sm10 5 6

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PROBLEMS (10.5 and 10.6) Solve Probs. 10.1 and 10.2, using the multiple-integration method.
SOLUTION (10.5)
''''EIv w=−
1
'''EIv wx C=− +
212
1
'' 2 EIv wx C x C=− + +
32
13
11
'62 EIv wx C x C=− + + (1)
43 2
1234
111
24 6 2
EIv wx C x C x C x C=− + + + + (2)
Boundary conditions:
1
'''(0) 0, 0
A
EIv V C=− = =
3
3
''(0) 0, '( ) 0: 6
A
EIv M EIv L C wL== = =
4
3
() 0,: 8EIv L C wL==−
Then, Eqs.(1) & (2) give the results of the solution of Prob. 10.1.
SOLUTION (10.6)
Segment AC
111
2 ''' 2 ''EIv V P EIv Px C=− = = +
2
112
1
2'
2
EIv Px C x C=++ (1)
32
1123
11
262 EIv Px C x C x C=+ ++ (2)
Segment BC
224
''' ''EIv P EIv Px C==+ Continued on next slide
L
A
B
y
x
w
A
C
B
2E
I
x
L/
2
L/
2
R
A
=P
y
P
EI
M
A
=PL

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
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Copyright Act without the permission of the copyright owner is unlawful.
2
245
1
'2 EIv Px C x C=++ (3)
32
2456
11
62
EIv Px C x C x C=+ ++ (4)
Boundary conditions:
11 12
2 ''(0) : , 2 '(0) 0 : 0EIv PL C PL EIv C=− =− = =
13
(0) 0: 0EIv C==
Conditions of continuity:
12 44
2 ''( ) ''( ) : ,
2222
L L PL PL
EIv EIv C C PL=−=+=−
22 2
12 55
33 3
'( ) '( ) : ,
22168 16
L L PL PL
vv CCPL=−=−+ =
33 3
12 66
5
() (): ,
2 2 96 96 24
L L PL PL PL
vv CC=−=−+ =−
Then, Eqs. (1) through (4) yield the results given in solution of Prob. 10.2.