBLEMS (*12.65 and *12.66) A load P is supported at joint B of a structure consisting of
three bars of equal axial rigidity AE (Fig. P12.65 and P12.66). Use Castigliano's theorem to determine the
force in each bar.
SOLUTION (*12.65)
Consider RD as redundant.
C
D
4
A
∑F
x
∂FBD
=1
∂RD
FBD =RD
FBC
3
3 ft
4.5 ft
B
P
4
∑F
3
FAB
y
B
= 0:
FBC = 1.25 P − 1.25RD
P
= 0 : FAB = −0.75P + 0.75 RD
where
∂FBC ∂RD = −1.25
Thus,
∂FAB ∂RD = 0.75
∂Fi
=0
∂RD
= (1.25 P − 1.25 RD )(7.5)(−1.25)
+ (−0.75 P + 0.75 RD )(4.5)(0.75) + RD (3)1 = 0
= −14.25P + 17.25 RD = 0
AEδ D = ∑ Fi Li
or
RD = 0.826 P
Then
FBD = 0.826 P
FAB = −0.131P
FBC = 0.22 P
Continued on next slide
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SOLUTION (*12.66)
RD =FAB=R
D
C
0.5L
0.6L
0.8L
vD=0
R
FBC
B
3
P
4
5
4
3
0.5L
B
P
A
Figure (a)
FAB
Figure (b)
Structure is statically indeterminate to the 1st degree. We have
δ D = ∂U ∂R = 0. That is
δD =
∂F
∂F
∂F
1
[ FAB LAB AB + FBC LBC BC + FBD LBD BD ] = 0
∂R
∂R
∂R
AE
Joint B.
Equilibrium results in:
FAB = 0.8R − 0.8P
FBC = 0.6 P − 0.6 R
FBD = R
(1)
(2)
Equation (1) is thus
δ D = (0.8R − 0.8P )(0.8L)(0.8) + (0.6 P − 0.6 R)(0.6 L)(−0.6)
+ R(0.5L)(1) = 0
Solving
R = 0.593P = FBD
Equations (2) give then
FAB = −0.32 P
FBC = 0.244 P
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
Copyright Act without the permission of the copyright owner is unlawful.
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