sm10 13 14

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 (10.13 and 10.14) Solve Probs. 10.10 and 10.11, using the multiple-integration
approach.
SOLUTION (10.13)
Segment AC
1'''' 0EIv =
11
'''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)
Segment BC
2'''' 0EIv =
25
'''EIv C=
256
''EIv C x C=+
2
2567
1
'2 EIv C x C x C=++ (3)
32
25 678
11
62
EIv C x C x C x C=+++ (4)
Boundary conditions:
11
'''(0) 0 : 0vC==
12
''(0) :vPLCPL==
25
'''( ) :
2
L
vPCP=− =−
26
3
''( ) :
22
L
vPLCPL==
22 2
277
13
'( ) 0 : 0,
22
v L PL PL C C PL=− + += =−
333 3
288
35
() 0: 0,
64 12
PL
v L PL PL C C PL=−+ −+= =
Continued on next slide
P
L/
2 C
A
P
L
B
y
L/
2
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.
Continuity conditions:
222
12 3 3
37
'( ) '( ) : ,
222 8 8
LLPL
vv CPLCPL= + =− =−
33 3 3
12 4 4
719
() (): ,
22816 12 48
L L PL PL PL
vv C C PL=−+==−
Then, Eqs.(1) through (4) yield the results given in solution of Prob. 10.10.
SOLUTION (10.14)
Segment AC
111
'''' '''EIv w EIv wx C=− =− +
2
112
1
'' 2 EIv wx C x C=− + +
32
1123
11
'62 EIv wx C x C x C=− + + + (1)
43 2
11234
111
24 6 2
EIv wx C x C x C x C=− + + + + (2)
Segment BC
225
'''' 0 '''EIv EIv C==
256
''EIv C x C=+
2
2567
1
'2 EIv C x C x C=++ (3)
32
25 678
11
62
EIv C x C x C x C=+++ (4)
Continued on next slide
2a
R
A=2wa
A
C
B
a
M
A
=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.
Boundary conditions:
11
'''(0) 2 : 2EIv V aw C aw=− = =
22
12
''(0) 2 : 2EIv aw C aw=− =−
13
'(0) 0 : 0EIv C==
14
(0) 0: 0EIv C==
Continuity conditions:
12 5
'''(2 ) '''(2 ) : 0vavaC==
12 6
''(2 ) ''(2 ) : 0vavaC==
3
12 7
4
'(2 ) '(2 ) : 3 vava waC=−=
44 4
12 88
82
(2 ) (2 ): 2 ,
33
v a v a wa wa C C wa=−=−+ =
Then, Eqs.(1) through (4) yield the results given in solution of Prob. 10.11