sm12 78 80

Telechargé par Gingermon
________________________________________________________________________
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
A
M
as redundant.
Due to symmetry: 1
2
AB
R
Rw== L
2
1
2
AA
M
Rx M wx=−
1
A
M
M
=
2
0
1
0( )(1
2
L
AAA
wx )
R
xM dx
EI
θ
== − −
or
233
0
3
2646
L
AA A
0
x
wx wL wL
RMx M=− + + = + +
Solving
2
12
AwL
M =
Statics: 2
12
BwL
M=
SOLUTION (12.79)
Consider
A
R
and A
M
as redundants.
Segment AC: 1AA
M
Rx M=−
Segment CB: 20AA
Rx M M=−
Continued on next slide
A
B
L
x
w
M
R
A
RB
A
MBB
A
B
L/2
x
R
A
RB
M
A
MB
L/2
y
M0
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,
0
i
Ai
A
M
EI M M
θ
==
or
2
0
02
0( )(1) ( )(1)
LL
AA AA
L
R
x M dx R x M M dx=−+
∫∫
2
22
0
02
22
LL
AA
AA
L
Rx Rx
M
xMxMx+ + + +
0
=−
After simplifying
(1)
0
2
AA
MMRL+− =
Similarly,
0
i
Ai
A
M
EIv M R
=
=
or
2
0
02
0( ) ( )
LL
AA AA
L
R
xMxdx RxM Mxdx=−+
∫∫
22
32 32
0
02
32 322
L
L
AA AA
L
Mx
Rx Mx Rx Mx
=+ ++ −
Simplifying,
0
390
28
AA
RL M M−−= (2)
Solving Eqs. (1) and (2):
0
03
1
42
AA
M
MM RL
=
=↑
Statics:
AB AB
R
RMM=− =
SOLUTION (*12.80)
Consider
A
R
as redundant.
R
A
aa
2E
I
P
P
EI
a
B
R
B =2RA-P
D
A
E
R
C =P-RA
C
a
x’
x
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:
11AA
M
Rx M R x=− =−
Segment BD:
22
()
AA
M
Rx Px a M R x=− + =−
Segment BE:
33
()'(') ', '
M
AA A
PRxPxa RxPa M R x=− = + ∂ ∂=
Segment EC:
44
()', '
M
AA
PRx M R x=− ∂ ∂=
Therefore
3
2
1
00
1
22
aa
AA
A6
M
RR
dx x dx
EI R EI EI
==
∫∫
a
22
2
2
11
[()](
22
aa
A
aa
A
M)
M
dx R x P x a x dx
EI R EI
=−+
∫∫
33
75
612
A
Ra Pa
EI EI
=−
22
2
3
3
11
(' ')
'
aa
A
aa
A
M
M
dx R x Pax dx
EI R EI
=−
∫∫
33
73
32
A
Ra Pa
EI EI
=−
22
4
4
00
11
(' ')
aa
A
A
M'
M
dx Pax R x dx
EI R EI
=−+
∫∫
33
33
A
Ra Pa
EI EI
=−
Since
0
i
Ai
A
M
vM
R
==
We have
1771 5 31
()(
6633 1223
A
RP+++ − ++ =
)0
Solving
9
16
A
R
P=↓
Statics:
17
81
BC
6
R
PRP=↑ = ↑
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%

sm12 78 80

Telechargé par Gingermon
La catégorie de ce document est-elle correcte?
Merci pour votre participation!

Faire une suggestion

Avez-vous trouvé des erreurs dans linterface ou les textes ? Ou savez-vous comment améliorer linterface utilisateur de StudyLib ? Nhésitez pas à envoyer vos suggestions. Cest très important pour nous !