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________________________________________________________________________
PROBLEMS (10.40 through 10.43) A beam is supported and loaded as shown in Figs. P10.40
through P10.43. Determine:
(a) The equation of the elastic curve.
(b) The slope at the free end.
(c) The deflection at the free end.
SOLUTION (10.40)
(a)
22
31
'' 22
EIv wax wa w x a=− −<−>
22 3
1
131
'226
EIv wax wa x w x a C=−−<−>+ (a)
322 4
12
13 1
64 24
EIv wax wa x w x a C x C=− −<−>++ (b)
Boundary conditions:
'
21
(0) 0: 0 (0) 0: 0vCv C== ==
Equation (a) and (b) become
'3
1(3 )
62
wax
EIv w x a a x=− < − > − − (1)
2
4
1(9 2 )
24 12
wax
EIv w x a a x=− < − > − − (2)
(b) Make x=2a in Eq.(1):
3
7
6
Bwa
EI
θ
=
(c) Make x=2a in Eq.(2):
4
41
24
Bwa
vEI
=↓
Continued on next slide
a
R
A=wa
A
C
B
a
2
3
2
A
M
wa=
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.
SOLUTION (10.41)
(a)
'' 2
L
EIv PL P x=−<−>
21
'22
PL
EIv PLx x C=−<−>+ (a)
23
12
1
262
PL
EIv PLx x C x C=−<−>++ (b)
Boundary conditions:
2
22
11
7
'( ) 0 : ,
88
PL
v L PL C C PL=−+ =−
23 3 3
22
719
() 0: 0,
248 8 48
PL PL PL
vL C C PL=−−+= =−
Equations (a) and (b) become
22
'[84 7]
82
PL
vLxPx L
EI
=−<−>− (1)
2323
[24 8 42 19 ]
48 2
PL
vLxPxLL
EI
=−<−>−+ (2)
(b) Make x=2a in Eq.(1):
2
7
8
APL
EI
θ
=
(c) Make x=0 in Eq.(2):
3
19
48
BPL
vEI
=↑
Continued on next slide
L/
2
P
A
C
B
L/
2
y
x
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.
SOLUTION (10.42)
(a)
22 2
11
'' 2 2 2
22
EIv wax wa wx w x a=−−+<−>
22 3 3
1
11
'2 2
66
EIv wax wa x wx w x a C=− − +<−>+ (a)
322 4 4
12
111
2
32424
EIv wax wa x wx w x a C x C=−−+<−>++ (b)
Boundary conditions:
12
'(0) 0: 0 (0) 0: 0vCvC== ==
Then, Eqs. (a) and (b) become
223 3
'[612 2]
6w
vaxaxxxa
EI
=−−+<−> (1)
3224 4
[8 24 2 ]
24
w
vaxaxxxa
EI
=−−+<−> (2)
(b) Make x=3a in Eq.(1):
3
4
3
Bwa
EI
θ
=
(c) Make x=3a in Eq.(2):
4
10
3
Bwa
vEI
=↓
Continued on next slide
R
A=2wa
A
C
B
M
A
=2wa2
w
x
2a
a
y
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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.
SOLUTION (10.43)
Refer to the Fig. P10.43:
2
01
0: 0
2
AB
MMwLRL=−− +=
∑
or 0
1
2
B
M
RwL
L
=+
00
1
0: 22
yA
M
M
wL
FRwLwL
LL
==−−=−
∑
Thus
222
211
() 22
AB
dv
EI M x R x wx R x L w x L
dx ==−+<−>+<−>
23 2 3
1
111 1
262 6
AB
dv
EI R x wx R x L w x L C
dx = − + <−>+ <−>+ (1)
34 3 4
12
111 1
6246 24
AB
EIv R x wx R x L w x L C x C=−+<−>+<−>++ (2)
Boundary conditions:
2
(0) 0: 0EIv C==
34
1
11
() 0: 0
624
A
EIv L R L wL C L=−+=
or
23 3
0
1624 246
A
M
L
RL wL wL
C=− + =− +
Continued on next slide
R
A
A
C
B
M
0
w
x
L
w
R
B
y
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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.
Equation (1) and (2) become then
(a) 3
22 2
00
1[()
22 3 2
MM
dv wL wx wL
x
xxL
dx EI L L
θ
== − − + + <−>
3
30]
3123
M
L
wwL
xL+<−>− +
34
33
00
1[()
62 4 2
Mx M
wL wx wL
vx xL
EI L L
=−−++<−>
3
40]
44
wwL
x
LxMLx+<−>− +
(b) 3
0
31
() ( 5 )
26 4
CLwL
M
L
EI
θθ
== −
(c) 42
0
31931
() ( )
224 163
CLwL
vv ML
EI
== −
1
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5
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