sm10 10 11

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 (10.10 and 10.11) A cantilever beam is loaded as depicted in Fig. P10.10 and
P10.11. Using the double-integration approach, determine:
(a) The equation of the deflection curve.
(b) The slope at the free end.
(c) The deflection at the free end.
SOLUTION (10.10)
(a)
1(0 2)
M
PL x L=≤≤
23()
22
L
M
PL Px x L=− ≤≤
Segment AC
1''EIv PL=
11
'EIv PLx C=+
2
112
1
2
EIv PLx C x C=++
Segment BC
23
'' 2 EIv PL Px=−
2
23
31
'22 EIv PLx Px C=−+
23
234
31
46
EIv PLx Px C x C=−++
Boundary conditions:
22 2
233
31
'( ) 0: 0,
22
v L PL PL C C PL=−+==−
333 3
244
31 5
() 0: 0,
46 12
v L PL PL PL C C PL=−−+==
Thus,
22
2'(3 2)
2P
vLxxL
EI
=−− (1)
23 2 3
2(9 2 12 5 )
12
P
vLxxLxL
EI
=−−+ (2)
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:
22 2
12 1 1
13 7
'( ) '( ) : ,
222 8 8
LL
v v PL C PL C PL= + =− =−
3
32
12 2 2
519
() (): ,
2 2 16 12 48
LL PL
v v PL C C PL=−+=− =−
Therefore,
2
1'(87)
8P
vLxL
EI
=−− (3)
22 3
1(24 42 19 )
48
P
vLxLxL
EI
=−+ (4)
(b) x=0 in Eq. (3):
22
77
88
APL PL
EI EI
θ =− =
(c) x=0 in Eq. (4):
3
19
48
APL
vEI =− ↑
SOLUTION (10.11)
(a)
22
11(4 4 ) (0 2)
2
M
wa axx xa=−+− ≤≤
20()
2
L
M
xL=≤≤
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.
Segment AC
22
1'' ( 4 4 )
2
w
EIv a ax x=−+ −
3
22
11
'(4 2 )
23
wx
EIv a x ax C=− + − +
4
22 3
112
2
(2 )
2312
wx
EIv a x ax C x C=− + − + +
Boundary conditions:
1112
'(0) 0 : 0, (0) 0 : 0vCvC== ==
Thus,
223
1'(126 )
6w
vaxaxx
EI
=−+− (1)
22 3 4
1(24 8 )
24
w
vaxaxx
EI
=−+− (2)
Segment BC
223234
'' 0, ' ,EIv EIv C EIv C x C== =+
Continuity conditions:
3
12 3
4
'(2 ) '(2 ) : 3 vava waC=−=
44 4
12 44
82
(2 ) (2 ): 2 ,
33
v a v a wa wa C C wa=−=−+ =−
Therefore,
3
24
'3wa
vEI =− (3)
3
2(4 2)
3
wa
vxa
EI
=−+ (4)
(b) x=3a in Eq. (3):
33
44
33
Bwa wa
EI EI
θ =− =
(c) x=3a in Eq. (4):
44
10 10
33
Bwa wa
vEI EI =− = ↓