SolidWorks Motion Tutorial 2013

8/10/2019 SolidWorks Motion Tutorial 2013

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IntroductiontoSolidModelingUsingSolidWorks2013 SolidWorksMotionTutorial Page1

Inthistutorial,wewilllearnthebasicsofperformingmotionanalysisusingSolidWorksMotion.

AlthoughthetutorialcanbecompletedbyanyonewithabasicknowledgeofSolidWorkspartsand

assemblies,wehaveprovidedenoughdetailsothatstudentswithanunderstandingofthephysicsof

mechanicswillbeabletorelatetheresultstothoseobtainedbyhandcalculations.

Wewillbelookingatthreedifferentanalyses:

1.

Rotationofawheel,inwhichwewilllearnhowtosetupamotionanalysisandseetheeffectsof

changingthemassmomentofinertiaonangularacceleration.

2.

Fourbarlinkage,inwhichwewillseehowplottingaquantitysuchasaccelerationovera

mechanismsfullrangeofmotionallowsustoidentifytheextremevaluesofthequantity.

3. Rolleronaramp,inwhichtheeffectsoffrictionwillbeevaluated.

1. RotationofaWheel

Beginbycreatingthethreepartmodelsdetailedbelow,orbydownloadingthepartsfromthebooks

website. Theeightholepatternoneachwheelisaddedtohelpvisualizationoftherotationofthepart.


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Ifyoudownloadedthepartfiles,thematerialsforthe

partsaredefined. Ifyoucreatedthemyourself,thenfor

eachpart,definethematerialbyrightclickingMaterial

intheFeatureManagerandselectingEditMaterial. The

MaterialEditorwillappear,asshownhere. SelectAlloy

SteelfromthelistofsteelsintheSolidWorksmaterials

library. ClickApplyandthenClose.

Tobegin,wewillanalyzeasimplemodelofawheel

subjectedtoatorque. FromNewtonsSecondLaw,we

knowthatthesumoftheforcesactingonabodyequals

themassofthebodytimestheaccelerationofthebody,or

1

Theaboveequationappliestobodiesundergoinglinearacceleration. Forrotatingbodies,Newtons

SecondLawcanbewrittenas:

2Where isthesumofthemomentsaboutanaxispassingthroughthebodyscenterofmass,isthemassmomentofinertiaofthebodyaboutthataxis,andistheangularaccelerationofthebody. Themomentofinertiaaboutanaxisisdefinedas:

3whereistheradialdistancefromtheaxis. Forsimpleshapes,themomentofinertiaisrelativelyeasytocalculate,asformulasforofbasicshapesaretabulatedinmanyreferencebooks. However,formorecomplexcomponents,calculationofcanbedifficult. SolidWorksallowsmassproperties,includingmomentsofinertia,tobedeterminedeasily.

OpenthepartWheel1. Fromthemainmenu,selectTools:MassProperties.

Themasspropertiesofthewheelarereportedinthepopupbox. Forthispart,themassis15.29

pounds,andthemomentofinertiaaboutthezaxis(labeledasLzzinSolidWorks)is105.36lbin2.

Notethatifyoucenteredthepartabouttheorigin,thentheproperties,labeledTakenatthecenterof

massandalignedwiththeoutputcoordinatesystemwillbeidenticaltothoselabeledPrincipal

moments...takenatthecenterofmass. Notethattheunitsofmassusedareactuallypoundsmass,

thatis,apartthatweighsonepoundhasamassofonepoundmass.Whenwemakeourcalculations

later,wewillhavetoconvertourvaluessothatweuseunitsofmassthatareconsistentwiththeother

unitsthatweareusing.


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ClosetheMassPropertieswindow. OpenthepartWheel2. Fromthemainmenu,selectTools:Mass

Properties.

Notethatalthoughthemassof15.45isalmostthesameasthatofWheel1,Wheel2smomentof

inertiais146.54lbin2,whichisalmost40%greaterthanthatofWheel1. Thereasonforthedifference

isthatmorematerialinWheel2isplacedneartheouterrim. Inthedefinitionofthemomentofinertia

shownasEqn.3,thecontributionofeachparticleofmassonthevalueofdependsonitsdistancefromtheaxissquared. Therefore,addingmassneartheouterrimofthewheelincreasesitsmomentof

inertiagreatly.

Openanewassembly. InsertthecomponentBase.

Sincethefirstcomponentinsertedintoanassemblyisfixed,itislogicaltoinsertthecomponent

representingthestationarycomponent(theframeorgroundcomponent)first.

InsertthepartWheel1intotheassembly. SelecttheMateTool.Addaconcentricmatebetween

thecenterholeofthewheelandtheholeinthebase. Besuretoselectthecylindricalfacesforthe

mateandnotedges.Addacoincidentmatebetweenthebackfaceofthewheelandthefrontfaceof

thebase.

Youshouldnowbeabletoclickanddragthewheel,withrotationabouttheaxisofthematedholesthe

onlymotionallowedbythemates.Theadditionofthesetwomateshasaddedarevolutejointtothe

assembly. Arevolutejointissimilartoahingeinthatitallowsonlyonedegreeoffreedom.

ClickontheMotionStudy1tabnearthelowerleft

corner,whichopenstheMotionManageracrossthe

lowerportionofthescreen.


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TheMotionManagercanbeusedtocreatesimulationsofvariouscomplexities:

Animation

allowsthesimulationofthemotionwhenvirtualmotorsareappliedtodriveoneor

moreofthecomponentsatspecifiedvelocities,

BasicMotionallowstheadditionofgravityandsprings,aswellascontactbetweencomponents,

tothemodel,and

MotionAnalysis(SolidWorksMotion)allowsforthecalculationofvelocities,accelerations,and

forcesforcomponentsduringthemotion. Italsoallowsforforcestobeappliedtothemodel.

ThefirsttwooptionsarealwaysavailableinSolidWorks. SolidWorksMotionisanaddinprogram,and

mustbeactivatedbeforeitcanbeused.

Fromthemainmenu,chooseTools:AddIns. Inthelistof

availableaddins,clickthecheckboxbesideSolidWorksMotion

toactivateit. ClickOK.


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SelectMotionAnalysisfromthesimulation

optionspulldownmenu.

SelecttheForceTool.

Wewillapplyatorque(moment)tothewheel.Wewill

setthetorquetohaveaconstantvalueof5inlb,and

willapplyitforadurationoftwoseconds.

IntheForcePropertyManager,selectTorqueandthenclickonthefrontfaceofthewheel.

Notethatthearrowshowsthatthetorquewillbeappliedinthecounterclockwisedirectionrelativeto

theZaxis(wesaythatthistorquesdirectionis+Z). Thearrowsdirectlybelowthefaceselectionbox

canbeusedtoreversethedirectionofthetorque,ifdesired.

ScrolldownintheForcePropertyManagerandsetthevalueto5inlb.

ScrollbacktothetopofthePropertyManagerandclickthecheck

marktoapplythetorque.


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IntheMotionManager,clickanddragthediamondshapedicon(termedakey)fromthedefaultfive

secondstothedesiredtwoseconds(00:00:02). SelecttheMotionStudyPropertiesTool,andchange

thecalculationratefromthedefaultof25to100framespersecond. Clickthecheckmark.

Usingalargernumberofframespersecondwillresultinsmootherplots,butwillrequiremore

calculationtime.

ClicktheCalculatorIcontoperformthesimulation.

Theanimationofthesimulationcanbeplayedbackwithoutrepeating

thecalculationsbyclickingthePlayfromStartkey. Thespeedofthe

playbackcanbecontrolledfromthepulldownmenubesidethePlay

controls.

Asnotedearlier,SolidWorksMotionprovidesquantitativeanalysis

resultsinadditiontoqualitativeanimationsofmotionmodels.Wewillcreateplotsoftheangular

accelerationandangularvelocityofthewheel.

SelecttheResultsandPlotsTool. InthePropertyManager,usethepulldownmenustoselect

Displacement/Velocity/Acceleration:AngularAcceleration:ZComponent. Clickonthefrontfaceof

thewheel,andclickthecheckmark.


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Aplotwillbecreatedoftheangularacceleration

versustime. Theplotcanbedraggedaroundthe

screenandresized. Itcanalsobeeditedbyright

clickingtheplotentitytobemodified,similartothe

editingofaMicrosoftExcelplot.

Weseethattheaccelerationisaconstantvalue,

about1050degreespersecondsquared. Sincethe

appliedtorqueisconstant,itmakessensethatthe

angularaccelerationisalsoconstant. Wecancheck

thevaluewithhandcalculations. Notethatwhilewe

canperformverycomplexanalyseswithSolidWorksMotion,checkingamodelbyapplyingsimpleloads

ormotionsandcheckingresultsbyhandisgoodpracticeandcanpreventmanyerrors.

Weearlierfoundthemassmomentofinertiatobe105.36lbin2. Sincethepoundisactuallyaunitof

force,notmass,weneedtoconvertweighttomassbydividingbythegravitationalacceleration( . Sinceweareusinginchesasourunitsoflength,wewilluseavalueof386.1in/s2: 105.36 lb in386.1 ins

0.2729 lb in s 4Sincethetorqueisequaltothemassmomentofinertiatimestheangularacceleration,wecanfindthe

angularaccelerationas:

5 in lb0.2729 lb in s 18.323 rads 5

Noticethatthenondimensionalquantityradiansappearsinouranswer. Sincewewantouranswerin

termsofdegrees,wemustmakeonemoreconversion:

18.323 rads 180 deg rad 1050 degs 6

ThisvalueagreeswithourSolidWorksMotionresult.

SelecttheResultsandPlotsTool. InthePropertyManager,usethepulldownmenustoselect

Displacement/Velocity/Acceleration:AngularVelocity:ZComponent. Clickonthefrontfaceofthe

wheel,andclickthecheckmark. Resizeandmovetheplotsothatbothplotscanbeseen,andformat

theplotasdesired.


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Asexpected,sincetheaccelerationisconstant,thevelocityincreaseslinearly. Thevelocityattheendof

twosecondsisseentobeabout2100degreespersecond. Thisresultcanbeverifiedwithasimplehand

calculation:

1050 degs 2 s 2100 degs 7Often,theangularvelocityisexpressedinrevolutionsperminute(rpm),commonlydenotedbythe

symbol:

2100deg

s 1 rev360 deg

60 s1 min 350 rpm 8

Wewillnowexperimentwithvariationsofthesimulation.

Movetheplotsoutoftheway,butdonotclosethem. Clickanddragthekeyatthetopofthe

simulationtreefrom2secondstofour,sothatthesimulationwillnowlastforfourseconds.Placethe

cursoronthelinecorrespondingtotheappliedtorque(Torque1)atthe2secondmark. (Ifdesired,you

canclickthe+andsignsattherightendofthetimelinetoscalethetimeline.) Rightclickandselect

Off.

Anewkeywillbeplacedatthatlocation. Thetorquewillnowbeappliedfortwoseconds,butthe

simulationwillcontinueforthefourseconds.


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PresstheCalculatoricontoperformthesimulation.

Theplotswillbeautomaticallyupdated. Notethattheangularaccelerationnowdropstozeroattwo

seconds,whiletheangularvelocitywillbeconstantaftertwoseconds. Sincethereisnofrictioninthe

model,thewheelwillcontinuetospinataconstantvelocitywithoutanytorqueapplied.

Intheprevioussimulations,thetorquewasappliedasaconstantvalue. Thatmeansthatthechangeof

theaccelerationrelativetotime(commonlyreferredtoasjerk)isinfiniteattime=0andattime=2

seconds. Amorerealisticapproximationistoassumethatthetorquebuildsupoversomeperiodof

time,andalsorampsdowngradually. Forexample,wewillassumethatittakestwosecondstoreach

thefullvalueoftorqueandtwosecondstorampdown.

Rightclickthekeyaddedtothetorqueattime=2secondsanddeleteit. Movethekeydefiningthe

durationofthesimulationtosixseconds.

Movethetimebarbacktozero. RightclickonTorque1andselectEditFeature.

ScrolldowninthePropertyManager,andselect

SegmentsasthetypeofForceFunction.Enterthethree

rowsas

shown

here,

with

Cubic

as

the

Segment

Type.


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Twographsaredisplayed:thetorqueasafunctionoftimeandthe

derivativeofthetorque. Youcanexperimentwithdifferentsegment

typestoseehowtheyaffectthetorque,butthecubiccurvewillwork

fineforthisexample.

ClickOK

and

then

the

check

mark

to

apply

the

torque.

Calculate

the

simulation.

Notethattheangularaccelerationcurveissmooth,andpeaksat1050deg/s2. Attheendofthesix

seconds,thewheelwillbeturningatabout4,200deg/s(700rpm).

NowletsseetheeffectofreplacingtheWheel1componentwithWheel2,whichhasahighermass

momentofinertia. Ofcourse,wecouldstartwithanewassembly,butitiseasiertoreplacethe

componentintheexistingassembly. Thiswillallowustoretainmostoftheassemblymatesand

simulationentities.

Clickthemodeltabatthebottomofthescreen. ClickonWheel1intheFeatureManagertoselectit.

Fromthemainmenu,selectFile:Replace. BrowsetofindWheel2. InthePropertyManager,clickthe

checkmarktoacceptthereplacementoffacesintheexistingmateswiththoseofthenewpart. Click

thecheckmarktomakethereplacement.


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Dependingonhowyoumodeledtheparts,itispossiblethaterrorswillbeencounteredwhenthe

programattemptstoreapplythemates. Ifthishappens,closetheerrormessages,deletebothmates,

andapplynewmatesmanually.

SwitchtotheMotionStudy. Rightclickeachtorque,selectEditFeature,andclickonthefrontfaceof

thewheeltodefinethedirection. Inthesimulationtree,rightclickoneachplot,andclickonthefront

faceofthewheeltodefinethecomponentforwhichvelocity/accelerationistobeplotted. Calculate

thesimulation.


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Notethatthemaximumangularaccelerationisabout755deg/s2,whichissignificantlylessthatofthe

simulationwiththeearlierwheel. Thisvaluecanbeverifiedfromtheratio:

9

105.36146.54 1050 degs 755 degs 10

2. FourBarLinkage

Inthisexercise,wewillmodela4barlinkagesimilartothatofChapter11ofthetext. Inthetext,we

were

able

to

qualitatively

simulate

the

motion

of

the

simulation

when

driven

by

a

constant

speed

motor. Inthisexercise,wewilladdaforceandalsoexploremoreofthequantitativeanalysistools

availablewithSolidWorksMotion.

Downloadorconstructthecomponentsofthelinkageshownonthenextpage,andassemblethemas

detailedinChapter11ofthetext.ThematerialshouldbeAlloySteelforalloftheparts.TheFramelink

shouldbeplacedintheassemblyfirst,sothatitisthefixedlink.

YoushouldbeabletoclickanddragtheCranklinkaroundafull360degreerotation.

NotethattheConnectorlinkhasthreeholes. Themotionofthethirdholecanfollowmanypaths,

dependingonthegeometryofthelinksandthepositionofthehole.


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Beforebeginningthesimulation,wewillsetthelinkstoapreciseorientation. Thiswillallowusto

compareourresultstohandcalculationsmoreeasily.

Addaperpendicularmatebetweenthetwofacesshownhere.

ExpandtheMatesgroupofthe

FeatureManager,andrightclickonthe

perpendicularmatejustadded. Select

Suppress.

Theperpendicularmatealignsthecranklinkatapreciselocation.However,

wewantthecranktobeabletorotate,sowehavesuppressedthemate.Wecouldhavedeletedthe

mate,butifweneedtorealignthecranklater,wecansimplyunsuppressthemateratherthan

recreatingit.


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SwitchtotheFrontView.Zoomoutsothattheviewlookssimilartotheoneshownhere.

TheMotionManagerusesthelastview/zoomofthemodelasthestartingviewforthesimulation.

MakesurethattheSolidWorksMotionaddinisactive. ClicktheMotionManagertab.

SetthetypeofanalysistoMotionAnalysis. SelecttheMotoricon. Inthe

PropertyManager,setthevelocityto60rpm. Clickonthefrontfaceofthe

Cranktoapplythemotor,andclickthecheckmark.


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Clickanddragthesimulationkeyfromthedefault

fivesecondstoonesecond(0:00:01).

Sincewesetthemotorsvelocityto60rpm,aone

secondsimulationwillincludeonefullrevolutionof

theCrank.

ClicktheMotionStudyPropertiesTool. UndertheSolidWorksMotiontab,set

thenumberofframesto100(framespersecond),andclickthecheckmark.

Thissettingwillproduceasmoothsimulation.

ChooseSolidWorksMotionfromthepull

downmenu,andpresstheCalculatoriconto

runthesimulation.

ClicktheResultsandPlotsTools. InthePropertyManager,setthetypeoftheresulttoDisplacement/

Velocity/Acceleration:TracePath. ClickontheedgeoftheopenholeoftheConnector.

Playbackthesimulationtoseetheopenholespathoverthefull

revolutionoftheCrank.

Ifdesired,youcanaddpathsfor

theothertwojointsthatundergo

motion.


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Thefourbarlinkagecanbedesignedtoproduceavarietyofmotionpaths,asillustratedbelow.

Wewillnowaddaforcetotheopenhole.

SelecttheForceTool. InthePropertyManager,thehighlightedboxpromptsyouforthelocationofthe

force. Clickontheedgeoftheopenhole,andtheforcewillbeappliedatthecenterofthehole.

Thedirectionboxisnowhighlighted. Rotateandzoominsothatyoucanselectthetopfaceofthe

Framepart. Theforcewillbeappliednormaltothisforce.Asyoucansee,theforceactsupwards.

Clickthearrowstoreversethedirectionof

theforce.


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ScrolldowninthePropertyManagerandsetthemagnitudeofthe

forceto20pounds. Clickthecheckmarktoapplytheforce.

Runthesimulation.

We

will

now

plot

the

torque

of

the

motor

that

is

required

to

produce

the60rpmmotionwiththe20lbloadapplied.

SelectResultsandPlots. Inthe

PropertyManager,specifyForces:Applied

Torque:ZComponent. Clickonthe

RotaryMotorintheMotionManagertoselect

it,andclickthecheckmarkinthe

PropertyManager.

Format

the

resulting

plot

as

desired.

Notethattheappliedtorquepeaksatabout51inlb. Att=0,thetorqueappearstobeabout 30 inlb

(thenegativesignsindicatesthedirectionisabouttheZaxis,orclockwisewhenviewedfromtheFront

View).

In

order

to

get

a

more

exact

value,

we

can

export

the

numerical

values

to

a

CSV

(comma

separatedvalues)filethatcanbereadinWordorExcel.

RightclickinthegraphandchooseExportCSV. Savethefiletoa

convenientlocation,andopenitinExcel.


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Attime=0,weseethatthemotortorqueis 29.2

inlb.

Handcalculationsforastaticanalysisofthe

mechanismareattached,whichshowavalueof

29.4inlb.

Itisimportantwhencomparingthesevaluesto

recognizetheassumptionsthatarepresentinthe

handcalculations:

1. Theweightsofthememberswerenot

includedintheforces,and

2. Theaccelerationsofthememberswereneglected.

The

first

assumption

is

common

in

machine

design,

as

the

weights

of

the

members

are

usually

small

in

comparisontotheappliedloads. Incivilengineering,thisisusuallynotthecase,astheweightsof

structuressuchasbuildingandbridgesareoftengreaterthantheappliedforces.

Thesecondassumptionwillbevalidonlyiftheaccelerationsarerelativelylow. Inourcase,theangular

velocityofthecrank(60rpm,oronerevolutionpersecond)producesaccelerationsinthemembersthat

aresmallenoughtobeignored. Letsaddgravitytothesimulationtoseeitseffect.

ClickontheGravityicon. InthePropertyManager,selectYasthedirection. Therewillbeanarrow

pointingdowninthelowerrightcornerofthegraphicsarea,showingthatthedirectioniscorrect.

Click

the

check

mark.

Run

the

simulation.

Thetorqueplotisalmostunchanged,withthepeak

torqueincreasingbyonlyoneinlb. Therefore,omittinggravityhadverylittleeffectonthecalculations.

Nowwewillincreasethevelocityofthemotortoseetheeffectonthetorque.


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Dragthekeyattheendofthetopbarinthe

MotionManagerfrom1secondto0.1second. Usethe

ZoomInToolinthelowerrightcornerofthe

MotionManagertospreadoutthetime

line,ifdesired.

Dragthesliderbarshowingthetimewithinthesimulation

backtozero.

Thisisanimportantstepbeforeeditingexistingmodel

items,aschangescanbeappliedatdifferenttimesteps.

Because

we

want

the

motors

speed

to

be

changed

from

thebeginningofthesimulation,itisimportanttosetthe

simulationtimeatzero.

RightclickontheRotaryMotorintheMotionManager. Inthe

PropertyManager,setthespeedto600rpm. Clickthecheckmark.

Sinceafullrevolutionwilloccurinonly0.1seconds,weneedtoincrease

theframerateofthesimulationtoachieveasmoothplot.

SelecttheMotionStudyProperties. InthePropertyManager,settheSolidWorksMotionframerateto

1000frames/second. Clickthecheckmark.

Runthesimulation. (ClickNoifyoureceiveamessageaskingifyouwant

toincreasethesimulationtime.)


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Thepeaktorquehasincreasedfrom51to180inlb,

demonstratingthatasthespeedisincreased,the

accelerationsofthemembersarethecriticalfactors

affectingthetorque.

Youcanverifythisconclusionfurtherbysuppressing

bothgravityandtheapplied20lbloadandrepeating

thesimulation. Thepeaktorqueisdecreasedonly

from180to151inlb,evenwithnoexternalloads

applied.

Toperformhandcalculationswiththeaccelerationsincluded,itisnecessarytofirstperformakinematic

analysistodeterminethetranslationalandangularaccelerationsofthemembers.Youcanthendraw

freebodydiagramsofthethreemovingmembersandapplythreeequationsofmotiontoeach:

F ma F ma M ITheresultisnineequationsthatmustbesolvedsimultaneouslytofindthenineunknownquantities(the

appliedtorqueandthetwocomponentsofforceateachofthefourpinjoints).

Theresultsapplytoonlyasinglepointintime. Thisisamajoradvantageofusingasimulationprogram

suchasSolidWorksMotion:sinceitisnotevidentatwhatpointinthemotionthattheforcesare

maximized,ouranalysisevaluatestheforcesoverthecompleterangeofthemechanismsmotionand

allowsustoidentifythecriticalconfiguration.


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3. RolleronaRamp

Inthisexercise,wewilladdcontactbetweentwobodies,andexperimentwithfrictionbetweenthe

bodies.Wewillbeginbycreatingtwonewpartsarampandaroller(skipthesestepsifyouhave

downloadedtheparts).

Openanewpart. IntheFrontPlane,sketchanddimensionthetriangleshownhere.

Extrudethetriangleusingthemidplaneoption,withathicknessof1.2inches.

IntheTopPlane,usingtheCorner

RectangleTool,drawarectangle.

Addamidpointrelationbetweenthe

leftedgeoftherectangleandthe

origin.Addthetwodimensions

shown,andextrudetherectangle

down0.5inches.

Modifythematerial/appearanceasdesired(shownhereas

Pine). SavethispartwiththenameRamp.

Openanewpart. Sketchanddimensiona

oneinchdiametercircleintheFrontPlane.

Extrudethecirclewiththemidplaneoption,

toatotalthicknessofoneinch. Setthe

materialofthepartasPVCRigid.Modify

thecolorofthepartasdesired(overriding

thedefaultcolorofthematerialselected).


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Openanewsketchonthefrontfaceofthecylinder.

Addanddimensionthecirclesandlinesasshownhere

(thepartisshowninwireframemodeforclarity). The

twodiagonallinesaresymmetricaboutthevertical

centerline.

ExtrudeacutwiththeThroughAlloption,withthesketchcontoursshownselected. Ifdesired,change

thecolorofthecutfeature.

Createacircularpatternoftheextrudedcutfeatures,withfive

equallyspacedcuts. SavethepartwiththenameRoller.

Openanewassembly. Inserttherampfirst,andplaceitattheoriginoftheassembly. Insertthe

Roller.


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Addtwomatesbetweentherampandtheroller.MatetheRightPlanesofbothparts,andadda

tangentmatebetweenthecylindricalsurfaceoftherollerandthesurfaceoftheramp.

Thebestwaytosetthecorrectheightoftherollerontherampisto

addamatedefiningthepositionoftheaxisoftheroller.

FromtheHeadsUpViewToolbar,selectView:TemporaryAxes.

Thiscommandturnsonthedisplayofaxesthatareassociatedwith

cylindricalfeatures.

Addadistancematebetweentherollersaxisandtheflat

surfaceatthebottomoftheramp. Setthedistanceas6.5

inches.

Sincetheradiusoftherolleris0.5

inches,theaxiswillbe0.5inches

abovetheflatsurfacewhenthe

surfaceoftherollercontactsthat

surface. Therefore,thevertical

distancetraveledbytherollerwill

be6.0inches. Also,notethatthe

distancetravelleddowntheramp

willbe12inches(6inchesdivided

bythesineoftherampangle,30

degrees).


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Turnoffthetemporaryaxisdisplay. SwitchtotheMotionStudy. SelectSolidWorksSimulationasthe

typeofanalysis.Addgravityinthe ydirection.

SelecttheContactTool. InthePropertyManager,youwillbepromptedto

selectthebodiesforwhichcontactcanoccur. Clickoneachofthetwo

parts.

Clear

the

check

boxes

labeled

Material

and

Friction.

Leave

the

otherpropertiesastheirdefaults.

Wewilladdfrictionlater,butourinitialsimulationwillbeeasiertoverifywithoutfriction. IntheElastic

Propertiessection,notethatthedefaultissetasImpact,withseveralotherproperties(stiffness,

exponent,etc.)specified. Ateachtimestep,theprogramwillcheckforinterferencebetweenthe

selectedbodies. Ifthereisinterference,thenthespecifiedparametersdefineanonlinearspringthat

actstopushthebodiesapart. Contactsaddconsiderablecomplexitytoasimulation. Ifthetimesteps

aretoolarge,thenthecontactmaynotberecognizedandthebodieswillbeallowedtopassthrough

eachother,oranumericalerrormayresult.

SelecttheMotionStudiesPropertyTool. Settheframerateto500andchecktheboxlabeledUse

PreciseContact. Clickthecheckmark.

Forsomesimulations,itmaybenecessarytolowerthesolutiontoleranceinordertogetthesimulation

torun. Forthisexample,thedefaulttoleranceshouldbefine.

Thematesthatweaddedbetweenthepartstopreciselylocatetherollerontherampwillprevent

motionoftheroller. Ratherthandeletethesemates,wecansuppressthemintheMotionManager.

RightclickoneachofthematesintheMotionManagerandselectSuppress. Runthesimulation.

Youwillseethattherollerreachesthebottomoftherampquickly.

Changethedurationofthesimulationto0.5seconds,andrunthesimulationagain.Createaplotof

themagnitudeofthelinearvelocityoftherollervs.time.

Therollerreachesthebottomoftherampinabout0.35seconds,andthevelocityatthebottomofthe

rampisabout68in/s.Thesevaluesagreewiththosecalculatedintheattachmentattheendofthis

document.


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Nowletsaddfriction.

Movethetimelineofthesimulationbacktozero.RightclickonthecontactintheMotionManager

tree,andselectEditFeature. ChecktheFrictionbox,andsetthecoefficientoffrictionto0.25.

Calculatethesimulation.

Theresultingvelocityplotshowsthevelocityatthebottomoftheramptobeabout54in/s. Thisvalue

agreeswiththatofthecalculationsshownintheattachment.

Toconfirmthattherollerisnotslipping,wecantrace

thepositionofasinglepointontheroller.

SelecttheResultsandPlotsTool. Definetheplotas

Displacement/Velocity/Acceleration:TracePath.

Clickonapointneartheouterrimoftheroller(not

onaface,butonasinglepoint). Clickthecheck

mark.

Thetracepathshowssharpcuspswherethepointsvelocityapproacheszero(itwillnot

becomeexactlyzerounlessthepointisonthe

outersurfaceoftheroller). Forcomparison,

repeattheanalysiswithalowerfriction

coefficient.


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Changethefrictioncoefficientto0.15andrecalculatethesimulation.

Thistime,thetracepathsshowssmoothcurves

whenthepointisneartherampssurface,

indicating

that

sliding

and

rolling

are

taking

placesimultaneously.

Intheattachment,itisshownthatthe

coefficientoffrictionrequiretopreventslipping

isabout0.21.

Itisinterestingtonotethatthefrictioncoefficienttopreventslippingandthetimerequiredtoreachthe

bottomoftheramparebothfunctionsoftheratioofthemomentofinertiatothemassoftheroller. (A

parametercalledtheradiusofgyrationisdefinedasthesquarerootofthemassmomentofinertia

dividedbythemass,andisafunctiononlyofthepartsgeometry.) Youcanconfirmthisbychangingthe

materialoftherollerandseeingthattheresultsofthesimulationareunchanged. However,ifyou

changethegeometryoftheroller(theeasiestwayisbysuppressingthecutoutregions),thenthe

resultswillchange.


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ATTACHMENT:VERIFICATIONCALCULATIONS

STATICANALYSISOFFOURBARLINKAGESUBJECTEDTO20LBAPPLIEDFORCE

FreebodydiagramofConnector:

NotethatmemberCDisa2forcemember,andsotheforceattheendisalignedalongthemembers

axis.

Applyequilibriumequations:

M5.714 in sin75.09 1.832 in cos75.09 11.427 in20 lb 05.522 in 0.4714 in 228.5 in lb

5.993 in 228.5 in lb

228.5 in lb5.993 in 38.13 lb

F 38.13 lbcos75.09 0 9.812 lb


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F 38.13 lbsin75.09 20 lb 0 16.85 lb

FreebodydiagramofCrank:

NotethatBxandByareshowninoppositedirectionsasinConnectorFBD.

SummomentsaboutA:

M 3 in 0 3 in9.812 lb 29.4 in lb


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ROLLERCALCULATIONS

NoFriction:

Freebodydiagram:

F W sin mF N W cos 0

Whereistherampangle(30degrees)Sincetheweightisequalthemass

mtimesthegravitationalacceleration

g,theaccelerationinthe

x

directionxwillbe: x g sin Theaccelerationisintegratedwithrespecttotimetofindthevelocityinthexdirection:

x g sin d g sin vxoWherevxoistheinitialvelocityinthexdirection. Thevelocityisintegratedtofindthedistancetravelledinthexdirection:

g sin v d g2 sin v xWherex0istheinitialposition. Ifwemeasurexfromthestartingposition,thenx0iszero. Iftheblockisinitiallyatrest,thenvxoisalsozero. Inoursimulation,theblockwillslideadistanceof12inchesbeforecontactingthebottomoftheramp(seethefigureonpage23).

Knowingthedistancetravelledinthexdirection,andenteringthenumericalvaluesofgas386.1in/s2andofsinof0.5(sinof30o),wecanfindthetimeittakestheblocktoslidetothebottom:


30/32


12 in 386.1in s2 0.5or 0.353 sSubstituting

this

value

into

Equation

4,

we

find

the

velocity

at

the

bottom

of

the

ramp:

x 386.1 ins2 0.50.353 s 68.1 ins Thisvelocitycanalsobefoundbyequatingthepotentialenergywhentherollerisatthetopoftheramp

(heightabovethedatumequals6inches)tothekineticenergywhentherollerisatthebottomofthe

ramp:

12 2

2 2386.1 ins6 in 68.1 ins FrictionIncluded:

FreeBodyDiagram:

Whiletherollerwithoutfrictionslidesandcanbetreatedasaparticle,therollerwithfriction

experiencesrigidbodyrotation. Theequationsofequilibriumare:

F W sin mF N W cos 0

M r I


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Ifthereisnoslipping,thentherelativevelocityoftheroller

relativetotherampiszeroatthepointwherethetwo

bodiesareincontact(pointO). Sincetherampis

stationary,thisleadstotheobservationthatthevelocityof

pointOisalsozero.

SincepointOisthecenterofrotationoftheroller,the

tangentialaccelerationofthecenteroftherollerxcanbewrittenas: x rSubstitutingthisexpressionintothefirstequilibriumequationandsolvingforthefrictionforce,

W sin mrSubstitutingthisexpressionintothethirdequilibriumequationandsolvingfortheangularacceleration

, W sin mr r IoWsinrI mr

WsinrI mr Themassandthemomentofintertia

IcanbeobtainedfromSolidWorks. Fortheroller,thevaluesare:

m 0.017163 lbI 0.002515 lb in

Sincepoundsareunitsofweight,notmass,theyquantitiesabovemustbedividedbygtoobtainthe

quantitiesinconsistentunits:

m 0. 017163 lb

386.1 in/s 4.4453 X 10 lb s

in

I0.002515 lb in386.1 in/s 6.5139 X 10 lb in s

Thevalueoftheangularaccelerationcannowbefound:

WsinrI mr 0. 017163 lb sin 300.5 in

6.5139e 6 lb in s 4.4453 5 lb s

in 0.5 in 243.4

rads


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Therefore,thelinearaccelerationinthexdirectionis:x r 0.5 in 243.4 rads2 121.7 ins2

Integrating

to

obtain

the

velocity

and

position

at

any

time:

x x d x vxo 121.7 ins2 v d 12 v x60.85 ins

Fortherollertotravel12inchesinthedirection,thetimerequiredis

12 in60.85 in s 0.444 sAndthevelocityatthebottomoftherampis:

x 121.7 ins2 0.444 s 54.0 ins Wecanalsocalculatethefrictionforce:

W sin mr 0. 017163 lbsin 30 4.4453 e 5 lb sin 0.5 243.4 rads 0.00318 lbFromthesecondequilibriumequation,thenormalforceis:

N Wcos 0.017163 lbcos30 0.1486 lbSincethemaximumfrictionforceisthecoefficientoffrictiontimesthenormalforce,thecoefficientoffrictionmustbeatleast:

0.00318 lb0.1486 lb 0.21Thisistheminimumcoefficientoffrictionrequiredfortherollertorollwithoutslipping.

Documents

SolidWorks Motion Tutorial 2013