sm10 150 153

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________________________________________________________________________
PROBLEMS (10.150 through 10.153) Determine all of the reactions for the propped cantilever
beam AB shown in Figs. P10.150 through P10.153.
SOLUTION (10.150)
Let
A
M
be redundant.
22
12
38 12
LwL wL
A==
211
()
22
AA
AMLML=− =−
24
/12
2
0()()
23243
A
BA
M
L
LLwL
EIt A A== + = −
or
2
1
8
A
M
wL=
Statics:
53
88
AB
R
wL R wL=↑ =↑
SOLUTION (10.151)
Assume
A
R
as redundant
10
AML
=
2
211
()
22
BB
ARLLRL=− =−
Continued on next slide
M
x
w
Parabola
A
1
-MA
wL2/8
L
tangent at A
3L/2
R
A RB
M
A
-wL/2
V
wL2/8
x
L/
2
A
wL/2
B
A
2
M
x
-RBL
L
tangent at A
2L/3
R
A RB
M
A
M
0
L/
2
A
B
A
2
M0
A
1

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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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/12 0
21 1
0()()
2332
BA B
LL
EIt A A R L M== + = −
or 0
3
2
BM
RL =↓
Statics:
0
3
0: 2
yA
M
FR
L
==↑
∑
0
1
0: 2
AA
M
MM==
∑
SOLUTION (10.152)
Consider
A
R
as redundant.
2
111
()
22
AA
ARLLRL==
23
00
21()
46 24
wL wL
AL=− =−
/12 0
2411
0()()
33330
AB A
LL
EIt A A R w L== + = −
or 0
1
10
A
R
wL=↑
Statics:
0
2
0: 5
yB
FRwL==↑
∑
2
0
1
0: 15
BB
M
MwL==
∑
Continued on next slide
tangent at B
M
B
A
1
x
B
M
A
-w0L2/6
A
2
R
A RB
RAL
2L
/
3
L
Cubic
4L/3
w0

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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 (10.153)
Choose
A
R
as redundant.
2
2
21()
22 8
LwL
Mw=− =−
2
11()()
22
AA
R
LRL
AL
EI EI
==
23
21()()
38 2 48
wL L wL
AEI EI =− =−
Requirement:
12
23
0()( )
328
AB LLL
tAA== + +
34
7
3 384
A
RL wL
EI EI
=−
Solving
7
128
AwL
R=↑
A
1
x
RAL
2L
/
3
M
2
B
A
Parabolic
Spandrel
C
A
2
3L/8
B
M/EI
Tangent at A
A
L/
2