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)
2
111
()
216 2 64
PL L PL
AEI EI == 2
211
()
24 2 16
PL L PL
AEI EI ==
3
/1 2
23
() ( )
3364
AB LLPL
tA A EI =+ =
2
/
13
64
BAB PL
t
LEI
θ ==
0
/BD B D
θ
θθ =−
Also
2
1()
22 4
BPx Px
x
EI EI
θ ==
and
22
33
,
64 4 4
PL Px L
x
EI EI
==
Therefore 23
max / 12
()
22 3 6
BD Px x Px
vt EI EI == =
33
33 3
6 64 128
PL PL
EI EI
==↓
Continued on next slide
C
x
A
θ
D
B
t
A
/B
A
A
1
A
P/
2
4E
I
L/
2
L/
2
EI
A
2
M/EI
B
P
L
/
16E
I
P
L
/
4EI
C
D
P
x
/
2EI
P/
2
P
tangent at D
B
θ
tan
g
ent a
t
B
A
B
tB/D
max
v
x

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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SOLUTION (10.139)
00 0 0
12
11
() ()
24 2 16 22 2 8
M
ML M ML
LL
AA
EI EI EI EI
== =−=−
0
/1 2
21
() ( )
3316
BA
M
L
LL
tA A EI
=+ =−
0
116
BAB
M
t
LEI
θ == (1)
00
/1()
2
BD B D Mx
x
EIL
θθθ =− =− (2)
From Eqs. (1) and (2): 22
L
x=
We have
23
00
/12
()
233
BD
M
xMx
tx
EIL EIL
=− =−
Hence,
2
0
max 48 2
DML
vv EI
== ↑
x
B
A
1
A
C
M
0x/4EIL
M/EI
A
2
L/
2
L/
2
tB/D
A
B
x
D
M
0/4EI
2E
I
M
0/L
M
0/L
EI
M0
tangent at D
B
θ
tan
g
ent at B
max
v
L/3
tA/B