sm10 22 23

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________________________________________________________________________
PROBLEMS (10.22 and 10.23) Redo Probs. 10.18 and 10.19, using the multiple-integration
method.
SOLUTION (10.22)
Segment AC
0
111
'''' 0 '''
M
EIv EIv C L ===−
0
12
'' M EIv x C
L
=− +
2
0
13
'2
M
EIv x C
L
=− + (1)
3
0
134
6
M
EIv x C x C
L
=− + + (2)
Segment BC
0
225
'''' 0 '''
M
EIv EIv C L ===−
0
26260
'' ''( ) 0,
M
EIv x C EIv L C M
L
=− + = =
2
0
207
'2
M
EIv x M x C
L
=− + +
32
0
2078
1
62
M
EIv x M x C x C
L
=− + + + (3)
Boundary and continuity conditions:
14
(0) 0: 0vC==
2
0
278
() 0: 0
3
ML
vL CLC=++=
12 307
'( ) '( ) :va va C MaC==+
12 3 078
1
() (): 2 va va Ca Ma Ca C==++
Continued on next slide
0
a
R
A=M0/L
A
C
B
R
B= M0/L
y
x
M
0
L b

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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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Copyright Act without the permission of the copyright owner is unlawful.
Solving,
2
00
30
32
M
LMa
CMa
L
=− + −
2
00
732
M
LMa
CL =− − 2
0
82
M
a
C=
Then, Eqs.(1) through (3) yield the results given in solution of Prob.10.18.
SOLUTION (10.23)
Segment AC
111
''' ''
22
PPx
EIv EIv C==+
2
12
1
'4 EIv Px C=+ (1)
3
123
1
12
EIv Px C x C=++ (2)
Segment BC
224
1
''' ''
22
P
EIv EIv Px C=− =− +
242
1
''(2 ) 0 : , '' 2 EIv a C Pa EIv Px Pa== =−+
2
25
1
'4 EIv Px Pax C=− + +
32
256
11
12 2
EIv Px Pax C x C=− + + + (3)
Boundary & continuity conditions:
13
(0) :
22
PPEI
vC
kk
=− =−
2
6
25
2
(2 ) 0: 23
C
va C Pa
a
==−−
2
12 2 5
1
'( ) '( ) : 2 va va C Pa C==+
3
12 6
1
() (): 62
PEI
va va C Pa k ==−
Continued on next slide
0
a
a
A
C
B
P
/2
y
x
P
P
/2
k

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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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Copyright Act without the permission of the copyright owner is unlawful.
Solving,
2
53
44
PEI
CPa
ak
=− + 2
21
44
PEI
CPa
ak
=− +
Then, Eqs.(1) through (3) yield the results given in solution of Prob.10.19.