材料力学例题及Examples of mechanics of materials and.docx

材料力学例题及Examples of mechanics of materials and.docx

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材料力学例题及Examplesofmechanicsofmaterialsand

材料力学例题及（Examplesofmechanicsofmaterialsand）

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Examplesofmechanicsofmaterialsandguidanceofproblemsolving

（secondtosixthchapters）

Thesecondchapter,stretching,compressionandshear

Inexample2-1,trytofigureouttheaxisofthestraightrodoffigurea

Solution:

thestraightrodissubjectedtoaxialexternalforceatA,B,CandDpoints

ABaxialforceisfirstsought

Thememberiscutoffatany1-1sectionofthesection

Examineleftsegment（Figure2-5b）

ApositiveaxialforceissetonthecrosssectionN1

ThustheequilibriumequationSX=0isobtained

N16=0

N1=6kN

N1wasoriginallyassumedthataforceiscorrect

Italsoshowsthattheaxialforceispositive

TheaxialforceofanysectionintheABsectionisequalto+6kN

BCaxialforceisagainsought

Thememberiscutoffatany2-2sectionoftheBCsection

Theleftsegmentisstillexamined（Figure2-5c）

TheaxialforceN2isstillsetonthecrosssection

BySX=0

618N2=0

N2=12kN

N2minusthatoriginallyassumedtensioniswrong（forpressure）

ItalsoshowsthattheaxialforceN2isnegative

TheaxialforceofanysectionintheBCsectionisequalto-12kN

Similarly,theaxialforceofanysectionintheCDsectionis-4kN

Drawinternalforcechart

ThehorizontalaxisXrepresentsthesectionpositionoftherod

RepresentstheaxialforceofthecrosssectionwiththeverticalaxisofX

Drawanaxlebarattheselectedscale

AsshowninFigure2-5（d）

Fromthis,itisshownthatthemaximumaxialforceoccursintheBCsegment

Problemsolvinginstruction:

whentheaxialforceiscalculatedbythecrosssectionmethod,thepositiveaxialforceisalwayssetonthesectionoftheincisionN

ThentheaxialforceNisobtainedbySX=0

IftheNiscorrect,thepositiveaxialforce（tension）isindicated

Ifnegative,itmeansnegativeaxialforce（pressure）

Example2-2testthefreesuspensionofthestraightbar（Figure2-6a）bylongitudinaluniformdistributionofloadQ（force/length）causedbystressandlongitudinaldeformation

ThebarlengthL,sectionalareaAandelasticmodulusEareallknown

Solution:

takeanarbitrarycrosssectionM-Matthelowerendoftherodatx

TheaxialforceofthissectionisN（x）=QX

Accordingtothisformula,theaxlecanbemadeasshowninfigure2-6b

ThestressoftheM-Mcrosssectioniss（x）=N（x）/A=qx/A

Obviously

ThesuspensionendhasmaximumaxialforceNmax=QLandmaximumpositivestress

Longitudinaldeformationofrod

Duetounequalaxialforcesoneachcrosssection

Theformula（2-4）cannotbeapplieddirectly

AndshouldstartfromalongsectionofDX

TakethemicrosegmentDXatx

Itslongitudinalelongationcanbewrittenas

Totalelongationofbars

Theelongationattheendofthefixedmemberduetodeadweightisinvestigated

Thebarselfweightisauniformlongitudinaldistributionforce

Atthistime,thedistributionforceofunitrodlengthisq=A*1*g

Here,Gistheunitvolumeofmaterial,thatis,bulkdensity

GetQintotheupperform

HereG=Algistheweightofthewholepole

Theupperformshowsthatthetotalelongationcausedbythedeadweightoftheequalstraightrodisequaltohalftheelongationofalltheweightconcentratedatthelowerend

Problemsolving:

arodwithvariableaxialforce

CalculationofaxialdeformationofbarsbyusingHooke'slaw

Deformationshallbecalculatedinsections

Thenthealgebraisaddedtothefullbar

Whentheaxialforceisacontinuousfunctionisrequiredbyintegralroddeformation

Example2-3.Figure2-7showsthattworoundsectionbarmaterialsarethesame

Strainenergyoftwobariscalculated

Andcomparethesize

Solution:

arod:

Brod:

Ratioofstrainenergyoftwobar:

Problemsolvinginstruction:

ascanbeseenfromthisexample

Underthesameforce

Thestrainenergyofthebarwithsmallrigidityislarge

Example2-4parallelbars1,2and3suspensionrigidbeamAB,asshowninfigure2-8a

LoadonthebeamG

Suchasrod1,2,3ofthecross-sectionalarea,lengthandmodulusofelasticityarethesame

TheyareA,landErespectively

TrytofindtheaxialforceN1,N2andN3ofthethreepoles

-

Solution:

undertheactionofloadG

ThebeamismovedtoAPhiPhiBposition（Figure2-8b）

Thenthereductionamountofrod1isDl1

Theelongationofrod2and3isDl2andDl3

TakethecrossmemberABastheseparatingbody

Asshowninfigure2-8c

ExceptforloadG

ThereareaxialforcesN1,N2,N3,andX

Duetotheassumptionthatthe1rodisshortened

2and3rodelongation

Therefore,N1shouldbesetaspressure

AndN2andN3aresetastension

（1）equilibriumequation

（a）

Thethreeequilibriumequationscontainfourunknownforces

Therefore,itisastaticallyindeterminateproblem

（2）deformationgeometryequationsbythedeformationdiagram2-8bshowsthattheB1B=2C1CPhiPhi

Thatis

or

（b）

（3）physicalequation

（c）

Substituting（c）form（b）

Thenitissolvedinconjunctionwith（a）

Available

Problemsolvinginstruction:

insolvingstaticallyindeterminateproblem:

assumingthattheaxialforceofeachrodistensionorpressure

Onthebasisoftheextensionorshorteningofeachrodinthedeformationdiagram

Thetwomustagree

Bycalculation,theaxialforceofthreebarsispositive

Instructionsareassetoutinthedeformationdiagram

Rod2and3elongate

Therod1isshortened

Examplesandproblemsolvinginstruction

Example2-5showninFigure3-6thescrewsaresubjectedtoaxialtensionF

Therelationbetweentheknownpermittedshearstress[t]andthetensileallowancestress[s]is:

[t]=0.6[s]

Therelationbetweenthepermittedcompressivestress[sbs]andthetensileallowancestress[s]is[sbs]=2[s]

TrysettingupD

D

ThereasonableratiobetweenTandthree

Solution:

（1）tensilestrengthofscrews

（2）extrudingstrengthofnut;

（3）shearstrengthofnut;

Got:

D:

D:

T=1.225:

1:

0.415

Problemsolvinginstruction:

payattentiontotheshearingsurfaceandextrusionsurfaceofthisproblem

Example2-6apalletisrivetedtoanuprightpostwith8rivets

Asshowninfigure3-7a

Therivetspacingisa

F=80kN

DistanceL=3A

Knownrivetdiameterd=20mm

Permittedshearstress=[t]=130MPa

Checktheshearstrengthofrivet

Solution:

thecenterCoftherivetgroupislocatedontheYaxisofthecolumn

WillforceFtothepointCtranslationgetaCpointytoforceFandacoupleFlclockwise

TheforceFcausedbyCisequaltotheshearforceresultingfromtheshearsurfaceofeachrivet

ItsvalueisF/8

AsshowninFigure3-7（c）

TheshearforcealongthenegativeYdirectionofthreerivets,1,2and8,isshowninthefigureF/8

TheFlalsocausedtheshearineveryrivet

ItisassumedthatthesheardirectionisorthogonaltotheconnectionoftherivetcentertotheC

Thesizeisdirectlyproportionaltothelengthoftheconnection

Figure3-7（b）showsthatFlinducedshearrivetrivet;1,3,5,7areQofshear1;2,4,6,8shearareQof2

ThesumofthemomentsoftheshearforceoftherivetsisequaltoFl.C

Thatis

Recycle

Substitutionformula

2ofthetotalshearrivetF/8F/4=Q2=3F/8

Thetotalshearforceofrivet1is

Sorivet1and3aremostdangerous

so

=115MPa<[t]

Problemsolving:

whencalculatingtheshearstrengthoftheconnectingcomponentoftherivetgroup

Tocorrectlyanalyzetheforceofeachrivet

Whentheexternalforcepassesthroughthecenteroftherivetcluster

Canbeapproximatedastheforceofeachrivetisthesame

Whentheexternalforcedoesnotpassthroughtherivetcentroid,therivetforceshouldbeanalyzedaccordingtotheactualforcecondition

Thethirdchapteristheexampleandtheinstructionofproblemsolving

Example3.1.TheRevn=300r/minoftheknowndriveshaft（Figure4-5（a））

ThepowerinputofthedrivingwheelAisP=400kW

TheoutputpowerofthethreedrivenwheelsisPB=120kW

PC=120kW

PD=160kW

TryHuazhoutorquediagram

Solution:

（1）calculatetorqueactingoneachwheelM

BecauseAisthedrivingwheel

Therefore,thesteeringofthemAisconsistentwiththesteeringoftheshaft;thetorqueonthedrivenwheelistheresistancetotherotationoftheshaft

Therefore,thetorqueofthedrivenwheelB,CandDisoppositetothatoftheaxle

（2）seekingthetorqueofeachshaft;

Askfor1-1sectiontorquefirst

Cutfromthissection

Retainrightsegment

Andsetoutonthesection

PositivetorqueMT1（Figure4-5（b））

ByequilibriumconditionSmx=0

Yes

MDmAMT1=0

haveto

HereMT1showsthatthecrosssectionofminustorqueisminus

TorqueacrossallsectionsoftheAandBwheels

Equalto12.74kNm

ItcanbeconcludedthatMT2=8.92kNm

MT3=10kNm

（3）drawatorquediagram

Thepositionofthecrosssectionisexpressedinabscissa

Torqueisexpressedinordinate

Torquechartsofthethree,BC,andCDsectionsoftheAB,,andaxleareselectedatselectedscales

Becausethetorqueisconstantineachsegment

Therefore,thetorquediagramconsistsofthreehorizontallines

AsshowninFigure4-5（c）

Themaximumtorque7.64kNmoccursinthemiddlesection

Problemsolving:

whentheaxlecrosssectiontorque

PositivetorqueisalwayssetatthecrosssectionMT

UseSmx=0forthistorque

IfMTisapositivetorque

Ifthesignisnegativetorque

Ifinthiscase,A,Bonthewheel

ThetorquediagramisshowninFigure4-5（d）

Fromthiswecansee

Themaximumvalueoftheinternalforcecanbereducedbyreasonablyarrangingtheload

Improvethecarryingcapacityofthemember

Example3.2itisknownthatthetransmissionsh