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________________________________________________________________________
PROBLEMS (10.146 through 10.149) A beam is supported and loaded as shown in Figs.
P10.146 through P10.149. Determine the reactions at the support.
SOLUTION (10.146)
Let
A
M
be redundant. Because of symmetry:
2
AB
wL
RR== ↑ and AB
M
M=
23
12
2
38 12 A
M
L
wL wL
AL A
EI EI EI
== =− 00
/0
AB B A
θθθ
=
−=
Thus 3
/12
012 A
AB
M
L
wL
AA EI EI
θ
== + = −
or
2
12
AwL
M= 2
12
BwL
M=
Continued on next slide
C
M/EI
x
w
Parabola
A
1
-MA/EI
A
2
wL2/16EI
A
L/
2
L
/2
L/
2
R
A RB
MB
M
A
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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 (10.147)
Let
A
R
and A
M
as redundants.
2
12
1()
22
AA A
R
LRL ML
AL A
EI EI EI
== =−
0
32
M
L
AEI
=−
2
123
022
o
A
AA
M
L
RL
AAA ML
θ
== + + = − −
or
0
2
AA
R
LM M−= (1)
We have 0:
B
v=
123
0: () () () 0
324
BA LLL
tAAA=++=
or
0
3
34
AA
R
LM M−= (2)
Solving Eqs.(1) and (2):
0
1
4
A
M
M= 0
3
2
AM
RL
=
↑
Statics:
BA BA
R
RMM=− =
Continued on next slide
C
M/EI
x
A
1
-M0/EI
A
2
A
L/
2
L/
2
L/
4
R
A RB
MB
M
A
B
A
C
A
3-
M
A
/E
I
M0
L/
3
R
A
L/EI
B
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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 (10.148)
Let
A
M
be redundant.
0
A
θ
=
11(2 )
2AA
AMaMa==
2
21(2 )
2
APaaPa=− =−
2
31
2
APa=−
/12
14 2
0[()()]
33
BBA aa
vt A A
EI
=== +
or
1
2
A
M
Pa=
Statics:
73
44
BA
RP R=↑ =↓
SOLUTION (10.149)
Consider
A
R
and A
M
as redundant.
00
/0
BA B A
θθθ
=−=
/0
BA
t=
Continued on next slide
C
x
tan
g
ent at B
B
M
A
2
A
2a
P
A
-Pa
a
4a/3
A
3
R
A RB
B
C
2a/3
M
A
MA
A
1
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2
111
()
22
AA
ARLLRL==
2A
AML
=
−
2
311
()
22
APbbPb=− =−
/123
0
BA
EI A A A
θ
=++=
or
22
20
AA
RL ML Pb−−= (1)
/1 2 3
() () () 0
323
BA LLb
EIt A A A=++=
or
323
30
AA
RL ML Pb−−= (2)
Solving Eqs.(1) and (2)
2
2
APab
ML
= 2
3(2 )
APb
R
aL
L
=
+↑
Statics:
2
2
0:
BB
Pa b
MM
L
==
∑
2
2
0: (2 )
yB
Pa b
FRPab
L
==−+↑
∑
x
M
B
C
y
A
1
C
x
tangent at A
B
M
A
3
A
P
A
-Pb
b
L/
2
A
2
R
A RB
b/3
-MA
MA
B
A
a
R
A
RAL
MA
L/
3
L
1
/
4
100%