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PROBLEMS (10.1 through 10.4) A cantilever beam is loaded as shown in Figs. P10.1 through
10.4. Using the double-integration method, determine:
(a) The equation of the elastic curve.
(b) The slope at the free end.
(c) The deflection at the free end.
SOLUTION (10.1)
(a) 2
1
"2
EIv wx=−
31
1
'6
EIv wx C=− +
412
1
24
EIv wx C x C=− + +
Boundary conditions:
3
11
'( ) 0: 6
vL C wL==
44 4
22
11 1
() 0: ,
24 6 8
v L wL wL C C wL=− + + =−
Thus,
33
'()
6w
vxL
EI
=− − (1)
43 4
(4 3)
24
w
vxLxL
EI
=− − + (2)
(b) x=0 in Eq. (1): 3
6
AwL
EI
θ
=
(c) x=0 in Eq. (2): 44
88
AwL wL
vEI EI
=− = ↓
SOLUTION (10.2)
()()
M
PL x Px L=− − = −
(a)
Segment AC
1
2''( )EIv P x L=− Continued on next slide
2
1
2
M
wx=−
y
w
A
x
V
A
C
B
2E
I
x
L/
2
L/
2
P
y
P
EI
P
L
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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.
2
11
1
2' ( )
2
EIv P x L C=−+
3
112
1
2()
6
EIv P x L C x C=−++
Segment CB
2'' ( )EIv P x L=−
2
23
1
'()
2
EIv P x L C=−+
3
234
1()
6
EIv P x L C x C=−++
Boundary conditions:
23
11 12
11
'(0) 0: , (0) 0:
26
v C PL v C PL==− ==
Thus
2
1'(2)
4P
vxxL
EI
=− (1)
32
1(3)
12
P
vxxL
EI
=− (2)
Boundary conditions:
22 2
12 33
31 5
'( ) '( ) : ,
22168 16
LL
v v PL PL C C PL=−=+=−
333 3
12 44
515 1
() (): ,
2 2 96 48 32 8
LL
v v PL PL PL C C PL=−=−−+=−
So,
22
25
'[()]
28
P
vxLL
EI
=−− (3)
3
32
215
[( ) ]
23 8 4
PL
vxLLx
EI
=−−+ (4)
(b) x=L in Eq. (3):
22
55
16 16
BPL PL
EI EI
θ
=− =
(c) x=L in Eq. (4):
33
33
16 16
BPL PL
EI EI
θ
=− = ↓
Continued on next slide
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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
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SOLUTION (10.3)
(a)
3
0
1
"6
EIv w x
L
=−
4
01
1
'24
EIv w x C
L
=− +
5
012
1
120
EIv w x C x C
L
=− + +
Boundary conditions:
3
10
1
'( ) 0: 24
vL C wL==
54 3
002 20
11 1
() 0: 0
120 24 24
vL wL wL C C wL=− + += =
Therefore
44
0
'()
24
w
vxL
EIL
=− (1)
54 5
0(5 4)
120
w
vxLxL
EIL
=− − + (2)
(b) x=0 in Eq. (1):
2
0
24
AwL
EI
θ
=
(c) x=0 in Eq. (2):
44
00
30 30
AwL wL
vEI EI
=− = ↓
SOLUTION (10.4)
(a)
Segment AC Continued on next slide
3
0
6
x
M
w
L
=−
A
x
0
x
w
L
y
2
0
2
xw
L
x/
3
V
R
A= wa
2
3
2
A
M
wa=
w
aC
A
B
y
a
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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2
13(2 3 ) (0 )
22
wa
M
wa wax x a x a=− + = − ≤ ≤
1'' (2 3 )
2
wa
EIv x a=−
2
11
'(3)
2
wa
EIv x ax C=−+
32
112
3
()
232
wa x
EIv ax C x C=−++
Boundary conditions:
1112
'(0) 0 : 0, (0) 0 : 0vCvC== ==
Thus
1'(3)
2
wax
vxa
EI
=− (1)
2
1(2 9 )
12
wax
vxa
EI
=− (2)
Segment BC
22 2
2(2 ) ( 4 4 )
22
ww
M
ax x ax a=− − = − + −
22
2'' ( 4 4 )
2
w
EIv x ax a=−+ −
322
23
'( 2 4)
23
wx
EIv ax a x C=−+ − +
4322
234
2
(2)
2123
wx
EIv ax a x C x C=−+ − + +
Continuity conditions:
33 3
12 33
71
'( ) '( ) : ,
66
v a v a wa wa C C wa=−=−+ =
44 4
12 44
713 1
() (): ,
12 24 24
v a v a wa wa C C wa=−=−+ =−
Thus
3223
2'(212242)
12
w
vxaxaxa
EI
=−+−− (3)
43 2234
2(824 4 )
24
w
vxaxaxaxa
EI
=−+− +− (4)
Continued on next slide
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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) x=2a in Eq. (3):
33
77
66
Bwa wa
EI EI
θ
=− =
(c) x=2a in Eq. (4):
44
41 41
21 21
Bwa wa
vEI EI
=− = ↓
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