sm10 90 93

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________________________________________________________________________
PROBLEMS (10.90 through 10.93) A beam is supported and loaded as shown in Figs. P10.90
through P10.93. What are the reactions at each support?
SOLUTION (10.90)
Use Table B.14 (case 1):
() () 0
AAPAR
vv v=+=
33
(2) ()() 0
323
A
CP RL
PL L
EI EI
θ =− − + =
333 0
24 16 3A
RL
PL PL
EI EI EI
=− − − = , 5
16
A
R
P
=
↑
Statics:
11 3
16 16
BB
R
PMP=↑ =
SOLUTION (10.91)
Continued on next slide
C
L/
2
L/
2
A
C
B
P
=
L
A
B
M
B
R
A
A
B
P
+
R
A
RB
A
M0
M
0
L
A
B
=
L
A
B
M
A
R
B
B
+
R
A
RB

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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Copyright Act without the permission of the copyright owner is unlawful.
Use Table B.14 (case 2 and 1):
() () 0
BBMBR
vv v=+=
23
00
3
0,
23 2
BB
ML M
RL R
EI EI L
=− − = = ↓
Statics:
00
31
22
AA
M
R
MM
L
=↑ =
SOLUTION (10.92)
Use Table B.14 (cases 11 and 9):
0
() () () 0
AB
AAMAMAM
θ
θθθ =− + + =
00
3624
AB
ML
ML ML
EI EI EI
=− + + = (1)
Similarly, 0
B
θ
= gives
00
6324
AB
ML
ML ML
EI EI EI
−+ = (2)
Solving Eqs.(1) and (2):
04
B
MM
=
04
A
MM
=
Statics:
00
33
22
AB
MM
RR
LL
=↓ =↑
Continued on next slide
M0
L/
2
L/
2
M0
C
M
A
A
B
=
M
A
M
B
-
R
A
RB
A
B
M
B
A
B
-
A
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 (10.93)
Use Table B.14 (case 3 and 1):
3
(2)
() () 0 () 6
BBwBR CwwL
vv v EI
θ =+= =−
343
(2) (2)
() 0
8623
B
RL
wL wL L
EI EI EI
=− − + =
344 0
128 96 3B
RL
wL wL
EI EI EI
=− − + =
Solving
7
128
B
R
wL=↑
Statics:
2
57 9
128 128
AA
R
wL M wL=↑= ↑
A
L/
2
A
B
=
A
B
M
A
R
B
B
+
R
A
RB
C
L/
2
C
w
w