SAMPLE PROBLEM 5/11
Determine the equivalent concentrated load(s) and external reactions for
the simply supported beam which is subjected to the distributed load shown.
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
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 Aand B. Using
principles of equilibrium, we have
Ans.
Ans.
R
A
⫽ 696 lb
R
A
(10) ⫺ 1200(5) ⫺ 480(2) ⫽ 0[ΣM
B
⫽ 0]
R
B
⫽ 984 lb
1200(5) ⫹ 480(8) ⫺ R
B
(10) ⫽ 0[ΣM
A
⫽ 0]
274 Chapter 5 Distributed Forces
A
8 m
B
w(x)
1000 N/m 202
4
N/m
x
w = w0 + kx3
5′8′
120 lb/ft 120 lb/f
t
160 lb/f
t
(120) (10) = 1200 lb
(160) (6) = 480 lb
1200 lb 480 lb
5′
A
RARB
AB
B
3′
1
–
2
AB
4′6′
120 lb/ft 280 lb/f
t
4.49 m
Ay
A
AxB
MA
10 050 N
y
x
Helpful Hint
Note that it is usually unnecessary
to reduce a given distributed load to
a single concentrated load.
SAMPLE PROBLEM 5/12
Determine the reaction at the support Aof the loaded cantilever beam.
Solution.
The constants in the load distribution are found to be w0⫽1000
N/m and k⫽2 N/m4. The load Ris then
The x-coordinate of the centroid of the area is found by
From the free-body diagram of the beam, we have
Ans.
Ans.
Note that Ax⫽0 by inspection.
A
y
⫽ 10 050 N[ΣF
y
⫽ 0]
M
A
⫽ 45 100 N 䡠m
M
A
⫺ (10 050)(4.49) ⫽ 0[ΣM
A
⫽ 0]
⫽ 1
10 050 (500x
2
⫹ 2
5x
5
)兩
8
0
⫽ 4.49 m
x ⫽
冕
xw dx
R ⫽ 1
10 050
冕
8
0
x(1000 ⫹ 2x
3
) dx
R ⫽
冕
w dx ⫽
冕
8
0
(1000 ⫹ 2x
3
) dx ⫽
冢
1000x ⫹ x
4
2
冣
冏
8
0
⫽ 10 050 N Helpful Hints
Use caution with the units of the
constants w0and k.
The student should recognize that
the calculation of Rand its location
is simply an application of centroids
as treated in Art. 5/3.
x
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