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________________________________________________________________________
PROBLEMS (*10.53 and *10.54) An overhanging beam is supported and loaded as shown in
Fig. P10.53 and P10.54. Determine the equation of the elastic curve and deflection at point A.
SOLUTION (*10.53)
Refer to the above figure:
22 2 0
'' ( )
222 2
wwwa a
EIv x a x x a w x=− + + < > + < >+ < − >
33 22 1
'()
664 2
wwwa a
EIv x a x x a w x C=− + + < > + < > + < − >+
2
44 3 2
12
()
24 24 12 2 2
wwwaawa
EIv x a x x x C x C=− + + < > + < > + < − > + + (1)
Using boundary conditions:
44
22
(0) 0: 0,
24 24
wa wa
vCC=− += =
444 4
4 3
11
23
() 0: 0,
324128 24 8
wa wa wa wa
va wa Ca C wa=− +++++= =
Equation (1) becomes
44 32 234
[( ) 2 12 9 ]
24 2
wa
vxaxaxaxaxa
EI
=−++<>+<>+<−>++
Let x=-a:
4
3
Cwa
vEI
=↓
SOLUTION (*10.54)
From conditions of equilibrium:
20
AB
RP R=↑ =
Using singularity functions:
22
M
Px P x a P x a=− + < − >− < − >
Using singularity functions:
2
222
dv
EI Px P x a P x a
dx =− + < − >− < − > (1)
Continued on next slide
R
A=wa/2
A
C
B
wa2
w
x
w
a
R
B
y
Equivalent loading
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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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22 2
1
11
2
22
dv
EI Px P x a P x a C
dx =− + < − > − < − > + (2)
33 3
12
11 1 2
63 6
EIv Px P x a P x a C x C
=
− + <−>− <− >+ + (3)
Boundary conditions:
3
12
1
() 0: 6
EIv a C a C Pa=+= (4)
2
12
(3 ) 0: 3 2EIv a aC C Pa=+= (5)
Eq.(5)-Eq.(4): 2
111
12
CPa=, then 3
23
4
CPa=−
Equation (3) is therefore
3323
2113
[]
63 6 124
Px xa xa axa
vEI
<−> <− >
=−+ − + −
At the free end (x=0), we have
33
33
44
CPa Pa
vEI EI
=− = ↓
1
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2
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