sm10 34 35

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.
________________________________________________________________________
PROBLEM (*10.34 and *10.35) An overhanging beam is loaded as shown in Figs. P10. 34 and
P10.35. 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 (*10.34)
(a)
Segment AB
2
11
31
'' 82 EIv M wLx wx== −
23
11
31
'16 6 EIv wLx wx C=−+ (a)
34
12
11
16 24
EIv wLx wx C x C=−++ (b)
Segment BC
2
22
39
'' ( )
828
wx
EIv M wLx wL x L== − + −
3
22
23
39
'()
16 6 16
wx
EIv wLx wL x L C=−+−+ (c)
34 3
234
3()
16 24 16
wLx wx
EIv wL x L C x C=−+ −++ (d)
Boundary & continuity conditions:
3
1211
(0)0: 0, ()0: 48
wL
vCvLC== ==−
12 132 4
'( ) '( ) : , ( ) 0 : 0vL vL C C vL C====
Continued on next slide
3wL/8
w
L
C
A
B
y
L/
2
x
9wL/8

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.
Then, Eqs.(a) through (d) become
233
1'(98)
48
w
vLxxL
EI
=−− (1)
233
1(3 2 )
48
wx
vLxxL
EI
=−− (2)
23 23
2'[9827()]
48
w
vLxxxLL
EI
=−−−− (3)
343 3
2[3 2 9 ( ) ]
48
w
vLxxLxLxL
EI
=−−+− (4)
(b) Let x=L/2 in Eq.(2):
4
192
MwL
vEI =↓
(c) Let x=3L/2 in Eq.(4):
44
2max
( ) 0.0078
128
CwL wL
vv EI EI ==− = ↓
To find
2max
()v, make 1'0v
=
in Eq.(1): 233 98 0Lx x L
−
−=
Solving, by trial and error: 0.422
m
x
L
=
Then Eq.(2) yields
4
1max
( ) 0.0054 wL vEI =↓
(d) Let x=3L/2 in Eq.(3):
3
1
12
CwL
EI
θ =
SOLUTION (*10.35)
(a)
Due to symmetry only one-half the beam need be considered.
Continued on next slide
3P/2
C
L/
2
E
y
x
L/
2
P
A
B
3P/2
P
L
D
P

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 CA
11
''EIv M Px==−
2
11
1
'2 EIv Px C=− + (a)
3
112
1
6
EIv Px C x C=− + + (b)
Segment AE
12 313
'' ( )
2224
PL
EIv M Px x Px PL==−+ −= −
2
23
13
'44 EIv Px PLx C=−+ (c)
32
234
13
12 8
EIv Px PLx C x C=−++ (d)
Boundary & continuity conditions
22 2
233
13 1
'( ) 0 : 0,
44 2
v L PL PL C C PL=−+==
2222 2
12 1 1
35
'( ) '( ) : ,
228 1682 16
L L PL PL PL PL
vv C CPL=−+=−+ =
33 3
122
513
'( ) 0 : 0,
24832 96
LPLPL PL
vCC=++= =
333 3
244
3
() 0: 0,
296324 6
LPLPLPL PL
vCC=−+= =−
Equations (a), (b), and (d):
22
1'(85)
16
P
vxL
EI
=−+ (1)
32 3
1(16 30 13 )
96
P
vxLxL
EI
=−+− (2)
3223
2(8 36 48 16 )
96
P
vxLxLxL
EI
=−+− (3)
(b) Making x=L in Eq.(3):
3
24
EPL
vEI =↑
Continued on next slide

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.
(c) Making x=0 in Eq.(2):
3
13
96
CPL
vEI =↓
Thus,
maxC
vv=
(d) Making x=0 in Eq.(1):
2
5
16
CD PL
EI
θθ =− =