sm12 53 55

Telechargé par Gingermon
________________________________________________________________________
PROBLEMS (12.53 through 12.55) For the beam and loading shown in Figs. P12.53 through
P12.55, employ Castigliano's theorem to determine:
(a) The deflection at point C.
(b) The slope at point C.
SOLUTION (12.53)
P
a
B
x’
A
C
L
x
C
Pa C
LL
+
Segment AB
1aC
M
Px x
LL
=−
Segment BC
2'
M
Px C=−
(a) For this case C=0:
00
11
( )( ) '( ') '
La
CPax ax
vdx
EI L L EI
=−+
∫∫
Pxxdx
23 3
200
'
33
La
Pa x P x
EIL EI
=+ 2()
3
Pa La
EI
=
+↓
(b) 12
,1
MM
x
CL C
∂∂
=− =−
∂∂
.
For C=0, we have :
00
11
( )( ) '( ') '
La
CPax x dx Px x dx
EI L L EI
θ
=−+
∫∫
32
200
'
32
La
Pa x P x
EIL EI
=+
(2 3 )
6
Pa La
EI
=+
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 (12.54)
x’
Q
x
C
0
33
MQ
aa
A
02
33
MQ
a
+
2a
B
M0
Segment AC
0
133
MQ
M
xx
a
=− +
Segment CB
0
2'2'
33
Mx Q
x
a
=− +
(a) We have
12
2' 0
33
MM
xx
Q
QQ
∂∂
==
∂∂ =
Thus,
200
00
'
11
'
33 3 3
aa
CMx Mx
xx
2'
xdx
EI a EI a
=− +
∫∫
vd
2
0
2
9Ma
EI
=↑
(b) In this case Q=0:
12
00
'
,
33
MM
x
x
M
aM
∂∂
=− =
∂∂a
So,
200
00
'
11
'
33 33
aa
CMx Mx
xx
dx dx
EI a a EI a a
θ
=+
∫∫
'
0
3
M
a
EI
=
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 (12.55)
Q
a
B
x’
A
C
L
x
0
MQa
LL
+
M0
0()
MQLa
LL
+
Segment AB
0
1MQa
M
xx
LL
=−
Segment BC
M
20 'MQx=−
(a) We have
12
'
MM
ax
x
QL Q
∂∂
=− =−
∂∂
, and 0Q
=
.
Hence,
00
00
11
() (')
La
CMx ax 'xMxdx
EI L L EI
=−+
∫∫
vd
2
0(2 3 )
6
Ma La
EI
=+
(b) Now Q=0:
12
00
,1
MM
x
ML M
∂∂
==
∂∂
.
Thus,
00
00
1(1) '
La
CMxxdx M dx
EI L L
θ
=+
∫∫
0(3
3
MLa
EI )=+
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.
1 / 3 100%

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