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EnginEEring mEchanics

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00_Front Matters.indd 1 4/27/09 10:02:19 AM




r. c. hiBBElEr

PRENTICE HALLUpper saddle river, n.J.


Published in 2010 byPrentice HallPearson Education South Asia Pte Ltd23/25 First Lok Yang Road, JurongSingapore 629733

Pearson Education offices in Asia: Bangkok, Beijing, Hong Kong, Jakarta, Kuala Lumpur, Manila, New Delhi, Seoul, Singapore, Taipei, Tokyo

Authorized adaptation from the United States edition, entitled Engineering Mechanics: Statics, Twelfth Edition, ISBN: 0136091830 by HIBBELER, RUSSELL C., published by Pearson Education, Inc, publishing as Prentice Hall, Copyright © 2010 R. C. Hibbeler.

All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording or by any information storage retrieval system, without permission from Pearson Education, Inc.

WORLD WIDE adaptation edition published by PEARSON EDUCATION SOUTH ASIA PTE LTD., Copyright © 2010.

Printed in Singapore

4 3 2 1 12 11 10 09

ISBN-13 978-981-06-8135-7ISBN-10 981-06-8135-6

Copyright © 2010 by R. C. Hibbeler. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

www.pearson-asia.com

Prentice Hallis an imprint of

TA new.indd 4 4/28/09 12:10 PM

Contents

What’s in This Package ix

Preface xi

Section-by-Section, Chapter-by-Chapter Summaries Part I with Review Questions and Answers 1

1 General Principles 3 MainGoalsofthisChapter 3 1.1 Mechanics 3 1.2 FundamentalConcepts 4 1.3 UnitsofMeasurement 5 1.4 TheInternationalSystemofUnits 5 1.5 NumericalCalculations 5 1.6 GeneralProcedureforAnalysis 6 HelpfulTipsandSuggestions 6 ReviewQuestions 7

2 Force Vectors 8 MainGoalsofthisChapter 8 2.1 ScalarsandVectors 8 2.2 VectorOperations 8 2.3 VectorAdditionofForces 10 2.4 AdditionofaSystemofCoplanarForces 10 2.5 CartesianVectors 11 2.6 AdditionandSubtractionofCartesianVectors 12 2.7 PositionVectors 12 2.8 ForceVectorDirectedAlongaLine 13 2.9 DotProduct 13 HelpfulTipsandSuggestions 14 ReviewQuestions 14


3 Equilibrium of a Particle 15 MainGoalsofthisChapter 15 3.1 ConditionfortheEquilibriumofaParticle 15 3.2 TheFree-BodyDiagram 15 3.3 CoplanarForceSystems 16 3.4 Three-DimensionalForceSystems 16 HelpfulTipsandSuggestions 17 ReviewQuestions 17

4 Force System Resultants 18 MainGoalsofthisChapter 18 4.1 MomentofaForce—ScalarFormulation 18 4.2 CrossProduct 19 4.3 MomentofaForce—VectorFormulation 20 4.4 PrincipleofMoments 21 4.5 MomentofaForceAboutaSpecifiedAxis 22 4.6 MomentofaCouple 22 4.7 SimplificationofaForceandCoupleSystem 22 4.8 FurtherSimplificationofaForceandCoupleSystem 24 4.9 ReductionofaSimpleDistributedLoading 24 HelpfulTipsandSuggestions 24 ReviewQuestions 24

5 Equilibrium of a Rigid Body 26 MainGoalsofthisChapter 26 5.1 ConditionsforRigid-BodyEquilibrium 26 5.2 Free-BodyDiagrams 26 5.3 EquationsofEquilibrium 27 5.4 Two-andThree-ForceMembers 28 5.5 Free-BodyDiagrams 28 5.6 EquationsofEquilibrium 29 5.7 ConstraintsandStaticalDeterminacy 29 HelpfulTipsandSuggestions 30 ReviewQuestions 30

6 Structural Analysis 31 MainGoalsofthisChapter 31 6.1 SimpleTrusses 31 6.2 TheMethodofJoints 32 6.3 ZeroForceMembers 32 6.4 TheMethodofSections 32 6.5 SpaceTrusses 33 6.6 FramesandMachines 34 HelpfulTipsandSuggestions 34 ReviewQuestions 35

7 Internal Forces 36 MainGoalsofthisChapter 36 7.1 InternalForcesDevelopedinStructuralMembers 36 7.2 ShearandMomentEquationsandDiagrams 37

vi Contents


7.3 RelationsBetweenDistributedLoad,ShearandMoment 39 7.4 Cables 39 HelpfulTipsandSuggestions 41 ReviewQuestions 41

8 Friction 42 MainGoalsofthisChapter 42 8.1 CharacteristicsofDryFriction 42 8.2 ProblemsInvolvingDryFriction 43 8.3 Wedges 43 8.4 FrictionalForcesonScrews 44 8.5 FrictionalForcesonFlatBelts 45 8.6 FrictionalForcesonCollarBearings,PivotBearings,andDisks 46 8.7 FrictionalForcesonJournalBearings 47 8.8 RollingResistance 47 HelpfulTipsandSuggestions 48 ReviewQuestions 48

9 Center of Gravity and Centroid 49 MainGoalsofthisChapter 49 9.1 CenterofGravity,CenterofMassandtheCentroidofaBody 49 9.2 CompositeBodies 50 9.3 TheoremsofPapusandGuldinus 51 9.4 ResultantofaGeneralDistributedLoading 51 9.5 FluidPressure 52 HelpfulTipsandSuggestions 53 ReviewQuestions 53

10 Moments of Inertia 54 MainGoalsofthisChapter 54 10.1 DefinitionofMomentsofInertiaforAreas 54 10.2 ParallelAxisTheoremforanArea 55 10.3 RadiusofGyrationofanArea 55 10.4 MomentsofInertiaforCompositeAreas 56 10.5 ProductofInertiaforanArea 56 10.6 MomentsofInertiaforanAreaAboutInclinedAxes 57 10.7 Mohr’sCircleforMomentsofInertia 58 10.8 MassMomentofInertia 59 HelpfulTipsandSuggestions 60 ReviewQuestions 60

11 Virtual Work 61 MainGoalsofthisChapter 61 11.1 DefinitionofWorkandVirtualWork 61 11.2 PrincipleofVirtualWorkforaParticleandaRigidBody 62 11.3 PrincipleofVirtualWorkforaSystemofConnectedRigidBodies 62 11.4 ConservativeForces 64 11.5 PotentialEnergy 64 11.6 PotentialEnergyCriterionforEquilibrium 64

Contents vii


11.7 StabilityofEquilibrium 65 ProcedureforSolvingProblems 66 ReviewQuestions 66

ANSWERS TO REVIEW QUESTIONS 67

Part II Free-Body Diagram Workbook 71

1 Basic Concepts in Statics 73 1.1 Equilibrium 73

2 Free-Body Diagrams: the Basics 75 2.1 Free-BodyDiagram:Particle 75 2.2 Free-BodyDiagram:RigidBody 78

3 Problems 83 3.1 Free-BodyDiagramsinParticleEquilibrium 85 3.2 Free-BodyDiagramsintheEquilibriumofaRigidBody 115

viii Contents


What’s in This Package TheStaticsStudyPackwasdesignedtohelpstudentsimprovetheirstudyskills.Itconsistsoftwostudycomponents—achapter-by-chapterreviewandafree-bodydiagramworkbook.

• Chapter-by-Chapter Review and Free-Body Diagram Workbook—PreparedbyPeterSchiavoneoftheUniversityofAlberta.Thisresourcecontainschapter-by-chapterStaticsreview,includingkeypoints,equations,andcheckupquestions.TheFree-BodyDiagramworkbookstepsstudentsthroughnumerousfree-bodydiagramproblemsthatincludefullexplanationsandsolutions.

• Companion Website—Locatedatwww.pearsoned-asia.com/hibbeler,theCompanionWebsiteincludesthefollowingresources:

• Video Solutions—Complete, step-by-step solutionwalkthroughsof representativehomeworkproblemsfrom each section in the Statics text. Developed by Professor Edward Berger of University ofVirginia,VideoSolutionsoffer:

• Fully-worked Solutions —Showingeverystepofrepresentativehomeworkproblems,tohelpstudentsmakevitalconnectionsbetweenconcepts.

• Self-paced Instruction —Studentscannavigateeachproblemandselect,play,rewind,fast-forward,stop,andjump-tosectionswithineachproblem’ssolution.

• 24/7 Access—Helpwheneverstudentsneeditwithover20hoursofhelpfulreview.


PrefaceThissupplementisdividedintotwoparts.PartIprovidesasection-by-section,chapter-by-chaptersummaryofthekeyconcepts,principlesandequationsfromR.C.Hibbeler’stext,Engineering Mechanics: Statics, Twelfth Edition.PartIIisaworkbookwhichexplainshowtodrawandusefree-bodydiagramswhensolvingproblemsinStatics.

Part I: Chapter-by-Chapter Summaries

Thispartofthesupplementprovidesasection-by-section,chapter-by-chaptersummaryofthekeyconcepts,principlesandequationsfromR.C.Hibbeler’stext, Engineering Mechanics: Statics, Twelfth Edition. Wefollowthesamesectionandchapterorderasthatusedinthetextandsummarizeimportantconceptsfromeachsectionineasy-to-understandlanguage.We end each chapter summary with a simple set of review questions designed to see if the student hasunderstoodthekeyconceptsandchapterobjectives.

Thissectionofthesupplementwillbeusefulbothasaquickreferenceguideforimportantconceptsandequationswhensolvingproblemsin,forexample,homeworkassignmentsorlaboratoriesandalsoasahandyreviewwhenpreparingforanyquiz,test,orexamination.

Part II: Free-Body Diagram Workbook

A thorough understanding of how to draw and use a free-body diagram is absolutely essential when solving problems in mechanics.

Thisworkbookconsistsmainlyofacollectionofproblemsintendedtogivethestudentpracticeindrawingandusingfree-bodydiagramswhensolvingproblemsinStatics.

Alltheproblemsarepresentedastutorialproblemswiththesolutiononlypartiallycomplete.Thestudentisthenexpectedtocompletethesolutionby“fillingintheblanks”inthespacesprovided.Thisgivesthestudenttheopportunitytobuild free-body diagrams in stagesandextract therelevant informationfromthemwhenformulatingequilibriumequations.Earlierproblemsprovidestudentswithpartiallydrawnfree-bodydiagramsandlotsofhintstocompletethesolution.Laterproblemsaremoreadvancedandaredesignedtochallengethestudentmore.Thecompletesolutiontoeachproblemcanbe foundon thebackof thepage.Theproblemsarechosen fromtwo-dimensional theoriesofparticleandrigidbodymechanics.Oncetheideasandconceptsdevelopedintheseproblemshavebeenunderstoodandpracticed,thestudentwillfindthattheycanbeextendedinarelativelystraightforwardmannertoaccommodatethecorrespondingthree-dimensionaltheories.

Theworkbookbeginswithabriefprimeronfree-bodydiagrams:wheretheyfit intothegeneralprocedureofsolvingproblemsinmechanicsandwhytheyaresoimportant.Nextfollowsafewexamplestoillustrateideasandthentheworkbookproblems.

For best results, the student should read the primer and then, beginning with the simpler problems, try tocompleteandunderstandthesolutiontoeachofthesubsequentproblems.Thestudentshouldavoidthetemptationtoimmediatelylookatthecompletedsolutiononthebackofthepage.Thissolutionshouldbeaccessedonlyasalastresort(afterthestudenthasstruggledtothepointofgivingup),ortocheckthestudent’sownsolutionafterthefact.Theideabehindthisisverysimple:we learn most when wedothe thing we are trying to learn—readingthroughsomeone


else’ssolutionisnotthesameasactuallyworkingthroughtheproblem.Intheformer,thestudentgainsinformation,inthelatterthestudentgainsknowledge.Forexample,howmanypeoplelearntoswimordriveacarbyreadinganinstructionmanual?

Consequently, since the workbook is based on doing, the student who persistently solves the problems in theworkbookwillultimatelygainathorough,usableknowledgeofhowtodrawandusefree-bodydiagrams.

Peter Schiavone

xii Preface


PART I

Section-by-Section, Chapter-by-Chapter

Summaries with Review Questions and

Answers

01_Gen Principles.indd 1 4/21/09 12:32:16 PM

MAIN GOALS OF THIS CHAPTER:

• TointroducethebasicideasofMechanics.

• TogiveaconcisestatementofNewton’slawsofmotionandgravitation.

• ToreviewtheprinciplesforapplyingtheSIsystemofunits.

• Toexaminestandardproceduresforperformingnumericalcalculations.

• Tooutlineageneralguideforsolvingproblems.

1.1 MECHANICS

Mechanics is that branch of the physical sciences concerned with the behavior of bodies subjected to the action offorces.Thesubjectofmechanicsisdividedintotwoparts:

• statics—thestudyofobjectsinequilibrium(objectseitheratrestormovingwithaconstantvelocity).

• dynamics—thestudyofobjectswithacceleratedmotion.

Althoughstaticscanbeconsideredasaspecialcaseofdynamics(inwhichtheaccelerationiszero),itdeservesspecialtreatmentsincemanyobjectsaredesignedwiththeintentionthattheyremaininequilibrium.

General Principles

1


� Chap. 1 General Principles

1.2 FUNDAMENTAL CONCEPTS

BASIC QUANTITIES

• Length, time, mass, force

IDEALIZATIONS

Mathematicalmodelsoridealizationsareusedinmechanicstosimplifythetheory.Themorecommonones,inorderofsophistication,are:

• Particle—aparticlehasamassbutasizethatcanbeneglected,i.e.thegeometryofthebodyisignored.Aparticleisoftenrepresentedbyapointinspace.

• Rigid Body—a rigid body has a mass and a size (shape) but it is assumed that any changes in shape can beneglected, i.e. the geometry of the body is taken into account but any deformations (changes in shape) areignored.Consequently,thematerialpropertiesofthebodycanbeignored.Arigidbodyisoftenrepresentedasa collectionofparticles inwhichall theparticles remainatafixeddistance fromeachotherbeforeandafterapplyingaload.

• Deformable or Elastic Body—adeformablebodyhasamass,asize(shape)andthedeformations(changes inshape)of thebodyare taken intoaccount.Hence, thematerialpropertiesof thebodymustbe considered indescribingthebehaviorofthebody.

• Concentrated Force—Aconcentratedforcerepresentstheeffectofaloadingwhichisassumedtoactatapointonabody.Thisidealizationrequiresthattheareaoverwhichtheloadisappliedisverysmallcomparedtotheoverallsizeofthebody,e.g.contactforcebetweenwheelandground.

NEWTON’S THREE LAWS OF MOTION

Newton’slawsapplytothemotionofaparticleasmeasuredfromanonaccelerating(inertial)referenceframe.

• First Law—aparticleoriginallyatrestormovinginastraightlinewithconstantvelocity,willremaininthisstateprovidedtheparticleisnotsubjectedtoanunbalancedforce.

• Second Law—aparticleacteduponbyanunbalancedforceFexperiencesanaccelerationa thathasthesamedirectionastheforceandamagnitudedirectlyproportionaltotheforce,i.e.

F=m a

• Third Law—Themutualforcesofactionandreactionbetweentwoparticlesareequal,oppositeandcollinear.

NEWTON’S LAW OF GRAVITATIONAL ATTRACTION

F =Gm m

r

1 2

2

F =forceofgravitationbetweentwoparticles G =universalconstantsofgravitation m1,m2 =massofeachofthetwoparticles r =distancebetweenthetwoparticles

MASS AND WEIGHT

• Massisa(scalar)propertyofmatterthatdoesnotchangefromonelocationtoanother.Inotherwords,massisanabsolutequantity

• Weight isaforce(andhencehasamagnitudeanddirection)whichreferstothegravitationalattractionoftheearthonaquantityofmassm.Weight isnotanabsolutequantity. Itsmagnitudedependson theelevationatwhichthemassislocated.WewritethemagnitudeofweightasW=mgwheregistermedtheaccelerationduetogravity.


1.3 UNITS OF MEASUREMENT

Thefourbasicquantitiesforce,mass,lengthandtimearerelatedbyNewton’s2ndlaw.Hence,theunitsusedtodefinethesequantitiesarenotindependent,i.e.threeofthefourunitsarecalledbase units(arbitrarilydefined)andthefourthunitaderived unit(derivedfromNewton’s2ndlaw).

SI UNITS (INTERNATIONAL SYSTEM OF UNITS)

• IntheSIsystem,theunitofforce,thenewton,isaderivedunit.Themeter,secondandkilogramarebaseunits.

• Onenewtonisequaltoaforcerequiredtogiveonekilogramofmassanaccelerationof1m/s2.

• Innewtons,theweightofabodyhasmagnitude

W=mgwhereg=9.81m/s2.

Thefollowingtablesummarizesthesystemofunits.

Name Length Time Mass Force

InternationalSystem(SI) meter(m) second(s) kilogram(kg) newton* N =

kg.m

s2

*Derived Unit

1.� THE INTERNATIONAL SYSTEM OF UNITS

PREFIXES

Whenanumericalquantityiseitherverylargeorverysmall,theunitsusedtodefineitssizemaybemodifiedbyusingaprefix.Forexample:

Exponential Form Prefix SI Symbol

1000000000 109 giga G

1000000 106 mega M

1000 103 kilo k

0.001 10–3 milli m

0.000001 10–6 micro µ

0.000000001 10–9 nano n

RULES FOR USE

Youshouldknow the rules for theproperuseof thevariousSI symbols.Theseareusedextensively inengineeringpracticethroughouttheworld.

1.5 NUMERICAL CALCULATIONS

Itisimportantthatthenumericalanswerstoanyproblemencounteredinengineeringpracticebereportedwithbothjustifiableaccuracyandappropriatesignificantfigures.

1.5 Numerical Calculations 5


� Chap. 1 General Principles

DIMENSIONAL HOMOGENEITY

Eachterminanyequationusedtodescribeaphysicalprocessmustbeexpressedinthesameunits,i.e.thetermsmustbedimensionally homogeneous.Algebraicmanipulationsofanequationcanbechecked,inpart,byverifyingthattheequationremainsdimensionallyhomogeneous.

SIGNIFICANT FIGURES

Theaccuracyofanumberisspecifiedbythenumberofsignificantfiguresitcontains.Asignificant figureisanydigit,including a zero, provided it is not used to specify the location of the decimal point for the number. For example,0.00546and2500expressedtothreesignificantfigureswouldbe5.46×(10–3)and2.50×(103),respectively(engineering notation).

ROUNDING OFF NUMBERS

Fornumericalcalculations,theaccuracyofthesolutionofaproblem(generally)canneverbebetterthantheaccuracyof the problem data. Consequently, a calculated result should always be rounded off to an appropriate number ofsignificantfigures.Toconveyappropriateaccuracy,therearerulesforroundingoffnumbers.Youshouldknowthese.

CALCULATIONS

Perform numerical calculations to several significant figures and then report the final answer to three significantfigures.

1.� GENERAL PROCEDURE FOR ANALYSIS

• Themosteffectivewaytolearnengineeringmechanicsistosolve problems.

• Youmustpresentyourworkinalogicalandorderlymannerasfollows:

– Readtheproblemcarefullyandtrytoestablishalinkbetweentheactualphysicalsituationandtheappropriatepartofthetheorystudies.

– Drawanynecessarydiagramsandtabulatetheproblemdata.

– Applytherelevantprinciples.

– Solve thenecessaryequationsalgebraically,as faraspractical, then,makingsure theyaredimensionallyhomogeneous,useaconsistentsetofunitsandcompletethesolutionnumerically.Reporttheanswerwithnomoresignificantfiguresthantheaccuracyofthegivendata.

– Studytheanswerandseeifitmakes sensephysically—inthecontextofthephysicalproblem.

HELPFUL TIPS AND SUGGESTIONS

• Thelanguageofengineeringmechanicsismathematics.Consequently,makesureyoureview/re-readthenecessarymathematical notation/concepts as they arise in your mechanics course (trying to review all of the necessarymathematicsatonceisnotrecommended—there’sjusttoomuchtodigestatonetime).Youshouldaimtoachievefluencyinbasicmathematicaltechniques/notationsothatyourlearningofmechanicsisnotdistractedbytryingtorememberthingswhichyourinstructorassumesyouknow,e.g.howtosolvelinearsystemsofalgebraicequations,howtoperformbasicvectoralgebra,differentiationandintegration,etc.

• Remember that in solving problems from engineering mechanics you are solving real practical problems andproducingrealdatawithphysicalsignificance.Thus,youareresponsibleformakingsureyourresultsarecorrect,consistentandwell-presented.Getintothehabitofdoingthisnowsothatitwillbecomesecondnaturebythetimeyougraduate.Intheworldofprofessionalengineeringyouhavearesponsibilitytoyourprofessionandtothemanypeoplewhowillusetheproductyouwillhelptodesign,manufactureorimplement.


REVIEW QUESTIONS: TRUE OR FALSE1?

1. ThesubjectcalledStaticsstudiesonlybodieswhichareatrest.

2. Aparticlehasamassbutnegligibleshape/size.

3. Arigidbodyhasamassbutnegligibleshape/size.

4. Newton’sthreelawsofmotioncanbeprovedmathematically.

5. Weightisapropertyofmatterthatdoesnotchangefromonelocationtoanother.

6. IntheSIsystemofunits,thenewtonisaderivedunit.

7. Whenperformingnumericalcalculations,thefinalanswershouldbereportedtothreesignificantfigures.

8. Inanequationit’spermittedtohavedifferenttermsexpressedindifferentunits.Thisisreferredtoasdimensionallyinhomogeneous.

11.F 2.T 3.F 4.F 5.F 6.T 7.T 8.F

Review QuestioNs: true or False? �


MAIN GOALS OF THIS CHAPTER

In this chapter we define scalars, vectors and vector operations and use them to analyze forces acting on objects. Specifically:

• To show how to add forces and resolve them into components.

• To express force and position in Cartesian vector form.

• To explain how to determine a vector’s magnitude and direction.

• To introduce the dot product and use it to find the angle between two vectors or the projection of one vector onto another.

2.1 SCALARS AND VECTORS

Most of the physical quantities in mechanics can be represented by either scalars or vectors:

• A scalar is a real number, e.g. mass, time, volume and length are represented by scalars.

• A vector has both magnitude and direction, e.g. force, velocity and acceleration are vectors.

2.2 VECTOR OPERATIONS

MULTIPLICATION OR DIVISION OF A VECTOR BY A SCALAR

• The product of a vector A and a scalar a is a vector a A with magnitude | a A | = | a | | A |. The direction is the same as that of A if a is positive and opposite to that of A if a is negative.

Force Vectors

2

02_Force Vectors.indd 8 4/21/09 12:36:15 PM

A

2 A

–1.5 A

0.5 A

Scalar multiplication and division

VECTOR ADDITION

• Two vectors A and B can be added to form a resultant vector R = A + B by using the parallelogram law. If the two vectors are collinear (both vectors have the same line of action), the resultant is formed by an algebraic or scalar addition.

A

B

A

B

A

B

B

A

Parallelogram law(b)

Triangle construction(c)

Triangle construction(d)(a)

Vector addition

R=B+A

A B

R

Addition of Collinear Vectors

R = A + B

R=A+B

R=A+B

RESOLUTION OF A VECTOR

• A vector may be resolved into components having known lines of action by using the parallelogram law.

2.2 Vector Operations �


10 Chap. 2 Force Vectors

Resolution of a vector

a

b

A

B

R

a

b

R

(a) (b)

ComponentsExtend parallel lines from the head of Rto form components

Resultant

2.3 VECTOR ADDITION OF FORCES

• A force is a vector quantity since it has a specified magnitude and direction. Consequently, forces are added together or resolved into components using the rules of vector algebra.

• Two common problems in statics involve either finding the resultant force given its components or resolving a known force into components.

• Often the magnitude of a resultant force can be determined from the law of cosines, while its direction is determined from the law of sines:

A

C

B

b

c

a

Sine law:

sin a sin b sin c A = B = C

Cosine law:

C = A2 + B2 – 2AB cos c

2.4 ADDITION OF A SYSTEM OF COPLANAR FORCES

In the plane, a force can be resolved into two rectangular components. There are two separate notations for doing this:

• Scalar Notation—we write the force F as (Fx , Fy) where Fx and Fy are the scalar components of the force F in the directions of the positive x- and y-axes, respectively. If Fx and Fy are negative, it means that | Fx | and | Fy | are directed along the negative x- and y-axes, respectively.

• Cartesian Vector Notation—we write the force F as

F = Fx i + Fy j,

where i and j represent the positive directions of the x- and y-axes, respectively.


COPLANAR FORCE RESULTANTS

• The resultant of several coplanar forces can easily be determined if an x, y-coordinate system is established and the forces are resolved along the axes. For example,

F1 = F1x i + F1y j, F2 = F2x i + F2y j, F3 = F3x i + F3y j,

then the resultant is given by

FR = F1 + F2 + F3

= (F1x + F2x + F3x) i + (F1y + F2y + F3y) j

= (FRx) i + (FRy) j.

• In the general case, the x and y components of the resultant of any number of coplanar forces can be represented symbolically by the algebraic sum of the x and y components of all the forces, i.e.

FRx = ∑Fx ,

FRy = ∑Fy .

• The magnitude and direction of the resultant force are given by:

| FR | = FR = F FRx Ry2 2+ (2.0)

q = tan–1 F

FRy

Rx

,

respectively.

2.5 CARTESIAN VECTORS

• A Cartesian coordinate system is often used to solve problems in three dimensions. The coordinate system is right-handed which means that the thumb of the right hand points in the direction of the positive z-axis when the right hand fingers are curled about this axis and directed from the positive x toward the positive y-axis.

z

y

x

Right-handed coordinate system

2.5 Cartesian Vectors 11



• The unit vector from a vector A is given by AA

where A 0 is the magnitude of vector A. The unit vector is

dimensionless and defines the direction of vector A.

• The positive directions of the x, y, z axes are defined by the Cartesian unit vectors i, j, k, respectively. Consequently, any vector A with scalar components Ax , Ay and Az can be written in the Cartesian vector form

A = Ax i + Ay j + Az k. (2.1)

• The magnitude of vector A is given by

| A | = A = A A Ax y z2 2 2+ + . (2.2)

• The direction of vector A is defined by the angles a, b and g measured between the tail of A and the positive x, y, z axes located at the tail of A.

A

z

y

x

Ay j

Azkz

ab

g

Ax i

• The angles a, b and g are found from their direction cosines

cos a = AA

x , cos b = A

Ay

, cos g = A

Az , cos2 a + cos2 b + cos2 g = 1. (2.3)

This means that only two of the angles a, b and g have to be specified—the third can be found from cos2 a + cos2 b + cos2 g = 1.

2.6 ADDITION AND SUBTRACTION OF CARTESIAN VECTORS

• To find the resultant of a concurrent force system, express each force as a Cartesian vector and add the i, j, k components of all the forces in the system.

2.7 POSITION VECTORS

• The position vector r is defined as a fixed vector which locates a point in space relative to another point. For example, from the origin of coordinates O, the point in space P (x, y, z) has position vector r = x i+ y j+ z k.

• More generally, the position vector may be directed from point A to point B in space. In this case, the position vector is again denoted by r (or sometimes rAB) and is given by

rAB = rB – rA , (2.4)


where rB and rA are the position vectors of A and B from the origin of coordinates O. For example, if A (xA, yA, zA) and B (xB, yB, zB) then

rA = xA i + yA j + zA k, rB = xB i + yB j + zB k, rAB = (xB – xA) i + (yB – yA) j + (zB – zA) k.

2.8 FORCE VECTOR DIRECTED ALONG A LINE

• A force F (with magnitude F ) acting in the direction of a line represented by a position vector r can be written in the form

F = F rr

= F u,

where u = rr

is a unit vector representing the direction of the line.

2.� DOT PRODUCT

• The dot product is used to determine

– The angle between two vectors.

– The projection of a vector in a specified direction.

• The dot product of two vectors A and B is defined as

A · B = AxBx + AyBy + AzBz = AB cos q, (2.5)

where A and B are the magnitudes of A and B, respectively, and q is the angle between the tails of A and B. Consequently,

q = cos–1 A B·AB

.

• The dot product is commutative (A · B = B · A), and distributive A· (B + D) = A · B + A · D. Also, if a ∈ :

a(A · B) = (aA) · B = A · (aB) = (A · B)a. (2.6)

• In some engineering applications, you must resolve a vector into components which are parallel and perpendicular (normal) to a given line (direction). The component of vector A in the direction specified by the unit vector u is given by

A|| = A · u = A cos q. (2.7)

This component is also referred to as the scalar projection of A onto the line with direction u or the component of vector A parallel to a line with direction u. Clearly, the vector A|| is defined by A|| = (A||) u.

• Once the parallel component has been determined, we can determine the component of A perpendicular (or normal) to a line with direction u by

A⊥ = A A2 2– || (2.8)

• Clearly, in terms of vectors,

A = A|| + A⊥.

2.9 Dot Product 13




• Be aware of the differences between vectors and scalars. For example, force, velocity and acceleration are vectors while speed, time and distance are scalars. If you are asked to find a vector (e.g. a force) you must report both magnitude and direction.

• Vector operations are essential in describing the basic principles of mechanics. Make sure you take the time to review basic vector algebra. It doesn’t take long but the payoff (in terms of your effectiveness in mechanics) is significant.

REVIEW QUESTIONS

1. How are the scalar components of a vector defined in terms of a Cartesian coordinate system?

2. If you know the scalar components of a vector, how can you determine its magnitude and direction?

3. Suppose you know the coordinates of two points A and B. How do you determine the scalar components of the position vector of point B relative to point A?

4. How do you identify a right-handed coordinate system?

5. What are the direction cosines of a vector? If you know them, how do you determine the components of the vector?

6. What is the definition of the dot product? Is the dot product a vector or a scalar?

7. If the dot product of two vectors is zero, what does that mean?

8. If you know the components of two vectors A and B, how can you determine their dot product?

9. Simplify

i. A · (B + D).

ii. (aA) · B where a is a scalar.

10. How can you use the dot product to determine the components of a vector parallel and perpendicular to a line?


MAIN GOALS OF THIS CHAPTER

In this chapter we:

• Introduce the concept of the free-body diagram for an object modelled as a particle (an object with mass but negligible shape/size—henceforth referred to simply as a particle).

• Show how to solve particle equilibrium problems using the equations of equilibrium.

3.1 CONDITION FOR THE EQUILIBRIUM OF A PARTICLE

A particle is in equilibrium provided it is at rest if originally at rest (static equilibrium) or has a constant velocity if originally in motion.

• To maintain equilibrium it is necessary and sufficient that the resultant force acting on a particle be equal to zero. In terms of Newton’s laws of motion, this is expressed mathematically as:

∑F = 0 , (3.0)

where ∑ F is the vector sum of all forces acting on the particle.

3.2 THE FREE-BODY DIAGRAM

To apply the equation of equilibrium (3.0), we must account for all the known and unknown forces ( ∑ F) which act on the particle. The easiest way to do this is to draw a free-body diagram.

• A free-body diagram is simply a sketch which shows the particle ‘free’ from its surroundings with all the forces that act on it. There are three main steps:

Draw Outlined Shape. Imagine the particle to be isolated or cut ‘free’ from its surroundings by drawing its outlined shape.

Show All Forces. Indicate on this sketch all the forces that act on the particle—it may help to carefully trace around the particle’s boundary, noting each force acting.

Identify Each Force. The forces which are known should be labeled with their proper magnitudes and directions. Letters are used to represent the magnitudes and directions of forces that are unknown.

Equilibrium of a Particle

3

03_Equilibrium.indd 15 4/21/09 12:38:01 PM

16 Chap. 3 Equilibrium of a Particle

* Connections—there are two types of connections often encountered in particle equilibrium problems:

(a) Springs—The magnitude of force exerted on a linear elastic spring with stiffness k, deformed a distance s measured from its unloaded position is

F = ks. (3.1)

Here s is determined from the difference in the spring’s deformed length and its undeformed length.

(b) Cables and Pulleys—Assume cables (or cords) have negligible weight and cannot stretch. Also, a cable can support only tension which always acts in the direction of the cable.

There are several examples and practice problems, as well as much more on drawing free-body diagrams in Part II of this study pack.

3.3 COPLANAR FORCE SYSTEMS

Coplanar force equilibrium problems for a particle can be solved using the following procedure.

1. Free-Body Diagram

• Establish the x, y axes in any suitable orientation.

• Label all the known and unknown force magnitudes and directions on the diagram.

• The sense of a force having an unknown magnitude can be assumed.

2. Equations of Equilibrium

• Resolve each force into its i (x) and j (y) components and apply the scalar equations of equilibrium

∑Fx = 0, ∑Fy = 0. (3.2)

(the algebraic sum of the x and y components of all the forces acting on the particle equal to zero).

• Components are positive if they are directed along a positive axis and negative if they are directed along a negative axis.

• If more than two unknowns exist and the problem involves a spring, apply F = ks to relate the spring force to the deformation s of the spring.

• If the solution yields a negative result, this indicates the sense of the force is the reverse of that shown on the free-body diagram.

3.4 THREE-DIMENSIONAL FORCE SYSTEMS

Three-dimensional force equilibrium problems for a particle can be solved using the following procedure.

1. Free-Body Diagram

• Establish the x, y, z axes in any suitable orientation.

• Label all the known and unknown force magnitudes and directions on the diagram.

• The sense of a force having an unknown magnitude can be assumed.


2. Equations of Equilibrium

• When it’s easy to do so, resolve each force into its i(x), j(y), and k(z) components and apply the scalar equations of equilibrium

∑Fx = 0, ∑Fy = 0, ∑Fz = 0. (3.3)

(the algebraic sum of the x , y and z components of all the forces acting on the particle equal to zero).

• If the three-dimensional geometry appears difficult, then first express each force as a Cartesian vector and substitute these vectors into the vector equation of equilibrium (3.0)

∑F = 0

and then set the i, j and k components equal to zero.

• If the solution yields a negative result, this indicates the sense of the force is the reverse of that shown on the free-body diagram.


• Since we must account for all the forces acting on the (object modelled as a) particle when applying the equations of equilibrium, the importance of first drawing a free-body diagram cannot be over-emphasized.

• One of the most common mistakes made in writing equilibrium conditions is forgetting to include all of the forces acting. When drawn carefully, a free-body diagram will make it easier for you to identify all the forces acting.

• Use Part II of this supplement to get lots of practice in drawing free-body diagrams and applying the equations of equilibrium for a particle.

REVIEW QUESTIONS

1. What is meant by ‘equilibrium of a particle’?

2. What do you know about the sum of the external forces acting on an object modelled as a particle in equilibrium?

3. What are the steps in drawing a free-body diagram?

4. What is a coplanar force system?

5. What is a three-dimensional system of forces?

6. What is the difference between equilibrium of coplanar and three-dimensional force systems?

7. What is the relation between the magnitude of the force exerted on a linear spring and the change in its length?

8. The following is the correct free-body diagram for the ring at E. True or False?

4

53

45° A

B

100 NC

ED

(a)

30°

G

30°

45°

yTEG

TEC

xE

100 N

(b)

REVIEW QUESTIONS 17



• Todiscusstheconceptofthemoment of a forceandshowhowtocalculateitintwoandthreedimensions.

• Toprovideamethodforfindingthemomentofaforceaboutaspecifiedaxis.

• Todefinethemomentofacouple.

• Topresentmethodsfordeterminingtheresultantsofnonconcurrentforcesystems.

• Toindicatehowtoreduceasimpledistributedloadingtoaresultantforcehavingaspecifiedlocation.

4.1 MOMENT OF A FORCE—SCALAR FORMULATION

• ThemomentMOofaforceFaboutanaxispassingthroughaspecificpointOprovidesameasureofthetendencyoftheforcetocausethebodytorotateabouttheaxis(sometimesreferredtoasatorque).Clearlythemomentisavectorandsohasboth magnitude and direction.

• Themagnitudeof themoment isdeterminedfromM0=Fd,whered is theperpendicularorshortestdistancefrompointOtothelineofactionoftheforceF.

• Using theright-handrule, thedirection (sense)of rotation is indicatedby thefingerswith the thumbdirectedalongthemomentaxisorlineofactionofthemoment.

Force System Resultants

4

04_Force System.indd 18 4/21/09 12:41:05 PM

Senseofrotation

O

Momentaxis

dF

MO

MO

F

d

O

(a)

(b)

• Ifasystemofforcesliesinthex –yplane,thenthemomentproducedbyeachforceaboutpointOwillbedirectedalongthez-axis.TheresultantmomentMROofthesystemcanbedeterminedbysimplyaddingthemomentsofalltheforcesalgebraicallysinceallthemomentvectorsarecollinear,i.e.

+MRO = ∑Fd.

Here thecounterclockwisecurlwrittenalongside theequation indicates that themomentofany forcewillbepositive if it is directed along the positive z-axis, whereas a negative moment is directed along the negativez-axis.

4.2 CROSS PRODUCT

• ThecrossproductoftwovectorsAandByieldsavectorCwrittenC=A×B.

ThemagnitudeofvectorCisgivenbyABsinqwhereqistheanglebetweenthetailsofAandB.

VectorChasadirectionwhichisperpendiculartotheplanecontainingAandBsuchthatCisspecifiedbytheright-handrule,i.e.curlingthefingersoftherighthandfromvectorA(cross)tovectorB,thethumbthenpointsinthedirectionofC.

4.2 Cross Product 19


20 Chap. 4 Force System Resultants

C = A × B

A

B

q

LAWS OF OPERATION

• A×BB×A,ratherA×B=–B×A.

• a (A×B)=(a A)×B=A×(a B)=(A×B) a.

• A×(B+D)=(A×B)+(A×D).

CARTESIAN VECTOR FORMULATION

• TofindthecrossproductofanytwoCartesianvectorsAandBweusethedeterminant

A×B=

i j k

A A A

B B B

x y z

x y z

(4.0)

• Thefollowingusefulresultscanbeobtainedbyapplyingtheright-handruleanddon’tneedtobememorized.

i×j=k, i×k=–j, i×i=0,

j×k=i, j×i=–k, j×j=0,

k×i=j, k×j=–i, k×k=0.

4.3 MOMENT OF A FORCE—VECTOR FORMULATION

Inthree-dimensionsitispreferabletousethevectorcrossproducttodeterminethemoment:

• ThemomentMOofaforceFaboutthemomentaxispassingthroughpointOandperpendiculartotheplanecontainingOandFcanberepresentedby

MO=r×F=

i j k

r r r

F F F

x y z

x y z

(4.1)

whererrepresentsapositionvectordrawnfromO to any pointlyingonthelineofactionofF.


O

Momentaxis

MO

rA

F

(a)

O

Momentaxis

d

MO

rAr

F

(b)

q q

PRINCIPLE OF TRANSMISSIBILITY

• Sincein(4.1),rcanextendfromOtoanypointonthelineofactionofF,Fisasliding vectorandcanactatanypointalongitslineofactionandcreatethesame momentaboutpointO.

RESULTANT MOMENT OF A SYSTEM OF FORCES

• If a body is acted upon by a system of n forces, the resultant moment about O is just the vector sum of theindividualmoments:

MRO=i

n

=∑

1

(ri×Fi ).

z

x

yO

r2

r1r3

F3 F1

F2

MROR

4.4 PRINCIPLE OF MOMENTS

• Theprinciple of moments (Varignon’s theorem)states that the moment of a force about a point is equal to the sum of the moments of the force’s components about the point.This is particularly convenient since it is ofteneasiertodeterminethemomentsofaforce’scomponentsratherthanthemomentoftheforceitself(e.g.intwodimensions).

4.4 Principle of Moments 21



4.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS

Recallthatwhenthemomentofaforceiscomputedaboutapoint,themomentanditsaxisarealwaysperpendiculartotheplanecontainingtheforceandthemomentarm.Insomeproblemsitisimportanttofindthecomponentofthismomentaboutaspecified axisthatpassesthroughthepoint.

• In termsof a scalar analysis, the momentof a force F abouta specifiedaxis canbedeterminedprovided theperpendiculardistancedafromboththeforcelineofactionandtheaxiscanbedetermined.ThenMa=Fda.

• Intermsofvectoranalysis,thecomponentMa=ua ·(r×F),whereuadefinesthedirectionoftheaxisandr isdirectedfromanypointontheaxis toanypointonthe lineofactionoftheforce.Thequantityua · (r×F) iscalledatriplescalarproductandcanbecomputedusingthedeterminant

ua·(r×F)=

u u u

r r r

F F F

a a a

x y z

x y z

x y z

OnceMaisdetermined,wecanexpressM aasaCartesianvector,namelyM a=Ma ua.

• IfMaiscalculatedasanegativescalarthenthesenseofdirectionofM aisoppositetoua.

4.6 MOMENT OF A COUPLE

• Acoupleisdefinedastwoparallelforcesthathavethesamemagnitude,oppositedirections,andareseparatedbyaperpendiculardistanced.Sincetheresultantforceiszero,theonlyeffectofacoupleistoproducearotationinaspecifieddirection.

F

–F

d

• Themomentproducedbyacoupleiscalledacouple momentwhichisafreevectorand,asaresult,itcausesthesameeffectofrotationonabodyregardlessofwherethecouplemomentisappliedtothebody.Consequently,thecouplemomentcanbecomputedaboutanypoint.Forconvenience,thispointisoftenchosenonthelineofactionofoneoftheforcesinthecouple.

• Thecouplemoment iseasilydeterminedfromthevector formulationM=r×Fwherer isdirectedfromanypointonthelineofactionofoneoftheforcestoanypointonthelineofactionoftheotherforceF.

• Aresultant couple momentissimplythevectorsumofallthecouplemomentsofthesystem.

4.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM

Aforcehastheeffectofbothtranslatingandrotatingabodyandtheamountbywhichitdoessodependsonwhereandhowthe force isapplied. It ispossible,however, toreplace a systemof forcesandcouplemomentsactingonabodywithanequivalent singleresultantforceandcouplemomentactingataspecifiedpointO.Hereequivalentmeansthat the system and the resultant each produce the same external effects of translation and rotation.There are twocasestoconsider:


• PointOis on the Line of Action of the Force—simplyslidetheforcealongitslineofactiontothepointO.

O

AF

=

(a)

O

AF

F

–F

O

A

= F

(b) (c)

• PointO is not on the Line of Action of the Force—movethe force to thepointOandaddacouplemomentanywheretothebody.

= =O

A

F

O

A

F

F

–F

O

A

Fr

M = r × F

P

(a) (b) (c)

Next,weshowhowtodeterminetheequivalent resultantsmentionedabove.

• TosimplifyanyforceandcouplemomentsystemtoaresultantforceactingatpointOandaresultantcouplemomentweusethefollowingequations

FR =∑F, (4.2)

(MR)O =∑MC+∑MO .

The first equation states that the resultant force of the system is equivalent to the sum of all the forces.Thesecondequationstatesthattheresultantcouplemomentofthesystemisequivalenttothesumofallthecouplemoments∑ MCplusthemoments∑ MOaboutpointOofalltheforces.

• Thefollowingtipsmayproveusefulwhenapplyingequations(4.2):

Establish the coordinate axes with the origin located at point O and the axes having a selectedorientation.

Iftheforcesystemliesinthex–yplane,andanycouplemomentsareperpendiculartothisplane,i.e.alongthez-axis,theequations(4.2)reducetothescalarequations:

(FR)x =∑Fx ,

(FR)y =∑Fy ,

(MR)O =∑MC+∑MO .

4.7 Simplification of a Force and Couple System 23



4.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM

• Incertainspecialcircumstances(whenthesystemofforcesiseitherconcurrent, coplanar or parallel),thesystemofforcesandcouplemomentsactingonarigidbodyreducesatpointOtoaresultantforceFRandaresultantcouplemoment(MR)Owhichareperpendicular to one another.Whenthisoccurs,itispossibletofurthersimplifytheforceandcouplemomentsystembymovingFRtoanotherpointP(onoroffthebody)sothatno resultant couple moment has to be applied to the body.Thatis,only the force resultantwillhavetobeappliedtothebody(atP).The locationofpointPmeasuredfrompointOcanalwaysbedeterminedprovidedFRand(MR)Oareknown.

• ReductiontoaWrench—Ageneral forceandcouplemomentsystemactingonabodywill reducetoasingleresultantforceFRandaresultantcouplemoment(MR)OatOwhicharenotperpendiculartooneanother.Inthiscasetheforceandcouplemomentsystemactingonabodycanbereducedtoawrench or screw:acombinationofacollinear force and couple moment.

4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING

Inmanysituationsaverylargesurfaceareaofabodymaybesubjectedtodistributed loadingssuchasthosecausedbywind,fluidsorsimplytheweightofmaterialsupportedoverthebody’ssurface.

• Distributed loadingsaredefinedbyusingaloadingfunctionw=w(x)thatindicatestheintensityoftheloadingalongthelengthofthemember.ThisintensityismeasuredinN/m.

• Theexternaleffectscausedbyacoplanardistributedloadactingonabodycanberepresentedbyasingle resultant force.

• Themagnitude of the resultant forceisequaltothetotalareaunderthedistributedloadingdiagramw=w(x).

• Thelocationoftheresultantforceisgivenbythefactthatitslineofactionpassesthroughthecentroidorgeometriccenterofthisarea.


• Beawareofthedifferencebetweenthecross productandthedot productoftwovectors.Theformerisavectorwhilethelatterisascalar.Also

A·B=B·A BUT A×BB×A (=–(A×B)).

• Theright-handruleisessentialinthecalculationofmoments.Youshouldbeabletoapplythisrulequicklyandaccurately.

REVIEW QUESTIONS

1. Whatismeantbyamoment?Isitavectororascalar?

2. What’sthemagnitudeofthemomentofaforceaboutapoint?

3. Howdoyoucalculatethesense(direction)ofamoment?

4. IfthelineofactionofaforcepassesthroughapointP,whatdoyouknowaboutthemomentoftheforceaboutP ?

5. IfyouknowthecomponentsoftwovectorsAandB,howdoyoudeterminetheircrossproduct?Isthecross-productavectororascalar?

6. IfA×B=0,whatdoesthismean?


7. WhenyouusetheequationMO=r×FtodeterminethemomentofaforceFaboutO,howdoyouchooser?

8. IfyouknowthecomponentsofthevectorMO=r×FhowcanyoudeterminetheproductofthemagnitudeofFandtheperpendiculardistancedfromOtothelineofactionofF?

9. HowdoyoufigureoutthesenseofthemomentofFaboutOusingtheformulaMO=r×F?

10. HowwouldyoucalculatethemomentexertedaboutapointObyacoupleconsistingofforcesFand–F?

11. True or false? The moment of a couple about O is different than the moment of the same couple aboutPO?

12. Whatismeantbyanequivalent system?

13. Howdowereplaceasystemofforcesandcouplemomentsactingonabodywithanequivalent singleresultantforceandcouplemomentactingataspecifiedpointO?Whataretherelevantequations?

14. Definewhatismeantbyawrench.Whendoesaforceandcouplemomentsystemactingonabodyreducetoawrench?

15. Howistheresultantforceexertedbyacoplanardistributedloadactingonabodydeterminedfromthefunctionw(x)?

REVIEW QUESTIONS 25



• Todeveloptheequationsofequilibriumforarigidbody.

• Tointroducetheconceptofthefree-bodydiagramforarigidbody.

• Toshowhowtosolverigidbodyequilibriumproblemsusingtheequationsofequilibrium.

5.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM

A rigid body is thenext levelof sophistication (after theparticle) in themodellingofanobject.Basicallywe‘add’size/shapetotheexistingmodelofaparticle.Consequently,themaindifferencebetweenaparticleandarigidbodyisthatarigidbodycansupportmoments.Toobtainequationsfortheequilibriumofarigidbody,therefore,weneedtosupplementtheequationsofparticleequilibriumwithanexpressionofmomentbalance.

• Thetwo equationsofequilibriumforarigid bodyare

∑F=0,

∑MO=0,

whereOisanarbitrarypoint.

EQUILIBRIUM IN TWO DIMENSIONS

5.2 FREE-BODY DIAGRAMS

• Noequilibriumproblemshouldbesolvedwithoutfirstdrawingthefree-bodydiagram,soastoaccountforalltheforcesand couple momentsthatactonthebody.

• PartIIofthisstudypackisdevotedtothedrawingoffree-bodydiagramsincluding,specifically,free-body diagrams for rigid body equilibrium in two dimensions.StudyPartIIofthisstudypackmakingspecialnoteofthefollowingimportantpoints:

Equilibrium of a Rigid Body

5


– Ifasupportprevents translationofabodyinaparticulardirection,thenthesupportexertsaforceonthebodyinthatdirection.

– Ifrotation is prevented,thenthesupportexertsacouple momentonthebody.

– StudyTable2.1inPartIIofthissupplement(orTable5-1ofthetext).

– Internalforcesarenevershownonthefree-bodydiagramsincetheyoccurinequalbutoppositecollinearpairsandthereforecancelout.

– Theweightofabodyisanexternalforceanditseffectisshownasasingleresultantforceactingthroughthebody’scenterofgravityG.

– Couple momentscanbeplacedanywhereonthefree-bodydiagramsincetheyarefree vectors.Forcescanactatanypointalongtheirlinesofactionsincetheyaresliding vectors.

5.3 EQUATIONS OF EQUILIBRIUM

• Whenthebodyissubjectedtoasystemofforceswhichalllieinthex–yplane,theforcescanberesolvedintotheirxandycomponents.Consequently,theconditionsforequilibriumintwodimensionscanbewritteninscalarformas:

∑Fx=0, (5.0)

∑Fy=0,

∑MO=0,

where∑ MOrepresentsthealgebraicsumofthecouplemomentsandmomentsofalltheforcecomponentsaboutanaxisperpendiculartothexy-planeandpassingthroughanarbitrarypointO(onoroffthebody).

TWO ALTERNATIVE SETS OF EQUILIBRIUM EQUATIONS

• ∑Fa=0,

∑MA=0,

∑MB=0,

Here,theonlyrequirementisthatalinepassingthroughpointsAandBisnotperpendiculartothea-axis.

• ∑MA=0,

∑MB=0,

∑MC=0.

Here,theonlyrequirementisthatpointsA,BandCdonotlieonthesameline.

5.3 Equations of Equilibrium 27


28 Chap. 5 Equilibrium of a Rigid Body

PROCEDURE FOR SOLVING COPLANAR FORCE EQUILIBRIUM PROBLEMS

• Free-Body Diagram

– Establishthex,ycoordinateaxesinanysuitableorientation.

– Drawanoutlinedshapeofthebody.

– Showalltheforcesandcouplemomentsactingonthebody.

– Labelalltheloadingsandspecifytheirdirectionsrelativetothexy-axes.

– Indicatethedimensionsofthebodynecessaryforcomputingthemomentsofforces.

• Equations of Equilibrium

– Applythemomentequationofequilibrium(∑ MO=0)aboutapointOthatliesattheintersectionofthelinesofactionoftwounknownforces.Inthisway,themomentsoftheseunknownsarezeroaboutO,andadirectsolutionforthethirdunknowncanbedetermined.

– Whenapplyingtheforceequilibriumequations(∑ Fx=0and∑ Fy=0),orientthexandyaxesalonglinesthatwillprovidethesimplestresolutionoftheforcesintotheirxandycomponents.

– Ifthesolutionoftheequilibriumequationsyieldsanegativescalarforaforceorcouplemomentmagnitude,itmeansthatthesenseisoppositetothatwhichwasassumedonthefree-bodydiagram.

5.4 TWO- AND THREE-FORCE MEMBERS

Thesolutiontosomeequilibriumproblemscanbesimplifiedifoneisabletorecognizemembersthataresubjectedtoonlytwoorthreeforces.

• Two-Force Members—Whenamemberissubjectedtono couple momentsandforcesappliedatonlytwopointsAandBonamember,thememberiscalledatwo-force member.Inthiscase,forthemembertobeinequilibrium,itisnecessarythattheresultantforcesatAandBmustbeequal, opposite and collinear.Thelineofactionofboth(resultant)forcesisknownsinceitalwayspassesthroughAandB.Hence only the force magnitude (remember both resultants are equal in magnitude!) needs to be determined or stated.

• Three-Force Members—Whenamemberissubjectedtoonly three forces,itisnecessarythattheforcesbeeither concurrent or parallelforthemembertobeinequilibrium.OncethepointofconcurrencyO(wherethelinesofactionoftheforcesintersect)isidentified,thennecessarily∑ MO=0.Iftwoofthethreeforcesareparallel,thepointofconcurrencyO,issaidtobeat“infinity”andthethirdforcemustbeparalleltotheothertwoforcestointersectatthis‘point’.

EQUILIBRIUM IN THREE DIMENSIONS

5.5 FREE-BODY DIAGRAMS

Thefirststepinsolvingthree-dimensionalequilibriumproblems,asinthecaseoftwodimensions,istodraw a free-body diagram.Thegeneralprocedurefordoingthisisthesameasthatoutlinedforthetwo-dimensionalcaseinSection5.2ofthetext.However,thereareafewsubtledifferencesofwhichyoushouldbeaware:

• It isnecessary tobe familiarwith thedifferent typesof reactive forcesandcouplemomentsactingatvarioustypesofsupportsandconnectionswhenmembersareviewedinthreedimensions.Itisimportanttorecognizethesymbolsusedtorepresenteachofthesesupportsandtounderstandclearlyhowtheforcesandcouplemomentsaredevelopedbyeachsupport.ThesearesummarizedinTable5-2ofthetext.Remember:

– As in the two-dimensional case, a force is developed by a support that restricts the translation of the attached member, whereas a couple moment is developed when rotation of the attached member is prevented.


5.6 EQUATIONS OF EQUILIBRIUM

When the body is subjected to a three-dimensional force system, equilibrium requires that the resultant force andresultantcouplemomentactingonthebodybeequaltozero.

• Invectorformthetwoequilibriumequationsare

∑F=0,

∑MO=0,

where∑ F is thevector sumofall theexternal forcesactingon thebodyand∑ MO is the sumof the couplemomentsandthemomentsofalltheforcesaboutanypointO(onoroffthebody).

• Writing

∑F=∑Fx i+∑Fy j+∑Fz k,

∑MO=∑Mx i+∑My j+∑Mz k.

Thesix scalarequilibriumequationsare

∑Fx=0, ∑Fy=0, ∑Fz=0, (5.1)

∑Mx=0, ∑My=0, ∑Mz=0.

5.7 CONSTRAINTS AND STATICAL DETERMINACY

Toensureequilibriumofarigidbody,itisnotonlynecessarytosatisfytheequationsofequilibrium,butthebodymustalsobeproperlyheldorconstrainedbyitssupports.

• Redundant Constraints. Whenabodyhasredundant supports,thatis,moresupportsthanarenecessarytoholdit inequilibrium, itbecomesstatically indeterminate.Thismeans that therewillbemoreunknown loadingsonthe body than equations of equilibrium available for their solution.The additional equations needed to solveindeterminateproblemsaregenerallyobtainedfromthedeformation conditionsatthepointsofsupport.Theseequationsinvolvemodellingthebodynotasarigidbodybutasadeformable body(thenextlevelofsophistication).Thisisdoneincoursesdealingwith‘mechanicsofmaterials’.

• Improper Constraints. Insomecases,theremaybeasmanyunknownforcesonthebodyasthereareequationsofequilibrium;however,instabilityofthebodymaydevelopbecauseofimproperconstrainingbythesupports.Whenthishappens,eitherthenumberofavailableequilibriumequationsisreducedbyone(makingthesystemindeterminate) or we will not be able to satisfy all the equilibrium equations. Proper constraining (avoidinginstabilityofabody)requires

1. Thelinesofactionofthereactiveforcesdonotintersectacommonaxis,and

2. Thereactiveforcesmustnotallbeparalleltooneanother.

Whentheminimumnumberofreactiveforcesisneededtoproperlyconstrainthebodyinquestion,theproblemwillbestatically determinateandthereforetheequationsofequilibriumcanbeusedtodetermineallthereactiveforces.

5.7 Constraints and Statistical Determinancy 29


30 Chap. 5 Equilibrium of a Rigid Body


• The first step in solving equilibrium problems is to draw a free-body diagram. Don’t try to skip this stage nomatterhowtrivialyouthinkitis!

• Makethefree-body diagramasclearandconciseaspossible.Itwillaidyourunderstandingoftheproblemanditwillhelpyouconstructtheequilibriumequations.

REVIEW QUESTIONS

1. Whyistherenomomentequilibriumequationforabodymodelledasaparticle?

2. Writedownthesixindependentscalarequilibriumequationsforarigidbodyinthreedimensions.Adapttheseequationstothetwo-dimensionalcaseexplainingwhytherearenowonlythreeindependentequations.

3. Whatdoesitmeanwhenanobjectissaidtohaveredundantsupports.

4. Howdoyouknowifanobjectisstaticallyindeterminateasaresultofredundantsupports?

5. Howdoyouavoidinstabilityofabodyduetoimproperconstraining?



• Toshowhowtodeterminetheforcesinthemembersofatrussusingthemethodofjointsandthemethodofsections.

• Toanalyzetheforcesactingonthemembersofframesandmachinescomposedofpin-connectedmembers.

6.1 SIMPLE TRUSSES

A truss is a structure composed of slender members joined together at their end points.The members are usuallywoodenstrutsormetalbars.Thejointconnectionsareusuallyformedbyboltingorweldingtheendsofthememberstoacommonplatecalledagusset plate,orbysimplypassingalargeboltorpinthrougheachofthemembers.

(b)(a)

• Planar Trusseslieinasingleplaneandareoftenusedtosupportroofsandbridges.

• Assumptions for Design.To design both the members and the connections of a truss, it is first necessary todetermine the force developed in each member when the truss is subjected to a given loading.The followingassumptionsallowustoconsidereachtrussmemberasatwo-force membersothattheforcesattheendsofthemembermustbedirectedalongtheaxisofthemember.

– All loadings are applied at the joints.

– The members are joined together by smooth pins.

Structural Analysis

6

06_Struct Analysis.indd 31 4/27/09 9:50:27 AM

32 Chap. 6 Structural Analysis

• Simple Trusses. Topreventcollapse,theformofatrussmustberigid.Thesimplestformwhichisrigidorstableisa triangle.Consequently,asimple truss isconstructedbystartingwithabasic triangularelement.Additionalelementsconsistingoftwomembersandajointareaddedtothetriangularelementtoformasimpletruss.

6.2 THE METHOD OF JOINTS

In order to analyze or design a truss, we must obtain the force in each of its members.To do this, we consider theequilibrium of a jointofthetruss.Thisisthebasisforthemethod of joints.

• Sincethetrussmembersareallstraighttwo-forcememberslyinginthesameplane,theforcesystemactingateach joint iscoplanar and concurrent.Consequently,momentequilibriumisautomaticallysatisfiedat the jointanditisonlynecessarytosatisfytwo independent scalar force equilibrium equations.

PROCEDURE FOR ANALYZING A (PLANAR) TRUSS USING THE METHOD OF JOINTS

• Drawthefree-bodydiagramofajointhavingatleastoneknownforceandatmosttwounknownforces.(Ifthisjoint is at one of the supports, it generally will be necessary to know the external reactions at the trusssupport).

• Establishthesenseofanunknownforcebyeither:

– Assuming that the unknown force is in tension and interpreting negative scalar results as members incompression.OR

– By inspection.

• Orientthe xandyaxessuchthattheforcesonthefree-bodydiagramcanbeeasilyresolvedintotheirxandy componentsandthenapplythetwoforceequilibriumequations∑ Fx=0and∑ Fy=0.Solveforthetwounknownmemberforcesandverifytheircorrectsense.

• Continuetoanalyzeeachoftheotherjointsasabove.

• Once the force inamember is found fromtheanalysisofa jointatoneof itsends, theresultcanbeused toanalyzetheforcesactingonthejointatitsotherend.Rememberthatamemberincompression pushesonthejointandamemberintension pullsonthejoint.

6.3 ZERO FORCE MEMBERS

Trussanalysisusing themethodof joints isgreatly simplified ifone isfirstable todetermine thosememberswhichsupport no loading.These zero-force members areused to increase stabilityof the trussduring constructionand toprovidesupportiftheappliedloadingischanged.

• Zero-forcemembersofatrussaregenerallydeterminedbyinspection of each of its joints.Asageneralrule:

– Ifonlytwomembersformatrussjointandnoexternalloadorsupportreactionisappliedtothejoint,themembersmustbezeroforcemembers.

– Ifthreemembersformatrussjointforwhichtwoofthemembersarecollinear,thethirdmemberisazeroforcememberprovidednoexternalforceorsupportreactionisappliedtothejoint.

6.4 THE METHOD OF SECTIONS

Themethod of sectionsisusedtodeterminetheloadingsactingwithinabody.Itisbasedontheprinciplethatif a body is in equilibrium then any part (section) of the body is also in equilibrium.


PROCEDURE FOR ANALYZING THE FORCES IN THE MEMBERS OF A TRUSS USING THE METHOD OF SECTIONS


– Make a decision as to how to ‘cut’ or section the truss through the members where forces are to bedetermined.

– Beforeisolatingtheappropriatesection,itmayfirstbenecessarytodeterminethetruss’externalreactions.Thenthreeequilibriumequationsareavailabletosolveformemberforcesatthecutsection.

– Drawthefree-bodydiagramofthatpartofthesectionedtrusswhichhastheleastnumberofforcesactingonit.

– Establishthesenseofanunknownmemberforcebyeither:

* Assuming that the unknown member force is in tension and interpreting negative scalar results asmembersincompression.OR

* By inspection.


– Momentsshouldbesummedaboutapointthatliesattheintersectionofthelinesofactionoftwounknownforces,sothatthethirdunknownforceisdetermineddirectlyfromthemomentequation.

– Iftwooftheunknownforcesareparallel,forcesmaybesummedperpendiculartothedirectionoftheseunknownstodeterminedirectlythethirdunknownforce.

6.5 SPACE TRUSSES

A space truss consists of members joined together at their ends to form a stable three-dimensional structure.Thesimplestelementofaspacetrussisatetrahedron,formedbyconnectingsixmemberstogether.

P

Either the method of joints or the method of sections can be used to determine the forces developed in themembersofasimplespacetruss:

• Method of Joints.Iftheforcesinallthemembersofthetrussmustbedetermined,themethodofjointsismostsuitablefortheanalysis.Solvethethreescalarequilibriumequations∑ Fx=0,∑ Fy=0and∑ Fz=0ateachjoint.Thesolutionofmanysimultaneousequationscanbeavoidediftheforceanalysisbeginsatajointhavingatleastoneknown forceandatmost three unknown forces.UseaCartesianvectoranalysis if the three-dimensionalgeometryoftheforce-systematthejointishardtovisualize.

6.5 Space Trusses 33


34 Chap. 6 Structural Analysis

• Method of Sections.Ifonlyafewmemberforcesaretobedetermined,themethodofsectionsismostsuitable.Whenanimaginarysectionispassedthroughatrussandthetrussisseparatedintotwoparts,theforcesystemactingononeofthepartsmustsatisfythesixscalarequilibriumequations∑ Fx=0,∑ Fy=0,∑ Fz=0,∑ Mx=0and∑ My=0and∑ Mz=0.Byproperchoiceofthesectionandaxesforsummingforcesandmoments,manyoftheunknownmemberforcesinaspacetrusscanbecomputeddirectlyusingasingleequilibriumequation.

6.6 FRAMES AND MACHINES

Frames and machines are two common types of structures which are often composed of pin-connected multiforcemembers. Frames are generally stationary and are used to support loads, while machines contain moving parts andare designed to transmit and alter the effect of forces. Once the forces at the joints are obtained (see below) it isthenpossibletodesignthesizeofthemembers,connectionsandsupportsusingthetheoryofmechanicsofmaterials(deformablebodies)andanappropriateengineeringdesigncode.

PROCEDURE FOR DETERMINING THE JOINT REACTIONS ON FRAMES OR MACHINES COMPOSED OF MULTIFORCE MEMBERS


– Drawthefree-bodydiagramoftheentirestructure,aportionofthestructure,oreachofitsmembers.Thechoiceshouldbemadesothatitleadstothemostdirectsolutionoftheproblem.

– Whenthefree-bodydiagramofagroupofmembersofastructureisdrawn,theforcesattheconnectedpartsofthisgroupareinternalforcesandarenotshownonthefree-bodydiagramofthegroup.

– Forcescommontotwomemberswhichareincontactactwithequalmagnitudebutoppositesenseontherespectivefree-bodydiagramsofthemembers.

– Two-forcemembers,regardlessoftheirshape,haveequalbutoppositecollinearforcesactingattheendsofthemember.

– Inmanycasesitispossibletotellbyinspectionthepropersenseoftheunknownforcesactingonamember;however,ifthisseemsdifficult,thesensecanbeassumed.

– Acouplemomentisafreevectorandcanactatanypointonthefree-bodydiagram.Also,aforcevectorisaslidingvectorandcanactatanypointalongitslineofaction.


– Count the number of unknowns and compare it to the total number of equilibrium equations that areavailable,Intwodimensions,therearethreeequilibriumequationsthatcanbewrittenforeachmember.

– Summomentsaboutapointthatliesattheintersectionofthelinesofactionofasmanyunknownforcesaspossible.

– Ifthesolutionofaforceorcouplemomentmagnitudeisnegative,itmeansthatthesenseisthereverseofthatshownonthefree-bodydiagrams.


• Theimportanceofdrawingandusingaclearandconcisefree-bodydiagramcannotbeoverstated.

• Asinmostmechanicsproblems,practiceisthekey.MakesureyoureadExamples6-9through6-13inthetextandattempttodrawtherequestedfree-bodydiagramsyourself.Whendoingso,makesuretheworkisneatandthatalltheforcesandcouplemomentsareproperlylabelled.


REVIEW QUESTIONS

1. Whatisatruss?

2. Whatassumptionsallowustoconsideratrussmemberasatwo-force member?

3. Whatisthemethodofjoints?

4. Howmanyindependentscalarequilibriumequationsareavailablefromthefree-bodydiagramofajoint?

5. Whatisthemethodofsections?

6. Whatmethodsareavailabletodeterminetheforcesdevelopedinthemembersofasimplespacetruss?

7. What’sthedifferencebetweenaframeandamachine?

8. Whenthefree-bodydiagramofagroupofmembersofastructureisdrawn,theforcesattheconnectedpartsofthisgrouparenotshownonthefree-bodydiagramofthegroup.Why?

REVIEW QUESTIONS 35



• Toshowhowtousethemethodofsectionsfordeterminingtheinternalloadingsinamember.

• Togeneralizethisprocedurebyformulatingequationsthatcanbeplottedsothattheydescribetheinternalshearandmomentthroughoutamember.

• Toanalyzetheforcesandstudythegeometryofcablessupportingaload.

7.1 INTERNAL FORCES DEVELOPED IN STRUCTURAL MEMBERS

Thedesignofanystructuralormechanicalmemberrequiresaninvestigationofboththeexternalloadsandreactionsactingonthememberandtheloadingactingwithinthemember—inordertobesurethe material can resist this loading.Theseinternalloadingscanbedeterminedusingthemethod of sections.

Theideaistocutan‘imaginarysection’throughthemembersothattheinternalloadings(ofinterest)atthesectionbecomeexternalonthefree-bodydiagramofthesection.

PROCEDURE FOR FINDING THE INTERNAL LOADINGS AT A SPECIFIC LOCATION IN A MEMBER USING THE METHOD OF SECTIONS

• Support Reactions

Before the member is ‘cut’ or ‘sectioned’, it may first be necessary to determine the member’s supportreactions,sothattheequilibriumequationsareusedonlytosolvefortheinternalloadingswhenthememberissectioned.

If the member is part of a frame or machine, the reactions at its connections are determined using themethodsoutlinedinSection6.6.

Internal Forces

7

07_Internal Forces.indd 36 4/21/09 12:43:18 PM


Keepalldistributed loadings, couplemomentsand forcesactingon themember in theirexact locations,thenpassanimaginarysectionthroughthemember,perpendiculartoitsaxisatthepointwheretheinternalloadingistobedetermined.

Afterthesectionismade,drawafree-bodydiagramofthesegmentthathastheleastnumberofloadsonit,andindicatethex,y,zcomponentsoftheforceandcouplemomentresultantsatthesection.

Ifthememberissubjectedtoacoplanar system of forces,onlyN(normalforce),V(shearforce),andM(bendingmoment)actatthesection.

V

N

M

Shearforce

Normalforce

Bendingmoment

C

Inthreedimensions,ageneralinternalforceandcouplemomentresultantwillactatthesection.

y

z

Ny

Normalforce

My

Torsionalmoment

Vx

Vz

Mx

x

C

Mz

Shearforcecomponents

Bendingmomentcomponents

Inmanycasesitmaybepossibletotellbyinspectionthepropersenseoftheunknownloadings;however,ifthisseemsdifficult,thesensecanbeassumed.


Momentsshouldbesummedatthesectionaboutaxespassingthroughthecentroidorgeometriccenterofthemember’scross-sectionalareainordertoeliminatetheunknownnormalandshearforcesandtherebyobtaindirectsolutionsforthemomentcomponents.

Ifthesolutionsoftheequilibriumequationsyieldsanegativescalar,theassumedsenseofthequantityisoppositetothatshownonthefree-bodydiagram.

7.2 SHEAR AND MOMENT EQUATIONS AND DIAGRAMS

Beams are designed to support loads perpendicular to their axes.The actual design of a beam requires a detailedknowledgeofthevariationoftheinternalshearforceVandbendingmomentMactingateach point alongtheaxisofthebeam.Afterthis,thetheoryofmechanicsofmaterialsisusedwithanappropriateengineeringdesigncodetodeterminethebeam’srequiredcross-sectionalarea.

7.2 Shear and Moment Equations and Diagrams 37


38 Chap. 7 Internal Forces

ThevariationsofVandMasfunctionsofthepositionxalongthebeam’saxiscanbeobtainedusingthemethodofsections(Section7.1).However,itisnecessarytosectionthebeamatanarbitrarydistancexfromoneendratherthanataspecifiedpoint.Iftheresultsareplotted,thegraphicalvariationsofVandMasfunctionsofxaretermedtheshear diagram and bending moment diagram,respectively.

Thesediagramscanbeconstructedasfollows:

• Support Reactions

Determineallthereactiveforcesandcouplemomentsactingonthebeamandresolvealltheforcesintocomponentsactingperpendicularandparalleltothebeam’saxis.

• Shear and Moment Functions

Specifyseparatecoordinatesxhavinganoriginatthebeam’s left endandextendingtoregionsofthebeambetween concentrated forces and/or couple moments, or where there is no discontinuity of distributedloading.

Sectionthebeamperpendiculartoitsaxisateachdistancexanddrawthefree-bodydiagramofoneofthesegments.BesureVandMareshownactingintheirpositivesenseinaccordancewiththefollowingsignconvention:

Positiveshear

Positivemoment

Beamsignconvention

M M

V

V

V

V

M M

TheshearVisobtainedbysummingforcesperpendiculartothebeam’saxis.

ThemomentMisobtainedbysummingmomentsaboutthesectionedendofthesegment.

• Shear and Moment Diagrams

Plotthesheardiagram(Vversusx)andthemomentdiagram(Mversusx).IfthecomputedvaluesofthefunctionsdescribingVandMarepositive,thevaluesareplottedabovethex-axis,whereasnegativevaluesareplottedbelowthex-axis.

Generally, it isconvenient toplot theshearandbending-momentdiagramsdirectlybelowthe free-bodydiagramofthebeam.SeeExamples7-7and7-8inthetext.


7.3 RELATIONS BETWEEN DISTRIBUTED LOAD, SHEAR AND MOMENT

Incaseswhereabeamissubjectedtoseveralconcentratedforces,couplemoments,anddistributedloads,themethodofconstructingtheshearandbendingmomentdiagramsmaybecomequitetedious.Inthissectionasimplermethodforconstructing thesediagrams ispresented—basedondifferential relations thatexistbetweenthe load, shear,andbendingmoment.Thefollowingarethemainpoints:

• Theslopeofthesheardiagramisequaltothenegativeoftheintensityofthedistributedloading,wherepositivedistributedloadingisdownward,i.e.

dVdx

=–w(x). (7.0)

• Ifaconcentratedforceactsdownwardonthebeam,theshearwilljumpdownwardbytheamountoftheforce.

• Thechangeintheshear∆Vbetweentwopointsisequaltothenegative of the areaunderthedistributed-loadingcurvebetweenthepoints.

• Theslopeofthemomentdiagramisequaltotheshear,i.e.

dMdx

=V. (7.1)

• Thechangeinthemoment∆Mbetweentwopointsisequaltotheareaunderthesheardiagrambetweenthetwopoints.

• Ifaclockwisecouplemomentactsonthebeam,theshearwillnotbeaffected,however, themomentdiagramwilljumpupwardbytheamountofthemoment.

• Pointsofzeroshearrepresentpointsofmaximum or minimum momentsince

dMdx

=0. (7.2)

7.4 CABLES

Flexiblecablesandchainsareusedinengineeringstructuresforsupportandtotransmitloadsfromonemembertoanother.Intheforceanalysisofsuchsystems,theweightofthecableitselfmaybeneglected(cableisreferredtoas‘weightless’)becauseitisoftensmallcomparedtotheloaditcarries.Inmodellingthecable,itisassumedthat:

1. Thecable isperfectly flexible (cableoffersno resistance tobending so the tensile forceacting in thecable isalwaystangenttothecableatpointsalongitslength).

2. Thecableisinextensible(cablehasaconstantlengthbeforeandafterloadisapplied—cablecanbetreatedasarigidbody).

CABLE SUBJECTED TO CONCENTRATED LOADS

• Whenacableofnegligibleweightsupportsseveralconcentratedloads,thecabletakestheformofseveralstraight-line segments,eachofwhich is subjected toaconstant tensile force.Theequilibriumanalysis isperformedbywritingdownasufficient numberofequilibriumequations(basedontheentirecableoranypartthereof)andequationsdescribingthegeometryofthecabletosolveforalltheunknownsleadingtoadescriptionofthetensionin(eachsegmentof)thecable.SeeExample7-13intext.

7.4 Cables 39


40 Chap. 7 Internal Forces

CABLE SUBJECTED TO A DISTRIBUTED LOAD

A

B

w = w(x)

∆xx

y

x

• Theequation

yF

x dx dxH

( ) ,= ( )∫∫1w (7.3)

determinesthecurveforthecabley=f(x).Here,FHmeasuresthehorizontalcomponentoftensileforceatany pointalongthecableandw(x)istheloadingfunctionmeasuredinthe x-direction.Inpractice,FHandthetwoconstantsofintegrationaredeterminedfromtheboundaryconditionsforthecable—seeExample7-14intext.

CABLE SUBJECTED TO ITS OWN WEIGHT

• Whentheweightofthecablebecomesimportantintheforceanalysis(e.g.cablesusedastransmissionlines),theloadingfunctionalongthecablebecomesafunctionofthearclengthsratherthantheprojectedlengthx.Theequationofthedeflectioncurveisgivenbyy=f(x)where

dydx F

s dsH

( ) ,= ∫1w (7.4)

xds

Fx ds

H

( )

.=

+ ( )

∫

∫1

12

212

w

(7.5)

The twoconstantsof integration from(7.5)are foundusing theboundaryconditions for thecable.First solve(7.5)forw(s)thenuse(7.4)togettheshapey=f(x)forthecable.SeeExample7-15intext.



• Useclearandconcisefree-bodydiagrams.

• Asinmostmechanicsproblems,practiceisthekey.MakesureyoureadExamples7-1through7-15inthetextbefore you attempt any corresponding problems.These examples will serve as templates with which to solveproblems.Drawtherequestedfree-bodydiagramsyourself.Whendoingso,makesuretheworkisneatandthatalltheforcesandcouplemomentsareproperlylabelled.

REVIEW QUESTIONS

1. Whatarethenormal(oraxial)force,shearforceandbendingmoment?

2. Howarethepositivedirections(sense)oftheshearforceVandbendingmomentMdefined?

3. Foraportionofabeamwhichissubjectedonlytoadistributedloadw,howaretheshearforceandbendingmomentdistributionsdeterminedfromequations(7.0)and(7.1)?

4. Whatdoesitmeanwhenacableisassumedtobe‘weightless’,‘inextensible’or‘perfectlyflexible’?

5. Ifacableissubjectedtoaloadthatisuniformlydistributedalongastraightlineanditsweightisnegligible,whatmathematicalcurvedescribesitsshape?

REVIEW QUESTIONS 41



• Tointroducetheconceptofdryfrictionandshowhowtoanalyzetheequilibriumofrigidbodiessubjectedtothisforce.

• Topresentspecificapplicationsoffrictionalforceanalysisonwedges,screws,beltsandbearings.

• Toinvestigatetheconceptofrolling resistance.

8.1 CHARACTERISTICS OF DRY FRICTION

Asaresultofexperimentsthatpertaintotheforegoingdiscussion,thefollowingruleswhichapplytobodiessubjectedtodryfrictionmaybestated:

• Thefrictionalforceactstangenttothecontactingsurfacesinadirectionopposed to the relative motionortendencyformotionofonesurfaceagainstanother.

• Themaximumstatic frictional forceFs that canbedeveloped is independentof theareaof contact,providedthenormalpressureisnotverylownorgreatenoughtoseverelydeformorcrushthecontactingsurfacesofthebodies.

• ThemaximumstaticfrictionalforceisgenerallygreaterthanthekineticfrictionalforceFkforanytwosurfacesofcontact.However,ifoneofthebodiesismovingwithavery low velocityoverthesurfaceofanother,FkbecomesapproximatelyequaltoFs,i.e.µ sµ k .

• When slipping at the surface of contact is impending (about to occur), the maximum static frictional force isproportionaltothenormalforceN,suchthatFs=µ s N.

• Whenslippingatthesurfaceofcontactisoccurring,thekineticfrictionalforceisproportionaltothenormalforceN,suchthatFk=µ k N.

Friction

8

08_Friction.indd 42 4/27/09 9:57:24 AM

8.2 PROBLEMS INVOLVING DRY FRICTION

Ifarigidbodyisinequilibriumwhenitissubjectedtoasystemofforceswhichincludestheeffectoffriction,theforcesystemmustsatisfynotonlytheequationsofequilibriumbutalsothelawsthatgovernthefrictionalforces.

Ingeneral,therearethreetypesofmechanicsproblemsinvolvingdryfriction.Theyareclassifiedoncethefree-bodydiagramsaredrawnandthe totalnumberofunknownsare identifiedandcomparedwith the totalnumberofavailableequilibriumequations:

• Equilibrium—Thetotalnumberofunknownsisequaltothetotalnumberofavailableequilibriumequations.Inthiscase,oncethefrictionalforcesaredetermined,checkthatF≤µ s Notherwiseslippingwilloccurandthebodywillnotremaininequilibrium.

• Impending Motion at All Points—Thetotalnumberofunknownswillequalthetotalnumberofavailableequilibriumequationsplusthetotalnumberofavailablefrictionalequationsorconditionalequationsfortipping.Asaresult,severalpossibilitiesformotionorimpendingmotionwillexistandtheproblemwillinvolveadeterminationofthekindofmotionwhichactuallyoccurs.

• Impending Motion at Some Points—Thetotalnumberofunknownswillbelessthanthetotalnumberofavailableequilibriumequationsplusthetotalnumberofavailablefrictionalequations,F=µ N.Ifmotionisimpendingatthepointsofcontact,thenFs=µ s Nwhereasifthebodyisslipping,thenFk=µ k N.

EQUILIBRIUM VERSUS FRICTIONAL EQUATIONS

WhenthefrictionalforceFisanequilibriumforce,i.e.F<µ N,wecanalwaysassumethesenseofthefrictionalforceF(sincethefrictionalforcealwaysactssoastoopposetherelativemotionorimpedethemotionofabodyoverthecontactingsurface).ThecorrectsenseisdeterminedaftersolvingtheequilibriumequationsforF.However,incaseswhereF=µ Nisused,wecannolongerassumethesenseofFsincetheequationF=µ Nrelatesonlythemagnitudesoftwoperpendicularvectors.Consequently,inthiscase,Fmustalwaysbeshownactingwithitscorrectsenseonthefree-bodydiagram.

PROCEDURE FOR ANALYSIS

• Free-Body Diagrams

Drawthenecessaryfree-bodydiagramsand,unless it isstatedintheproblemthat impendingmotionorslippingoccurs,alwaysshowthefrictionalforcesasunknowns,i.e.do not assume F=µ N.

Determine the number of unknowns and compare this with the number of available equilibriumequations.

Iftherearemoreunknownsthanequilibriumequations,itisnecessarytoapplythefrictionalequationsatsome,ifnotall,pointsofcontacttoobtaintheextraequationsneededforcompletesolution.

IftheequationF=µ Nistobeused,itwillbenecessarytoshowFactingintheproperdirectiononafree-bodydiagram.

• Equations of Equilibrium and Friction

Applytheequationsofequilibriumandthenecessaryfrictionalequations(orconditionalequationsiftippingispossible)andsolvefortheunknowns.

Iftheprobleminvolvesathree-dimensionalforcesystemsuchthatitbecomesdifficulttoobtaintheforcecomponentsorthenecessarymomentarms,applythevector equations of equilibrium.

8.3 WEDGES

Awedge isa simplemachinewhich isoftenused to transformanapplied force intomuch larger forces,directedatapproximatelyrightanglestotheappliedforce.Alsowedgescanbeusedtogivesmalldisplacementsoradjustmentsto heavy loads.The analysis of problems involving wedges proceeds as above, i.e. we draw free-body diagrams ofthewedgeandanyothercontactingbodiesand formulate theappropriateequilibriumand frictionalequations.SeeExample8-7intext.

8.2 Problems Involving Dry Friction 43

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44 Chap. 8 Friction

8.4 FRICTIONAL FORCES ON SCREWS

Ascrewmaybethoughtofsimplyasaninclinedplaneorwedgewrappedaroundacylinder.Inmostcasesscrewsareusedasfasteners;however,inmanyapplications,theyareincorporatedtotransmitpowerormotionfromonepartofthemachinetoanother.

Beforeproceedingtosolveproblemsinvolvingfrictionalforcesonscrews,eachofthefollowingcasesshouldbethoroughlyunderstood:

• Frictional Analysis with Upward Screw Motion. Themomentnecessarytocauseupwardimpendingmotionofthescrewis

M=Wrtan(q+φ), φ=φs=tan–1µ s . (8.0)

Ifφisreplacedbyφk=tan–1µ k,weobtainasmallervalueofMnecessarytomaintainuniformupwardmotionofthescrew.

W

Upwardscrewmotion

N

F

R

n

S

q

q φ

• Frictional Analysis with Downward Screw Motion (whenthesurfaceof thescrewisveryslippery:q>φ).Themomentnecessarytocausedownwardimpendingmotionofthescrewis

M ′=Wrtan(q–φ), φ=φ s. (8.1)

Ifφ is replaced byφk = tan–1 µ k, we obtain a (smaller) value of M necessary to maintain uniform downwardmotionofthescrew.

W

Downwardscrewmotion( > )

S ′

n

Rq

q φ

q φ

08_Friction.indd 44 4/27/09 9:57:26 AM

• Frictional Analysis with Downward Screw Motion (when the surface of the screw is very rough: q < φ).Themomentnecessarytocausedownwardimpendingmotionofthescrewis

M ′=Wrtan(q–φ), φ=φ s . (8.2)

Ifφ is replaced byφk = tan–1 µ k, we obtain a (smaller) value of M necessary to maintain uniform downwardmotionofthescrew.

W

Downwardscrewmotion(<)

S″

R

n

q φ

q

q φ

• Frictional Analysis with a Self-Locking Screw. IfthemomentM(oritseffectS)isremoved,thescrewwillremainself-locking,i.e.itwillsupporttheloadWbyfrictionforcesalone(providedφ≥q).

W

Self-lockingscrew(=)(onthevergeofrotatingdownward)

R

qφn

q

q φ

=

8.5 FRICTIONAL FORCES ON FLAT BELTS

Wheneverbeltdrivesaredesigned,itisnecessarytodeterminethefrictionalforcesdevelopedbetweenthebeltanditscontactingsurface.

• ThetensionT2inthebeltrequiredtopullthebeltcounterclockwiseoverthesurfaceandtherebyovercomeboththe frictional forcesat the surfaceof contactand theknown tensionT1 (motionor impendingmotionofbeltrelativetosurface)is:

T2=T1 e µ β,

where µ is the coefficient of static or kinetic friction between the belt and the surface of contact, and β is inradians.

8.5 Frictional Forces on Flat Belts 45

08_Friction.indd 45 4/27/09 9:57:26 AM

46 Chap. 8 Friction

Motionorimpendingmotionofbeltrelativetosurface

r

T2

T1

β

q

8.6 FRICTIONAL FORCES ON COLLAR BEARINGS, PIVOT BEARINGS, AND DISKS

Pivotandcollarbearingsarecommonlyusedinmachinestosupportanaxialloadonarotatingshaft.

R

Pivotbearing

(a)

M

P

Collarbearing

(b)

R1

R2

M

P

• Themagnitudeofthemomentrequiredforimpendingrotationoftheshaftisgivenby

M=23

µ s PR R

R R23

13

22

12

–

–

.

08_Friction.indd 46 4/27/09 9:57:26 AM

z

rR1R2

dF

dNdAp

M

P

q

• Thefrictionalmomentdevelopedattheendoftheshaft,whenitisrotatingatconstantspeedcanbefoundbysubstitutingµ kforµ sintheexpressionforM.

8.7 FRICTIONAL FORCES ON JOURNAL BEARINGS

Whenashaftoraxleissubjectedtolateralloads,ajournal bearingiscommonlyusedforsupport.

• Themomentneededtomaintainconstantrotationoftheshaftisgivenby

M=Rrsinφk, (8.3)

whereφk istheangleofkineticfrictiondefinedbytanφk=µ kandR isthemagnitudeofthebearingreactiveforceactingatA.

M

P

rA

N

R

F

(b)

φ A

zRotation

(a)

k

φ k

8.8 ROLLING RESISTANCE

• The force P necessary to initiate and maintain rolling (of a cylinder with weight W and radius r) at constantvelocityhasmagnitude

PWa

r.

Here,thedistanceaisreferredtoasthecoefficient of rolling resistance.

8.8 Rolling Resistance 47

08_Friction.indd 47 4/27/09 9:57:26 AM

48 Chap. 8 Friction

r

W

P

A

a

N

q


• Useclearandconcisefree-bodydiagrams.

• Asinmostmechanicsproblems,practiceisthekey.MakesureyoureadExamples7-1through7-15inthetextbefore you attempt any corresponding problems.These examples will serve as templates with which to solveproblems.Drawtherequestedfree-bodydiagramsyourself.Whendoingso,makesuretheworkisneatandthatalltheforcesandcouplemomentsareproperlylabelled.

REVIEW QUESTIONS

1. Ifrelativeslippingoftwodrysurfacesincontactisimpending,whatcanyousayaboutthefrictionalforcestheyexertoneachother?

2. Iftwodrysurfacesincontactareslidingrelativetoeachother,whatcanyousayaboutthefrictionalforcestheyexertoneachother?

3. Whatarethecharacteristicsofanequilibriumprobleminvolvingdryfriction?

4. What are the characteristics of a problem involving dry friction when motion is impending at all points ofcontact?

5. What’sthefirststepinsolvingamechanicsprobleminvolvingdryfriction?

6. Ifascrewissubjectedtoalargeaxialload,whatistheequationwhichwillgivethemomentnecessarytorotatethescrewandcauseittomoveinadirectionoppositetotheaxialload?

7. Ifascrewissubjectedtoalargeaxialload,whatistheequationwhichwillgivethemomentnecessarytorotatethescrewataconstant rateandcauseittomoveinthedirectionoftheaxialload?

8. IftheshaftofajournalbearingissubjectedtoalateralloadwithmagnitudeP,howdoyoufindthemomentnecessarytomaintainconstantrotationoftheshaft?

08_Friction.indd 48 4/27/09 9:57:27 AM


• Todiscusstheconceptofthecenter of gravity,center of mass,andthecentroid.

• Toshowhowtodeterminethe locationof thecenterofgravityandcentroidforasystemofdiscreteparticlesandabodyofarbitraryshape.

• TousethetheoremsofPappusandGuldinusforfindingtheareaandvolumeforasurfaceofrevolution.

• Topresentamethodforfindingtheresultantofageneraldistributedloadingandshowhowitappliestofindingtheresultantofafluid.

9.1 CENTER OF GRAVITY, CENTER OF MASS AND THE CENTROID OF A BODY

• Thecenter of gravity Gisapointwhichlocatestheresultant weight of a system of particles.ThiscoordinatesofthecenterofgravityGofasystemofparticlesisgivenby

x=∑ x W

∑W, y=

∑ y W

∑W, z=

∑ z W

∑W, (9.0)

wherex,y,zrepresentthecoordinatesofeachparticleofthesystemand∑ Wistheresultantsumoftheweightsofalltheparticlesinthesystem.

• Thecenter of massofasystemofparticlesisobtainedbysubstitutingW=mg(assuminggforeveryparticleisconstant)into(9.0).Consequently,thecenterofmasshascoordinates

x=∑ x m

∑ m, y=

∑ y m

∑ m, z=

∑ z m

∑ m,

where∑ Wistheresultantsumofthemassesofalltheparticlesinthesystem.Notethatthelocationofthecenterofgravitycoincideswiththatofthecenterofmass.However,thecenter of mass is independent of gravityandsocanbeusedinsituationswhenparticlesarenotundertheinfluenceofagravitationalattraction.

Center of Gravity and Centroid

9

09_Gravity.indd 49 4/27/09 9:59:39 AM

50 Chap. 9 Center of Gravity and Centroid

• Thecoordinatesofthecenter of gravity Gofabodyaregivenby

x= x dW

dW, y=

y dW

dW, z=

z dW

dW, (9.1)

wherex,y,zrepresentthecoordinatesofanarbitrarypointinthebody.

• Thecenter of massofarigidbodyisobtainedbysubstitutingdW=ginto(9.1)andcancellinggfromboththenumeratorsanddenominators.Thisyields(9.1)withWreplacedbym.

CENTROID

• ThecentroidC (x, y, z) isapointwhichdefines thegeometric center of a body.Thispointcoincideswith thecenter of mass orthecenter of gravityonly if thematerialcomposingthebodyisuniform or homogeneous(inwhichcasebothγandρareconstantthroughoutthebody).

• Formulas used to locate the center of gravity or the centroid simply represent a balance between the sum ofmomentsofallthepartsofthesystemandthemomentofthe‘resultant’forthesystem.Therearethreecasestoconsider:

– Volume. C (x,y,z)isgivenby

x=V

x dV

V dV, y=

V y dV

V dV, z=

V z dV

V dV, (9.2)

– Area. C (x,y,z)isgivenby

x=A

x dA

A dA, y=

A y dA

A dA, z=

A z dA

A dA, (9.3)

– Line. C (x,y,z)isgivenby

x=L

x dL

L dL, y=

L y dL

L dL, z=

L z dL

L dL. (9.4)

• Thecentroidwilllieonanyaxisofsymmetryofthebody.Also,thecentroidmaybelocatedoffthebody,e.g.inthecaseofaringwherethecentroidisatthecenter.

9.2 COMPOSITE BODIES

Acompositebodyconsistsofaseriesofconnected‘simpler’shapedbodies.


Thelocationofthecenterofgravityofacompositebodycanbedeterminedusingthefollowingprocedure:

• Composite Parts

Using a sketch, divide the body or object into a finite number of composite parts that have simplershapes.

Ifacompositeparthasahole,thenconsiderthecompositepartwithouttheholeandconsidertheholeasanadditionalcompositeparthavingnegativeweightorsize.

09_Gravity.indd 50 4/27/09 9:59:40 AM

• Moment Arms

Establishthecoordinateaxesonthesketchanddeterminethecoordinates(x,y,z)ofthecenterofgravityofeachcompositepart.

• Summations

• Determinex,y,zbyapplyingthecenterofgravityequations:

x=∑ x W

∑W, y=

∑ y W

∑W, z=

∑ z W

∑W, (9.5)

where∑ Wisthesumoftheweightsofallthecompositepartsofthebody(totalweightofthebody).

• Ifanobjectissymmetricalaboutanaxis,thecentroidoftheobjectliesonthisaxis.

CENTROID FOR A COMPOSITE

Whenthe(composite)bodyhasconstantdensityorspecificweight,thecenterofgravitycoincideswiththecentroidofthebodywhich,forlines,areasandvolumes,canbefoundusingrelationsanalogousto(9.5)withtheW’sreplacedbyL’s,A’sandV’s,respectively[asin(9.2)–(9.4)].Centroidsforcommonshapesoflines,areas,shellsandvolumesthatoftenmakeupacompositebodyaregiveninthetableontheinsidebackcoverofthetext.

9.3 THEOREMS OF PAPUS AND GULDINUS

Thefollowingtwotheorems(ofPapusandGuldinus)areusedtofindthesurface area and volumeofanyobjectofrevolution:

• Surface Area. TheareaAofasurfaceofrevolutionequalstheproductofthelengthofthegeneratingcurveandthedistancetravelledbythecentroidofthecurveingeneratingthesurfacearea.Thatis:

A=q r L,

whereq istheangleofrevolution(radians), r istheperpendiculardistancefromtheaxisofrevolutiontothecentroidofthegeneratingcurveandListhetotallengthofthegeneratingcurve.

• Volume. ThevolumeVofabodyofrevolutionequalstheproductofthegeneratingareaandthedistancetraveledbythecentroidoftheareaingeneratingthevolume.Thatis:

V=q r A,

whereq istheangleofrevolution(radians), r istheperpendiculardistancefromtheaxisofrevolutiontothecentroidofthegeneratingareaandAisthegeneratingarea.

• Composite Shapes. Thesetwotheoremsmayalsobeappliedtolinesorareasthatmaybecomposedofaseriesof compositeparts. In this case, the total surfaceareaorvolumegenerated is the sumof the surfaceareasofvolumesgeneratedbyeachofthecompositeparts:

A=q ∑ r L, V=q ∑ r A,

wherer isthedistancefromtheaxisofrevolutiontothecentroidofeachcompositepart(rememberthateachpartundergoesthesameangleofrevolutionq).

9.4 RESULTANT OF A GENERAL DISTRIBUTED LOADING

In Section 4.9 we discussed the method used to simplify a distributed loading which is uniform along an axis of arectangularsurface.Here,wegeneralizethismethodtoincludesurfaceswhichhaveanarbitraryshapeandaresubjectedtoavariableloaddistribution.

– Pressure Distribution over a Surface. Consideraflatplate subjected to the loading functionp (x,y)Pa (Pa=1N/m2).Theentireloadingontheplatecanbesimplifiedtoasingle resultant forceFR

9.4 Resultant of a General Distributed Loading 51

09_Gravity.indd 51 4/27/09 9:59:40 AM

52 Chap. 9 Center of Gravity and Centroid

• Magnitude of Resultant Force

FR=V dV,

i.e.totalvolumeunderthedistributedloadingdiagram.

• Location of Resultant Force. Thelocation (x,y)ofFRisgivenby

x=V

x dV

V dV, y=

V y dV

V dV. (9.6)

Inotherwords,

LineofActionofFRpassesthroughthegeometriccenterorcentroidofthevolumeunderthedistributedloadingdiagram,i.e.

x y

yx

(a)

dF

p

dVdA

p = p(x, y)

x y

yx

(b)

FR

9.5 FLUID PRESSURE

AccordingtoPascal’s lawafluidatrestcreatesapressurepatapointthatisthesame in alldirections.Themagnitudeofp(forceperunitarea)isgivenby p=γ z=ρ gz, (9.7)

whereγ isspecificweight,ρismassdensityandzisthedepthofthepointfromthefluidsurface.Equation(9.7)isvalidonly for incompressible fluids (inwhichpressureand temperaturevariationsdonotproduceany significantdensityvariations).

UsingEquation(9.7)andtheresultsofSection9.4, it ispossible todeterminetheresultant forcecausedbyaliquidpressuredistributionandspecifyitslocationonthesurfaceofasubmergedplate.Threecasesareconsidered:

• Flat Plate of Constant Width. Theeasiestofthethreecases.Consideraflatrectangularplateofconstantwidthsubmergedinaliquidwithspecificweightγ .ThemagnitudeoftheresultantforceFRresultingfromthedistributionofpressureovertheplate’ssurfaceisequaltothetrapezoidalvolumehavinganintensityofp1=γ z1atdepthz1andp2=γ z2atdepthz2.ThelineofactionofFRpassesthroughthevolume’s centroid C [seeEquation(9.6)]—thisisnotthecentroidoftheplatebutratherthecenter of pressure P of the plate.

• Curved Plate of Constant Width. ThecalculationofthemagnitudeofFRanditslocationPismorecomplicatedfora(general)curvedplatethanaflatplate.Thereisasimplification,however,whentheplatehasconstantwidth.ThismethodrequiresseparatecalculationsforthehorizontalandverticalcomponentsofFR.

• Flat Plate of Variable Width. TheloadingcausedbythepressuredistributionactingonthesurfaceofasubmergedplatehavingavariablewidthhasresultantFRwithmagnitude given by the volume described by the plate area as its base and linearly varying pressure distribution as its height. FromEquation(9.6),thecentroidofV againdefinesthepointthroughwhichFRacts,i.e.thecenterofpressureP,whichliesonthesurfaceoftheplatejustbelowC,hascoordinatesP (x,y ′)definedbytheequations

x= V x dV

V dV, y=

V y ′ dV

V dV.

09_Gravity.indd 52 4/27/09 9:59:41 AM

Notethatthispointisnotthecentroidoftheplate’sarea.

y

x

y′

z

Liquidsurface

γ FR dF

C x

P(x, y ′)

p = z

dy′

z


• Findingthelocationofthecenterofmassorcentroidinvolvesmultipleintegration(e.g.overvolumesorsurfaces).TheseintegrationscanoftenbereducedtosingleintegrationsusingtheprocedureoutlinedattheendofSection9.1.StudythisprocedurethenreadandperformExamples9-1through9-8ofthetext.

REVIEW QUESTIONS

1. TrueorFalse?

(a) Thelocationofthecenterofmasscoincideswiththatofthecenterofgravity.

(b) Thecentroidofabodyalwayscoincideswiththebody’scenterofmass.

(c) Thecentroidofabodyisalwayslocatedonthebodyinquestion.

(d) Theformulaforthecentroidforthesurfaceareaofaplateinvolvesintegralsoverthesamesurfacearea.

2. Ifthetotalweightofanobjectistoberepresentedbyasingleequivalentforce,wheremustthisforceact?

3. What’sthedifferencebetweenthecenterofgravityandthecenterofmass?

4. Howisthespecificweightofabodydefined?

5. What’stherelationshipbetweenthemassdensityρandthespecificweightγ ofabody?

6. Whatdoesitmeanwhenwesayabodyishomogeneous?

7. Ifanobjectishomogeneous,whatdoyouknowaboutthepositionofitscenterofmass?

8. Intheanalysisofthelocationofthecenterofgravityofacompositebody,howdoyoudealwithaholeinthebody?

9. WhatarethetheoremsofPapusandGuldinususedfor?

10. Showthatthecentroidforthevolumeofabodycoincideswiththecenter of massonlyifthematerialcomposingthebodyisuniform or homogeneous.

REVIEW QUESTIONS 53

09_Gravity.indd 53 4/27/09 9:59:41 AM


• Todevelopamethodfordeterminingthemoment of inertiaforanarea.

• Tointroducetheproductofinertiaandshowhowtodeterminethemaximumandminimummomentsofinertiaofanarea.

• Todiscussthemassmomentofinertia.

10.1 DEFINITION OF MOMENTS OF INERTIA FOR AREAS

• Themomentsofinertia(secondmoments)oftheareaAaboutthexandyaxesaregiven,respectively,by

I y dAxA

,= ∫ 2 (10.0)

I x dAyA

.= ∫ 2 (10.1)

• ThemomentofinertiaoftheareaAaboutthepoleOorthez-axis(alsoknownasthepolar moment of inertia)is

J r dA I IOA

x y= = +∫ 2 (10.2)

wherer2=x2+y2istheperpendiculardistancefromthepole(z-axis)totheelementdA.

Ox

y

y

x

r

dA

A

Moments of Inertia

10

10_Moments.indd 54 4/21/09 12:50:04 PM

• ClearlyIx ,IyandJOarealwayspositiveandhaveunitsoflengthraisedtothefourthpower.

• Theterminology‘momentofinertia’isactuallyamisnomerinthiscontext—ithasbeenadoptedbecauseofthesimilaritywithintegralsofthesameformrelatedtomass.

10.2 PARALLEL AXIS THEOREM FOR AN AREA

• SupposethecentroidofanareaislocatedatC(x ′,y ′,z ′).The moment of inertia of an area A about an axis is equal to the moment of inertia of the area about a parallel axispassing through the area’s centroidplus the product of the area and the square of the perpendicular distance between the axes, e.g.

I I Ad I I Ad J J Ax x y y y x O C, ,= + = + = +′ ′2 2 dd 2, (10.3)

where I I Jx y C′ ′, and representmomentsofinertiaoftheareaaboutacorrespondingparallelaxispassingthroughthearea’scentroid.

Ox

y

d

dx

dy

x′

y′

x′

y′dA

C

10.3 RADIUS OF GYRATION OF AN AREA

• Providedtheareasandmomentsofinertiaareknown,theradii of gyrationofaplanarareaaredeterminedfromtheformulas

kI

Ak

I

Ak

J

Axx

yy

OO, ,= = = (unitsoflength). (10.4)

• Itisclearfromequations(10.0)–(10.1)thatfindingmomentsofinertiarequirestheevaluationofarea integrals.IfonechoosestodescribetheareaelementdAwithdifferentialsizeintwodirections(e.g.dA=dx dy)thenadouble integrationmustbeperformed.Mostoften,however,it iseasiertoperformonlyasingleintegrationbyspecifying the differential element dAhavingadifferentialsizeorthicknessinonlyonedirection.Theprocedureisasfollows:

10.3 Radius of Gyration of an Area 55

10_Moments.indd 55 4/21/09 12:50:04 PM

56 Chap. 10 Moments of Inertia


• MostoftentheelementdAcanberectangularwithafinite lengthanddifferentialwidth.

• Theelementshouldbelocatedsothatitintersectstheboundaryoftheareaatthearbitrarypoint(x,y).Thereare two ways to orient the element dA with respect to the axis about which the moment of inertia is to bedetermined:

Case 1:Thelengthoftheelementcanbeorientedparalleltotheaxis.Inthiscase,Equations(10.0)–(10.1)canbeuseddirectlysinceall partsoftheelementlieatthesamemoment-armdistancefromtheaxis.

Case 2:The length of the element can be oriented perpendicular to the axis. Here neither of Equations(10.0)–(10.1)canbeuseddirectlysinceallpartsoftheelementwillnotlieatthesamemoment-armdistancefromtheaxis.Instead,itisnecessarytofirstcalculatethemomentofinertiaoftheelement(separately)andthenintegratethisresultovertheareaAtoobtaintheappropriateareamomentofinertia.

• SeeExamples10-1to10-4intext.

10.4 MOMENTS OF INERTIA FOR COMPOSITE AREAS

Provided the moment of inertia of each of the simpler areas (making up the composite area) is known, or can bedeterminedaboutacommonaxis,thenthemomentofinertiaofthecompositeareaequalsthealgebraic sumofthemomentsofinertiaofallitsparts.


Themomentofinertiaofacompositeareaaboutareferenceaxiscanbedeterminedusingthefollowingprocedure:

• Composite Parts

– Usingasketch,dividetheareaintoitscompositepartsandindicatetheperpendiculardistancefromthecentroidofeachparttothereferenceaxis.

• Parallel-Axis Theorem

– Themomentofinertiaofeachpartshouldbedeterminedaboutitscentroidalaxis,whichisparalleltothereferenceaxis—use the table given on the inside back cover of the text.

– Ifthecentroidalaxisdoesnotcoincidewiththereferenceaxis,usetheparallelaxistheoremtodeterminethemomentofinertiaofthepartaboutthereferenceaxis.

• Summation

– Themomentofinertiaoftheentireareaaboutthereferenceaxisisfoundbysummingtheresultsofthecompositeparts.

– Ifacompositeparthasa‘hole’, itsmomentofinertiaisfoundby‘subtracting’themomentofinertiafortheholefromthemomentofinertiaoftheentirepartincludingthehole.

10.5 PRODUCT OF INERTIA FOR AN AREA

Ingeneral,themomentofinertiaforanareaisdifferentforeveryaxisaboutwhichitiscomputed.Insomeapplications,itisnecessarytoknowtheorientationofthoseaxeswhichgive,respectively,themaximumandminimummomentsofinertiaforthearea(seeSection10.6).Essentialtothisistheideaofaproduct of inertia for an area.

• TheproductofinertiafortheareaAis

I xy dAxyA

= ∫ .

• Theproductofinertiamaybenegative,positiveorzero(unlikemomentofinertia).ForexampleIxywillbezeroifxoryisanaxisofsymmetryfortheareaA.

• The sign of Ixy depends on the quadrant where the areaA is located. In fact, if the area is rotated from onequadranttoanother,thesignofIxywillchange.

10_Moments.indd 56 4/21/09 12:50:04 PM

• Parallel Axis Theorem for Product of Inertia of an Area A

I I Ad dxy x y x y= +′ ′

Itisimportantthatthealgebraic signsfordxanddybemaintainedwhenapplyingthisresult.

x

y

x′

y′

dx

dy

C

dA y′

x′

10.6 MOMENTS OF INERTIA FOR AN AREA ABOUT INCLINED AXES

• Themomentsandproductsofinertiaforanareawithrespecttoasetofinclined(atanangleq)uandυaxesaregivenby(assumingq,Ix ,IyandIxyareknown):

II I I I

Iux y x y

xy

–cos – sin=

++

2 22 2q q ,

II I I I

Ix y x yxyυ q q–

–cos – sin=

+2 2

2 2 , (10.5)

II I

Iux y

xyυ q q–

sin cos= +2

2 2 .

• Thepolarmomentofinertiaaboutthez-axispassingthroughthepointOisindependentoftheorientationofthe uandυaxes,i.e.

JO=Iu+Iυ=Ix+Iy.

PRINCIPAL MOMENTS OF INERTIA

• Whentheangleqin(10.5)takesthevalueq=qpdefinedby

tan–

( – ) /2

2q p

xy

x y

I

I I= , (10.6)

theaxesuandυarecalledtheprincipal axes of the areasincetheyidentifytheorientationoftheaxesuandυaboutwhichthemomentsofinertiaIuandIυaremaximum or minimum.Inthiscase,theyarecalledprincipal moments of inertiaandaregivenby

II I I Ix y x y

maxmin

–=

+±

+

2 2

2

II xy2 . (10.7)

Dependingonthesignchosen,thisresultgivesthemaximumorminimummomentofinertiaforthearea.

• Theproduct of inertia with respect to the principal axes is zero.Henceanysymmetricalaxisrepresentsaprincipalaxisofinertiaforthearea.

10.6 Moments of Inertia For an Area About Inclined Axes 57

10_Moments.indd 57 4/21/09 12:50:04 PM


10.7 MOHR’S CIRCLE FOR MOMENTS OF INERTIA

Equations10.5 to10.7haveagraphical solutionwhich isconvenient touseandeasy to remember—this solution iscalledaMohr’s circle.


Mohr’scircleprovidesaconvenientmeansfortransformingIx ,IyandIxyintotheprincipalmomentsofinertiausingthefollowingprocedure:

• DetermineIx ,Iy ,Ixy .Establishthex,yaxesforthearea,withtheorigin locatedatthepointPof interestanddetermineIx ,IyandIxy .

x

y

u

υ

p1

Axis for minor principalmoment of inertia, Imin

Axis for major principalmoment of inertia, Imax

P θ

• Construct the Circle

– Constructarectangularcoordinatesystemsuchthattheabscissarepresentsthemomentof inertiaIandtheordinaterepresentstheproductofinertiaIxy .

– DeterminethecenterofthecircleO,whichislocatedatadistanceI Ix y+

2fromtheoriginandplotthe

referencepointAhavingcoordinates(Ix ,Ixy ).Bydefinition,Ixisalwayspositive,whereasIxywillbeeitherpositiveornegative.

– ConnectthereferencepointAwiththecenterofthecircleanddeterminethedistanceOAbytrigonometry.Thisdistancerepresentstheradiusofthecircle.Finallydrawthecircle.

I

Ixy

O

Imax

Imin

Ix

Ixy

A

Ix – Iy

2

2 p1

Ix + Iy

2

( )2

+IxyR =

q

Ix – Iy

22

10_Moments.indd 58 4/21/09 12:50:05 PM

• Principal Moments of Inertia. ThepointswherethecircleintersectstheabscissagivethevaluesoftheprincipalmomentsofinertiaIminandImax.Noticethattheproduct of inertia will be zero at these points.

• Principal Axes. To find the direction of the major principal axis, determine, by trigonometry, the angle 2qp1,measuredfromtheradiusOAtothepositiveI-axis.ThisanglerepresentstwicetheanglefromthexaxisoftheareainquestiontotheaxisofmaximummomentofinertiaImax.Boththeangleonthecircle,2qp1andtheangletotheaxisonthearea,qp1,mustbemeasuredinthesamesense.TheaxisforminimummomentofinertiaIminisperpendiculartotheaxisforImax.

10.8 MASS MOMENT OF INERTIA

• Themassmomentofinertiaaboutthez-axisisgivenby

I r dmm

= ∫ 2 , (10.8)

whereristheperpendiculardistancefromtheaxistothearbitraryelementdm.

r

dm

z

• Whentheaxispassesthroughthebody’smasscenterG,themomentofinertiaisdenotedbyIG .Themassmomentofinertiaisalwayspositiveandhasunitskg.m2.

• Ifthebodyconsistsofamaterialhavingavariabledensityρ(x,y,z),thebody’smomentofinertiaiscomputedusingvolumeintegrationas

I r dVV

= ∫ 2 ρ .

Thisintegralisgenerallycomputedasa triple integral.Theintegrationprocesscan,however,besimplifiedtoasingle integrationprovidedthechosenelementalvolumehasadifferentialsizeorthicknessinonlyonedirection.Shellordiskelementsareoftenusedforthispurpose.

PARALLEL AXIS THEOREM

• Ifthemomentof inertiaofthebodyaboutanaxispassingthroughthebody’smasscenter isknown,thenthemomentofinertiaaboutanyotherparallelaxiscanbecomputedfrom

I=IG+md 2, (10.9)

wheremisthemassofthebodyanddistheperpendiculardistancebetweentheaxes.

10.8 Mass Moment of Inertia 59

10_Moments.indd 59 4/21/09 12:50:05 PM


RADIUS OF GYRATION

• TheradiusofgyrationkhasunitsoflengthandisrelatedtothemassmandmomentofinertiaIofthebodyby

I=mk2 or k=1m

.

COMPOSITE BODIES

• Ifabodyisconstructedfromanumberofsimpleshapessuchasdisks,spheresandrods,themomentofinertiaofthebodyaboutanyaxiszcanbedeterminedbyaddingalgebraicallythemomentsofinertiaofallthecompositeshapescomputedaboutthez-axis.Acompositepartmustbeconsideredasanegativequantityifithasalreadybeenincludedwithinanotherpart,e.g.a‘hole’subtractedfromasolidplate.


• BecarefulwhenusingtheParallelAxisTheorem(10.9).Itisapplicableonlytothesituationwhenthemomentofinertiaofthebodyaboutanaxispassing through the body’s mass centerisknown.ItcannotbeappliedintheformI=IB+md 2whereBisanarbitrarypoint.

REVIEW QUESTIONS

1. TrueorFalse?Theareamomentsofinertia(10.0)–(10.2)canbenegative.

2. Itisoftenthecasethatthemomentofinertiaforanareaisknownaboutanaxispassingthroughitscentroid.Whatcanthenbesaidaboutthemomentofinertiaofthesameareaaboutacorrespondingparallelaxis?

3. Howaretheradiiofgyrationforaplanarareadetermined?

4. Ifacompositebodyhasa‘hole’,howwouldyoufindmomentofinertiaofthebody?

5. Findthemomentofinertiaofthefollowingcompositeareaaboutthex-axis.

x

100mm

75mm

75mm

25mm

(a)

x

100mm

75mm

75mm

25mm

–

(b)

6. Whataretheprincipal axes of an areaandtheprincipal moments of inertia?

7. WhatareMohr’s circlesusedfor?

8. Definethemassmomentofinertiaanddescribehowitcanbefound.

10_Moments.indd 60 4/21/09 12:50:05 PM


• Tointroducetheprincipleofvirtual workandshowhowitappliestodeterminingtheequilibriumconfigurationofaseriesofpin-connectedmembers.

• Toestablishthepotentialenergyfunctionandusethepotential-energymethodtoinvestigatethetypeofequilibriumorstabilityofarigidbodyorconfiguration.

11.1 DEFINITION OF WORK AND VIRTUAL WORK

WORK OF A FORCE

• AforceFdoesworkonlywhenitundergoesadisplacementinthedirectionoftheforce.

• Workisascalar quantitydefinedbythedotproduct

dU=F·dr (11.0) =Fcosqds ,

wheredUistheincrementofworkdonewhentheforceFisdisplaceddr,qistheanglebetweenthetailsofdr andF,anddsisthemagnitudeofdr.

• Positive workisdonewhentheforceanditsdisplacementhavethesamesense.Otherwisenegative workisdone.

• IntheSIsystem,thebasicunitofworkisaJoule(J)(1J=1N·m).

WORK OF A COUPLE

• The two forces of a couple do work when the couple rotates about an axis perpendicular to the plane of thecouple.

• WorkdonebyacoupleMisascalar quantitydefinedbythedotproduct

dU=M·dq, (11.1)

Virtual Work

11

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62 Chap. 11 Virtual Work

wheredUistheincrementofworkdonewhenthecoupleMis‘displaced’d(adifferentialrotationofthebodyaboutanaxisperpendiculartotheplaneofthecouple).

• Theresultantwork ispositivewhen thesenseofM is thesameas thatofdandnegativewhen theyhaveanopposite sense.Thedirectionandsenseofdareagaindefinedbytheright-hand rule.Hence,ifmovementofthebodyoccursinthesame plane,thelineofactionofdwillbeparalleltothelineofactionofMandEquation(11.1)becomes

dU=Mdq (11.2)

VIRTUAL WORK

• Avirtualmovement(displacementorrotation)isanimaginarymovementwhichisassumedanddoes not actually exist.Avirtual displacementisadifferentialthatisgiveninthepositivedirectionofthepositioncoordinateandisdenotedbythesymbold s.Similarly,avirtual rotationisdenotedbyd q.

• Thevirtual workdonebyaforceundergoingavirtualdisplacementd sis

d U=Fcosd q s. (11.3)

• Thevirtual workdonebyacoupleundergoingavirtualrotationd qintheplaneofthecoupleforcesis

d U=Md q. (11.4)

11.2 PRINCIPLE OF VIRTUAL WORK FOR A PARTICLE AND A RIGID BODY

• Particle. Ifaparticleundergoesanimaginaryorvirtualdisplacementd r,thenthevirtualwork(d U)donebytheforcesystemmustbezeroforequilibrium,i.e.

d U=0. (11.5)

Inotherwordswecanwrite three independent virtual work equationscorrespondingto thethreeequationsofequilibrium:

∑Fxd x=0, ∑Fyd y=0, ∑Fzd z=0.

• Rigid Body. Asinthecaseofaparticlewecanalsowriteasetofthreevirtualworkequations(11.5)forarigidbodysubjectedtoacoplanarforcesystem(twoinvolvingvirtualtranslationsinthexandydirectionsandanotheravirtualrotationaboutanaxisperpendiculartothex–yplaneandpassingthroughanarbitrarypointO).

• NOTE.Asinthecaseofaparticle,noaddedadvantageisgainedbysolvingrigidbodyequilibriumproblemsusingtheprincipleofvirtualworksinceforeachapplicationofthevirtualworkequation,thevirtualdisplacementorrotation,commontoeveryterm,factorsout,leavinganequationthatcouldhavebeenobtainedinamoredirectmannerbysimplyapplyingtheequationsofequilibrium.

11.3 PRINCIPLE OF VIRTUAL WORK FOR A SYSTEM OF CONNECTED RIGID BODIES

Themethodofvirtualworkismostsuitableforsolvingequilibriumproblemsthatinvolveasystemofseveralconnectedrigidbodies.

• Degrees of Freedom.Ann–degree-of-freedom systemrequiresn independent coordinates qntospecifythelocationofallitsmembers.


• Principle of Virtual Work.The principle of virtual work for a system of rigid bodies whose connections arefrictionlessmaybestatedasfollows:

Asystemofconnectedrigidbodiesisinequilibriumprovidedthevirtualworkdonebyalltheexternalforcesandcouplesactingonthesystemiszeroforeachindependentvirtualdisplacementofthesystem.

Mathematically,wewrite

d U=0, (11.6)

where d U represents the virtual work of all the external forces and couples acting on the system during anyindependentvirtualdisplacement.

• Thismeansthatforann–degree-of-freedomsystemitispossibletowritenindependentvirtualworkequations,one foreveryvirtualdisplacement takenalongeachof the independent coordinateaxes,while the remainingn–1remainingindependentcoordinatesareheldfixed.

The following procedure shows how to use the equation of virtual work to solve problems involving a system offrictionlessconnectedrigidbodieshavingasingle degree of freedom.


– Drawthefree-bodydiagramoftheentiresystemofconnectedbodiesanddefinetheindependentcoordinateq.

– Sketchthe‘deflectedposition’ofthesystemonthefree-bodydiagramwhenthesystemundergoesapositivevirtualdisplacementd q.

• Virtual Displacements

– Indicate position coordinates si , measured from a fixed point on the free-body diagram to each of the inumberof‘active’forcesandcouples,i.e.thosethatdowork.

– Eachcoordinateaxisshouldbeparalleltothelineofactionofthe‘active’forcetowhichitisdirected,sothatthevirtualworkalongthecoordinateaxiscanbecalculated.

– Relate each of the position coordinates si to the independent coordinate q; then differentiate theseexpressionstoexpressthevirtualdisplacementsd siintermsofd q.

• Virtual Work Equation

– Writethevirtualworkequation(11.6)forthesystemassumingthat,whetherpossibleornot,allthepositioncoordinatessiundergopositivevirtualdisplacementsd si .

– Usingtherelationsford si ,expresstheworkofeach‘active’forceandcoupleintheequationintermsofthesingleindependentvirtualdisplacementd q.

– Factor out the common displacement from all the terms and solve for the unknown force, couple orequilibriumposition,q.

– If thesystemcontainsndegreesoffreedom,n independentcoordinatesqnmustbespecified.Followtheabove procedure and let only one of the independent coordinates undergo a virtual displacement whiletheremainingn–1coordinatesareheld fixed.Inthisway,nvirtual-workequationscanbewritten,oneforeachindependentcoordinate.

• EXAMPLES.TheaboveprocedureisillustratedinExamples11-1through11-4inthetext.Note that had these examples been solved using the equations of equilibrium, it would have been necessary to dismember the links and apply three scalar equations to each link. The principle of virtual work, by means of calculus, has eliminated this task so that the answer is obtained directly.

11.3 Principle of Virtual Work For a System of Connected Rigid Bodies 63



11.4 CONSERVATIVE FORCES

• IfaforceFisdisplacedoverapathwithfinitelengthStheworkdonebytheforceisgivenbytheintegral

U dU F dsS S

cos .= =∫ ∫ q

Ifthisintegralisindependent of its path(dependsonlyontheinitialandfinallocationsofitspath),theforceiscalledaconservative force.

• EXAMPLES OF CONSERVATIVE FORCES:Weight, Elastic Springs.

• EXAMPLES OF NONCONSERVATIVE FORCES:Friction:theworkdonebythefrictionalforcedepends on the path: the longer thepath, thegreater thework.Theworkdone isdissipated fromthebody in the formofheat.

11.5 POTENTIAL ENERGY

Whenaconservative forceactsonabody, it gives thebody thecapacity todowork.This capacity isknownas thebody’spotential energyanddependsonthelocationofthebody.

• Gravitational Potential Energy. Measuring y positive upward, the gravitational potential energy of a body’sweightWis

Vg=Wy. (11.7)

• Elastic Potential Energy. TheelasticpotentialenergyVe thataspringproducesonanattachedbody,whenthespringiselongatedorcompressedfromanundeformedposition(s=0)toafinalpositionsis

Ve=12

ks 2. (11.8)

• Potential Function.Inthegeneralcase,ifabodyissubjectedtobothgravitationalandelasticforces,thepotentialenergy(function)Vofthebodycanbeexpressedasthealgebraicsum

V=Vg+Ve,

wheremeasurementofVdependsonthelocationofthebodywithrespecttoaselecteddatuminaccordancewithEquations(11.7)and(11.8).

11.6 POTENTIAL ENERGY CRITERION FOR EQUILIBRIUM

• System Having One Degree of Freedom (q). Whenafrictionlessconnectedsystemofrigidbodiesisinequilibrium,werequirethatthepotentialenergy(function)Vofthebodysatisfies

dVdq

=0. (11.9)

• System Having n Degrees of Freedom (q1,…, qn).When a frictionless connected system of rigid bodies is inequilibrium,werequirethatthepotentialenergy(function)Vofthebodysatisfies

∂∂

Vq1

=0,∂∂

Vq2

=0, …,∂∂

Vqn

=0.

Inotherwords,itispossibletowrite n independent equations for a system having n degrees of freedom.


11.7 STABILITY OF EQUILIBRIUM

Oncetheequilibriumconfigurationforabodyorasystemofconnectedbodiesisdefined,itisimportanttoinvestigatethe‘type’ofequilibriumorthestabilityoftheconfiguration.

• Types of Equilibrium

1. Stable Equilibrium. Asmalldisplacementofthesystemcausesthesystemtoreturntoitsoriginalposition.Potentialenergyofthesystemisataminimuminthiscase.

2. Neutral Equilibrium.Asmalldisplacementofthesystemcausesthesystemtoremaininitsdisplacedstate.Potentialenergyofthesystemremainsconstantinthiscase.

3. Unstable Equilibrium.Asmalldisplacementofthesystemcausesthesystemtomovefartherawayfromitsoriginalposition.Originalpotentialenergyofthesystemisamaximuminthiscase.

• System Having One Degree of Freedom (q).Werequirethatthepotentialenergy(function)Vofthebodysatisfiesthefollowingconditionsineachcase:

1. Stable Equilibrium

dVdq

d V

dq, .= >0 0

2

2 (11.10)

2. Neutral Equilibrium

dVdq

d V

dq

d V

dq= = = … =

2

2

3

30. (11.11)

3. Unstable Equilibrium

dVdq

d V

dq, .= <0 0

2

2 (11.12)

• System Having Two Degrees of Freedom (q1, q2).Things become much more complicated as the number ofdegreesoffreedomofthesystemincreases.However,forasystemfortwodegreesoffreedom,wecansay:

1. Equilibrium and Stabilityoccuratapoint(q1eq ,q2eq)when

∂∂

= ∂∂

=Vq

Vq1 2

0 ,

∂

∂ ∂

∂∂

∂∂

2

1 2

22

12

2

22

Vq q

V

q

V

q–

< ,0

∂∂

+ ∂∂

>

2

12

2

22

0V

q

V

q.

2. Equilibrium and Instabilityoccurwhen

∂∂

= ∂∂

=Vq

Vq1 2

0 ,

∂∂ ∂

∂∂

∂∂

2

1 2

22

12

2

22

Vq q

V

q

V

q–

< ,0

∂∂

+ ∂∂

<

2

12

2

22

0V

q

V

q.

11.7 Stability of Equilibrium 65



PROCEDURE FOR SOLVING PROBLEMS

Usingpotential-energymethods,theequilibriumpositionsandthestabilityofabodyorasystemofconnectedbodieshavingasingle degree of freedom qcanbeobtainedusingthefollowingprocedure.

• Potential Function

– Sketchthesystemsothatitislocatedatsomearbitrary positionspecifiedbytheindependentcoordinateq.

– Establishahorizontaldatumthroughafixed pointandexpressthegravitational potential energy VgintermsoftheweightWofeachmemberanditsverticaldistanceyfromthedatum,Vg=Wy.

– Express the elastic potential energy Ve of the system in terms of the stretch or compression, s, of any

connectingspringandthespringstiffnessk,Ve=12

ks 2.

– FormulatethepotentialfunctionV=Vg+Veandexpresstheposition coordinates y andsintermsoftheindependentcoordinateq.

• Equilibrium Position

– TheequilibriumpositionisdeterminedfromEquation(11.9),i.e.dVdq

=0.

• Stability

– StabilityattheequilibriumpositionisdeterminedfromEquations(11.10)–(11.12).

REVIEW QUESTIONS

1. WhatistheworkdonebyaforceFwhenitspointofapplicationisdisplaceddr?

2. WhatistheworkdonebyacoupleMwhentheobjectonwhichitactsrotatesthroughanangled ?

3. Whatdoestheprincipleofvirtualworksaywhenanobjectinequilibriumissubjectedtoavirtualtranslationorrotation?

4. Whatismeantbya‘conservativeforce’?

5. Whatisthepotentialenergyofabodyandhowisitrelatedtotheconceptof‘conservativeforce’?

6. What is thepotentialenergycriterionforequilibriumfora frictionlessconnectedsystemofrigidbodieswithonedegreeoffreedom?

7. Whatdoesitmeanwhenanequilibriumpositionofabodyisstableorunstable?

8. How do you know when an equilibrium position of a system having one degree of freedom is stable orunstable?


Chapter 1:

1. F 2. T 3. F 4. F 5. F 6. T 7. T 8. F

Chapter 2:

1. See(2.1). 2. See(2.0)(intheplane)or(2.2)and(2.3). 3. See(2.4).

4. SeeSection2.5. 5. See(2.3). 6. See(2.5).

7. Vectorsareperpendicular.

8. See(2.5). 9. See(2.6). 10. See(2.7)and(2.8).

Chapter 3:

1. SeeSection3.1. 2. See(3.0). 3. SeeSection3.2.

4. Linesofactionoftheforceslieinaplane.

5. Linesofactionoftheforceslieinthree-dimensionalspace.

6. Onemoreequation—see(3.2)and(3.3).

7. See(3.1). 8. True.

Chapter 4:

1. SeeSection4.1. 2. SeeSection4.1. 3. Right-handrule.

4. Nomoment. 5. See(4.0). 6. Vectorsareparallel.

7. r representsapositionvectordrawnfrom O to any pointlyingonthelineofactionofF.

8. |MO|=|F||r|sinq=Frsinq=Fd.

9. Right-handrule,i.e.curlingthefingersoftherighthandfromvectorr(cross)tovectorF,thethumbthenpointsinthedirectionofMO.

10. SeeSection4.6. 11. False. 12. SeeSection4.7. 13. SeeSection4.7and(4.2).

14. SeeSection4.8. 15. SeeSection4.9.

Chapter 5:

1. Aparticlehasnosize/shapeandsocannotsupportrotation,onlytranslation.

2. See(5.1)and(5.0).

3. SeeSection5.7:theobjecthasmoresupportsthanarenecessarytoholditinequilibrium.

ANSWERS TO REVIEW QUESTIONS

12_Ans to Rev Qns.indd 67 4/21/09 12:58:16 PM

68 ANSWERS TO REVIEW QUESTIONS

4. SeeSection5.7:whentherearemoreunknownloadingsonthebodythanequationsofequilibriumavailablefortheirsolution.

5. ReviewSection5.7.

Chapter 6:

1. SeeSection6.1:Atrussisastructurecomposedofslendermembersjoinedtogetherattheirendpoints.

2. SeeSection6.1.

3. SeeSection6.2:Inordertoanalyzeordesignatruss,wemustobtaintheforceineachofitsmembers.Todothis,weconsidertheequilibrium of a jointofthetruss.Thisisthebasisofthemethodofjoints.

4. Two.

5. SeeSection6.4.

6. SeeSection6.5—eitherthemethod of jointsormethod of sections.

7. SeeSection6.6.

8. Theforcesattheconnectedpartsofthegroupareinternalforcesandarenotshownonthefree-bodydiagramof the group.

Chapter 7:

1. SeeSection7.1:NormalforceNactsparallel tothebeam’saxis.ShearforceVactsnormaltothebeam’saxis.BendingmomentMisacouplemomentwhichcausesthebeamtobend.

2. SeeSection7.2.Followthesignconventionshowninthefigure.

3. ByintegrationtoobtainVandMasfunctionsofx.

4. See Section 7.4. Note that no cable is truly ‘weightless,’ ‘inextensible’ or ‘perfectly flexible’. These terms aresimplificationstoaidthemodeling.

5. Weuseequation(7.3)andintegrate(notingthatwisconstant).Weobtain

y xF

xc x c

H

( ) = + +

12

2

1 2w .

This is a parabola. If the origin of the x – y coordinate system is chosen so that y = 0,dydx

= 0 at x = 0, weobtain

y xxFH

( ) = w 2

2.

Chapter 8:

1. SeeSection8.1:Fs=µ s N.

2. SeeSection8.1:Fk=µ k N.

3. SeeSection8.2:Thetotalnumberofunknownsisequaltothetotalnumberofavailableequilibriumequations.Inthiscase,oncethefrictionalforcesaredetermined,checkthatF≤µsNotherwiseslippingwilloccurandthebodywillnotremaininequilibrium.

4. SeeSection8.2:Thetotalnumberofunknownswillequal thetotalnumberofavailableequilibriumequationsplusthetotalnumberofavailablefrictionalequationsorconditionalequationsfortipping.

5. Drawafree-bodydiagram!

6. Equation(8.0).

7. Equation(8.1)or(8.2)withφ=φ k .

8. Equation(8.3)withRreplacedbyP(sincethereactiveforceRisequalinmagnitudetotheloadP).


ANSWERS TO REVIEW QUESTIONS 69

Chapter 9:

1. (i) T. (ii) F(materialcomprisingthebodymustbehomogeneous). (iii) F. (iv) T—see(9.3).

2. SeeSection9.1—atthecenterofgravity(orcenterofmass).

3. The location of the centerofgravitycoincideswith thatof the centerofmass.However, thecenter of mass is independent of gravityandsocanbeusedinsituationswhenparticlesarenotundertheinfluenceofagravitationalattraction—seeSection9.1.

4. Specificweightisγorweightperunitvolume.

5. γ=ρ g—seeSection9.1.

6. Abodywhosemassdensityisconstantthroughoutitsvolumeissaidtobehomogeneous—seeSection9.1.

7. Centerofmasscoincideswithcentroidandcenterofgravity—seeSection9.1.

8. SeeSection9.2.Ifacompositeparthasahole,thenconsiderthecompositepartwithouttheholeandconsidertheholeasanadditionalcompositeparthavingnegativeweightorsize.

9. Tofindthesurfacearea and volume ofanyobjectofrevolution—seeSection9.3.

10. From(9.1)withdW=ρ gdV,theformulaforthecenterofmassofabodyisgivenby

x=V

x ρgdV

V ρgdV, y=

V y ρgdV

V ρgdV, z=

V z ρgdV

V ρgdV.

Ifthebodyishomogeneousρisconstant.Thusweobtain:

x=V

x dV

V dV, y=

V y dV

V dV, z=

V z dV

V dV,

whichisexactly(9.2).

Chapter 10:

1. SeeSection10.1: False.

2. SeeSection10.2: It’s equal to the moment of inertia about the axis passing through the area’s centroidplus the product of the area and the square of the perpendicular distance between the axes..

3. SeeSection10.3,Equation(10.4).

4. SeeSection10.4:Ifacompositeparthasa‘hole’,itsmomentofinertiaisfoundby‘subtracting’themomentof inertiafortheholefromthemomentofinertiaoftheentirepartincludingthehole.

5. Thecompositeareaisdeterminedbysubtractingthecirclefromtherectangle.Thecentroidofeachareaislocatedinthefigure.

• Moment of inertia for Circle: UsingEquation(10.3)

Ix=Ix′+ Ad y2

=14

p(25)4+p(25)2(75)2=11.4(106)mm4.

• Moment of inertia for Rectangle: UsingEquation(10.3)

Ix=Ix′+ Ad y2

=1

12100(150)3+100(150)2(75)2=112.5(106)mm4.


70 ANSWERS TO REVIEW QUESTIONS

• Summation. Themomentofinertiaforthecompositeareaisthus

Ix=–11.4(106)+112.5(106)

=101(106)mm4.

6. See Section 10.6: The principal axes of the area identify the orientation of the axes u and υ about which themomentsofinertiaIuandIυaremaximum or minimum.(theprincipal moments of inertia).

7. SeeSection10.7:Mohr’scircleprovidesaconvenient(graphical)meansfortransformingIx ,IyandIxy intothe principalmomentsofinertia.

8. SeeEquation(10.8).Thisintegralisgenerallycomputedasavolumeandhencetriple integral(usingtherelationdm=ρ dV).Theintegrationprocesscan,however,besimplifiedtoasingle integrationprovidedthechosenelementalvolumehasadifferential sizeor thickness inonlyonedirection.Shellordiskelementsareoftenusedfor thispurpose.ThedetailedprocedureisgiveninSection10.8ofthetext.

Chapter 11:

1. See(11.0). 2. See(11.1).

4. Thatthevirtualwork(δU)donebytheforcesystemmustbezero—see(11.5).

5. SeeSection11.4—thattheworkdonebytheforceoverafinitepathisindependentofthepathitself.

6. SeeSection11.5. Whenaconservativeforceactsonabody,itgivesthebodythecapacitytodowork.Thiscapacityisknownasthebody’spotential energyanddependsonthelocationofthebody.

7. Equation(11.9).

8. SeeSection11.7: Stable Equilibrium:asmalldisplacementofthesystemcausesthesystemtoreturntoitsoriginalposition.Potentialenergyofthesystemisataminimuminthiscase;Unstable Equilibrium.Asmalldisplacementofthesystemcausesthesystemtomovefartherawayfromitsoriginalposition.Originalpotentialenergyofthesystemisamaximuminthiscase.

9. TestusingEquation(11.10)forstabilityandEquation(11.12)forinstability.


PART II

Free-Body Diagram Workbook

13_Basic Concepts.indd 71 4/21/09 12:59:15 PM

Statics is a branch of mechanics that deals with the study of bodies that are at rest (if originally at rest) or move with constant velocity (if originally in motion) that is, bodies which are in (static) equilibrium.

In mechanics, real bodies (e.g. planets, cars, planes, tables, crates, etc.) are represented or modeled using certain idealizations which simplify application of the relevant theory. In this book we refer to only two such models:

• Particle. A particle has a mass but a size/shape that can be neglected. For example, the size of an aircraft is insignificant when compared to the size of the earth and therefore the aircraft can be modeled as a particle when studying its three-dimensional motion in space.

• Rigid Body. A rigid body represents the next level of sophistication after the particle. That is, a rigid body is a collection of particles which has a size/shape but this size/shape cannot change. In other words, when a body is modeled as a rigid body, we assume that any deformations (changes in shape) are relatively small and can be neglected. For example, the actual deformations occurring in most structures and machines are relatively small so that the rigid body assumption is suitable in these cases.

1.1 Equilibrium

Equilibrium of a Particle

A particle is in equilibrium provided it is at rest if originally at rest or has a constant velocity if originally in motion. To maintain equilibrium, it is necessary and sufficient to satisfy Newton’s first law of motion which requires the resultant force acting on the particle or rigid body to be zero. In other words

∑F = 0 , (1.1)

where ∑ F is the vector sum of all the external forces acting on the particle. Successful application of the equations of equilibrium (1.1) requires a complete specification of all the known

and unknown external forces (∑ F) that act on the object. The best way to account for these is to draw the object’s free-body diagram.

Basic Concepts in Statics

1


74 Chap. 1 Basic Concepts in Statics

Equilibrium of a Rigid Body

A rigid body will be in equilibrium provided the sum of all the external forces acting on the body is equal to zero and the sum of the external moments taken about a point is equal to zero. In other words

∑F = 0 , (1.2) ∑MO = 0 , (1.3)

where ∑ F is the vector sum of all the external forces acting on the rigid body and ∑ MO is the sum of the external moments about an arbitrary point O.

Successful application of the equations of equilibrium (1.2) and (1.3) requires a complete specification of all the known and unknown external forces (∑ F) and moments (∑ MO) that act on the object. The best way to account for these is again to draw the object’s free-body diagram.


2.1 Free-Body Diagram: Particle

The equilibrium equation (1.1) is used to determine unknown forces acting on an object (modeled as a particle) in equilibrium. The first step in doing this is to draw the free-body diagram of the object to identify the external forces acting on it. The object’s free-body diagram is simply a sketch of the object freed from its surroundings showing all the (external) forces that act on it. The diagram focuses your attention on the object of interest and helps you identify all the external forces acting. For example:

TAB (Force of cable acting on crate)

W (Weight or gravity acting on crate)

B

A

Isolate the crate and part of the cable AB

Free-Body diagram of Crate

Figure 1

Note that once the crate is separated or freed from the system, forces which were previously internal to the system become external to the crate. For example, in Figure 1, such a force is the force of the cable AB acting on the crate.

Next, we present a formal procedure for drawing free-body diagrams for a particle.

Free-Body Diagrams: the Basics

2

14_FreeBody Diag.indd 75 4/21/09 1:01:01 PM

76 Chap. 2 Free-Body Diagrams: the Basics

2.1.1 Procedure for Drawing a Free-Body Diagram: Particle

1. Identify the object you wish to isolate. This choice is often dictated by the particular forces you wish to determine.

2. Draw the outlined shape of the isolated object. Imagine the object to be isolated or cut free from the system of which it is a part.

3. Show all external forces acting on the isolated object. Indicate on this sketch all the external forces that act on the object. These forces can be active forces, which tend to set the object in motion, or they can be reactive forces which are the result of the constraints or supports that prevent motion. This stage is crucial: it may help to trace around the object’s boundary, carefully noting each external force acting on it. Don’t forget to include the weight of the object (unless it is being intentionally neglected).

4. Identify and label each external force acting on the (isolated) object. The forces that are known should be labeled with their known magnitudes and directions. Use letters to represent the magnitudes and arrows to represent the directions of forces that are unknown.

5. The direction of a force having an unknown magnitude can be assumed.

Example 2.1

The crate in Figure 2 has a weight of 20 kN. Draw free-body diagrams of the crate, the cord BD and the ring at B. Assume that the cords and the ring at B have negligible mass.

45°

BC

A

D

Figure 2

Solution

Free-Body Diagram for the Crate. Imagine the crate to be isolated from its surroundings, then, by inspection, there are only two external forces acting on the crate, namely, the weight of 20 kN and the force of the cord BD.

FD (Force of cord acting on crate)

20 kN (Weight or gravity acting on crate)

Figure 3


Free-Body Diagram for the Cord BD. Imagine the cord to be isolated from its surroundings, then, by inspection, there are only two external forces acting on the cord, namely, the force of the crate FD and the force FB caused by the ring. These forces both tend to pull on the cord so that the cord is in tension. Notice that FD shown in this free-body diagram (Figure 4) is equal and opposite to that shown in Figure 3 (a consequence of Newton’s third law).

FD (Force of ring acting on cord)

FD (Force of crate acting on cord)

B

D

Figure 4

Free-Body Diagram for the ring at B. Imagine the ring to be isolated from its surroundings, then, by inspection, there are actually three external forces acting on the ring, all caused by the attached cords. Notice that FB shown in this free-body diagram (Figure 5) is equal and opposite to that shown in Figure 4 (a consequence of Newton’s third law).

B

45°

FA (Force of cord BA acting on ring)

FC (Force of cord BC acting on ring)

FD (Force of cord BD acting on ring)

Figure 5

2.1.2 Using the Free-Body Diagram: Equilibrium

The free-body diagram is used to identify the unknown forces acting on the particle when applying the equilibrium equation (1.1) to the particle. The procedure for solving equilibrium problems for a particle once the free-body diagram for the particle is established, is therefore as follows:

1. Establish the x, y-axes in any suitable orientation.

2. Apply the equilibrium equation (1.1) in component form in each direction:

∑Fx = 0 and Fy = 0. (2.1)

3. Components are positive if they are directed along a positive axis and negative if they are directed along a negative axis.

4. If more than two unknowns exist and the problem involves a spring, apply F = ks to relate the magnitude of the spring force F to the deformation of the spring s (here, k is the spring constant).

5. If the solution yields a negative result, this indicates the sense of the force is the reverse of that shown/assumed on the free-body diagram.

2.1 Free-Body Diagram: Particle 77



Example 2.2

In Example 2.1, the free-body diagrams established in Figures 3–5 give us a ‘pictorial representation’ of all the information we need to apply the equilibrium equations (2.1) to find the various unknown forces. In fact, taking the positive x-direction to be horizontal (→ +) and the positive y-direction to be vertical (↑ +), the equilibrium equations (2.1) when applied to each of the objects (regarded as particles) are:

For the Crate: ↑ + ∑ Fy = 0: FD – 20 = 0 (See Figure 3)

FD = 20 kN (2.2)

For the Cord BD: ↑ + ∑ Fy = 0: FB – FD = 0 (See Figure 4)

FB = FD (2.3)

For the Ring: ↑ + ∑ Fy = 0: FA sin 45 – FB = 0 (See Figure 5) (2.4)

→ + ∑ Fx = 0: FC – FA cos 45 = 0 (See Figure 5) (2.5)

Equations (2.2)–(2.5) are now 4 equations which can be solved for the 4 unknowns FA, FB, FC and FD. That is: FB = 20 kN; FD = 20 kN, FA = 28.28 kN, FC = 20 kN. The directions of each of these forces is shown in the free-body diagrams above (Figures 3–5).

2.2 Free-Body Diagram: Rigid Body

The equilibrium equations (1.2) and (1.3) are used to determine unknown forces and moments acting on an object (modeled as a rigid body) in equilibrium. The first step in doing this is again to draw the free-body diagram of the object to identify all of the external forces and moments acting on it. The procedure for drawing a free-body diagram in this case is much the same as that for a particle with the main exception that now, because the object has ‘size/shape’, it can support also external couple moments and moments of external forces.

2.2.1 Procedure for Drawing a Free-Body Diagram: Rigid Body

1. Imagine the body to be isolated or ‘cut free’ from its constraints and connections and sketch its outlined shape.

2. Identify all the external forces and couple moments that act on the body. Those generally encountered are:

(a) Applied loadings

(b) Reactions occurring at the supports or at points of contact with other bodies (See Table 2.1)

(c) The weight of the body (applied at the body’s center of gravity G)

3. The forces and couple moments that are known should be labeled with their proper magnitudes and directions. Letters are used to represent the magnitudes and direction angles of forces and couple moments that are unknown. Establish an x, y-coordinate system so that these unknowns, e.g. Ax , By, etc. can be identified. Indicate the dimensions of the body necessary for computing the moments of external forces. In particular, if a force or couple moment has a known line of action but unknown magnitude, the arrowhead which defines the sense of the vector can be assumed. The correctness of the assumed sense will become apparent after solving the equilibrium equations for the unknown magnitude. By definition, the magnitude of a vector is always positive, so that if the solution yields a negative scalar, the minus sign indicates that the vector’s sense is opposite to that which was originally assumed.


Table 2.1. Supports for Rigid Bodies Subjected to Two-Dimensional Force Systems

(1)

cable

Types of Connection Reaction Number of Unknowns

One unknown. The reaction is a tension force which acts away from the member in the direction of the cable.

One unknown. The reaction is a force which acts along the axis of the link.

One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact.



One unknown. The reaction is a force which acts perpendicular to the slot.

One unknown. The reaction is a force which acts perpendicular to the rod.

Two unknowns. The reactions are the couple moment and the force which acts perpendicular to the rod.

Three unknowns. The reactions are the couple moment and the two force components, or the couple moment and the magnitude and direction of the resultant force.

Two unknowns. The reactions are two components of force, or the magnitude and direction of the resultant force. Note that and are not necessarily equal [usually not, unless the rod shown is a link as in (2)].

F

(2)

weightless link F

(3)

roller F

q

qq or

q

(4)

roller or pin in confined smooth slot

(5)

rocker

(6)

smooth contacting surface

F

F

q

F

F

(8)

smooth pin or hinge

(9)

member fixed connected to collar on smooth rod

Fy

M

orq φFx

F

(10)

fixed support

Fy

Fx

F

orM M

(7)

member pin connected to collar on smooth rod

or

q

or

q

q

q

qF

qq

qF

q

q

q q

F

φφ

φφ

q

2.2 Free-Body Diagram: Rigid Body 79



2.2.2 Important Points

• No equilibrium problem should be solved without first drawing the free-body diagram, so as to account for all the external forces and moments that act on the body.

• If a support prevents translation of a body in a particular direction, then the support exerts a force on the body in that direction.

• If rotation is prevented then the support exerts a couple moment on the body.

• Internal forces are never shown on the free-body diagram since they occur in equal but opposite collinear pairs and therefore cancel each other out.

• The weight of a body is an external force and its effect is shown as a single resultant force acting through the body’s center of gravity G.

• Couple moments can be placed anywhere on the free-body diagram since they are free vectors. Forces can act at any point along their lines of action since they are sliding vectors.

Example 2.3

Draw the free-body diagram of the beam, which is pin-connected at A and rocker-supported at B. Neglect the weight of the beam.

5 mB

A

8 m 4 m

500 N800 N · m

Figure 6

Solution

The free-body diagram of the beam is shown in Figure 7. From Table 2.1, since the support at A is a pin-connection, there are two reactions acting on the beam at A denoted by Ax and Ay. In addition, there is one reaction acting on the beam at the rocker support at B. We denote this reaction by the force F which acts perpendicular to the surface at B, the point of contact (see Table 2.1). The magnitudes of these vectors are unknown and their sense has been assumed (the correctness of the assumed sense will become apparent after solving the equilibrium equations for the unknown magnitude, i.e. if application of the equilibrium equations to the beam yields a negative result for F, this indicates the sense of the force is the reverse of that shown/assumed on the free-body diagram). The weight of the beam has been neglected. 2.2.3 Using the Free-Body Diagram: Equilibrium

The equilibrium equations (1.2) and (1.3) can be written in component form as:

∑Fx = 0, (2.6)

∑Fy = 0, (2.7)

∑MO = 0. (2.8)


5 m

A

Ay

Ax

B

8 m 4 m

500 N800 N · m

F

y

x22.62°

22.62°

Effect of Applied Force

Acting on Beam

Effect of Rocker SupportActing on Beam

Effect of Applied CoupleMoment Acting on Beam

Effect of Pin SupportActing on Beam

Figure 7

Here, ∑ Fx and ∑ Fy represent, respectively, the algebraic sums of the x and y components of all the external forces acting on the body and ∑ MO represents the algebraic sum of the couple moments and the moments of all the external force components about an axis perpendicular to the x – y plane and passing through the arbitrary point O, which may lie either on or off the body. The procedure for solving equilibrium problems for a rigid body once the free-body diagram for the body is established, is as follows:

• Apply the moment equation of equilibrium (2.8), about a point (O) that lies at the intersection of the lines of action of two unknown forces. In this way, the moments of these unknowns are zero about O and a direct solution for the third unknown can be determined.

• When applying the force equilibrium equations (2.6) and (2.7), orient the x and y-axes along lines that will provide the simplest resolution of the forces into their x and y components.

• If the solution of the equilibrium equations yields a negative scalar for a force or couple moment magnitude, this indicates that the sense is opposite to that which was assumed on the free-body diagram.

Example 2.4

A force of magnitude 150 kN acts on the end of the beam as shown. Find the magnitude and direction of the reaction at pin A and the tension in the cable.

3 m

8 m

C

34

5

A

1.5 m

2 m

150 kN

B

Figure 8

2.2 Free-Body Diagram: Rigid Body 81



Solution

Free-Body Diagram. The first thing to do is to draw the free-body diagram of the beam in order to identify all the external forces and moments acting on the beam.

3 m

8 m

C

345

A

1.5 m

2 m

150 kN

B

Ax

Ay

T

Tx

y

Figure 9

Equations of Equilibrium. The free-body diagram of the beam suggests we can sum moments about the point A to eliminate the moment contribution of the reaction forces Ax and Ay acting on the beam. This will allow us to obtain a direct solution for the third unknown, i.e. the cable tension T. Taking counterclockwise as positive when computing moments, we have:

+ ∑MA = 0: – (3/5)T (2 m) – (4/5)T (3 m) + 150 kN (10 m) = 0

– 3.6T + 150 kN (10 m) = 0

T = 416.7 kN Ans.

Summing forces to obtain Ax and Ay, using the result for T, we have

→ + ∑Fx = 0: – Ax + (4/5)(416.7 kN) = 0

Ax = 333.3 kN ←

↑ + ∑Fy = 0: (3/5)(416.7 kN) – 150 kN – Ay = 0

Ay = 100 kN ↓

Thus, the reaction force FA at pin A has magnitude FA given by:

FA = ( . (333 3 100kN) kN)2 2+

= 348.0 kN

and direction given by

q = tan–1 [(–100 kN) / (–333.3 kN)] = 196.7°

counterclockwise from the positive x-axis or 16.7° .


3.1 Free-BodyDiagramsinParticleEquilibrium 85

3.2 Free-BodyDiagramsintheEquilibriumofaRigidBody 115

Problems

3

15_Problems.indd 83 4/21/09 1:04:21 PM

15_Problems.indd 84 4/21/09 1:04:21 PM

3.1Free-BodyDiagramsinParticleEquilibrium

Problem 3.1 The sling is used to support a drum having a weight of 4500 N ( 450 kg). Draw a free-body diagram for the knot at A. Take q = 20°.

BA

D

C

qq

Solution

1. The knot at A has negligible size so that it can be modelled as a particle.

2. Imagine the knot at A to be separated or detached from the system.

3. The (detached) knot at A is subjected to three external forces. They are caused by:

i. ii.

iii.

4. Draw the free-body diagram of the (detached) knot showing all these forces labeled with their magnitudes and directions.

15_Problems.indd 85 4/21/09 1:04:22 PM

86 Chap. 3 Problems

Problem 3.1 The sling is used to support a drum having a weight of 4500 N ( 450 kg). Draw a free-body diagram for the knot at A. Take q = 20°.

BA

D

C

qq

Solution




i. CORD AB ii. CORD AC

iii. CORD AB (weight of drum)


4500 N

15_Problems.indd 86 4/21/09 1:04:22 PM

Problem 3.2 The spring ABC has a stiffness of 500 N/m and an unstretched length of 6 m. A horizontal force F is applied to the cord which is attached to the small pulley B so that the displacement of the pulley from the wall is d = 1.5 m. Draw a free-body diagram for the small pulley B.

F

B

C

d

A

k = 500 N/m

k = 500 N/m

6 m

Solution

1. The pulley B has negligible size so that it can be modelled as a particle.

2. Imagine the pulley B to be separated or detached from the system.

3. The (detached) pulley B is subjected to three external forces. They are caused by:

i. ii.

iii.

4. Draw the free-body diagram of the (detached) pulley showing all these forces labeled with their magnitudes and directions. You should also include any other available information, e.g. lengths, angles, etc. — which will help when formulating the equilibrium equations for the pulley.

3.1 Free-Body Diagrams in Particle Equilibrium 87

15_Problems.indd 87 4/21/09 1:04:23 PM

88 Chap. 3 Problems

Problem 3.2 The spring ABC has a stiffness of 500 N/m and an unstretched length of 6 m. A horizontal force F is applied to the cord which is attached to the small pulley B so that the displacement of the pulley from the wall is d = 1.5 m. Draw a free-body diagram for the small pulley B.

F

B

C

d

A

k = 500 N/m

k = 500 N/m

6 m

Solution

1. The pulley B has negligible size so that it can be modelled as a particle.

2. Imagine the pulley B to be separated or detached from the system.

3. The (detached) pulley B is subjected to three external forces. They are caused by:

i. Force F ii. Spring AB

iii. Spring BC


15_Problems.indd 88 4/21/09 1:04:24 PM

Problem 3.3 The 2-kg block is held in equilibrium by the system of springs. Draw a free-body diagram for the ring at A.

3 m

3 m 4 m

kAD = 40 N/m

kAB = 30 N/m

kAC = 20 N/m

C B

A

D

Solution

1. The ring at A has negligible size so that it can be modelled as a particle.

2. Imagine the ring at A to be separated or detached from the system.

3. The (detached) ring at A is subjected to three external forces. They are caused by:

i. ii.

iii.

4. Draw the free-body diagram of the (detached) ring showing all these forces labeled with their magnitudes and directions. You should also include any other available information, e.g. lengths, angles, etc. — which will help when formulating the equilibrium equations for the ring.


15_Problems.indd 89 4/21/09 1:04:25 PM

90 Chap. 3 Problems

Problem 3.3 The 2-kg block is held in equilibrium by the system of springs. Draw a free-body diagram for the ring at A.

3 m

3 m 4 m

kAD = 40 N/m

kAB = 30 N/m

kAC = 20 N/m

C B

A

D

Solution




i. Spring AD (weight of block) ii. Spring AC

iii. Spring AB

4. Draw the free-body diagram of the (detached) ring showing all these forces labeled with their magnitudes and directions. You should also include any other available information, e.g. lengths, angles, etc. — which will help when formulating the equilibrium equations for the ring.

15_Problems.indd 90 4/21/09 1:04:25 PM

Problem 3.4 The motor at B winds up the cord attached to the 30-kg crate with a constant speed. The force in cord CD supports the pulley C and the angle q represents the equilibrium state. Draw the free-body diagram of the pulley C. Neglect the size of the pulley.

125

13

B

A

C

D

q

Solution

1. The pulley C has negligible size so that it can be modelled as a particle.

2. Imagine the pulley C to be separated or detached from the system.

3. The (detached) pulley C is subjected to three external forces. They are caused by:

i. ii.

iii.



15_Problems.indd 91 4/21/09 1:04:26 PM

92 Chap. 3 Problems

Problem 3.4 The motor at B winds up the cord attached to the 30-kg crate with a constant speed. The force in cord CD supports the pulley C and the angle q represents the equilibrium state. Draw the free-body diagram of the pulley C. Neglect the size of the pulley.

125

13

B

A

C

D

q

Solution

1. The pulley C has negligible size so that it can be modelled as a particle.

2. Imagine the pulley C to be separated or detached from the system.

3. The (detached) pulley C is subjected to three external forces. They are caused by:

i. CORD CD ii. CORD CB

iii. CORD CA (weight of crate)


(30g)N (30g)N (30g)N , g = .8I m/s2P

15_Problems.indd 92 4/21/09 1:04:27 PM

Problem 3.5 The following system is held in equilibrium by the mass supported at A and the angle q of the connecting cord. Draw the free-body diagram for the connecting knot D.

C

E F

A

B

D

30 kg

60° q

40 kg

Solution

1. The knot D has negligible size so that it can be modelled as a particle.

2. Imagine the knot D to be separated or detached from the system.

3. The (detached) knot D is subjected to three external forces. They are caused by:

i. ii.

iii.

4. Draw the free-body diagram of the (detached) knot showing all these forces labeled with their magnitudes and directions. You should also include any other available information, e.g. lengths, angles, etc. — which will help when formulating the equilibrium equations for the knot.


15_Problems.indd 93 4/21/09 1:04:28 PM

94 Chap. 3 Problems

Problem 3.5 The following system is held in equilibrium by the mass supported at A and the angle q of the connecting cord. Draw the free-body diagram for the connecting knot D.

C

E F

A

B

D

30 kg

60° q

40 kg

Solution

1. The knot D has negligible size so that it can be modelled as a particle.

2. Imagine the knot D to be separated or detached from the system.

3. The (detached) knot D is subjected to three external forces. They are caused by:

i. CORD DE (weight of C) ii. CORD DF (weight of A)

iii. CORD DB (weight of B)

4. Draw the free-body diagram of the (detached) knot showing all these forces labeled with their magnitudes and directions. You should also include any other available information, e.g. lengths, angles, etc. — which will help when formulating the equilibrium equations for the knot.

15_Problems.indd 94 4/21/09 1:04:29 PM

Problem 3.6 The 500-N ( 50-kg) crate is hoisted using the ropes AB and AC. Each rope can withstand a maximum tension of 2500 N before it breaks. Rope AB always remains horizontal. Draw the free-body diagram for the ring at A and determine the smallest angle q to which the crate can be hoisted.

FBA

C q

Solution



3. The (detached) ring A is subjected to three external forces. They are caused by:

i. ii.

iii.

4. Draw the free-body diagram of the (detached) ring showing all these forces labeled with their magnitudes and directions.

5. Establish an xy-axes system on the free-body diagram and write down the equilibrium equations in each of the x and y-directions

+ ↑ Fy = 0 :

+ Fx = 0 : →

6. Solve for the angle q :


15_Problems.indd 95 4/21/09 1:04:30 PM

96 Chap. 3 Problems

Problem 3.6 The 500-N ( 50-kg) crate is hoisted using the ropes AB and AC. Each rope can withstand a maximum tension of 2500 N before it breaks. Rope AB always remains horizontal. Draw the free-body diagram for the ring at A and determine the smallest angle q to which the crate can be hoisted.

FBA

C q

Solution



3. The (detached) ring A is subjected to three external forces. They are caused by:

i. CORD AC ii. CORD AB

iii. CORD AD (weight of crate)

4. Draw the free-body diagram of the (detached) ring showing all these forces labeled with their magnitudes and directions.


+ ↑ Fy = 0 : TAC sin q – 500 = 0

+ Fx = 0 : TAB – TAC cos q = 0 →

6. Solve for the angle q :

Assume TAC =2500 N ⇒ q = 11.54° and TAB = 2449.49 N < 2500 N (O.K!) Ans.

500 N

500 N

15_Problems.indd 96 4/21/09 1:04:30 PM

Problem 3.7 The block has a weight of 20 N and is being hoisted at uniform velocity. The system is held in equilibrium at angle q by the appropriate force in each cord. Draw the free-body diagram for the small pulley.

T

q

30°

B

A

Solution

1. The pulley has negligible size so that it can be modelled as a particle.

2. Imagine the pulley to be separated or detached from the system.

3. The (detached) pulley is subjected to three external forces. They are caused by:

i. ii.

iii.

4. Draw the free-body diagram of the (detached) pulley showing all these forces labeled with their magnitudes and directions. You should also include any other available information, e.g. lengths, angles, etc. — which will help when formulating the equilibrium equations for the knot.


15_Problems.indd 97 4/21/09 1:04:31 PM

98 Chap. 3 Problems

Problem 3.7 The block has a weight of 20 N and is being hoisted at uniform velocity. The system is held in equilibrium at angle q by the appropriate force in each cord. Draw the free-body diagram for the small pulley.

T

q

30°

B

A

Solution

1. The pulley has negligible size so that it can be modelled as a particle.

2. Imagine the pulley to be separated or detached from the system.

3. The (detached) pulley is subjected to three external forces. They are caused by:

i. Cord AB ii. Force T

iii. Weight of block

4. Draw the free-body diagram of the (detached) pulley showing all these forces labeled with their magnitudes and directions. You should also include any other available information, e.g. lengths, angles, etc. — which will help when formulating the equilibrium equations for the knot.

(= 20 N) 20 N20 N

15_Problems.indd 98 4/21/09 1:04:32 PM

Problem 3.8 Blocks D and F weigh 5 kN ( 500 kg) each and block E weighs 8 kN ( 800 kg). The system is in equilibrium at a given sag s. Draw the free-body diagram for the connecting ring at A and find s. Neglect the size of the pulleys.

FD E

A

CB

4 m 4 m

s

Solution


2. Imagine the ring to be separated or detached from the system.

3. The (detached) ring is subjected to three external forces. They are caused by:

i. ii.

iii.

4. Draw the free-body diagram of the (detached) ring showing all these forces labeled with their magnitudes and directions. Include also any other information which may help when formulating the equilibrium equations for the ring.

5. Establish an xy-axes system on the free-body diagram and write down the equilibrium equations in the y-direction only (this is all that is required to solve this problem):

+ ↑ Fy = 0 :

6. Solve for the sag s :


15_Problems.indd 99 4/21/09 1:04:33 PM

100 Chap. 3 Problems

Problem 3.8 Blocks D and F weigh 5 kN ( 500 kg) each and block E weighs 8 kN ( 800 kg). The system is in equilibrium at a given sag s. Draw the free-body diagram for the connecting ring at A and find s. Neglect the size of the pulleys.

FD E

A

CB

4 m 4 m

s

Solution


2. Imagine the ring to be separated or detached from the system.

3. The (detached) ring is subjected to three external forces. They are caused by:

i. CORD AB (weight of D) ii. CORD AC (weight of F)

iii. CORD AE (weight of E)



+ ↑ Fy = 0 : 2(5) sin q – 8 = 0 ⇒ q = 53.13°

6. Solve for the sag s :

tan q = s4

⇒ s = 4 tan 53.13° = 5.33 m Ans.

5 kN5 kN

8 kN4 m

15_Problems.indd 100 4/21/09 1:04:34 PM

Problem 3.9 A vertical force P is applied to the ends of the cord AB and spring AC. The spring has an unstretched length of 2 m and the system is in equilibrium at angle q. Draw the free-body diagram of the connecting knot at A and write down the equilibrium equations for the knot at A.

2 m

k

2 m

q

A

B C

P

Solution




i. ii.

iii.



+ ↑ Fy = 0 :

+ Fx = 0 : →


15_Problems.indd 101 4/21/09 1:04:35 PM


Problem 3.9 A vertical force P is applied to the ends of the cord AB and spring AC. The spring has an unstretched length of 2 m and the system is in equilibrium at angle q. Draw the free-body diagram of the connecting knot at A and write down the equilibrium equations for the knot at A.

2 m

k

2 m

q

A

B C

P

Solution




i. CORD AB ii. SPRING AC

iii. Force P



+ Fx = 0 : Fs cos φ – T cos q = 0 →

+ ↑ Fy = 0 : T cos q + Fs sin φ – P = 0

15_Problems.indd 102 4/21/09 1:04:35 PM

Problem 3.10 The sling BAC is used to lift the 10-kN load with constant velocity. By drawing the free-body diagram for the ring at A, determine the magnitude of the force in the sling as a function of the angle q.

B

A

q

10 kN

Cq

Solution




i. ii.

iii.



+ ↑ Fy = 0 :

6. Solve for the magnitude of the force in the sling:


15_Problems.indd 103 4/21/09 1:04:36 PM


Problem 3.10 The sling BAC is used to lift the 10-kN load with constant velocity. By drawing the free-body diagram for the ring at A, determine the magnitude of the force in the sling as a function of the angle q.

B

A

q

10 kN

Cq

Solution




i. CORD AB ii. CORD AC

iii. CORD AD (weight of load)



+ ↑ Fy = 0 : 10 – 2T cos q = 0

6. Solve for the magnitude of the force in the sling:

T = 5cosq Ans.

10 kN

10 kN

15_Problems.indd 104 4/21/09 1:04:36 PM

Problem 3.11 When y is zero, the springs sustain a force of 300 N. The applied vertical forces F and –F pull the point A away from B a distance of y = 1 m. The cords CAD and CBD are attached to the rings at C and D. Draw the free-body diagrams for point A and ring C.

F

k = 40 N/M k = 400 N/M

1 m

1 m

1 m

1 m

–F

y

A

B

DC

Solution

1. Imagine A and C to be separated or detached from the system.

2. Each of A and C is subjected to three external forces. For A, they are caused by:

i. ii.

iii.

For C, they are caused by:

i. ii.

iii.

3. Draw the free-body diagrams of A and C showing all these forces labeled with their magnitudes and directions. You should also include any other available information, e.g. lengths, angles, etc. — which will help when formulating the equilibrium equations.


15_Problems.indd 105 4/21/09 1:04:37 PM


Problem 3.11 When y is zero, the springs sustain a force of 300 N. The applied vertical forces F and –F pull the point A away from B a distance of y = 1 m. The cords CAD and CBD are attached to the rings at C and D. Draw the free-body diagrams for point A and ring C.

F

k = 40 N/M k = 400 N/M

1 m

1 m

1 m

1 m

–F

y

A

B

DC

Solution

1. Imagine A and C to be separated or detached from the system.

2. Each of A and C is subjected to three external forces. For A, they are caused by:

i. CORD AC ii. CORD AD iii. Force F


i. CORD AC ii. CORD CB iii. Spring attached at C

3. Draw the free-body diagrams of A and C showing all these forces labeled with their magnitudes and directions. You should also include any other available information, e.g. lengths, angles, etc. — which will help when formulating the equilibrium equations.

15_Problems.indd 106 4/21/09 1:04:39 PM

Problem 3.12 By drawing a free-body diagram for the ring at A, determine the maximum weight W that can be supported in the position shown if each cable AC and AB can support a maximum tension of 300 N before it fails.

A

C30°

B

512 13

Solution




i. ii.

iii.

4. Draw the free-body diagram of the (detached) ring showing all these forces labeled with their magnitudes and directions. Include any other relevant information, e.g. lengths, angles, etc.


+ ↑ Fy = 0 :

+ Fx = 0 : →

6. Set the tension in AB to the maximum of 600 lb and solve for the maximum weight W:


15_Problems.indd 107 4/21/09 1:04:40 PM


Problem 3.12 By drawing a free-body diagram for the ring at A, determine the maximum weight W that can be supported in the position shown if each cable AC and AB can support a maximum tension of 3000 N before it fails.

A

C30°

B

512 13

Solution




i. CABLE AB ii. CABLE AB

iii. Weight of ball



+ Fx = 0 : FAB 513( ) + FAC sin 30° = 0 →

+ ↑ Fy = 0 : FAB 1213( ) + FAC cos 30° – W = 0

6. Set FAB, the tension in AB, to the maximum of 3000 N and solve for the maximum weight W:

FAC = 2308 N(< 3000 N!!), W = 4768 N ↓ Ans.

15_Problems.indd 108 4/21/09 1:04:41 PM

Problem 3.13 The cords suspend the two small buckets in the equilibrium position shown. Draw the free-body diagrams for each of the points F and C.

40°

65°C

A

F

E

B

D

20°

20°

Solution

1. Imagine the points F and C to be separated or detached from the system.

2. Each of F and C is subjected to three external forces. For F, they are caused by:

i. ii.

iii.


i. ii.

iii.

3. Draw the free-body diagrams of F and C showing all these forces labeled with their magnitudes and directions. You should also include any other available information, e.g. lengths, angles, etc. — which will help when formulating the equilibrium equations at these points.


15_Problems.indd 109 4/21/09 1:04:42 PM


Problem 3.13 The cords suspend the two small buckets in the equilibrium position shown. Draw the free-body diagrams for each of the points F and C.

40°

65°C

A

F

E

B

D

20°

20°

Solution

1. Imagine the points F and C to be separated or detached from the system.

2. Each of F and C is subjected to three external forces. For F, they are caused by:

i. CABLE FE ii. CABLE FC

iii. Weight of A


i. CABLE CF ii. CABLE CD

iii. Weight of B

3. Draw the free-body diagrams of F and C showing all these forces labeled with their magnitudes and directions. You should also include any other available information, e.g. lengths, angles, etc. — which will help when formulating the equilibrium equations at these points.

15_Problems.indd 110 4/21/09 1:04:43 PM

Problem 3.14The 30-kg pipe is supported at A by a system of five cords. Draw the free-body diagrams for the rings at A and B when the system is in equilibrium.

34

60° A

B

D

C

E

5

Solution

1. Imagine A and B to be separated or detached from the system.

2. Each of A and B is subjected to three external forces. For A, they are caused by:

i. ii.

iii.

For B, they are caused by:

i. ii.

iii.

3. Draw the free-body diagrams of A and B showing all these forces labeled with their magnitudes and directions. You should also include any other available information, e.g. lengths, angles, etc. — which will help when formulating the equilibrium equations.


15_Problems.indd 111 4/21/09 1:04:44 PM


Problem 3.14The 30-kg pipe is supported at A by a system of five cords. Draw the free-body diagrams for the rings at A and B when the system is in equilibrium.

34

60° A

B

D

C

E

5

Solution

1. Imagine A and B to be separated or detached from the system.

2. Each of A and B is subjected to three external forces. For A, they are caused by:

i. CABLE AB ii. CABLE AE

iii. Weight of Pipe

For B, they are caused by:

i. CABLE BC ii. CABLE BD

iii. CABLE BA

3. Draw the free-body diagrams of A and B showing all these forces labeled with their magnitudes and directions. You should also include any other available information, e.g. lengths, angles, etc. — which will help when formulating the equilibrium equations.

15_Problems.indd 112 4/21/09 1:04:46 PM

Problem 3.15The cord AB has a length of 5 m and is attached to the end B of the spring having a stiffness k =10 kN/M. The other end of the spring is attached to a roller C so that the spring remains horizontal as it stretches. If a 10-kN ( 1 tonne) weight is suspended from B, use the free-body diagram for the ring at B to determine the necessary unstretched length of the spring, so that q = 40° for equilibrium.

Bk = 10 kN/M

C

A q

5 m

5 m

5 m

Solution

1. Imagine the ring at B to be separated or detached from the system.

2. The (detached) ring at B is subjected to three external forces caused by:

i. ii.

iii.



+ ↑ Fy = 0 :

+ Fx = 0 : →

5. Determine the stretch in the spring BC and solve for the necessary unstretched length:


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Problem 3.15The cord AB has a length of 5 m and is attached to the end B of the spring having a stiffness k =10 kN/M. The other end of the spring is attached to a roller C so that the spring remains horizontal as it stretches. If a 10-kN ( 1 tonne) weight is suspended from B, use the free-body diagram for the ring at B to determine the necessary unstretched length of the spring, so that q = 40° for equilibrium.

Bk = 10 kN/M

C

A q

5 m

5 m

5 m

Solution

1. Imagine the ring at B to be separated or detached from the system.

2. The (detached) ring at B is subjected to three external forces caused by:

i. CABLE AB ii. SPRING BC

iii. 10 lb Weight



+ ↑ Fy = 0 : FAB sin 40° – 10 = 0 ⇒ FAB = 15.557 kN

+ Fx = 0 : FBC – FAB cos 40° = 0 ⇒ FBC = 11.918 kN →

5. Determine the stretch in the spring BC and solve for the necessary unstretched length l :

FBC = kx ⇒ x = 11 91810. = 1.1918 m (stretch in BC)

BC = 5 +(5 – 5 cos 40°) = 6.17 m

l = 6.17 – 1.1918 = 4.98 m Ans.

10 kN10 kN

15_Problems.indd 114 4/21/09 1:04:48 PM

3.2Free-BodyDiagramsintheEquilibriumofaRigidBody

Problem 3.16Draw the free-body diagram of the 50-kg uniform pipe, which is supported by the smooth contacts at A and B.

30°

40°

10°

A B

Solution

1. Imagine the pipe to be separated or detached from the system.

2. The supports at A and B are smooth contacts. Use Table 2.1 (6) to determine the number and types of reactions acting on the pipe at A and B.

3. The pipe is subjected to three external forces (don’t forget the weight!). They are caused by:

i. ii.

iii.

4. Draw the free-body diagram of the (detached) pipe showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the pipe (the correct sense will always emerge from the equilibrium equations for the pipe). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the pipe.

3.2 Free-Body Diagrams in the Equilibrium of a Rigid Body 115

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Problem 3.16Draw the free-body diagram of the 50-kg uniform pipe, which is supported by the smooth contacts at A and B.

30°

40°

10°

A B

Solution


2. The supports at A and B are smooth contacts. Use Table 2.1 (6) to determine the number and types of reactions acting on the pipe at A and B.

3. The pipe is subjected to three external forces (don’t forget the weight!). They are caused by:

i. The reaction at A ii. The reaction at B

iii. The weight of the pipe

4. Draw the free-body diagram of the (detached) pipe showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the pipe (the correct sense will always emerge from the equilibrium equations for the pipe). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the pipe.

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Problem 3.17Draw the free-body diagram of the hand punch, which is pinned at A and bears down on the smooth surface at B. Neglect the weight of the punch.

F = 40 N

AB

0.45 m

0.6 m0.06 m

Solution

1. Imagine the hand punch to be separated or detached from the system.

2. The support at B is a smooth contact. The punch is (smoothly) pin-connected at A. Use Table 2.1 (6) and (8) to determine the number and types of reactions acting on the pipe at A and B.

3. The punch is subjected to four external forces. They are caused by:

i. ii.

iii. iv.

4. Draw the free-body diagram of the (detached) punch showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the punch (the correct sense will always emerge from the equilibrium equations for the punch). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the punch.


15_Problems.indd 117 4/21/09 1:04:50 PM


Problem 3.17Draw the free-body diagram of the hand punch, which is pinned at A and bears down on the smooth surface at B. Neglect the weight of the punch.

F = 40 N

AB

0.45 m

0.6 m0.06 m

Solution

1. Imagine the hand punch to be separated or detached from the system.

2. The support at B is a smooth contact. The punch is (smoothly) pin-connected at A. Use Table 2.1 (6) and (8) to determine the number and types of reactions acting on the pipe at A and B.

3. The punch is subjected to four external forces. They are caused by:

i. The force F ii. The reaction at B

iii. & iv. The two reactions at A

4. Draw the free-body diagram of the (detached) punch showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the punch (the correct sense will always emerge from the equilibrium equations for the punch). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the punch.

40 N 0.45 m

0.06 m

0.6 m

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Problem 3.18Draw the free-body diagram of the jib crane AB, which is pin-connected at A and supported by member (link) BC. Neglect the weight of the crane.

8 kN

3 m

0.4 m

A

C

B

3

4

5

4 m

Solution

1. Imagine the jib crane AB to be separated or detached from the system.

2. There is a link support at B and the jib crane is (smoothly) pinned at A. Use Table 2.1 (2) and (8) to determine the number and types of reactions acting on the jib crane at A and B.

3. The jib crane is subjected to four external forces. They are caused by:

i. ii.

iii. iv.

4. Draw the free-body diagram of the (detached) crane showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the crane (the correct sense will always emerge from the equilibrium equations for the crane). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the jib crane.


15_Problems.indd 119 4/21/09 1:04:52 PM


Problem 3.18Draw the free-body diagram of the jib crane AB, which is pin-connected at A and supported by member (link) BC. Neglect the weight of the crane.

8 kN

3 m

0.4 m

A

C

B

3

4

5

4 m

Solution

1. Imagine the jib crane AB to be separated or detached from the system.

2. There is a link support at B and the jib crane is (smoothly) pinned at A. Use Table 2.1 (2) and (8) to determine the number and types of reactions acting on the jib crane at A and B.

3. The jib crane is subjected to four external forces. They are caused by:

i. & ii. The reactions at A iii. The reaction at B

iv. The 8 kN load

4. Draw the free-body diagram of the (detached) crane showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the crane (the correct sense will always emerge from the equilibrium equations for the crane). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the jib crane.

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Problem 3.19Draw the free-body diagram of the dumpster D of the truck, which has a weight of 25 kN ( 2.5 tonnes) and a center of gravity at G. It is supported by a pin at A and a pin-connected hydraulic cylinder BC (short link).

1.5 m

3 m1 m

20°30°B

A

DG

C

Solution

1. Imagine the dumpster D to be separated or detached from the truck.

2. There is a pin support at A and the dumpster is supported by a short link support at B. Use Table 2.1 (2) and (8) to determine the number and types of reactions acting on the dumpster at A and B.

3. The dumpster is subjected to four external forces. They are caused by:

i. ii.

iii. iv.

4. Draw the free-body diagram of the (detached) dumpster showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the dumpster (the correct sense will always emerge from the equilibrium equations for the dumpster). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the dumpster.


15_Problems.indd 121 4/21/09 1:04:54 PM


Problem 3.19Draw the free-body diagram of the dumpster D of the truck, which has a weight of 25 kN ( 2.5 tonnes) and a center of gravity at G. It is supported by a pin at A and a pin-connected hydraulic cylinder BC (short link).

1.5 m

3 m1 m

20°30°B

A

DG

C

Solution

1. Imagine the dumpster D to be separated or detached from the truck.

2. There is a pin support at A and the dumpster is supported by a short link support at B. Use Table 2.1 (2) and (8) to determine the number and types of reactions acting on the dumpster at A and B.

3. The dumpster is subjected to four external forces. They are caused by:


iv. The weight of the dumpster

4. Draw the free-body diagram of the (detached) dumpster showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the dumpster (the correct sense will always emerge from the equilibrium equations for the dumpster). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the dumpster.

25 kN

25 kN

15_Problems.indd 122 4/21/09 1:04:54 PM

Problem 3.20Draw the free-body diagram of the link CAB, which is pin-connected at A and rests on the smooth cam at B. Neglect the weight of the link.

425 N

50 mm

80 mm

30°C

AB

Solution

1. Imagine the link CAB to be separated or detached from the mechanism.

2. There is a pin connection at A and the link rests on the smooth surface at B. Use Table 2.1 (6) and (8) to determine the number and types of reactions acting on the link at A and B.

3. The link is subjected to four external forces. They are caused by:

i. ii.

iii. iv.

4. Draw the free-body diagram of the (detached) link showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the link (the correct sense will always emerge from the equilibrium equations for the link). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the link.


15_Problems.indd 123 4/21/09 1:04:55 PM


Problem 3.20Draw the free-body diagram of the link CAB, which is pin-connected at A and rests on the smooth cam at B. Neglect the weight of the link.

425 N

50 mm

80 mm

30°C

AB

Solution

1. Imagine the link CAB to be separated or detached from the mechanism.

2. There is a pin connection at A and the link rests on the smooth surface at B. Use Table 2.1 (6) and (8) to determine the number and types of reactions acting on the link at A and B.

3. The link is subjected to four external forces. They are caused by:


iv. The 425 N load at C

4. Draw the free-body diagram of the (detached) link showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the link (the correct sense will always emerge from the equilibrium equations for the link). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the link.

15_Problems.indd 124 4/21/09 1:04:56 PM

Problem 3.21Draw the free-body diagram of the uniform pipe which has a mass of 100 kg and a center of mass at G. The supports A, B and C are smooth.

C

1.75 m 0.1 m

1.25 m

0.5 m 0.2 m

30°A

G

B

Solution


2. The pipe rests on smooth surfaces at A, B and C. Use Table 2.1 (6) to determine the number and types of reactions acting on the pipe at A, B and C.

3. The pipe is subjected to four external forces. They are caused by:

i. ii.

iii. iv.

4. Draw the free-body diagram of the (detached) pipe showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the pipe (the correct sense will always emerge from the equilibrium equations for the pipe). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations for the pipe.


15_Problems.indd 125 4/21/09 1:04:57 PM


Problem 3.21Draw the free-body diagram of the uniform pipe which has a mass of 100 kg and a center of mass at G. The supports A, B and C are smooth.

C

1.75 m 0.1 m

1.25 m

0.5 m 0.2 m

30°A

G

B

Solution


2. The pipe rests on smooth surfaces at A, B and C. Use Table 2.1 (6) to determine the number and types of reactions acting on the pipe at A, B and C.

3. The pipe is subjected to four external forces. They are caused by:

i. The reaction at A iii. The reaction at B

iv. The reaction at C iv. Pipe’s weight

4. Draw the free-body diagram of the (detached) pipe showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the pipe (the correct sense will always emerge from the equilibrium equations for the pipe). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations for the pipe.

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Problem 3.22Draw the free-body diagram of the beam, which is pin-supported at A and rests on the smooth incline at B. Neglect the weight of the beam.

A

4

1 m 1 m 1 m 1 m

8 kN8 kN6 kN 6 kN

4 kN

0.4 m

35

B

0.2 m

0.2 m

Solution

1. Imagine the beam to be separated or detached from the system.

2. There is a pin connection at A and the beam rests on the smooth (inclined) surface at B. Use Table 2.1 (6) and (8) to determine the number and types of reactions acting on the beam at A and B.

3. In addition to the forces shown in the figure, the beam is subjected to three external forces. They are caused by:

i. ii.

iii.

4. Draw the free-body diagram of the (detached) beam showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the beam (the correct sense will always emerge from the equilibrium equations for the beam). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the beam.


15_Problems.indd 127 4/21/09 1:04:59 PM


Problem 3.22Draw the free-body diagram of the beam, which is pin-supported at A and rests on the smooth incline at B. Neglect the weight of the beam.

A

4

1 m 1 m 1 m 1 m

8 kN8 kN6 kN 6 kN

4 kN

0.4 m

35

B

0.2 m

0.2 m

Solution


2. There is a pin connection at A and the beam rests on the smooth (inclined) surface at B. Use Table 2.1 (6) and (8) to determine the number and types of reactions acting on the beam at A and B.

3. In addition to the forces shown in the figure, the beam is subjected to three external forces. They are caused by:


4. Draw the free-body diagram of the (detached) beam showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the beam (the correct sense will always emerge from the equilibrium equations for the beam). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the beam.

4 kN 8 kN 8 kN 6 kN 6 kN

I m I m I m I m 0.2 m 0.2 m

15_Problems.indd 128 4/21/09 1:04:59 PM

Problem 3.23Draw the free-body diagram of the member ABC, which is supported by a pin at A and a horizontal short link BD. Neglect the weight of ABC.

1 m

ADB

C

0.5 m

30°

50 kg

Solution

1. Imagine the member ABC to be separated or detached from the system.

2. There is a pin support at A and the member is supported by a horizontal short link at B. Use Table 2.1 (2) and (8) to determine the number and types of reactions acting on the member at A and B.

3. The member is subjected to four external forces. They are caused by:

i. ii.

iii. iv.

4. Draw the free-body diagram of the (detached) member showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the member (the correct sense will always emerge from the equilibrium equations for the member). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the member.


15_Problems.indd 129 4/21/09 1:05:00 PM


Problem 3.23Draw the free-body diagram of the member ABC, which is supported by a pin at A and a horizontal short link BD. Neglect the weight of ABC.

1 m

ADB

C

0.5 m

30°

50 kg

Solution

1. Imagine the member ABC to be separated or detached from the system.

2. There is a pin support at A and the member is supported by a horizontal short link at B. Use Table 2.1 (2) and (8) to determine the number and types of reactions acting on the member at A and B.

3. The member is subjected to four external forces. They are caused by:


iv. The weight at C

4. Draw the free-body diagram of the (detached) member showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the member (the correct sense will always emerge from the equilibrium equations for the member). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the member.

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Problem 3.24Draw the free-body diagram of the beam. The support at B is smooth. Neglect the weight of the beam.

8 m 4 m

A

B

800 N m ·

5 m

500 N

Solution


2. There is a pin support at A and a smooth contact support at B. Use Table 2.1 (6) and (8) to determine the number and types of reactions acting on the member at A and B.

3. In addition to those shown in the figure, the member is subjected to three external forces. They are caused by:

i. ii.

iii.

4. Draw the free-body diagram of the (detached) member showing all these forces and any external applied couple moments labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the member (the correct sense will always emerge from the equilibrium equations for the member). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the member.


15_Problems.indd 131 4/21/09 1:05:02 PM


Problem 3.24Draw the free-body diagram of the beam. The support at B is smooth. Neglect the weight of the beam.

8 m 4 m

A

B

800 N m ·

5 m

500 N

Solution


2. There is a pin support at A and a smooth contact support at B. Use Table 2.1 (6) and (8) to determine the number and types of reactions acting on the member at A and B.

3. In addition to those shown in the figure, the member is subjected to three external forces. They are caused by:


4. Draw the free-body diagram of the (detached) member showing all these forces and any external applied couple moments labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the member (the correct sense will always emerge from the equilibrium equations for the member). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the member.

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Problem 3.25Draw the free-body diagram of the vehicle, which has a mass of 5 Mg and center of mass at G. The tires are free to roll, so rolling resistance can be neglected.

435

1.75 m

CB

G

T

1 m 0.5 m

0.3 m

A

20°

0.6 m

Solution

1. Imagine the vehicle to be separated or detached from the system.

2. There are smooth contacts at A and B. Use Table 2.1 (6) to determine the number and types of reactions acting on the vehicle at A and B.

3. The vehicle is subjected to four external forces. They are caused by:

i. ii.

iii. iv.

4. Draw the free-body diagram of the (detached) vehicle showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the vehicle (the correct sense will always emerge from the equilibrium equations for the vehicle). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the vehicle.


15_Problems.indd 133 4/21/09 1:05:04 PM


Problem 3.25Draw the free-body diagram of the vehicle, which has a mass of 5 Mg and center of mass at G. The tires are free to roll, so rolling resistance can be neglected.

435

1.75 m

CB

G

T

1 m 0.5 m

0.3 m

A

20°

0.6 m

Solution

1. Imagine the vehicle to be separated or detached from the system.

2. There are smooth contacts at A and B. Use Table 2.1 (6) to determine the number and types of reactions acting on the vehicle at A and B.

3. The vehicle is subjected to four external forces. They are caused by:

i. The reaction at A iii. The reaction at B

iii. Car’s weight iv. Force T

4. Draw the free-body diagram of the (detached) vehicle showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the vehicle (the correct sense will always emerge from the equilibrium equations for the vehicle). Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the vehicle.

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Problem 3.26Draw a free-body diagram of the crane boom ABC, which has a mass of 45 kg, center of gravity at G, and supports a load of 30 kg. The boom is pin-connected to the frame at B and connected to a vertical chain CD. The chain supporting the load is attached to the boom at A.

AG B

C

D

0.4 m

0.6 m0.8 m1.25 m

Solution

1. Imagine the boom to be separated or detached from the system.

2. There is a pin connection at B and a vertical chain (cable) support at C. Use Table 2.1 (1) and (8) to determine the number and types of reactions acting on the boom at B and C.

3. The boom is subjected to five external forces. They are caused by:

i. ii.

iii. iv.

v.

4. Draw the free-body diagram of the (detached) boom showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the boom. Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the boom.


15_Problems.indd 135 4/21/09 1:05:06 PM


Problem 3.26Draw a free-body diagram of the crane boom ABC, which has a mass of 45 kg, center of gravity at G, and supports a load of 30 kg. The boom is pin-connected to the frame at B and connected to a vertical chain CD. The chain supporting the load is attached to the boom at A.

AG B

C

D

0.4 m

0.6 m0.8 m1.25 m

Solution

1. Imagine the boom to be separated or detached from the system.

2. There is a pin connection at B and a vertical chain (cable) support at C. Use Table 2.1 (1) and (8) to determine the number and types of reactions acting on the boom at B and C.

3. The boom is subjected to five external forces. They are caused by:

i. & ii. The reactions at B iii. The reaction at C

iv. Weight of boom v. Load at A

4. Draw the free-body diagram of the (detached) boom showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the boom. Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the boom.

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Problem 3.27Draw a free-body diagram of the beam. Neglect the thickness and weight of the beam.

400 N

15°600 N

8 m

3

4

5

4 m

AB

Solution


2. There is a pin connection at A and a rocker support at B. Use Table 2.1 to determine the number and types of reactions acting on the beam at A and B.

3. The beam is subjected to five external forces. They are caused by:

i. ii.

iii. iv.

v.

4. Draw the free-body diagram of the (detached) beam showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the beam. Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the beam.


15_Problems.indd 137 4/21/09 1:05:07 PM


Problem 3.27Draw a free-body diagram of the beam. Neglect the thickness and weight of the beam.

400 N

15°600 N

8 m

3

4

5

4 m

AB

Solution


2. There is a pin connection at A and a rocker support at B. Use Table 2.1 to determine the number and types of reactions acting on the beam at A and B.

3. In addition to those shown in the figure, the beam is subjected to three external forces. They are caused by:



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Problem 3.28The link shown in the figure is pin-connected at A and rests against a smooth support at B. Draw the free-body diagram for link ABC and use it to compute the horizontal and vertical components of reaction at pin A. Neglect the weight of the link.

0.75 m

30°

1 m0.5 m

60 N

90 N · m

A

B

Solution

1. Imagine the link ABC to be separated or detached from the system.

2. There is a pin connection at A and a smooth support at B. Use Table 2.1 to identify the reactions acting on the link at A and B.

3. The link is subjected to four external forces and one external applied couple moment.

4. Draw the free-body diagram of the (detached) link showing all these forces and couple moments labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the link. Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the link.

5. Sum moments about A and write down the moment equilibrium equation.

+ MA = 0 :

6. Establish an xy-axes system on the free-body diagram and write down the force equilibrium equations in each of the x and y-directions

+ Fx = 0 : →

+ ↑ Fy = 0 :

7. Solve the three equations for the required reaction components at pin A:


15_Problems.indd 139 4/21/09 1:05:09 PM


Problem 3.28The link shown in the figure is pin-connected at A and rests against a smooth support at B. Draw the free-body diagram for link ABC and use it to compute the horizontal and vertical components of reaction at pin A. Neglect the weight of the link.

0.75 m

30°

1 m0.5 m

60 N

90 N · m

A

B

Solution

1. Imagine the link ABC to be separated or detached from the system.

2. There is a pin connection at A and a smooth support at B. Use Table 2.1 to identify the reactions acting on the link at A and B.

3. The link is subjected to four external forces and one external applied couple moment.

4. Draw the free-body diagram of the (detached) link showing all these forces and couple moments labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the link. Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the link.


+ MA = 0 : – 90N.m – 60N (1m) + NB (0.75m) = 0


+ Fx = 0 : Ax – NB sin 30° N = 0 →

+ ↑ Fy = 0 : Ay – NB cos 30° N – 60N = 0

7. Solve the three equations for the required reaction components at pin A:

NB = 200N, Ax = 100N, Ay = 233N Ans.

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Problem 3.29A force of 150 kN acts on the end of the beam. Using the free-body diagram for the beam, find the magnitude and direction of the reaction at pin A and the tension in the cable. Neglect the weight of the beam.

3 m

8 m

C

34

5

A

1.5 m

2 m

150 kN

B

Solution


2. There is a pin connection at A and a cable support at B. Use Table 2.1 to identify the reactions acting on the beam at A and B.

3. The beam is subjected to four external forces.



+ MA = 0 :

You should obtain the cable tension directly from this equation.


+ Fx = 0 : →

+ ↑ Fy = 0 :

7. Solve the two equations for the magnitude and direction of the (resultant) force at pin A:


15_Problems.indd 141 4/21/09 1:05:10 PM


Problem 3.29A force of 150 kN acts on the end of the beam. Using the free-body diagram for the beam, find the magnitude and direction of the reaction at pin A and the tension in the cable. Neglect the weight of the beam.

3 m

8 m

C

34

5

A

1.5 m

2 m

150 kN

B

Solution


2. There is a pin connection at A and a cable support at B. Use Table 2.1 to identify the reactions acting on the beam at A and B.

3. The beam is subjected to four external forces.



+ MA = 0 : – 35

T( ) (2m) – 45

T( ) (3m) + 150 kN(10m) = 0

You should obtain the cable tension directly from this equation:

T = 416.7 kN. Ans.


+ Fx = 0 : – Ax + 45( ) (416.7 kN) = 0 →

+ ↑ Fy = 0 : 35( ) 416.7 kN – 150 kN – Ay = 0

7. Solve the two equations for the magnitude and direction of the (resultant) force at pin A:

Ax = 333.3 kN ← , Ay = 100 kN ↓

Thus, magnitude of force at A is ( . ) ( )333 3 1002 2+ = 348.0 kN

Direction is q = tan–1 –– .

100333 3

= 196.7° 16.7o Ans.

1.5 m

3 m

2 m8 m

150 kN

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Problem 3.30The oil-drilling rig shown has a mass of 24 Mg and mass center at G. If the rig is pin-connected at its base, use a free-body diagram of the rig to determine the tension in the hoisting cable and the magnitude of the hoisting force at A when the rig is in the position shown.

1.25 m5 m 3 m 10 m

A3

4

560°

B

G

Solution

1. Imagine the rig to be separated or detached from the system.

2. There is a pin connection at A and a cable support at B. Use Table 2.1 to identify the reactions acting on the rig at A and B. Note that since the hoisting cable is continuous and passes over the pulley, the cable is subjected to the same tension T throughout its length.

3. The rig is subjected to five external forces.

4. Draw the free-body diagram of the (detached) rig showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the rig. Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the rig.


+ MA = 0 :

You should obtain the cable tension T directly from this equation:


+ Fx = 0 : →

+ ↑ Fy = 0 :

7. Solve the two equations for the magnitude of the (resultant) force at pin A:


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Problem 3.30The oil-drilling rig shown has a mass of 24 Mg and mass center at G. If the rig is pin-connected at its base, use a free-body diagram of the rig to determine the tension in the hoisting cable and the magnitude of the hoisting force at A when the rig is in the position shown.

1.25 m5 m 3 m 10 m

A3

4

560°

B

G

Solution

1. Imagine the rig to be separated or detached from the system.

2. There is a pin connection at A and a cable support at B. Use Table 2.1 to identify the reactions acting on the rig at A and B. Note that since the hoisting cable is continuous and passes over the pulley, the cable is subjected to the same tension T throughout its length.

3. The rig is subjected to five external forces.

4. Draw the free-body diagram of the (detached) rig showing all these forces labeled with their magnitudes and directions. Assume the sense of the vectors representing the reactions acting on the rig. Include any other relevant information, e.g. lengths, angles, etc. — which may help when formulating the equilibrium equations (including the moment equation) for the rig.


+ MA = 0 : (235.4 kN)(10m) – 35( ) T(13m) + 4

5T( ) (1.25m) – T sin 60°(18m) + (T cos 60°)(1.25m) = 0

You should obtain the cable tension T directly from this equation: T =108.2 kN


+ Fx = 0 : Ax – 108.2 45( ) kN – 108.2 cos 60° kN = 0 →

+ ↑ Fy = 0 : Ay – 235.4 kN + 108.2 35( ) kN – 108.2 sin 60° kN = 0

7. Solve the two equations for the magnitude of the (resultant) force at pin A:

Ax =140.6 kN, Ay = 76.8 kN

Thus, magnitude of force at A is ( . ) ( . )140 6 76 82 2+ =160 kN Ans.

15_Problems.indd 144 4/21/09 1:05:13 PM

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