ME 270 – Fall 2019 Final exam review material

TOPIC Fall14 Fall15 Fall16 Fall17 Spring14 Spring15 Spring16 Spring17 Spring181 vectoroperations2 equilibriumofparticles 1a 1a 1b 1a 1a3 equilibriumofrigidbodies 3,5a 2,5a 2,4a 1b,5a 1a,2 1a,2,4a 1b,2,5a 1b,2,5a 5a4 equivalentsystems 2c 3a 5a5 centroids 1c 4d 3a6 hydrostaticloads 2a 1d 3d 1c7 friction 2b 1c 3c 2 1d,3 1d 1c 1c 2a,2b8 trusses 4 1b 1c 1b 1d,3a,3b 3 1a9 framesandmachines 3 1 1d 1c 3 3c 1d 1b10 SF/BMdiagrams 5b 4a,5b 4b 5b 5b 4b 4a,5b 4a,5b 3b,5b

axialstress 1f 3c 3e 3eshearstress-directshear 1d 1h 3d 4c 5a 3c 1cshearstress-torsion 1b 4b 3e 4 4d 5b 4b 4b 3c,4flexuralstress-bending 5c,5d 5c,5d,5f 4c,4e 5e 5d 4c,4d,4e 5c,5d 5c,5d 3d,5c

12 2ndareamoments 4c,4d 3b,4e 3a,3b,5d 4b,5c 5c,5d 4c,4d 4c,4d 1d,3aOther 1a 4a

ME270:TopicsofQuestionsfromPastFinalExams

11

1.VECTOROPERATIONSForcesasvectors

!F = FuAB = F

!rAB!rAB

where !rAB pointsFROMpointATOpointB.

Anglesbetweenvectors

cosθ =

!a i!b!a!b

Equivalently,youcoulduseacrossproducttofindtheangle;however,thatisnotrecommendedascrossproductsrequiremorecalculations,ingeneral.CalculationsofmomentsFor3Dforces,ALWAYSusethecrossproducttofindmoments:

!MO = !rOA ×

!FA

where !rOA pointsFROMpointOTOpointA

For2Dforces,useeitherthecrossproductapproachaboveorusethemomentarmapproach:

!

MO = d!FA

withdirectionfoundbytheright-handrule.3Dprojections

!FCD =

!F i uCD( )uCD =projectionof

!F ontolineCD

A

B

!F

!rAB

!a

!b

A

!FA

!rOA

O

d

D !F

C

!F i uCD

uCD

A

B

!F

!rAB

!a

!b

A

!FA

!rOA

O

d

D !F

C

!F i uCD

uCD

A

B

!F

!rAB

!a

!b

A

!FA

!rOA

O

d

D !F

C

!F i uCD

uCD

A

B

!F

!rAB

!a

!b

A

!FA

!rOA

O

d

D !F

C

!F i uCD

uCD

equivalent

SystemI SystemII

y

x

y

x

w( x ) F d

Bequivalent

SystemI SystemII

B

F

M

A A

L L

2.EQUILIBRIUMOFPARTICLESForeachFBD:

!F∑ =!0 (givestwoequationsin2D,andthreeequationsin3D)

3.EQUILIBRIUMOFRIGIDBODIESForeachFBD:

!F∑ =!0 (givestwoequationsin2D,andthreeequationsin3D)

!

M A∑ =!0 (givestwoequationsin2D,andthreeequationsin3D)

4.EQUIVALENTSYSTEMS–GivenloadinginI.FindequivalentloadFandcoupleMatgivenpointA.

!F∑( )I

=!F∑( )II

!

M B∑( )I=

!M B∑( )II

forANYpointBofyourchoice.ThisconceptisNOTrelatedtoequilibrium!Lineloads–Givenw(x).FindequivalentforceFandlocationd.

Fy∑( )II= Fy∑( )I

⇒ F = w x( )dx0

L

∫

M A∑( )II= M A∑( )I

⇒ Fd = xw x( )dx0

L

∫ ⇒ d = xw x( )dx0

L

∫⎛

⎝⎜⎜

⎞

⎠⎟⎟

/ F

equivalent

SystemI SystemII

y

x

y

x

w( x ) F d

Bequivalent

SystemI SystemII

B

F

M

A A

Lech

a

b

x

y

x2 y( )

x1 y( )

dy

12

x2 + x1( )

y

a

b

x

y

y2 x( )

y1 x( )

dx

12

y1 + y2( )

x

a

b

x

y

x2 y( )

x1 y( )

dy

12

x2 + x1( )

y

a

b

x

y

y2 x( )

y1 x( )

dx

12

y1 + y2( )

x

5.CENTROIDSCentroids–compositeparts A = Ai∑

xC = 1

Axi Ai∑

yC = 1

Ayi Ai∑

Thecoordinatesforallcomponentsshouldbereferencedtothesameorigin.Centroids–byintegrationUsingdifferentialareaparalleltoy-axis:

dA = y1 x( )− y2 x( )⎡⎣ ⎤⎦dx ⇒ A = y1 x( )− y2 x( )⎡⎣ ⎤⎦dx

0

a

∫

xC = 1

Ax dA∫ = 1

Ax y1 x( )− y2 x( )⎡⎣ ⎤⎦dx

0

a

∫

yC = 1A

y dA∫ = 1A

12

y1 x( ) + y2 x( )⎡⎣ ⎤⎦ y1 x( )− y2 x( )⎡⎣ ⎤⎦dx0

a

∫

= 12A

y12 x( )− y2

2 x( )⎡⎣

⎤⎦dx

0

a

∫

Alternately,usingdifferentialareaparalleltox-axis:

dA = x2 y( )− x1 y( )⎡⎣ ⎤⎦dy ⇒ A = x2 y( )− x1 y( )⎡⎣ ⎤⎦dy

0

b

∫

xC = 1A

x dA∫ = 1A

12

x2 y( ) + x1 y( )⎡⎣ ⎤⎦ x2 y( )− x1 y( )⎡⎣ ⎤⎦dy0

b

∫

= 12A

x22 y( )− x1

2 y( )⎡⎣

⎤⎦dy

0

b

∫

yC = 1

Ay dA∫ = 1

Ay x1 y( )− x2 y( )⎡⎣ ⎤⎦dy

0

b

∫

LeoB

lect

6.HYDROSTATICLOADSBuoyancyThebuoyancyforceonanobjectinafluidisequaltotheweightofthefluiddisplacedbythebodyinthefluid: FB = ρgV whereVisthevolumeoffluiddisplacedand ρ isthedensityofthefluid.NotethatifthemaximumpossiblebuoyancyforceisatleastaslargeastheweightWofthebody,thenthebodywillfloatonitsown;thatis,forabodytofloat: FB,max ≥W .HydrostaticpressureandloadsThehydrostaticpressureatagivendepthhinafluidisgivenby: p = ρgh where ρ isthedensityofthefluid.

• Thehydrostaticloadonaflat,verticalsurface(havingadepthvalueofbintothepage)canbedescribedbyalinearlyvaryinglineloadof:

w h( ) = ρghb

• Foraninclinedflatsurfaceoranycurvedsurface,drawanFBDofsurfacealongwiththefluidabove(orbelow)thesurface.Forthis,theloadsonthisFBDduetothefluidinclude:thehydrostaticloadonallsurfacesandtheweightWofthefluidincludedintheFBD(wheretheweightactsatthecentroidCofthefluidincludedintheFBD).

W

FB

fluidofdensityρ

free surface

ρgbH

W

C

H

h

free surface

h

ρgbH

W

C

H

Hw(h)

free surface

h

fluid

ρgbHG

7.FRICTIONFriction:general

• Fornoslidingoftheblockonthesurface: f ≤ µS N • Thefrictionforceisequalto f = µS N ONLYforimpendingslipping.

• Ifyouuse f = µS N ,youMUSTknowthedirectionofthefrictionforce(opposesimpendingmotion,IM)anddrawitappropriatelyinyourFBDs.Noexceptions!

Tippingvs.slipping

• ThenormalforceactingontheblockisNOTlocateddirectlyunderthecenterofmassoftheblock.

• Ingeneral, f ≠ µS N .• Fortippingvs.slippingproblemsolving:

i. Determinethedirectionofimpendingslippingmotion(IM)oftheblock.UseacarefulthoughtprocesstodeterminethedirectionofIMinaproblem.DrawthefrictionforceftoopposetheIM.

ii. Determinethecorneraboutwhichimpendingtippingwilloccur.LocatethenormalforceNatadistanceofdfromthatcorner.

iii. CompleteyourFBD.iv. Writedownthegeneralsetofequilibrium

equations.v. Assumetippingfortheblockbysettingd=0(but

NOT f = µS N ).Solvefortheloadingrequiredtocausetipping, Ptip .

vi. Assumeslippingfortheblockbysetting f = µS N (butNOTd=0).Solvefortheloadingrequiredtocauseslipping, Pslip .

vii. Choosetheappropriateloading Ptip or Pslip .Thiscanbethemostdifficultstepinthesolution.Carefullyreadthequestion.Lookforcluesinthewordingsuchas“maxorminvaluesofloadingtocreatemotion”,orthe“maxorminvaluesofloadingtopreventmotion”.

g

P

G

roughsurface

P

Gcornerofimpendingtipping

f

N

impendingmotion(IM)

d

P

N f

Pimpendingmotion(IM)

BeltfrictionFundamentalequation:

T1T2

= eµβ

Forthisequation:• YouneedtoknowwhichofthetensionforcesisthelargestBEFOREsolvingtheproblem.Inordertousetheabovebeltfrictionequation, T1 > T2 .

• TheaboveequationisvalidforONLYimpendingslipping(with µ = µS )andslipping(with µ = µk ).Itisnotvalidforanyothersituation.

WedgeswithfrictionInsolvingproblemswithwedges:

• YoumustknowthedirectionofimpendingslippingBEFOREdrawingyourFBDs.InyourFBDs,drawthefrictionforcestoopposethedirectionofimpendingmotion.

• Youcanuse f = µS N onlyonsurfacesforwhichimpendingslippingwilloccur.Formanysituations,youdonotknowthisinadvance.Tosolve,youmayneedtoassumeslippingatasurface,andthecheckyourassumptionaftersolvingtheproblem.

• Solvingwedgeproblemswithfrictioncanbecomeatediousprocessduetothenumberofsurfacesonwhichfrictionoccurs.Makeagoodplaninsolvingthesealgebraicequations.

directionofimpendingoractualmotion

C D

T1 T2

T1 > T2

note idealpulley is frictionless

8.TRUSSESItisgenerallyrecommendedthatyousolveforallreactionsonthetrussstructurebeforeattemptingtosolveforinternalloads.• Methodofjoints

Withthismethod:o DrawFBDsofindividualjointsinthetruss.o Itisrecommendedthatyoudrawallreactionsforcesonthejointsdueto

trussmembersusingthesameassumption(tensionorcompression).Forexample,ifyouchoosetodrawtheminTENSION,thenthememberforcespointOUTWARDfromthejoint.Ifyoufindanegativesolutionforanymemberforce,thenthatsaysthatyouroriginalassumptionwasincorrect.(Again,ifyouassumedamemberwasintension,findinganegativevaluefortheloadsaysthatthememberisincompression.)

o FromeachFBD,youcanwritedowntwoequilibriumequations.o Thismethodcanbecometedious,especiallyfortrusseswithalargenumber

ofjoints.

• MethodofsectionsWiththismethod:o Carefullyconsiderwhichmemberforcesforwhichyouwanttoknow

memberloads.o Makemathematicalcutsthroughthestructureexposingnomorethanthree

unknownmemberloads,generallythememberloadsthatyoudesiretoknow.

o FromeachFBD,youcanwritedownthreeequilibriumequations.o Itisrecommendedthatyoudrawallmemberforcesatthecutsusingthe

sameassumption(tensionorcompression).Forexample,ifyouchoosetodrawtheminTENSION,thenthememberforcespointOUTWARDatthecut.Ifyoufindanegativesolutionforanymemberforce,thenthatsaysthatyouroriginalassumptionwasincorrect.(Again,ifyouassumedamemberwasintension,findinganegativevaluefortheloadsaysthatthememberisincompression.)

o FromeachFBD,youcanwritedownthreeequilibriumequations.

o Thisisapowerfulmethodtousewhenyouneedtoknowonlyalimitednumberofmemberloads

• Zero-forcemembersBeforestartingtheanalysisofatruss,lookformembersthatcarrynoload.Todetermineifamemberiscarryingnoload,lookatthefreebodydiagramofajointtowhichthememberisattached.Ifallotherloadsatthejointarecollinear,thenthatmemberloadiszero.Recognizingzero-forcemembersishelpfulwhendecidingoncutstomakewiththemethodofsections.

F1

F2

F3

F3 =0

F1F2

F1 = F2 =0

9.FRAMESANDMACHINES• Generallystartouttheanalysisofframesandmachinesbyfirstdetermining

theexternalreactions.• Next,drawafreebodydiagramofeachmemberintheframe/machine.• NotethatmostmembersinframesandmachinesareNOTtwo-force

members.Asaresult,theloadingsatjointsduetoanontwo-forcememberareNOTalignedwiththemember;forthese,drawthereactionsashavingbothx-andy-components,withpossiblycouples.

• Makeaplanforyoursolution.Avoidsolvingforinternalreactionsthatyoudonotneed.Chooseyourmomentequationsaboutpointsthatallowyoutosolveoneequationforoneunknown,wheneverpossible.

10.SHEARFORCEANDBENDINGMOMENTDIAGRAMSRecall that from equilibrium relations we have for a line load p x( ) on a beam:

p x( ) = dVdx

⇒ V x2( ) =V x1( ) + p x( )dxx1

x2

∫ = shear force at x = x2

V x( ) = dMdx

⇒ M x2( ) = M x1( ) + V x( )dxx1

x2

∫ = bending moment at x = x2

and for a concentrated force PA at x = xA and a concentrated moment M B at

x = xB :

V xA

+( ) =V xA−( ) + PA

M xB

+( ) = M xB−( )− M B (note the “-“ sign in this expression)

Notes:

• V and M are areas under the p(x) and V(x) curves, respectively. Whenever possible1, find these areas by “inspection” to save effort and time.

• Be very aware of the sign conventions established in class for V and M when constructing your shear force and bending moment diagrams.

• Use the boundary conditions at the right end to CHECK your values of V and M at that end.

• From above, recall that if p(x) has a zero crossing, the shear force V will have either a maximum or minimum at that point.

• Also from above, recall that if V(x) has a zero crossing, the bending moment M will have either a maximum or minimum at that point.

• For non-linear curves for either V or M, you need to be able to establish whether these curves are convex or concave. Do so by considering if the slope is increasing or decreasing with x.

• It is strongly advised to always work from left to right (in the positive x-direction) when constructing your V and M curves.

1When the integrand is either zero, constant or linearly varying, we can easily find the areas by inspection. If the integrand is quadratic or higher order, this is generally not possible. In that case, you need to resort to formal integration over such a segment.

Between

2ends

f distribution

At theiscontinuity

11.STRESSESANDSTRESSDISTRIBUTIONS

Stressesduetodirectshear,axial,torsionalandbendingloads

Proper6es:•  shearstress,,whereAistheareaoverwhichtheshearforceVacts

•  approximatelyconstantovercross-sec5on

τ =V /A

directshearloads

P

τt

d

A = πtd

P

V

d

A = π d / 2( )2punching shearinpin

Examples:•  Punchingacircularholew/forceP:

•  Asingle-sidedshearforceonpin:

•  Adouble-sidedshearforceonpin:

τ = P

πtd

τ = P

πt d2 /4( )

τ = P /2

πt d2 /4( )

Stressesinstructuralmembersduetoaxial,torsionalandbendingloads

Proper6es:•  normal(axial)stress,,whereAisthecross-sec5onalarea•  constantovercross-sec5on

σ = F /A

Example:•  FBDofeachconnector•  Solveforaxialloadineachmemberusingequilibrium•  Determinetheaxialstressineachmemberwithaboveequa5on

4F F

F

σF

axialloads

Proper6es:•  shearstress,,whereJisthepolarareamoment

•  linearwithradialposi5onovercross-sec5on

τ = Tρ / J

Example:•  FBDofeachconnector•  Solvefortorqueineachmemberusingequilibrium

•  Determinetheshearstressineachmemberwithaboveequa5on

4T T

T

T

τ

ρ

Stressesinstructuralmembersduetoaxial,torsionalandbendingloads

torsionalloads

me270-cmk

Proper6es:•  normal(axial)stress,,whereyismeasuredfromcentroidandIisthe2ndareamoment•  linearwithdepthdimensionofbeam

σ = −My / I

M

σ

M

M

y

neutral

Example:•  Determinereac5ons(equilibrium)•  Drawshearforce(V)andbendingmoment(M)diagrams(equilibrium)

•  Determinethenormalstresswiththeaboveequa5onforwhereV=0

P

w0

bendingloads

surface

Fs OYLGallow02

FS Gallow

12.Calculationofsecondareamoments

Definition–secondareamoment:Thesecondareamomentofabeamcrosssectionaboutthez-axiswithCbeingthecentroidofthecrosssection(apointontheneutralplaneofthebeam)isdefinedas:

IC = y2 dA

A∫

Parallelaxistheorem:ThesecondareamomentsaboutthecentroidCandanarbitrarypointBontheplaneofsymmetryarerelatedthroughtheparallelaxistheorem: IB = IC + Ad2 wheredisthedistancebetweenCandB.IMPORTANTNOTE:ThesecondareamomentaboutthecentroidCmustappearontheright-handsideofthisequation.AwaytorememberthisisthatthecentroidalsecondareamomenthasitssmallestvalueaboutCascomparedtoanyotherpointonthecrosssection.Calculationofthesecondareamoment–byintegrationUsingthedifferentialareashownabove, dA = b Y( )dY ,thesecondareamomentaboutpointOisfoundfrom:

IO = Y 2 dA

A∫ = Y 2b Y( )dY

0

H

∫

Forthecentroidalsecondareamoment,usetheparallelaxistheorem: IO = IC + AYC

2 ⇒ IC = IO − AYC2

C

se#ngupintegral

dYb Y( )

dA

YC

Z

Y

Hd

C

B

Yz

Z

centroid

arbitrary point

plane of symmetry

y

beamcrosssec2on

O

C

usingcompositeparts

C2

C1section1

section 2

ZO

Y

dC2

dC1

Calculationofthesecondareamoment–bycompositepartsConsiderdividingthecrosssectionshownaboveintotwosections:1and2.Thentofindthesecondareamomentofthecrosssection:

1. Calculate(orlookupinatable)thecentroidalsecondareamomentforeachsection,

IC1

and IC2

,alongwiththecrosssectionalareas, A1 and A2 .2. DeterminethelocationofthecentroidCofthecompositearea;thatis,find

thedistances dC1 and dC2 fromCtothesectionalcentroids C1 and C2 ,respectively.

3. Usetheparallelaxistheoremtodeterminethecontributionsofthetwosectionstothecentroidalsecondareamoment:

IC( )1 = IC1+ A1dC1

2

IC( )2 = IC2+ A2dC2

2

4. Addtogetherthecontributionstothecentroidalsecondareamoment:

IC = IC( )1 + IC( )2

IMPORTANTNOTE:Thecontributionstothesecondareamomentfromthedifferentsectionsmustbereferencedtothesamepoint(here,C)BEFOREaddingtogetherinStep4.