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,
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 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,

7. 4.10 Reduction of a Simple Distributed Loading
Location of Resultant Force
Solving,
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, = 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,
For resultant force

13. 4.10 Reduction of a Simple Distributed Loading
Solution
For location of line of action,
Checking,

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
View Free Body Diagram
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

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,
For location of resultant force,

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

21. Chapter Summary Moment of a Force
A force produces a turning effect about the point O that does not lie on its line of action
In scalar form, moment magnitude, MO = Fd, where d is the moment arm or perpendicular distance from point O to its line of action of the force
Direction of the moment is defined by right hand rule
For easy solving,
- resolve the force components into x and y components

22. Chapter Summary Moment of a Force
- determine moment of each component about the point
- sum the results
Vector cross product are used in 3D problems
MO = r X F
where r is a position vector that extends from point O to any point on the line of action of F

23. Chapter Summary Moment about a Specified Axis
Projection of the moment onto the axis is obtained to determine the moment of a force about an arbitrary axis provided that the distance perpendicular to both its line of action and the axis can be determined
If distance is unknown, use vector triple product
Ma = ua·r X F
where ua is a unit vector that specifies the direction of the axis and r is the position vector that is directed from any point on the axis to any point on its line of action

24. Chapter Summary Couple Moment
A couple consists of two equal but opposite forces that act a perpendicular distance d apart
Couple tend to produce rotation without translation
Moment of a couple is determined from M = Fd and direction is established using the right-hand rule
If vector cross product is used to determine the couple moment, M = r X F, r extends from any point on the line of action of one of the forces to any point on the line of action of the force F

25. Chapter Summary Reduction of a Force and Couple System
Any system of forces and couples can be reduced to a single resultant force and a single resultant couple moment acting at a point
Resultant force = sum of all the forces in the system
Resultant couple moment = sum of all the forces and the couple moments about the point
Only concurrent, coplanar or parallel force system can be simplified into a single resultant force

26. Chapter Summary Reduction of a Force and Couple System
For concurrent, coplanar or parallel force systems,
- find the location of the resultant force about a point
- equate the moment of the resultant force about the point to moment of the forces and couples in the system about the same point
Repeating the above steps for other force system will yield a wrench, which consists of resultant force and a resultant collinear moment

27. Chapter Summary Distributed Loading
A simple distributed loading can be replaced by a resultant force, which is equivalent to the area under the loading curve
Resultant has a line of action that passes through the centroid or geometric center of the are or volume under the loading diagram

28. Chapter Review

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33. Chapter Review