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________________________________________________________________________
PROBLEM (*10.30 through 10.33) A beam supported and loaded as shown in Figs. P10.30
through P10.33, using a direct-integration method, determine:
(a) The equation of the elastic curve.
(b) The deflection at midspan.
(c) The magnitude and location of the maximum deflection.
(d) The slope at the right end.
SOLUTION (*10.30)
(a)
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:
2
''(0) 0: 0vC==
0
1
''( ) 0: 6
wL
vL C==
4
(0) 0: 0vC==
3
0
37
() 0: 360
wL
vL C==−
Equations (1) and (2) become
4224
0
' ( 15 30 7 )
360
w
vxLxL
EIL
=−+− (3)
4224
0(3 10 7 )
360
wx
vxLxL
EIL
=−+− (4)
(b) Make x=L/2 in Eq.(4):
4
0
5
768
CwL
vEI
=↓
Continued on next slide
y
A
wo
L
x
m
B
R
A
R
B
x
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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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(c) Set '0v= in Eq.(3):
8
1 0.5193
15
m
x
LL=− =
Then Eq.(2) gives
4
0
max max
( ) 0.00652 wL
vvx EI
== ↓
(d) Make x=L in Eq.(1):
3
0
45
BwL
EI
θ
=
SOLUTION (10.31)
(a)
Due to asymmetric but equal deflection configuration, only segment AC need be
considered.
0
'' M
EIv x
L
=
2
01
'2
M
EIv x C
L
=+
3
012
1
62
M
EIv x C x C
L
=++
Boundary conditions:
2
(0) 0: 0vC==
0
1
() 0:
212
M
L
L
vC==−
Then,
22
0
'(12)
12
M
vxL
EIL
=− (1)
22
0(4 )
12
Mx
vxL
EIL
=− (2)
Continued on next slide
L/
2
L/
2
M0/L
A
C
B
M0/L
y
x
M
0
xm
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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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(b) Let x=L/2 in Eq.(2):
0
C
v= as required
(c) Set '0v= in Eq.(1):
0.2887
12
mL
x
L==
Then, Eq.(2) gives
22
00
max max
( ) 0.00802
36 12
ML ML
vvx EI
EI
==− = ↓
or, in segment BC:
2
0
max 0.00802
M
L
vEI
=↑ ( 0.7113
m
x
L
=
)
(d) Make x=0 in Eq.(1):
2
0
24
AB
M
L
EI
θθ
=− =
SOLUTION (10.32)
(a)
00
'' M
EIv x M
L
=− +
2
001
'2
M
EIv x M x C
L
=− + + (a)
32
0012
1
62
M
EIv x M x C x C
L
=− + + + (b)
Boundary conditions:
210
1
(0) 0: 0, ( ) 0: 3
vCvLCML== ==
Continued on next slide
A
R
A
=M0/L
y
00
M
M
xM
L
=− +
M
0
x
V
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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.
Then Eqs.(a) and Eqs.(b) become
22
0
'(362)
6M
vxLxL
EIL
=− − + (1)
22
0(32)
6
Mx
vxLxL
EIL
=− − + (2)
(b) Let x=L/2 in Eq.(b):
2
0
16
MML
vEI
=↓
(c) For '0v=, Eq.(1) yields
22
3620xLxL−+= or 0.4227
m
x
L
=
.
Substitute this into Eq.(2):
2
0
max ( ) 0.0642
mML
vvx EI
== ↓
(d) Making x=L in Eq.(1):
0
6
B
M
L
EI
θ
=
SOLUTION (10.33)
(a)
Segment AC
22
11
111
'' 282
EIv M wLx wL wx== − −
22 3
11
4
'(2 )
83
w
EIv Lx L x x C=−−+ (a)
3224
112
2
()
83 2 3
wLx Lx x
EIv C x C=−−++ (b)
Segment BC
2'' 0EIv =
13 3 4
'EIv C EIv C x C==+ (c)
Continued on next slide
L/
2
R
A=wL/2
A
C
B
L/
2
M
A
=wL2/8
y
x
w
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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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Boundary and continuity condition:
121 1
(0) 0: 0 '(0) 0: 0vCv C== ==
3
12 3
'( ) '( ):
22 48
LL wL
vv C==−
4
12 4
() ():
2 2 384
LL wL
vv C==−
Then Eqs.(b) and (c) become
222
1(4 3 2 )
48
wx
vLxLx
EI
=−− (1)
3
2'48
wx
vEI
=− (2)
3
2(8 )
384
wx
vxL
EI
=−+ (3)
(b) Let x=L/2 in Eq.(3)
4
3
384
CwL
vEI
=↓
(c) Let x=L in Eq.(3)
4
7
384
BwL
vEI
=↓
(d) We have
3
48
BCwL
EI
θθ
==
1
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5
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