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SECTION 9.11
Representation of Loads on Beams by Discontinuity Functions
615
Representation of Loads on Beams by Discontinuity Functions
Problem 9.111 through 9.1112
A beam and its loading are shown in
the figure. Using discontinuity functions, write the expression for the
intensity
q
(
x
) of the equivalent distributed load acting on the beam
(include the reactions in the expression for the equivalent load).
Solution 9.111
Cantilever beam
P
a
L
b
x
B
D
A
y
F
ROM EQUILIBRIUM
:
R
A
5
PM
A
5
Pa
U
SE
T
ABLE
92.
52
P
K
x
L
2
1
1
Pa
K
x
L
2
2
1
P
K
x
2
a
L
2
1
q
(
x
)
R
A
K
x
L
2
1
1
M
A
K
x
L
2
2
1
P
K
x
2
a
L
2
1
P
a
L
b
x
B
D
A
y
R
A
M
A
Problem 9.112
Solution 9.112
Cantilever beam
q
a
L
b
x
B
D
A
y
F
ROM EQUILIBRIUM
:
R
A
5
qb
U
SE
T
ABLE
92.
1
q
K
x
2
a
L
0
2
q
K
x
2
L
L
0
qb
K
x
L
2
1
1
qb
2
(2
a
1
b
)
K
x
L
2
2
q
(
x
)
R
A
K
x
L
2
1
1
M
A
K
x
L
2
2
1
q
K
x
2
a
L
0
2
q
K
x
2
L
L
0
M
A
5
qb
2
a
1
b
)
q
a
L
b
x
B
D
A
y
M
A
R
A
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View Full Document Problem 9.113
Solution 9.113
Cantilever beam
616
CHAPTER 9
Deflections of Beams
q
= 2 k/ft
P
= 4 k
3 ft
6 ft
x
B
AD
y
a
5
6 ft
5
72 in.
b
5
3 ft
5
36 in.
L
5
9 ft
5
108 in.
q
5
2 k
/
ft
5
1
6
k
/
in.
F
ROM EQUILIBRIUM
:
R
A
5
16 k
M
A
5
864 kin.
U
SE
T
ABLE
92.
Units: kips, inches
(Units:
x
5
in.,
q
5
k
/
in.)
1
4
K
x
2
108
L
2
1
52
16
K
x
L
2
1
1
864
K
x
L
2
2
1
1
6
K
x
L
0
2
1
6
K
x
2
72
L
0
1
P
K
x
2
L
L
2
1
q
(
x
)
R
A
K
x
L
2
1
1
M
A
K
x
L
2
2
1
q
K
x
L
0
2
q
K
x
2
a
L
0
q
= 2 k/ft
P
= 4 k
b
5
3 ft
a
5
6 ft
x
B
y
M
A
R
A
Problem 9.114
Solution 9.114
Simple beam
b
L
a
x
B
D
A
y
P
F
ROM EQUILIBRIUM
:
U
SE
T
ABLE
92.
2
Pa
L
K
x
2
L
L
2
1
Pb
L
K
x
L
2
1
1
P
K
x
2
a
L
2
1
q
(
x
)
R
A
K
x
L
2
1
1
P
K
x
2
a
L
2
1
2
R
B
K
x
2
L
L
2
1
R
B
5
Pa
L
R
A
5
Pb
L
b
L
a
x
B
D
A
y
P
R
A
R
B
Problem 9.115
Solution 9.115
Simple beam
SECTION 9.11
Representation of Loads on Beams by Discontinuity Functions
617
b
L
a
x
B
D
A
y
M
0
F
ROM EQUILIBRIUM
:
U
SE
T
ABLE
92.
1
M
0
L
K
x
2
L
L
2
1
52
M
0
L
K
x
L
2
1
1
M
0
K
x
2
a
L
2
2
q
(
x
)
R
A
K
x
L
2
1
1
M
0
K
x
2
a
L
2
2
1
R
B
K
x
2
L
L
2
1
R
B
5
M
0
L
(downward)
R
A
5
M
0
L
b
L
a
x
B
D
A
y
M
0
R
A
R
B
Problem 9.116
Solution 9.116
Simple beam
a
L
x
B
D
a
A
y
P
E
P
F
ROM EQUILIBRIUM
:
R
A
5
R
B
5
P
U
SE
T
ABLE
92.
2
P
K
x
2
L
L
2
1
P
K
x
L
2
1
1
P
K
x
2
a
L
2
1
1
P
K
x
2
L
1
a
L
2
1
2
R
B
K
x
2
L
L
2
1
q
(
x
)
R
A
K
x
L
2
1
1
P
K
x
2
a
L
2
1
1
P
K
x
2
L
1
a
L
2
1
a
L
x
B
D
a
A
y
P
E
P
R
A
R
B
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View Full Document 618
CHAPTER 9
Deflections of Beams
Problem 9.117
Solution 9.117
Simple beam
10 ft
16 ft
x
B
D
A
y
M
0
= 20 kft
P
= 18 k
M
0
5
20 kft
5
240 kin.
P
5
18 k
a
5
16 ft
5
192 in.
b
5
10 ft
5
120 in.
L
5
26 ft
5
312 in.
F
ROM EQUILIBRIUM
:
R
A
5
7.692 k
R
B
5
10.308 k
U
SE
T
ABLE
92.
Units: kips, inches
(Units:
x
5
in.,
q
5
k
/
in.)
2
10.308
K
x
2
312
L
2
1
52
7.692
K
x
L
2
1
1
240
K
x
L
2
2
1
18
K
x
2
192
L
2
1
2
R
B
K
x
2
L
L
2
1
q
(
x
)
R
A
K
x
L
2
1
1
M
0
K
x
L
2
2
1
P
K
x
2
a
L
2
1
b
= 10 ft
a
= 16 ft
x
B
D
A
y
M
0
= 20 kft
P
= 18 k
R
A
R
B
Problem 9.118
Solution 9.118
Simple beam
a
L
q
x
B
D
A
y
F
ROM EQUILIBRIUM
:
U
SE
T
ABLE
92.
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This note was uploaded on 03/19/2008 for the course E M 316 taught by Professor Korkolis during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Korkolis

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