sm10 20 21

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.
________________________________________________________________________
PROBLEMS (10.20 and 10.21) Redo Probs. 10.16 and 10.17, using the multiple-integration
method.
SOLUTION (10.20)
2
00 0 01
'''' , ''' 2
xx
EIv w w EIv w x w C
LL
=− + =− + +
3
2
0012
1
'' 26
x
EIv w x w C x C
L
=− + + +
4
32
00123
11
'6242
x
EIv w x w C x C x C
L
=− + + + + (1)
5
432
001234
111
24 120 6 2
x
EIv w x w C x C x C x C
L
=− + + + + + (2)
Boundary conditions:
00
1
'''(0) :
33
wL wL
vV C=− = =
24
''(0) 0: 0 (0) 0: 0vCvC== ==
444 3
000 0
33
() 0: 0,
24 120 18 45
wL wL wL wL
vL CL C=−+++= =−
Then, Eqs.(1) and (2) yield the results given in solution of Prob.10.16.
Continued on next slide
x
y
A
wo
B
0
6
wL
0
3
wL
000
()
w
x
Lx w w
LL
−= −
x
L
-
x

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 (10.21)
Segment AC
1 1 1 112
'''' 0 ''' ''EIv EIv C 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)
Segment BC
20 205
'''' '''EIv w EIv w x C=− =− +
2
2056
1
'' 2 EIv w x C x C=− + +
32
20567
11
'62 EIv w x C x C x C=− + + +
432
205678
111
24 6 2
EIv w x C x C x C x C=− + + + + (3)
Boundary and continuity conditions:
14
(0) 0: 0EIv C==
11
11
'''(0) :
88
EIv V wL C wL=− = =
12
''(0) 0 : 0EIv C==
2055
335
'''( ) : ,
888
EIv L wL w L C wL C wL=− − + = =
2
26
1
''( ) 0 : 8 EIv L C wL==−
Continued on next slide
L
/2
wL/8
A
C
B
L
/23wL/8
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.
and 33
3
12 3 773
1
'( ) '( ) : ,
2 2 64 192 48
LLwL wL
vv C wLCCC=+=−+=+ (4)
44 4
12 88
() (): ,
2 2 384 192 384
L L wL wL wL
vv CC==+=−
3
27
() 0: 384
wL
vL C==
Equation (4): 3
37
128
wL
C=−
Then, Eqs.(1) through (3) yield the results given in the solution of Prob. 10.17.