sm12 65 66

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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.
1
BD
D
F
R
∂
=
∂
C
0:
y
F
=
∑
1.25 1.25
BC D
FPR
=
−
0: 0.75 0.75
x
AB D
FF P==−+
∑R
where
1.25 0.75
BC D AB D
FR FR∂∂ =− ∂∂=
Thus,
0
i
Dii
D
F
AE FL R
δ
∂
==
∂
∑
=− (1.25 1.25 )(7.5)( 1.25)
D
PR −
+ + =
+ =
+− ( 0.75 0.75 )(4.5)(0.75) (3)1 0
DD
PR R
=− 14.25 17.25 0
D
PR
or
0.826
D
R
P=
Then
0.826 0.131 0.22
BD AB BC
FPF PFP
=
=− =
Continued on next slide
A
B
D
3 ft 4
3
P
F
BC
FAB
B
3
4
P
F
B
DR =
D
4.5 ft
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SOLUTION (*12.66)
4
3
3
4
F
A
B
FBC
R
D =FAB=R
C
A
B
0.8L
0.5L
vD=
=0 0
D
P
0.6L
R
Figure (a)
B
0.5L
Figure (b)
P
5
Structure is statically indeterminate to the 1st degree. We have
0.
DUR
δ
=∂ ∂ = That is
1[]
BC
AB BD
DABABBCBC BDBD
F
FF
FL FL FL
AE R R R
δ
0
∂
∂
∂
=++
∂∂∂
= (1)
Joint B.
Equilibrium results in:
0.8 0.8 0.6 0.6
AB BC BD
FRPFPRFR
=
−=−= (2)
Equation (1) is thus
(0.8 0.8 )(0.8 )(0.8) (0.6 0.6 )(0.6 )( 0.6)
DRPL PRL
δ
=− +− −
(0.5)(1)RL 0
+
=
Solving
0.593 BD
R
PF==
Equations (2) give then
0.32 0.244
AB BC
FPFP
=
−=
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.