sm10 134 137

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________________________________________________________________________
PROBLEMS (10.134 through 10.137) A beam is supported and loaded as shown in Figs.
P10.134 and P10.137. Determine the maximum deflection between the supports A and B.
SOLUTION (10.134)
2
118 2
()
22 3 3
Pa a Pa
AEI EI
=− =−
2
214
()
24 3 6
Pa a Pa
AEI EI
==
/1 2
4128 1 4
[( )][(]
333 3 3 3
BA aaa a
tA Aa
=+++ +
32 3
44 7 3
27 54 2
Pa Pa Pa
EI EI EI
=− + =−
2
/
13
48
ABA
Pa
t
aEI
θ =− =
2
/31
:(),3
8222
OA O A Pa Pa x
x
xa
EI EI a
θθθ =− = =
Thus, 2
max / 132
[](3)
22 2 3
OA Pa a
vv a
EI a
⋅
== 3
3
4Pa EI
=
↑
Continued on next slide
x
x
B
A
1
A
C
B
M/EI
A
2
a
a
2a
P
a
/
4EI
- Pa/2EI
tA/O O
A
θ
P/
4
x
E
3
a
E
P/
4
D
y
max
v
8a/3
t
B/A
P
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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SOLUTION (10.135)
22
115 15
(3 )
22 4
wa wa
Aa
EI EI
==
23
212 2
(2 )
2wa wa
Aa
EI EI
=− =−
23
31()
32 6
wa wa
Aa
EI EI
=− =−
4
/1 2 3
257
() ( ) ()
3424
AB aawa
tAaA A EI =+ + =
3
/
157
372
BAB
wa
t
aEI
θ ==
02 2
/15 1
[()]
28
BO A O wax wa x a EI
θθθ =− = − −
Thus,
33 2
57 5 1 ( ) , 1.472
72 12 2
wa wax wa x a x a=−− =
Then,
2
max / 15 2 1 2
() ()()[()]
26 32 3
OBO x
v v t wax waxa xa a
EI EI
== = − − −+
Making 1.472
x
a= into the above
4
max 0.74
Owa
vv EI == ↓
Continued on next slide
tangent at O
x
B
θ
tan
g
ent at B
a
B
t
A
/B
A
A
1
a
5wa/6
A
C
B
7wa
/
6
B
y
M/EI
-wa2/2EI
5wa2/2EI
A
2
M/EI
A
3
5
6wa
x
EI
x
wa
(
x-a
)
wa
a
-wa2/2EI
tB/O
D
max
v
C
w

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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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SOLUTION (10.136)
23
00
11()
22 4
wa wa
Aa
EI EI
==
23
00
21()
46 24
wa wa
Aa
EI EI
=− =−
44 4
00 0
/1 2 2
24
() ()
3563015
BA wa wa wa
aa
tA A EI EI EI
=−=−=
4
0
max 2
15
AC wa
vt EI
== ↓
Continued on next slide
w0
x
x
A
tA/C
a
A
B
y
M/EI
a
B
A
1
A
C
2a
/
3
tan
g
ent at
C
4a
/
5
-w0a2/6EI
max
v
C
C
A
2
w0a/2
w0a/2
w0a2/2EI
Cubic

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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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SOLUTION (10.137)
22
12
13 3 1
() ()
24 2 16 24 2 16
PL L PL PL L PL
AA
EI EI EI EI
−
== =−=
33 3
/1 2 5
()()
2 6 3 8 48 48
BA LL L PL PL PL
tA A EI EI EI =++ =− =
2
/
15
48
ABA PL
t
LEI
θ =− =− (1)
0
/
AD A D
θ
θθ =−
Also
2
33
()
22
APx Px
x
EI EI
θ =− =− (2)
Equations (1) and (2) yield 5
6
x
L=
Then,
max / 13(5 6) 5 2 5
[]
22 636
AD PLL L
vt L
EI
==
33
55 0.0259
432 PL PL EI EI
== ↓
A
θ
P
L
x
x
B
A
1
A
C
B
M/EI
A
2
P/
2
L/
2
L/
2
-PL/4EI
tA/D
tan
g
ent a
t
A
A
B
3P
/
2
y
x
P
3PL/4EI
tB/A
D
max
v
3Px/2EI