a)While analyzing THE distributed load upon this beam, it is best practice to divide it into a minimum of _____ "simpler" distributed loads. (This will greatly reduce efforts in determining the reactions.) b)The highest order distributed load upon this beam is x to the power of _____. c)The uniform distribution has an intensity of _____ lb/ft. d)The resultant force of the uniform distribution has a magnitude of _____ lb. e)The centroid of the uniform distribution is located at x=_____ft as measured from A. f)The resultant force of the nonuniform distribution has a magnitude of _____ lb. g)The centroid of the nonuniform distribution is located at x=_____ft as measured from A.
a)While analyzing THE distributed load upon this beam, it is best practice to divide it into a minimum of _____ "simpler" distributed loads. (This will greatly reduce efforts in determining the reactions.) b)The highest order distributed load upon this beam is x to the power of _____. c)The uniform distribution has an intensity of _____ lb/ft. d)The resultant force of the uniform distribution has a magnitude of _____ lb. e)The centroid of the uniform distribution is located at x=_____ft as measured from A. f)The resultant force of the nonuniform distribution has a magnitude of _____ lb. g)The centroid of the nonuniform distribution is located at x=_____ft as measured from A.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
100%
a)While analyzing THE distributed load upon this beam, it is best practice to divide it into a minimum of _____ "simpler" distributed loads. (This will greatly reduce efforts in determining the reactions.)
b)The highest order distributed load upon this beam is x to the power of _____.
c)The uniform distribution has an intensity of _____ lb/ft.
d)The resultant force of the uniform distribution has a magnitude of _____ lb.
e)The centroid of the uniform distribution is located at x=_____ft as measured from A.
f)The resultant force of the nonuniform distribution has a magnitude of _____ lb.
g)The centroid of the nonuniform distribution is located at x=_____ft as measured from A.
![1 Note that it is usually unnecessary
- 4'
274
Chapter 5 Distributed Forces
Determine the equivalent concentrated load(s) and external reactions for
the simply supported beam which is subjected to the distributed load shown.
120 lb/ft
A
B
SAMPLE PROBLEM 5/11
Helpful Hint
Solution. The area associated with the load distribution is divided into the
rectangular and triangular areas shown. The concentrated-load values are deter-
mined by computing the areas, and these loads are located at the centroids of the
1 respective areas.
Once the concentrated loads are determined, they are placed on the free-
body diagram of the beam along with the external reactions at A and B. Using
principles of equilibrium, we have
to reduce a given distributed load to
a single concentrated load.
(160) (6) = 480 lb
%3D
160 lbt
5'
120 lb/t
1200(5) + 480(8) – RB(10) = 0
120 lb/ft
EMA = 0]
Ans.
RB = 984 lb
(120) (10) = 1200 lb
RĄ(10) – 1200(5) – 480(2) = 0
1200 lb
480 lb
[ZMB = 0]
5' 3
Ans.
RA = 696 lb
RA
|RB
SAMPLE PROBLEM 5/12
wx)
Determine the reaction at the support A of the loaded cantilever beam.
w = wo+ kx³
2024
N/m
1000 N/m
Solution. The constants in the load distribution are found to be wo = 1000
0 N/m and k = 2 N/m*. The load R is then
8 m
x4
1000x +
8
= 10 050 N
R =
w dx =
(1000 + 2x) dx =
Helpful Hints
2 The x-coordinate of the centroid of the area is found by
1 Use caution with the units of the
constants wo and k.
xw dx
x(1000 + 2x3) dx
2 The student should recognize that
the calculation of R and its location x
10 050
(500x2 +
is simply an application of centroids
as treated in Art. 5/3.
10 050
= 4.49 m
From the free-body diagram of the beam, we have
[ZMA = 0]
10 050 N
MA- (10 050)(4.49) = 0
4.49 m
MA =
= 45 100 N-m
MA
[2F, = 0]
Ans.
Ay
A,
Note that A, = 0 by inspection.
= 10 050 N
Ans.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6fa7ac50-e43d-4267-956c-145cd19bd43c%2F09d67b32-99bf-44a6-81b9-01bb916466b8%2Fqsil3ta_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1 Note that it is usually unnecessary
- 4'
274
Chapter 5 Distributed Forces
Determine the equivalent concentrated load(s) and external reactions for
the simply supported beam which is subjected to the distributed load shown.
120 lb/ft
A
B
SAMPLE PROBLEM 5/11
Helpful Hint
Solution. The area associated with the load distribution is divided into the
rectangular and triangular areas shown. The concentrated-load values are deter-
mined by computing the areas, and these loads are located at the centroids of the
1 respective areas.
Once the concentrated loads are determined, they are placed on the free-
body diagram of the beam along with the external reactions at A and B. Using
principles of equilibrium, we have
to reduce a given distributed load to
a single concentrated load.
(160) (6) = 480 lb
%3D
160 lbt
5'
120 lb/t
1200(5) + 480(8) – RB(10) = 0
120 lb/ft
EMA = 0]
Ans.
RB = 984 lb
(120) (10) = 1200 lb
RĄ(10) – 1200(5) – 480(2) = 0
1200 lb
480 lb
[ZMB = 0]
5' 3
Ans.
RA = 696 lb
RA
|RB
SAMPLE PROBLEM 5/12
wx)
Determine the reaction at the support A of the loaded cantilever beam.
w = wo+ kx³
2024
N/m
1000 N/m
Solution. The constants in the load distribution are found to be wo = 1000
0 N/m and k = 2 N/m*. The load R is then
8 m
x4
1000x +
8
= 10 050 N
R =
w dx =
(1000 + 2x) dx =
Helpful Hints
2 The x-coordinate of the centroid of the area is found by
1 Use caution with the units of the
constants wo and k.
xw dx
x(1000 + 2x3) dx
2 The student should recognize that
the calculation of R and its location x
10 050
(500x2 +
is simply an application of centroids
as treated in Art. 5/3.
10 050
= 4.49 m
From the free-body diagram of the beam, we have
[ZMA = 0]
10 050 N
MA- (10 050)(4.49) = 0
4.49 m
MA =
= 45 100 N-m
MA
[2F, = 0]
Ans.
Ay
A,
Note that A, = 0 by inspection.
= 10 050 N
Ans.
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