sm10 96 97

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________________________________________________________________________
PROBLEM (10.96 and 10.97) A beam is supported and loaded as shown in Figs. P10.96 and
P10.97. Determine all of the reactions.
SOLUTION (10.96)
From symmetry:
AB
R
R= AB
M
M=
Statics: 1
4
AB
R
RwL== ↑
Use Table B14 (cases 4.1 and 2):
() () () 0
BBwBRBM
vv v v=++ =
Here 43
4
00 0
(2) 2(2) 2(2)
() ()
30 30 24 2
Bw wL wL wL L
vEI EI EI
=− + +
4
11
192 wL EI
=−
33
() 3 () 2
BR B BM B
vRLEI v MLEI==−
Thus, 432
0
11 0
192 3 2
BB
BwL RL ML
vEI EI EI
=− + − =
Solving
22
00
55
96 96
BA
M
wL M wL==
Continued on next slide
A
B
=
M
A
M
B
+
R
A
RB
w0
A
B
M
B
R
B
w0
A
B
+
A
B
2w0
L
3
2( 2)
() 24
Cw wL
EI
θ =
C
+
A
B
w0
C
A
B
2w0
C

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SOLUTION (10.97)
Using Table B.14 (cases 2, 3 and 1):
234
00
283
A
AML RL
wL
vEI EI EI
=− − + =
or
0
33
28
AM
R
wL
L
=+↑
Then, the reactions B
R
and B
M
are found from equilibrium.
L/
2
A
B
=
A
M0
R
A
B
+
R
A
RB
C
L
B
w
w
M
B
M0