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________________________________________________________________________
PROBLEM (10.7 and 10.8) Redo Probs. 10.3 and 10.4, using the multiple-integration method.
SOLUTION (10.7)
2
001
'''' ''' 2
xx
EIv w EIv w C
LL
=− =− +
3
01 2
'' 6
x
EIv w C x C
L
=− + +
42
01 23
1
'24 2
x
EIv w C x C x C
L
=− + + + (1)
532
01 2 34
11
120 6 2
x
EIv w C x C x C x C
L
=− + + + + (2)
Boundary conditions:
1
'''(0) 0: 0EIv C==
3
230
1
''(0) 0: 0, '( ) 0 : 24
EIv C EIv L C w L== ==
44 4
004 40
11 1
() 0: 0,
120 24 30
EIvL wL wL C C wL
L
=− + += =
Then, Eqs.(1) and (2) yield the results given in solution of Prob. 10.3.
SOLUTION (10.8)
Continued on next slide
y
A
w0
(
x/L
)
w0
L
x
B
a
R
A=wa
A
C
B
a
MA=3
2wa2
y
x
w
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 AC
111
'''' 0 '''EIv EIv C==
112
''EIv C x C=+
2
1123
1
'2
EIv C x C x C=++ (1)
32
11 234
11
62
EIv C x C x C x C=+++ (2)
Boundary conditions:
11
'''(0) :
A
EIv V wa C wa=− = =
22
12
33
''(0) :
22
EIv wa C wa=− =−
13 14
'(0) 0: 0, (0) 0, 0EIv C EIv C== ==
Segment BC
225
'''' '''EIv w EIv wx C=− =− +
2
256
1
'' 2
EIv wx C x C=− + +
32
2567
11
'62
EIv wx C x C x C=− + + + (3)
432
25678
111
24 6 2
EIv wx C x C x C x C=− + + + + (4)
Continuity conditions:
12 1 55
'''( ) '''( ) : , 2va va C waC C wa==+=
22
22
12 66
''( ) ''( ) : 2 , 2
22
wa wa
v a v a wa C C wa=−=++ =
3
333 3
22 77
1
'( ) '( ) : 2 ,
66
wa
v a v a wa wa wa C C wa=−=++ =
12
444
44
88
() ():
71
,
12 24 3 6 24
va va
wa wa wa
wa C C wa
=
=+++ =
Then, Eqs.(1) through (4) yield the results given in solution of Prob.10.4.
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