sm10 138 139

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PROBLEM (*10.138 and 10.139) A simple beam AB with two different moments of inertia
carries a load as shown in Figs. P10.138 and P10.139. What is the largest deflection between the supports A
and B?
SOLUTION (*10.138)
P
EI
4EI
A
L/2
C
L/2
P/2
B
P/2
M/EI
PL/4EI
Px/2EI
PL/16EI
A
A2
A1
C D
x
θA
A
B
vmax
tA/B
D
θB
x
B
tB/D
tangent at D
tangent at B
A1 =
1 PL L
1 PL2
( )=
2 16 EI 2 64 EI
A2 =
1 PL L
1 PL2
( )=
2 4 EI 2 16 EI
L
2L
3 PL3
t A / B = A1 ( ) + A2 ( ) =
3
3
64 EI
2
1
3 PL
θB = tA/ B =
L
64 EI
θB / D = θB − θD0
Also
θB =
1 Px
Px 2
( x) =
2 2 EI
4 EI
and
3 PL2 Px 2
=
,
64 EI 4 EI
x=
3L
4
Therefore
1 Px 2 2 x
Px 3
( )=
2 2 EI 3
6 EI
3
P 3 3L
3 PL3
=
=
↓
6 EI 64
128 EI
vmax = t B / D =
Continued on next slide
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SOLUTION (10.139)
A
M0
2EI
L/2
EI
B
L/2
C
M0/L
M0/L
x
M0/4EI
A1
M/EI
L/3
x
A2
M0x/4EIL
tangent at B
tA/B
A
D
vmax
tangent at D
θB
tB/D
B
M L
1 M0 L M0L
1 M0 L
( )=
( )=− 0
A2 = −
2 4 EI 2 16 EI
2 2 EI 2
8EI
2L
1 M0L
L
t B / A = A1 ( ) + A2 ( ) = −
3
3
16 EI
M0
1
θ B = t AB =
L
16 EI
1 M0x
θ B / D = θB − θD0 = −
( x)
2 EIL
L
From Eqs. (1) and (2): x =
2 2
We have
M x3
1 M 0 x2 2
( x) = − 0
tB / D = −
2 EIL 3
3EIL
Hence,
M 0 L2
vmax = vD =
↑
48 2 EI
A1 =
(1)
(2)
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
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