sm12 81 83

________________________________________________________________________
PROBLEMS (12.81 through 12.83) A beam is supported and loaded as shown in Figs. P12.81
through P12.83. Apply Castigliano's theorem to determine the reactions.
SOLUTION (12.81)
Consider
B
R
as redundant.
2a
P
aC
x
A
B
R
A
RB
M
A
Segment BC:
1
M
Px=− 10
B
MR∂∂=
Segment AB:
22
()
BB
M
Px R x a M R x a=− + − ∂ ∂ = −
Thus,
3
0
0 ( )(0) [ ( )]( )
aa
BB
a
EIv Px dx PxRxa xadx== − + − + − −
∫∫
322 2
[2
a
BB B
aPx R x R ax Pax R a dx=−+ − + +
∫ ]
3
332
22
33 2
a
BBB
a
RxPx Pax
R
ax R xa=− + − + +
14 3
38
B
PR=− +
from which
7
4
B
R
P=↑
Statics:
13
24
AA
M
Pa R P=↓
=
Continued on next slide
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 (12.82)
Consider
A
R
and A
M
as redundants.
a
P
C b
A
B
L
x
R
A
M
A
y
Segment AC:
11
11
AA AA
MM
xM x
RM
∂
MR ∂
=− = =−
∂∂
Segment BC:
2() AA
M
Rx M Px a=−−−
22
1
AA
MRx MM∂∂= ∂∂=−
Thus,
0
0( ) [ ()]
aL
AAA AA
a
EIv RxMxdx RxM Pxaxdx== − + − − −
∫∫
3 2 33 22 33
32 3 2 3
AAA A
a a La La La
RM R M P
−
−−
=− + − −
22
0
2
La
Pa −
+
=
Simplifying,
32 2
11 (3 2 ) 0
32 6
AA
P
RL ML a bb−−+
= (1)
Similarly
0
0 ( )( 1) [( ( )]( 1)
aL
AAA AA
a
EI R x M dx R x M P x a dx θ == − − + − − − −
∫∫
222
()
22
AAA A
aLa L
RMaR MLaP
22
2
a
−
−
=− + − + − +
Continued on next slide
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.

()Pa L a 0
−
−=
This reduces to
2
11
0
22
AA
RL ML Pb−− =
2
(2)
Solving Eqs.(1) and (2):
22
23
(3 )
AA
Pab Pb
M
Ra
LL
b
=
=+↑
Statics:
22
32
(3)
BB
Pa Pa b
RabM
LL
=+↑ =
SOLUTION (12.83)
Consider
A
R
as redundant.
3
6
Awx
MR xL =−
A
M
Rx
∂
∂=
We have
45
3
200
0
0( )
63
LA
AA
wx wL
RL
EIv R x dx
LL
== − = −
∫30
Solving
0
10
AwL
R=↑
Statics:
2
0
0
2
31
BB
wL
RwL M=↑ =
5
y
wo
A
B
L
x
R
A
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.