sm12 81 83

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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 RB as redundant.
P
x
A
MA
a
B
2a
RA
C
RB
Segment BC:
M 1 = − Px
∂M 1 ∂RB = 0
Segment AB:
M 2 = − Px + RB ( x − a )
∂M 2 ∂RB = x − a
Thus,
a
3a
EIvB = 0 = ∫ (− Px)(0)dx + ∫ [− Px + RB ( x − a)]( x − a)dx
0
3a
a
= ∫ [− Px + RB x − 2 RB ax + Pax + RB a 2 ]dx
2
2
a
3a
Px3 RB x 3
Pax 2
=−
+
− RB ax 2 +
+ RB xa 2
3
3
2
a
=−
14
3
P + RB
3
8
from which
RB =
7
P↑
4
Statics:
MA =
1
Pa
2
RA =
3
P↓
4
Continued on next slide
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SOLUTION (12.82)
Consider RA and M A as redundants.
y
MA
A
a
P
b
C L
x
B
RA
Segment AC:
M 1 = RA x − M A
∂M 1
=x
∂RA
∂M 1
= −1
∂M A
Segment BC:
M 2 = RA x − M A − P ( x − a )
∂M 2 ∂RA = x
∂M 2 ∂M A = −1
Thus,
a
L
EIv A = 0 = ∫ ( RA x − M A ) xdx + ∫ [ RA x − M A − P ( x − a )]xdx
a
0
a
a
L − a3
L2 − a 2
L3 − a 3
= RA − M A + RA
− MA
−P
3
2
3
2
3
L2 − a 2
+ Pa
=0
2
3
2
3
Simplifying,
1
1
P
RA L3 − M A L2 − (3a + 2b)b 2 = 0
3
2
6
(1)
Similarly
a
L
EIθ A = 0 = ∫ ( RA x − M A )(−1)dx + ∫ [( RA x − M A − P( x − a)](−1)dx
a
0
a
L −a
L2 − a 2
= − RA + M A a − RA
+ M A ( L − a) + P
2
2
2
2
2
2
Continued on next slide
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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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− Pa( L − a) = 0
This reduces to
1
1
RA L2 − M A L − Pb 2 = 0
2
2
(2)
Solving Eqs.(1) and (2):
Pab 2
Pb 2
MA = 2
RA = 3 (3a + b) ↑
L
L
Statics:
Pa 2
Pa 2b
RB = 3 (a + 3b) ↑
MB = 2
L
L
SOLUTION (12.83)
Consider RA as redundant.
y
wo
A
L
RA
B
wx 3
6L
∂M ∂RA = x
M = RA x −
x
We have
EIv A = 0 = ∫
L
0
w0 x 4
RA L3 w0 L5
( RA x −
)dx =
−
6L
3
30 L
2
Solving
RA =
w0 L
↑
10
RB =
2
w0 L ↑
3
Statics:
MB =
w0 L2
15
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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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