sm10 98 101

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________________________________________________________________________
PROBLEMS (10.98 through 10.101) A beam is supported and loaded as shown in Fig. P10.98
and P10.101. What are the reactions at each support?
SOLUTION (10.98)
Using Table B.14 (cases 10 and 8):
323
(2 )
( ) [(20) 2(3 )(2 ) (3 ) ]
24
Bw wa
vaaa
EI
=− − +
4
11
12 wa
EI
=−
222
(2 )
() [(2) (3) ]
6(3 )
B
BR Ra a
vaaa
aEI
=−+
4
4
9wa
EI
=
Boundary condition,
() () 0
BBwBR
vv v=+=, gives:
33
16
B
R
wa=↑
Statics:
13 1
16 8
AC
R
wa R wa=↑ =↑
SOLUTION (10.99)
Continued on next slide
B
RB
R
B
Actual
Loading
A
2a
A
=
R
A
RC
C
a
w
C
Equivalent loadings
B
A
C
+
B
w
R
B
Actual
Loading
A
L
A
=
R
A
RC
C
L
C
Equivalent loading
B
A
C
+
B
M0
M0
R
B

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 (cases 7 and 9):
322
0
(2 ) [(2)]0
48 6(2 )
B
BML
RL
vLL
EI L EI
=− − − =
0
10
64
BM
RL =− + =
or
0
3
2
BM
RL =↓
Statics:
0
1
0: 4
AC
M
MR
L
==↑
∑
0
5
0: 4
yA
M
FR
L
==↑
∑
SOLUTION (10.100)
Using Table B.14 (case 10, 1, and 9):
() () 0
CCwCR
vv v=+= (1)
where,
34
() ()() ()
24 24
Cw Bw wL L
vLL
EI EI
θ ==−=
33
()
() ()() ()
33 3
CC C
CR BR
R
LRLL RL
vL L
EI EI EI
θ =+= +
3
2
3C
R
L
EI
=
Equation (1) yields
1
16
C
R
wL=↓
Continued on next slide
R
C
B
R
B
Actual
Loading
A
L
A
=
R
A
RC
C
L
w
C
Equivalent loadings
B
A
C
+
B
w

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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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Statics:
57
816
BA
R
wL R wL=↑ = ↑
SOLUTION (10.101)
Using Table B.14 (case 3 and 1):
() ()
BBwBR
vv v=+
34
83
B
R
L
wL
EI EI
=− + (1)
Since
B
B
R
vk =− (2)
Equations (1) and (2) give
3
33
1
BwL B
REI
kL
=↑
+
(Interestingly, for a rigid spring (k→∞): 38
B
R
wL
=
and for a free end
(0k=): 0.
B
R=)
Statics:
AB
RwLR=−↑
2
1
2
AB
M
wL R L=−
where B
R
is given above.
k
M
A
B
RB
R
B
Actual
Loading
A
L
A
=
R
A
w
Equivalent loading
B
A
+
B
w