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,

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

LoadingLarge 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

4.10 Reduction of a Simple Distributed

Loading

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

4.10 Reduction of a Simple Distributed

Loading

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

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

LoadingLoad-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

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)

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

Loading

AdAdxxwFFFL A

RR )(;

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

LoadingLocation 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,

x

L

RORo dxxxwFxMM )(;

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

LoadingLocation 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)

A

A

L

L

dA

xdA

dxxw

dxxxw

x)(

)(

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

LoadingLocation 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

4.10 Reduction of a Simple Distributed

Loading

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

4.10 Reduction of a Simple Distributed

Loading

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

x

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

LoadingExample 4.20Determine the magnitude and location of the equivalent resultant force acting on the shaft

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

LoadingSolutionFor the colored differential area element,

For resultant force

N

x

dxxdAF

FF

dxxwdxdA

AR

R

160

30

32

603

60

60

;

60

332

0

3

2

0

2

2

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

LoadingSolutionFor location of line of action,

Checking,

mmax

mNmabA

m

xdxxx

dA

xdA

x

A

A

5.1)2(43

43

1603

)/240(23

5.1

160

40

42

60

160

460

160

)60(44

2

0

42

0

2

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

LoadingExample 4.21A distributed loading of p = 800x Pa acts over the top surface of the beam. Determine the magnitude and location of the equivalent force.

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

LoadingSolution 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 triangleFR = ½(9m)(1440N/m) = 6480 N = 6.48 kN

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

LoadingSolutionResultant 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 triangleFR = V = ½(7200N/m2)(0.2m)

= 6.48 kN

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

LoadingExample 4.22The granular material exerts the distributed loading on the beam. Determine the magnitude and location of the equivalent resultant of this load

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

LoadingSolution 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

mmxmmx 5.4)9(21

,3)9(31

21

View Free Body Diagram

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

LoadingSolution 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,

mx

x

MM

kNxF

FF

ORo

R

R

4

)450(5.4)225(3)675(

;

675450225

;

Solution*Note: Trapezoidal area can be divided into two

triangular areas,F1 = ½(9m)(100kN/m) = 450kN

F2 = ½(9m)(50kN/m) = 225kN

4.10 Reduction of a Simple Distributed

Loading

4.10 Reduction of a Simple Distributed

Loading

mmxmmx 3)9(31

,3)9(31

21

Chapter SummaryChapter 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

Chapter SummaryChapter 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

Chapter SummaryChapter 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 Fwhere 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

Chapter SummaryChapter 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

Chapter SummaryChapter 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

Chapter SummaryChapter 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

Chapter SummaryChapter 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

Chapter ReviewChapter Review

Chapter ReviewChapter Review

Chapter ReviewChapter Review

Chapter ReviewChapter Review

Chapter ReviewChapter Review

Chapter ReviewChapter Review