sm12 78 80

________________________________________________________________________
PROBLEMS (12.78 through *12.80) A beam is supported and loaded as shown in Figs. P12.78
through P12.80. Use Castigliano's theorem to determine the reactions.
SOLUTION (12.78)
Consider M A as redundant.
Due to symmetry: RA = RB =
x
MA
MB
B
L
RA
1
EI
1
M = RA x − M A − wx 2
2
∂M
= −1
∂M A
w
A
θA = 0 =
1
wL
2
B
RB
∫
L
0
( RA x − M A −
wx 2
)(−1)dx
2
or
x2
wx3
0 = − RA + M A x +
2
6
L
=−
0
wL3
wL3
+ MA +
4
6
Solving
wL2
12
wL2
Statics: M B =
12
MA =
SOLUTION (12.79)
Consider RA and M A as redundants.
y
MB
M0
MA
A
L/2
L/2
RA
Segment AC:
Segment CB:
x
B
RB
M 1 = RA x − M A
M 2 = − RA x − M A − M 0
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.
Thus,
EIθ A = ∫ M i
∂M i
=0
∂M A
or
0=∫
L2
0
L
( RA x − M A )(−1)dx + ∫ ( RA x − M A − M 0 )(−1)dx
L2
L2
L
R x2
R x2
= − A + M Ax + − A + M Ax + M0x
2
2
0
L2
After simplifying
2 M A + M 0 − RA L = 0
Similarly,
(1)
∂M i
=0
∂RA
EIv A = ∫ M i
or
0=∫
L2
0
L
( RA x − M A ) xdx + ∫ ( RA x − M A − M 0 ) xdx
L2
R x3 M x
= A + A
3
2
2 L2
0
R x3 M x 2 M x 2
+ A + A − 0
3
2
2
L
L2
Simplifying,
3
9
RA L − M A − M 0 = 0
2
8
Solving Eqs. (1) and (2):
3M 0
1
M A = M0
RA =
↑
4
2L
Statics:
RA = − RB
MA = MB
(2)
SOLUTION (*12.80)
Consider RA as redundant.
A
RA
P
P
x
B
2EI
a
D
a
x’
EI
a
E
RB =2RA-P
a
C
RC =P-RA
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.
Segment AD:
M 1 = − RA x
∂M 1 ∂RA = − x
Segment BD:
M 2 = − RA x + P ( x − a )
∂M 2 ∂RA = − x
Segment BE:
M 3 = ( P − RA ) x '− P( x '− a ) = − RA x '+ Pa,
∂M 3 ∂RA = − x '
Segment EC:
M 4 = ( P − RA ) x ',
∂M 4 ∂RA = − x '
Therefore
RA a 2
RA a 3
1 a ∂M 1
=
=
dx
x
dx
2 EI ∫0 ∂RA
2 EI ∫0
6 EI
∂M 2
1 2a
1 2a
M2
dx =
[− RA x + P( x − a)](− x)dx
∫
∂RA
2 EI a
2 EI ∫a
7 RA a 3 5Pa 3
−
6 EI 12 EI
∂M 3
1 2a
1 2a
=
M
dx
( RA x '2 − Pax ')dx '
3
∫
∫
a
a
∂RA
EI
EI
=
7 RA a 3 3Pa 3
=
−
3EI
2 EI
a
a
∂M 4
1
1
M4
dx =
(− Pax '2 + RA x '2 )dx '
∫
∫
0
0
EI
∂RA
EI
RA a 3 Pa 3
−
3EI 3EI
=
Since
vA = ∑ ∫ M i
∂M i
=0
∂RA
We have
1 7 7 1
5 3 1
RA ( + + + ) − P ( + + ) = 0
6 6 3 3
12 2 3
Solving
RA =
9
P↓
16
Statics:
1
RB = P ↑
8
RC =
7
P↑
16
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.