ch 4 force system resultant

8/10/2019 Ch 4 Force System Resultant

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Chapter 4:Force System Resultants


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To discuss the concept of the moment of a force andshow 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 ofnon-concurrent force systems.

To indicate how to reduce a simple distributed loadingto a resultant force having a specified location.


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Moment of a Force Scalar FormationCross ProductMoment of Force Vector FormulationPrinciple of MomentsMoment of a Force about a Specified Axis


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Moment of a CoupleEquivalent SystemResultants of a Force and Couple System

Further Reduction of a Force and CoupleSystemReduction of a Simple Distributed Loading


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Moment of a force about a point or axis ameasure of the tendency of the force tocause a body to rotate about the point or axisCase 1Consider horizontal force Fx,which acts perpendicular to

the handle of the wrench andis located d y from the point O


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Fx tends to turn the pipe about the z axisThe larger the force or the distance d y , the greater theturning effect

Torque tendency ofrotation caused by Fxor simple moment ( Mo) z


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Moment axis (z) is perpendicular to shadedplane (x-y)Fx and d y lies on the shaded plane (x-y)

Moment axis (z) intersectsthe plane at point O


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Case 2Apply force Fz to the wrench

Pipe does not rotate about z axis

Tendency to rotate about x axisThe pipe may not actuallyrotate Fz creates tendencyfor rotation so moment(Mo) x is produced


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Case 2Moment axis (x) is perpendicular to shadedplane (y-z)

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


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Case 3Apply force Fy to the wrench

No moment is produced about point O

Lack of tendency to rotateas line of action passesthrough O


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In GeneralConsider the force F and the point O which lies inthe shaded planeThe moment MO about point O,or about an axis passingthrough O and perpendicular

to the plane, is a vector quantityMoment MO has its specifiedmagnitude and direction


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MagnitudeFor magnitude of MO,

MO = Fd where d = moment arm or perpendicular distancefrom the axis at point O to its line of action of theforce

Units for moment is N.m


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DirectionDirection of MO is specified by using righthand rule- fingers of the right hand are curled tofollow the sense of rotation when forcerotates about point O


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Direction- Thumb points along the moment axisto give the direction and sense of themoment vector- Moment vector is upwards andperpendicular to the shaded plane


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DirectionMO is shown by a vector arrowwith a curl to distinguish it fromforce vectorExample (Fig b)

MO is represented by the counterclockwise

curl, which indicates the action of F


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DirectionMoment acts about an axis perpendicularto the plane containing F and d Moment axis intersectsthe plane at point O


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Resultant Moment of a System ofCoplanar Forces

A clockwise curl is written along the equation toindicate that a positive moment if directed along the+ z axis and negativealong the z axis


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Moment of a force does not always cause rotationForce F tends to rotate the beam clockwise about Awith moment

MA = Fd AForce F tends to rotate the beam counterclockwiseabout B with moment

MB = Fd BHence support at A preventsthe rotation


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Example 4.1For each case, determine the moment of theforce about point O


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SolutionLine of action is extended as a dashed line toestablish moment arm d

Tendency to rotate is indicated and the orbit isshown as a colored curl

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

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

CW m N m N M b

CW m N m N M a

o

o


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Solution

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

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

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

CCW mkN mmkN M e

CCW m N m N M d

CW m N mm N M c

o

o

o


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Example 4.2Determine the moments ofthe 800N force acting on the

frame about points A, B, Cand D.


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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(

CCW m N m N M

mkN m N M

CW m N m N M

CW m N m N M

D

C

B

A


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Example 4.3Determine the resultant moment of the fourforces acting on the rod about point O


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SolutionAssume positive moments acts in the +k

direction, CCW

)(.334

.334

)30cos34)(40(

)30sin3)(20()0)(60()2)(50(

CW m N

m N

mm N

m N m N m N M

Fd M

Ro

Ro

View Free Body Diagram


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SolutionMethod 1MO = 400sin30

N(0.2m)-400cos30 N(0.4m)

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

As a Cartesian vector,M

O= {-98.6 k}N.m

View Free Body Diagram


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SolutionMethod 2:

Express as Cartesian vectorr = {0.4 i 0.2 j }NF = {400sin30 i 400cos30 j }N

= {200.0 i 346.4 j }N

For moment,

m N k

k ji

F X r M O

.6.98

04.3460.200

02.04.0


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Frictional forces (floor)on the blades of themachine creates a

moment Mc that tendsto turn itAn equal and oppositemoment must be

applied by the operatorto prevent turningCouple moment Mc = Fd is applied on the handle


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Example 4.10A couple acts on the gear teeth. Replace itby an equivalent couple having a pair offorces that cat through points A and B.


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SolutionMagnitude of coupleM = Fd = (40)(0.6) = 24N.m

Direction out of the page sinceforces tend to rotate CWM is a free vector and canbe placed anywhere


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SolutionTo preserve CCW motion,vertical forces acting through

points A and B must be directed asshownFor magnitude of each force,

M = Fd

24N.m = F (0.2m)F = 120N


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Example 4.11Determine the moment of the couple actingon the member.


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SolutionResolve each force into horizontal and verticalcomponents

F x = 4/5(150kN) = 120kNF y = 3/5(150kN) = 90kNPrinciple of Moment about point D,M = 120kN(0m) 90kN(2m)

+ 90kN(5m) + 120kN(1m)= 390kN (CCW)


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SolutionPrinciple of Moment about point A,M = 90kN(3m) + 120kN(1m)

= 390kN (CCW)*Note:- Same results if take momentabout point B- Couple can be replaced bytwo couples as seen in figure


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Solution- Same results for couple replaced bytwo couples

- M is a free vector and acts on anypoint on the member-External effects such as supportreactions on the member, will be thesame if the member supports thecouple or the couple moment


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Moment of a ForceA force produces a turning effect about thepoint O that does not lie on its line ofaction

In scalar form, moment magnitude, MO = Fd , where d is the moment arm orperpendicular distance from point O to itsline of action of the forceDirection of the moment is defined by righthand ruleFor easy solving,- resolve the force components into x and ycomponents


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Moment about a Specified AxisProjection of the moment onto the axis isobtained to determine the moment of a forceabout an arbitrary axis provided that thedistance perpendicular to both its line of actionand the axis can be determinedIf distance is unknown, use vector triple product

Ma = u a r X Fwhere u a is a unit vector that specifies thedirection of the axis and r is the position vectorthat is directed from any point on the axis toany point on its line of action


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Couple MomentA couple consists of two equal but oppositeforces that act a perpendicular distance d apartCouple tend to produce rotation withouttranslationMoment of a couple is determined from M = Fdand direction is established using the right-handrule

If vector cross product is used to determine thecouple moment, M = r X F, r extends from anypoint on the line of action of one of the forcesto any point on the line of action of the force F


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Reduction of a Force and Couple SystemFor concurrent, coplanar or parallel forcesystems,

- find the location of the resultant force about apoint- equate the moment of the resultant force aboutthe point to moment of the forces and couples in

the system about the same pointRepeating the above steps for other force systemwill yield a wrench, which consists of resultantforce and a resultant collinear moment


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ch 4 force system resultant

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