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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:
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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.