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________________________________________________________________________
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
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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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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
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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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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
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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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(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
=− =
1
/
4
100%