sm12 16 19

________________________________________________________________________
PROBLEMS (12.16 through 12.19) A beam is supported and loaded as shown in Figs. P12.16
through P12.19. What is the strain energy of the beam caused by bending?
SOLUTION (12.16)
0
3
AC M
M
x
a
=−
0
3
BC M
M
u
a
=
AC BC
UU U=+
22
00
2
00
() ()
33
22
aa
MMu
x
dx du
aa
EI EI
−
=+
∫∫
22
33
00
28
()
2(9)3 3 6
M
Ma
aa
EI a EI
=+=
SOLUTION (12.17)
Part AC: 1
2
M
Px=
2
12
02
AC M
Ud
EI
=∫x
2
22
3
3
x
22
00
88
L
L
PP
xdx
EI EI
∫
== 23
192
PL
EI
=
U=
By summery U and total energy is
AC BC
23
96
AB AC BC PL
UU UEI =+=
Continued on next slide
P
/2
B
A
P
C
L/
2
P
/2
L/
2
M0
0
3
M
a
a
u
B
C
2a
x
A
0
3
M
a
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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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________________________________________________________________________
SOLUTION (12.18)
0BC
M
M
=
0
AB M
M
x
L
=
AB BC
UU U=+
22
00
00
1
2
La
M
x
dF M dx
EI L
⎡⎤
⎛⎞
=+
⎢⎥
⎜⎟
⎝⎠
⎢⎥
⎣⎦
∫∫
22
3
00
2
1
()(
232
MM
LL
aa
EI L EI
=+=
)
3
+
=
SOLUTION (12.19)
The free-body diagrams
of beam AB and the portion
AO is shown in Fig. (a).
The condition
M
∑ gives
00
23
0()
6
w
M
Lx x
L
=−
Therefore
2
02
L
M
dx
U
EI
=∫
232
0
0
1{[ ( )] }
26
LwLx x dx
EI L
=−
∫
242 6 24
020[( 2 ) ]
72
L
wLx x Lx dx
EIL
=+−
∫
or
25
0
945
wL
U
EI
=
w0
0
3
wL
B
A
0
6
wL 2L/3
L/
3
0
x
w
L
M
x
/3
A
O
x
0
1
2()
xw
L
x
0
6
wL
Figure (a)
0
2
wL
M0
0
M
L
a
B
C
L
x
A
u
M
0
L
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.