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