sm12 84 86

________________________________________________________________________
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
B
R
as redundant.
2
1
2
B
M
Rx wx=−
B
M
Rx
∂
∂=
So,
334
2
0
0( )
23
LB
BB RL
wx wL
EIv R x dx== − = −
∫8
or
3
8
B
R
wL=↑
Statics:
2
51
88
AA
R
wL M wL=↑ =
SOLUTION (12.85)
Consider
B
R
as redundant.
0B
M
Rx M
=
−+
B
M
Rx
∂
∂=−
Therefore
23
00
0
11
0( )32
L
BB B
EIv R x M x dx R L M L== − = −
∫2
Solving
0
3
2
BM
RL =↓
Statics:
00
31
22
AA
M
R
MM
L
=↑ =
Continued on next slide
A
B
x
w
M
RB
L
R
A
A
A
B
x
RB
L
M
R
A
M
A
0
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.

SOLUTION (*12.86)
Consider
C
R
as redundant.
C
L/
2
B
x’
A
L
x
w
RB
1
22
AC
wL
RR=−
R
C
Segment BC:
1'
C
M
Rx= 1'
C
M
Rx∂∂=
Segment AB:
2
2()
22 2
C
R
wL wx
Mx=− − 22
C
x
MR
∂
∂=−
Thus,
2
23
22
00
0''(
444
LL
C
CC Rx
wLx wx
EIv R x dx dx== + − + +
∫∫ )
33
44
8121216
CC
RL RL
wL wL
=−++
from which
1
6
C
R
wL=↓
Statics:
53
12 4
AB
R
wL R wL=↑ =↑
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.