1. 4.10 Reduction of a Simple
Distributed Loading
Large surface area of a body may be
subjected to distributed loadings such as
those caused by wind, fluids, or weight of
material supported over body’s surface
Intensity of these loadings at each point
on the surface is defined as the pressure p
Pressure is measured in pascals (Pa)
1 Pa = 1N/m2

2. 4.10 Reduction of a Simple
Distributed Loading
Most common case of distributed pressure loading is
uniform loading along one axis of a flat rectangular
body
Direction of the intensity of the pressure load is
indicated by arrows shown on the load-intensity
diagram
Entire loading on the plate is a
system of parallel forces,
infinite in number, each
acting on a separate
differential area of the plate

3. 4.10 Reduction of a Simple
Distributed Loading
Loading function p = p(x) Pa, is a function of x
since pressure is uniform along the y axis
Multiply the loading function by the width
w = p(x)N/m2]a m = w(x) N/m
Loading function is a measure
of load distribution along line
y = 0, which is in the symmetry
of the loading
Measured as force per unit
length rather than per unit area

4. 4.10 Reduction of a Simple
Distributed Loading
Load-intensity diagram for w = w(x) can
be represented by a system of coplanar
parallel
This system of forces can be simplified into
a single resultant force FR
and its location can be
specified

5. 4.10 Reduction of a
Simple Distributed Loading
Magnitude of Resultant Force
FR = ∑F
Integration is used for infinite number of parallel
forces dF acting along the plate
For entire plate length,
+ ↓ FR = ΣF ; FR = ∫ w( x)dx = ∫ dA = A
L A
Magnitude of resultant force is equal to the total
area A under the loading diagram w = w(x)

6. 4.10 Reduction of a Simple
Distributed Loading
Location of Resultant Force
MR = ∑MO
Location x of the line of action of FR can be
determined by equating the moments of the
force resultant and the force distribution about
point O
dF produces a moment of
xdF = x w(x) dx about O
For the entire plate,
M Ro = ΣM O ; x FR = ∫ xw( x)dx
L

7. 4.10 Reduction of a
Simple Distributed Loading
Location of Resultant Force
Solving,
∫ xw( x)dx ∫ xdA
x= L
= A
∫ w( x)dx
L
∫ dA
A
Resultant force has a line of action which passes
through the centroid C (geometric center) of the
area defined by the distributed loading diagram
w(x)

8. 4.10 Reduction of a Simple
Distributed Loading
Location of Resultant Force
Consider 3D pressure loading p(x), the resultant
force has a magnitude equal to the volume under
the distributed-loading curve p = p(x) and a line of
action which passes through the centroid (geometric
center) of this volume
Distribution diagram can be
in any form of shapes such
as rectangle, triangle etc

9. 4.10 Reduction of a Simple
Distributed Loading
Beam supporting this stack of lumber is subjected
to a uniform distributed loading, and so the load-
intensity diagram has a rectangular shape
If the load-intensity is wo, resultant is determined
from the are of the rectangle
FR = wob

10. 4.10 Reduction of a Simple
Distributed Loading
Line of action passes through the centroid or
center of the rectangle,x = a + b/2
Resultant is equivalent to the distributed load
Both loadings produce same “external” effects or
support reactions on the beam

11. 4.10 Reduction of a Simple
Distributed Loading
Example 4.20
Determine the magnitude and location of the
equivalent resultant force acting on the shaft

12. 4.10 Reduction of a Simple
Distributed Loading
Solution
For the colored differential area element,
dA = wdx = 60 x 2dx
For resultant force
FR = ΣF ;
2
FR = ∫ dA = ∫ 60 x 2dx
A 0
x 3 2
 23 03 
= 60  = 60 − 
 3 0 3 3
= 160 N

13. 4.10 Reduction of a Simple
Distributed Loading
Solution
For location of line of action,
2 2
 x4   24 0 4 
∫ xdA ∫ x(60 x )dx 60 4  60 4 − 4 
2

0
x=A =0 = =  
∫
dA 160 160 160
A
= 1.5m
Checking,
ab 2m(240 N / m)
A= = = 160
3 3
3 3
x = a = (2m) = 1.5m
4 4

14. 4.10 Reduction of a Simple
Distributed Loading
Example 4.21
A distributed loading of p = 800x Pa acts
over the top surface of the beam. Determine
the magnitude and location of the equivalent
force.

15. 4.10 Reduction of a Simple
Distributed Loading
Solution
Loading function of p = 800x Pa indicates that
the load intensity varies uniformly from p = 0 at
x = 0 to p = 7200Pa at x = 9m
For loading,
w = (800x N/m2)(0.2m) = (160x) N/m
Magnitude of resultant force
= area under the triangle
FR = ½(9m)(1440N/m)
= 6480 N = 6.48 kN

16. 4.10 Reduction of a Simple
Distributed Loading
Solution
Resultant force acts through the centroid of
the volume of the loading diagram p = p(x)
FR intersects the x-y plane at point (6m, 0)
Magnitude of resultant force
= volume under the triangle
FR = V = ½(7200N/m2)(0.2m)
= 6.48 kN

17. 4.10 Reduction of a Simple
Distributed Loading
Example 4.22
The granular material exerts the distributed
loading on the beam. Determine the magnitude
and location of the equivalent resultant of this
load

18. 4.10 Reduction of a Simple
Distributed Loading
Solution
Area of loading diagram is trapezoid
Magnitude of each force = associated area
F1 = ½(9m)(50kN/m) = 225kN
F2 = ½(9m)(100kN/m) = 450kN
Line of these parallel forces act
through the centroid of associated
areas and insect beams at
1 1
x1 = (9m) = 3m, x2 = (9m) = 4.5m
3 2

19. 4.10 Reduction of a Simple
Distributed Loading
Solution
Two parallel Forces F1 and F2 can be reduced to a
single resultant force FR
For magnitude of resultant force,
+ ↓ FR = ΣF ;
FR = 225 + 450 x = 675kN
For location of resultant force,
M Ro = ΣM O ;
x (675) = 3(225) + 4.5(450)
x = 4m

20. 4.10 Reduction of a Simple
Distributed Loading
Solution
*Note:
Trapezoidal area can be divided into two triangular
areas,
F1 = ½(9m)(100kN/m) = 450kN
F2 = ½(9m)(50kN/m) = 225kN
1 1
x1 = (9m) = 3m, x2 = (9m) = 3m
3 3