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 (10.48 through *10.51) A simple beam loaded as shown in Figs. P10.48 through
P10.51. Determine:
(a) The equation of the elastic curve.
(b) The angle of rotation at the end A.
(c) The deflection at the midspan.
SOLUTION (10.48)
(a)
2
1
'' 822
wL
EIv wLx x=−<−>
23
1
1
'16 6 2
wL
EIv wLx x C=−<−>+ (a)
34
12
1
48 24 2
wL
EIv wLx x C x C=−<−>++ (b)
Using boundary conditions:
2
(0) 0: 0vC==
4
43
11
17
() 0: 0,
48 384 384
wL
v L wLx C L C wL=−+==−
Equations (a) and (b) become
233
' [24 64 7 ]
384 2
wL
vLxxL
EI
=−<−>− (1)
343
[8 16 7 ]
384 2
wL
vLxxLx
EI
=−<−>− (2)
(b) Make x=0 in Eq.(1):
3
7
384
AwL
EI
θ
=
(c) Make x=L/2 in Eq.(2):
4
5
768
CwL
vEI
=↓
Continued on next slide
x
L/
2
wL/8
A
C
B
L/
2
y
3wL/8
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 (10.49)
(a)
0
00
'' M
EIv x M x a
L
=−<−>
2
001
'2
M
EIv x M x a C
L
=+<−>+ (a)
32
00 12
62
MM
EIv x x a C x C
L
=+<−>++ (b)
Boundary conditions:
2
(0) 0: 0vC== 22
0
1
() 0: (3 )
6
M
vL C b L
L
==− −
Equations (a) and (b) become
222
0
'[36 (3)]
6M
vxLxabL
EIL
=−−<−>−− (1)
3222
0[3 (3 )]
6M
vxLxabLx
EIL
=−+<−>−− (2)
(b) Make x=0 in Eq.(1):
22
0(3)
6
AMLb
EIL
θ
=−
(c) Make x=L/2 in Eq.(2):
3222
093
(33 )
68 2
MML
vaLLabL
EIL
=−+−
SOLUTION (10.50)
(a)
Continued on next slide
L
M0
b
a
M0/L
A
C
B
M0/L
y
x
a
a
A
C
B
P
/2
y
x
P
P
/2
k
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.
'' 2
P
EIv x P x a=−<−>
22
1
'42
PP
EIv x x a C=−<−>+ (a)
33
12
12 6
PP
EIv x x a C x C=−<−>++ (b)
Boundary conditions:
2
(0) :
22
PPEI
vC
kk
=− =−
33 2
11
8
(2 ) 0: 2 ,
12 6 2 4 4
Pa Pa PEI Pa PEI
va Ca C
kak
=−+− =−+
Equations (a) and (b) become
222
'[2 ]
4PEI
vxxaa
EI ak
=−<−>−+ (1)
332
6
[2 3( ) ]
12
PEIEI
vxxaax
EI ak k
=−<−>+−+− (2)
(b) Make x=0 in Eq.(1):
2
()
4
APEI
a
EI ak
θ
=−+
(c) Make x=a in Eq.(2):
33
(2 )
12
CPEI
va
EI k
=+↓
SOLUTION (*10.51)
(a)
Continued on next slide
w0L/4
A
w
o
B
L/
2
w0L/4
L/
2
C
0
4( )
2
wL
x
L
−
0
2
x
w
L
=
x
w0L/4
A
B
w0L/4
C
2w0
y
2w0
x
(a) Actual load (b) Equivalent load
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.
Refer to Fig.(b):
223
00
04
12 1 1
()[ ]
42 32 23 2
wL w
x
xLL
Mx w x x
LL
=− + <−><−>
Thus,
33
00 0
2
'' 43 3 2
wL w w L
EIv x x x
LL
=−+<−>
24 4
000 1
'8126 2
wL w w L
EIv x x x C
LL
=−+<−>+ (a)
35 5
000 12
24 60 30 2
wL w w L
EIv x x x C x C
LL
=−+<−>++ (b)
Boundary conditions:
3
210
5
(0) 0: 0 ( ) 0: 192
vCvLCwL== ==−
Equations (a) and (b) become
22 4 4 2
0
' [24 16 32 5 ]
192 2
wL
vLxxxL
EIL
=−+<−>− (1)
23 5 4 3
0[8 3.2 6.4 5 ]
192 2
wL
vLxxxLx
EIL
=−+<−>− (2)
(b) Substitute x=0 into Eq.(1):
2
0
5
192
AwL
EI
θ
=
(c) Substitute x=L/2 into Eq.(2):
4
0
max 120
CwL
vv EI
== ↓
1
/
4
100%