Engineering Mechanics: Statics Chapter 4: Force System Resultants Chapter 4: Force System...

Engineering Mechanics: StaticsEngineering Mechanics: Statics

Chapter 4: Force System

Resultants

Chapter 4: Force System

Resultants

Chapter ObjectivesChapter Objectives

To discuss the concept of the moment of a force and show how to calculate it in two and three dimensions.

To provide a method for finding the moment of a force about a specified axis.

To define the moment of a couple. To present methods for determining the resultants of

non-concurrent force systems. To indicate how to reduce a simple distributed

loading to a resultant force having a specified location.

Chapter OutlineChapter Outline

Moment of a Force – Scalar Formation

Cross ProductMoment of Force – Vector

FormulationPrinciple of MomentsMoment of a Force about a Specified

Axis

Chapter OutlineChapter Outline

Moment of a CoupleEquivalent SystemResultants of a Force and Couple

SystemReduction of a Simple Distributed

Loading

Moment of a force about a point or axis – a measure of the tendency of the force to cause a body to rotate about the point or axisCase 1Consider horizontal force Fx,

which acts perpendicular to the handle of the wrench and is located dy from the point O

4.1 Moment of a Force – Scalar

Formation


Formation

Fx tends to turn the pipe about the z axis

The larger the force or the distance dy, the greater the turning effect

Torque – tendency of rotation caused by Fx

or simple moment (Mo) z

4.1 Moment of a Force

– Scalar Formation



Moment axis (z) is perpendicular to shaded plane (x-y)

Fx and dy lies on the shaded plane (x-y)Moment axis (z) intersects

the plane at point O





Case 2Apply force Fz to the wrenchPipe does not rotate about z axisTendency to rotate about x axisThe pipe may not actually

rotate Fz creates tendency

for rotation so moment (Mo) x is produced





Case 2Moment axis (x) is perpendicular to

shaded plane (y-z)

Fz and dy lies on the shaded plane (y-z)








– Scalar FormationCase 3Apply force Fy to the wrenchNo moment is produced about point OLack of tendency to rotate

as line of action passes through O




– Scalar FormationIn General Consider the force F and the point O which lies

in the shaded plane The moment MO about point O,

or about an axis passingthrough O and perpendicularto the plane, is a vector quantity

Moment MO has its specified

magnitude and direction


Formation


FormationMagnitudeFor magnitude of MO,

MO = Fdwhere d = moment arm or perpendicular distance from the axis at point O to its line of action of the force

Units for moment is N.m


Formation


FormationDirectionDirection of MO is specified by

using “right hand rule”- fingers of the right hand are curled to follow the sense of rotation when force rotates about point O


Formation


FormationDirection

- Thumb points along the moment axis to give the direction and sense of the moment vector- Moment vector is upwards and perpendicular to the shaded plane


Formation


FormationDirectionMO is shown by a vector arrow

with a curl to distinguish it from force vectorExample (Fig b)MO is represented by the

counterclockwise curl, which indicates the action of F


Formation


FormationDirectionArrowhead shows the sense of

rotation caused by FUsing the right hand rule, the

direction and sense of the moment vector points out of the page

In 2D problems, moment of the force is found about a point O


Formation


FormationDirectionMoment acts about an axis

perpendicular to the plane containing F and d

Moment axis intersects the plane at point O


Formation


FormationResultant Moment of a System of Coplanar ForcesResultant moment, MRo = addition of the

moments of all the forces algebraically since all moment forces are collinear

MRo = ∑Fd

taking clockwise to be positive


Formation


FormationResultant Moment of a System of Coplanar ForcesA clockwise curl is written along the

equation to indicate that a positive moment if directed along the + z axis and negative along the – z axis




– Scalar Formation Moment of a force does not always cause rotation

Force F tends to rotate the beam clockwise about A with moment

MA = FdA

Force F tends to rotate the beam counterclockwise about B with moment

MB = FdB

Hence support at A prevents the rotation


Formation


FormationExample 4.1For each case, determine the moment of

the force about point O


Formation


FormationSolution Line of action is extended as a dashed line

to establish moment arm d Tendency to rotate is indicated and the

orbit is shown as a colored curl

)(.5.37)75.0)(50()(

)(.200)2)(100()(

CWmNmNMb

CWmNmNMa

o

o


Formation


FormationSolution

)(.0.21)14)(7()(

)(.4.42)45sin1)(60()(

)(.229)30cos24)(40()(

CCWmkNmmkNMe

CCWmNmNMd

CWmNmmNMc

o

o

o


Formation


FormationExample 4.2Determine the moments of the 800N force acting on the frame about points A, B, C and D.

4.1 Moment of a Force – Scalar Formation

4.1 Moment of a Force – Scalar Formation

SolutionScalar Analysis

Line of action of F passes through C )(.400)5.0)(800(

.0)0)(800(

)(.1200)5.1)(800(

)(.2000)5.2)(800(

CCWmNmNM

mkNmNM

CWmNmNM

CWmNmNM

D

C

B

A

4.2 Cross Product4.2 Cross Product

Cross product of two vectors A and B yields C, which is written as

C = A X BRead as “C equals A cross B”


Magnitude Magnitude of C is defined as the

product of the magnitudes of A and B and the sine of the angle θ between their tails

For angle θ, 0° ≤ θ ≤ 180°Therefore,

C = AB sinθ


Direction Vector C has a direction that is

perpendicular to the plane containing A and B such that C is specified by the right hand rule- Curling the fingers of the righthand form vector A (cross) to vector B- Thumb points in the direction of vector C


Expressing vector C when magnitude and direction are known

C = A X B = (AB sinθ)uC

where scalar AB sinθ defines the magnitude of vector C unit vector uC

defines the direction of vector C


Laws of Operations1. Commutative law is not valid

A X B ≠ B X ARather,

A X B = - B X A Shown by the right hand rule Cross product A X B yields a vector opposite

in direction to CB X A = -C


Laws of Operations2. Multiplication by a Scalar

a( A X B ) = (aA) X B = A X (aB) = ( A X B )a

3. Distributive Law A X ( B + D ) = ( A X B ) + ( A X D )

Proper order of the cross product must be maintained since they are not commutative


Cartesian Vector FormulationUse C = AB sinθ on pair of

Cartesian unit vectorsExampleFor i X j, (i)(j)(sin90°) = (1)(1)(1) = 1


Laws of Operations In a similar manner,

i X j = k i X k = -j i X i = 0j X k = i j X i = -k j X j = 0k X i = j k X j = -i k X k = 0

Use the circle for the results. Crossing CCW yield positive and CW yields negative results


Laws of Operations Consider cross product of vector A and B

A X B = (Axi + Ayj + Azk) X (Bxi + Byj + Bzk)

= AxBx (i X i) + AxBy (i X j) + AxBz (i X k)

+ AyBx (j X i) + AyBy (j X j) + AyBz (j X k) + AzBx (k X i) +AzBy (k X j) +AzBz (k X k)

= (AyBz – AzBy)i – (AxBz - AzBx)j + (AxBy –

AyBx)k


Laws of Operations In determinant form,

zyx

zyx

BBB

AAA

kji

BXA

4.3 Moment of Force - Vector Formulation4.3 Moment of Force - Vector Formulation

Moment of force F about point O can be expressed using cross product

MO = r X F

where r represents position vector from O to any pointlying on the line of action of F


MagnitudeFor magnitude of cross product,

MO = rF sinθwhere θ is the angle measured between tails of r and F

Treat r as a sliding vector. Since d = r sinθ,

MO = rF sinθ = F (rsinθ) = Fd


DirectionDirection and sense of MO are determined

by right-hand rule - Extend r to the dashed position - Curl fingers from r towards F- Direction of MO is the same

as the direction of the thumb


Direction*Note:

- “curl” of the fingers indicates the sense of rotation- Maintain proper order of r and F since cross product is not commutative


Principle of TransmissibilityFor force F applied at any point A,

moment created about O is MO = rA x F

F has the properties of a sliding vector and therefore act at any point along its line of action and still create the same moment about O


Cartesian Vector Formulation For force expressed in Cartesian

form,

where rx, ry, rz represent the x, y, zcomponents of the position vectorand Fx, Fy, Fz represent that of the force vector

zyx

zyxO

FFF

rrr

kji

FXrM


Cartesian Vector Formulation With the determinant expended,

MO = (ryFz – rzFy)i – (rxFz - rzFx)j + (rxFy – yFx)k

MO is always perpendicular to

the plane containing r and F Computation of moment by cross

product is better than scalar for 3D problems


Cartesian Vector FormulationResultant moment of forces about

point O can be determined by vector addition

MRo = ∑(r x F)


Moment of force F about point A, pulling on cable BC at any point along its line of action, will remain constant

Given the perpendicular distance from A to cable is rd

MA = rdF In 3D problems,

MA = rBC x F


Example 4.4 The pole is subjected to a 60N force that

is directed from C to B. Determine the magnitude of the moment created by this force about the support at A.


Solution Either one of the two position vectors can

be used for the solution, since MA = rB x F or MA = rC x F

Position vectors are represented as rB = {1i + 3j + 2k} m and

rC = {3i + 4j} m Force F has magnitude 60N

and is directed from C to B


Solution

Substitute into determinant formulation

kji

kji

FXrM

Nkji

kjiN

uNF

BA

F

)]40(3)20(1[)]40(2)40(1[)]20(2)40(3[

402040

231

402040

)2()1()2(

)092)493)31()60(

)60(

222


SolutionOr

Substitute into determinant formulation

For magnitude,

mN

M

mNkjiM

kji

kji

FXrM

A

A

CA

.224

)100()120()160(

.100120160

)]40(4)20(3[)]40(0)40(3[)]20(0)40(4[

402040

043

222

4.4 Principles of Moments4.4 Principles of Moments

Also known as Varignon’s Theorem“Moment of a force about a point is equal to the sum of the moments of the forces’ components about the point”

For F = F1 + F2,

MO = r X F1 + r X F2

= r X (F1 + F2)

= r X F


The guy cable exerts a force F on the pole and creates a moment about the base at A

MA = Fd If the force is replaced

by Fx and Fy at point B where the cable acts on the pole, the sum of moment about point A yields the same resultant moment


Fy create zero moment about A

MA = Fxh Apply principle of

transmissibility and slide the force where line of action intersects the ground at C, Fx create zero moment about A

MA = Fyb


Example 4.6The force F acts at the end of the angle bracket. Determine the moment of the

force about point O.


SolutionMethod 1MO = 400sin30°N(0.2m)-400cos30°N(0.4m)

= -98.6N.m = 98.6N.m (CCW)

As a Cartesian vector, MO = {-98.6k}N.m


SolutionMethod 2: Express as Cartesian vector

r = {0.4i – 0.2j}NF = {400sin30°i – 400cos30°j}N = {200.0i – 346.4j}N

For moment,

mNk

kji

FXrMO

.6.98

04.3460.200

02.04.0

4.6 Moment of a Couple4.6 Moment of a Couple

Couple - two parallel forces - same magnitude but opposite direction- separated by perpendicular distance d

Resultant force = 0 Tendency to rotate in specified direction Couple moment = sum of

moments of both couple forces about any arbitrary point


Example Position vectors rA and rA are directed from

O to A and B, lying on the line of action of F and

–F Couple moment about O

M = rA X (-F) + rA X (F) Couple moment about A

M = r X Fsince moment of –F about A = 0


A couple moment is a free vector- It can act at any point since M depends only on the position vector r directed between forces and not position vectors rA and rB, directed from O to the forces

- Unlike moment of force, it do not require a definite point or axis


Scalar FormulationMagnitude of couple

moment M = Fd

Direction and sense are determined by right hand rule

In all cases, M acts perpendicular to plane containing the forces


Vector Formulation For couple moment,

M = r X F If moments are taken about point

A, moment of –F is zero about this point

r is crossed with the force to which it is directed


Equivalent CouplesTwo couples are equivalent if they

produce the same momentSince moment produced by the

couple is always perpendicular to the plane containing the forces, forces of equal couples either lie on the same plane or plane parallel to one another


Resultant Couple MomentCouple moments are free vectors

and may be applied to any point P and added vectorially

For resultant moment of two couples at point P,

MR = M1 + M2

For more than 2 moments,MR = ∑(r X F)


Frictional forces (floor) on the blades of the machine creates a moment Mc that tends to turn it

An equal and opposite moment must be applied by the operator to prevent turning

Couple moment Mc = Fd is applied on the handle


Example 4.10A couple acts on the gear teeth. Replace it by an equivalent couple having a pair of forces that cat through points A and B.


Solution Magnitude of couple

M = Fd = (40)(0.6) = 24N.m Direction out of the page since

forces tend to rotate CW M is a free vector and can

be placed anywhere


Solution To preserve CCW motion,

vertical forces acting through points A and B must be directed as shown

For magnitude of each force, M = Fd24N.m = F(0.2m)F = 120N

4.7 Equivalent System4.7 Equivalent System

A force has the effect of both translating and rotating a body

The extent of the effect depends on how and where the force is applied

We can simplify a system of forces and moments into a single resultant and moment acting at a specified point O

A system of forces and moments is then equivalent to the single resultant force and moment acting at a specified point O


Point O is on the Line of Action Consider body subjected to force F applied to

point A Apply force to point O without altering

external effects on body- Apply equal but opposite forces F and –F at O


Point O is on the Line of Action- Two forces indicated by the slash across them can be cancelled, leaving force at point O- An equivalent system has be maintained between each of the diagrams, shown by the equal signs


Point O is on the Line of Action- Force has been simply transmitted along its line of action from point A to point O- External effects remain unchanged after force is moved- Internal effects depend on location of F


Point O is Not on the Line of ActionF is to be moved to point ) without altering

the external effects on the bodyApply equal and opposite forces at point OThe two forces indicated by a slash across

them, form a couple that has a moment perpendicular to F


Point O is Not on the Line of ActionThe moment is defined by cross product

M = r X FCouple moment is free vector and can be

applied to any point P on the body

4.8 Resultants of a Force and Couple

System


System Consider a rigid body Since O does not lies on the line of

action, an equivalent effect is produced if the forces are moved to point O and the corresponding moments are

M1 = r1 X F1 and M2 = r2 X F2

For resultant forces and moments,FR = F1 + F2 and MR = M1 +

M2


System


System Equivalency is maintained thus each

force and couple system cause the same external effects

Both magnitude and direction of FR do not depend on the location of point O

MRo depends on location of point O since M1 and M2 are determined using position vectors r1 and r2

MRo is a free vector and can acts on any point on the body


System


System Simplifying any force and couple

system, FR = ∑F

MR = ∑MC + ∑MO

If the force system lies on the x-y plane and any couple moments are perpendicular to this plane,

FRx = ∑Fx

FRy = ∑Fy

MRo = ∑MC + ∑MO


System


SystemProcedure for Analysis When applying the following equations,

FR = ∑F

MR = ∑MC + ∑MO

FRx = ∑Fx

FRy = ∑Fy

MRo = ∑MC + ∑MO

Establish the coordinate axes with the origin located at the point O and the axes having a selected orientation

4.8 Resultants of a Force and Couple System


Procedure for Analysis Force Summation For coplanar force system, resolve each force

into x and y components If the component is directed along the positive

x or y axis, it represent a positive scalar If the component is directed along the

negative x or y axis, it represent a negative scalar

In 3D problems, represent forces as Cartesian vector before force summation



Procedure for Analysis Moment Summation For moment of coplanar force system

about point O, use Principle of Moment Determine the moments of each

components rather than of the force itself In 3D problems, use vector cross product

to determine moment of each force Position vectors extend from point O to any

point on the line of action of each force



Example 4.14 Replace the forces acting on the brace by an equivalent resultant force and couple moment acting at point A.



Solution Force Summation For x and y components of resultant force,

NN

NNF

FF

NN

NNF

FF

Ry

yRy

Rx

xRx

8.8828.882

54sin400600

;

8.3828.382

54cos400100

;



Solution For magnitude of resultant force

For direction of resultant force

6.66

8.382

8.882tantan

962

)8.882()8.382()()(

11

2222

Rx

Ry

RyRxR

F

F

N

FFF



Solution Moment Summation Summation of moments about point A,

When MRA and FR act on point A, they will produce the same external effect or reactions at the support

)(.551.551

)3.0)(54cos400(

)8.0)(54sin400()4.0(600)0(100

;

CWmNmN

mN

mNmNNM

MM

RA

ARA

4.9 Reduction of a Simple Distributed

Loading


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


Loading


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


Loading


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


Loading


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)


Loading


Loading

AdAdxxwFFFL A

RR )(;


Loading


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 )(;


Loading


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)(

)(


Loading


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


Loading


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


Loading


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


Loading


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


Loading


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


Loading


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


Loading


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.


Loading


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


Loading


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


Loading


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


Loading


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


Loading


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


Loading


Loading

mmxmmx 3)9(31

,3)9(31

21

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