ME 201
Engineering Mechanics: Statics
Chapter 4 – Part F
4.9 Reduction of a Simple Distributed
Loading
Distributed Loads
Thus far we’ve been working with loads that
are concentrated at a point:
Many times in engineering we need to be
concerned with another type of loading
referred to as distributed loading:
6 kN/m
10 N/m
Distributed Loads
Instead of being concentrated at a point, a
distributed load is spread out over a distance
It can be thought of as a collection of smaller
loads
Distributed Loads
To compute the resultant, FR, of a distributed
load, consider the following:
To find FR, we need to sum an infinite
number of small forces
Distributed Loads
Consider a small differential element, dF
with a width of dx
and a height of w(x)
w
dF
w(x)
x
dx
x
Distributed Loads
w
dF
w(x)
x
dx
x
The area of the element is
dF w( x)dx
Since infinite number of forces, need to integrate
to find FR
FR w( x)dx dA A
l
A
Distributed Loads
Where does FR act?
Can be determined by equating the moments of
the force resultant and the force distribution
FR w( x)dx
l
xFR xw( x)dx
l
xw( x)dx xw( x)dx xdA
x
l
l
FR
w( x)dx
l
This is also the centroid of the area
A
dA
A
Centroids
For simple shapes, centroid can be found in a table
(see back cover of textbook)
Where is the centroid for these common shapes?
h
h
b
b
b
xc
2
h
yc
2
b
xc
3
h
yc
3
Class Exercise
Given:
w=100x N/m
Find:
FR, x
600 N/m
A
B
6m
Example Problem Solution
Given:
w=100x N/m
Find:
FR, x
Solution:
FBD
FR
x
600 N/m
A
B
6m
FR
A
x
B
6m
1
FR 600 N / m 6 m
2
1800 Nm or 1.8 kNm
2
x 6m
3
4m
Example Problem
Given:
trapezoid
Find:
FR, x
w=60x2 N/m
2m
Example Problem Solution
Given:
trapezoid
Find:
FR, x
Solution:
FR integral
w=60x2 N/m
2m
FR w( x)dx
L
2
60 x dx
2
0
x3 2
60 |
3 0
160 N
Example Problem Solution
Given:
trapezoid
Find:
FR, x
Solution:
FR integral
x integral
w=60x2 N/m
2m
xw( x) dx
x
w( x)dx
x60 x dx
60 x dx
x
60 |
4
L
L
2
2
0
2
2
0
4
2
0
3
x
60
3
2
|
0
240
160
1.5 m
0
