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________________________________________________________________________
PROBLEMS (10.44 through 10.47) For the beam loaded and supported as shown in Figs.
P10.44 through P10.47, determine:
(a) The equation of the deflection curve.
(b) The slope at end B for the given numerical values:
P = 4 kips, a = 3 ft, L = 10 ft, E = 10 x 106 psi, I = 72 in.4 , w=2 kips/ft
SOLUTION (10.44)
(a)
'' 3 3EIv Px Pa P x a=− −<−>
22
1
11
'3 3
22
EIv Px Pax P x a C=−−<−>+ (a)
32 3
12
13 1 3
62 6
EIv Px Pax P x a C x C=− −<−>++ (b)
Boundary conditions:
12
'(0) 0: 0 (0) 0: 0vCvC== ==
Equations (a) and (b) become
22
'[31833]
6P
vxaxxa
EI
=−−<−> (1)
32 3
[9 3]
6P
vxaxxa
EI
=−−<−> (2)
(b) Make x=4a in Eq.(1):
2
9
2
BPa
EI
θ
=
32 3
6
9(4 10 )(36) 32.4 10
2(10 10 )(72) rad
−
×
==×
×.
SOLUTION (10.45)
(a)
Continued on next slide
3a
P
A
C
B
a
y
x
R
A=P
M
A
=3Pa
P
L
P
D
A
C
B
y
x
P
P
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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
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''EIv Px P x a P x L a= − <−>−<−+>
222
1
'22 2
Px P P
EIv x a x L a C= − <−>− <−+>+ (a)
333
12
66 6
Px P P
EIv x a x L a C x C= − <−>− <−+>+ + (b)
Boundary conditions:
2
11
'( ) 0 : ( ) 0, ( )
2822 2
LPLPL Pa
vaCCaL=−−+==−
2
(0) 0: 0vC==
Equations (a) and (b) become
22 2
'[ ()]
2P
vxxaxLaaaL
EI
= −<−>−<−+>+ − (1)
33 3
[3()]
6P
vxxaxLaaxaL
EI
= −<−>−<−+>+ − (2)
(b) Make x=L in Eq.(1):
()
2
BPa La
EI
θ
=−
33
6
4 10 (3 12)(10 3)12 8.4 10
2(10 10 )(72) rad
−
×× −
==×
×.
SOLUTION (10.46)
(a)
0
'' 2 2
4
Px
EIv P x a Pa x a= −<− >+ <− >
221
'22
82
Px P
EIv x a Pa x a C= − <− >+ <− >+ (a)
332
12
22
24 6 2
Px P Pa
EIv x a x a C x C= − <− >+ <− >+ + (b)
Continued on next slide
P
L
Pa2a
2a
4
AP
R=
A
C
B
34
BP
R=
y
x
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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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Boundary conditions:
2
(0) 0: 0vC==
2
333
11
64 8 5
(4 ) 0: 2 4 0,
24 6 6
Pa
v a Pa Pa Pa aC C=−++==−
Equations (a) and (b) become
22 2
'[312224220]
24
P
vxxaaxaa
EI
= − <− >+ <− >− (1)
33 22
[4 2 12 2 20]
24
P
vxxaaxaax
EI
=−<−>+<−>− (2)
(b) Make 4a in Eq.(1):
2
7
6
BPa
EI
θ
=
32 3
6
7(4 10 )(3 12) 8.4 10
6(10 10 )(72) rad
−
××
==×
×
SOLUTION (10.47)
0
22
w
dV L
wx
dx L
==− <−>
2
02A
w
dM L
VxR
dx L
==− <−>+
3
0
() 32
A
wL
M
xxRx
L
=− < − > +
At 3
0
,()0:0 ()
32
A
wL
x
LML L RL
L
===−−+
or
0
24
AwL
R=
Therefore
23
00
2() 3224
wwL
dv L
EI M x x x
dx L
==−<−>+
Continued on next slide
R
A
A
C
L/2 M(x)
x
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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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42
00
1
12 2 48
wwL
dv L
EI x x C
dx L
=− < − > + +
53
00
12
60 2 144
wwL
L
EIv x x C x C
L
=− < − > + + +
Boundary conditions:
2
(0) 0: 0EIv C==
3
4
5
00
11
37
() 0: () 0,
60 2 144 5760
wwL
LwL
EIv L C L C
L
=− ++= =−
(a) 24
53
0137
[]
60 5 144 5760
wLL L
vxxx
EIL
=−<−>+ −
(b) 24
42
0137
[]
12 5 48 5760
w
dv L L L
xx
dx EIL
=−<−>+ −
3
0
37
'(0) 5760
AwL
vEI
θ
==
Substituting the given data
33
6
37(2 10 )(10 12) 0.03083 1.77
5760(10 10 )(72) o
Arad
θ
××
===
×
1
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