sm12 84 86

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PROBLEMS (12.84 through *12.86) A beam is supported and loaded as shown in Figs. P12.84
through P12.86. Employ Castigliano's theorem to determine the reactions.
SOLUTION (12.84)
Consider RB as redundant.
x
w
MA
B
L
A
1
M = RB x − wx 2
2
∂M ∂RB = x
RB
RA
So,
L
EIvB = 0 = ∫ ( RB x 2 −
0
R L3 wL4
wx 3
)dx = B −
2
3
8
or
3
RB = wL ↑
8
Statics:
5
RA = wL ↑
8
1
M A = wL2
8
SOLUTION (12.85)
Consider RB as redundant.
M0
MA
A
B
L
x
RA
M = − RB x + M 0
∂M ∂RB = − x
RB
Therefore
L
1
1
EIvB = 0 = ∫ ( RB x 2 − M 0 x)dx = RB L3 − M 0 L2
0
3
2
Solving
RB =
3 M0
↓
2 L
RA =
3 M0
↑
2 L
Statics:
MA =
1
M0
2
Continued on next slide
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SOLUTION (*12.86)
Consider RC as redundant.
w
x’
A
B
L
x
RA =
L/2
RB
C
RC
wL 1
− RC
2
2
Segment BC:
M 1 = RC x '
∂M 1 ∂RC = x '
Segment AB:
x
wL RC
wx 2
∂M 2 ∂RC = −
M2 = (
− )x −
2
2
2
2
Thus,
wLx 2 RC x 2 wx 3
)dx
+
+
0
0
4
4
4
R L3 wL4 RC L3 wL4
= C −
+
+
8
12
12
16
EIvC = 0 = ∫
L2
L
RC x '2 dx ' + ∫ (−
from which
RC =
1
wL ↓
6
RA =
5
wL ↑
12
Statics:
RB =
3
wL ↑
4
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
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