山农双语材力

MechanicsofMaterialsChapter2Axialtensionandcompression§2–1Conceptsandpracticalexamples§2–2Internalforce、methodofsectionandstress§2–3Deformation

andstrain§2–4Internalforceandstressesonthecrosssection§2–5Stressesontheinclinedsection§2–6Mechanicalpropertiesofmaterialsinaxialtension§2–7Mechanicalpropertiesofmaterialsinaxialcompression§2–8Calculationoffailure,safetyfactorandstrength

§2–9Deformationofaxialtensionandcompression§2–10Strainenergyofaxialtensionandcompression§2–11

Staticallyindeterminateproblems§2–12

Temperaturestressandassemblystress§2–13StressConcentration§2–14Calculationofshearingandextrusion3

Chapter1Axialtensionandcompression§2–1

ConceptsandpracticalexamplesofaxialtensionandcompressionCharacteristicoftheexternalforce1、ConceptsCharacteristicofthedeformation5compressiveforcetensileforce7Practicalexamplesinengineering2、911F

12BACBF1BC2BA

1、Internalforce

Internalforceistheresultantofinternalforces,whichisactingmutuallybetweentwoneighbourpartsinsidethebody,causedbytheexternalforces.

§2–2Internalforce、methodofsectionandstress132、Methodofsection·axialforce1).Basicstepsofthemethodofsection：

Cutoff（截）Takeout（取）Substitute（代）Equilibrium（平）152).Axialforce—internalforceoftherodinaxialtensionorcompression，designatedbyFN.Suchas：DetermineFNbythemethodofsection.APPPAFNSimplesketchAPPCutoff:Takeout:Substitute：Equilibrium：17FN-P=0FN=Px3).Positiveandnegativesign

of

theaxialforce:19Tensileforceispositive.Compressiveforceisnegative.

>0<0①Reflectthevarietyoftheaxialforcealongtheaxis.②Findoutvalueofthemaximumaxialforceandthepositionofthesectioninwhichthemaximumaxialforceact.Determinethepositionofthecriticalsection.3、Diagramoftheaxialforce

xFNP+meaning19Example1Theforceswithmagnitudes

5P、8P、4PandPact

respectivelyatpointsA、B、C、Doftherod.Theirdirectionsareshowninthefigure.Trytoplotthediagramoftheaxialforceoftherod.Solution：DeterminetheinternalforceFN1insegmentOA.Takethefreebodyasshowninthefigure.ABCDPAPBPCPDOABCDPAPBPCPDFN121Similarly，wegettheinternalforcesinsegmentAB、BC、CD.Theyarerespectively：FN2=–3P

FN3=5PFN4=PThediagramoftheaxialforceisshownintherightfigure.BCDPBPCPDFN2CDPCPDFN3DPDFN4Nx2P3P5PP++–23Simplemethod

：Fromthelefttotheright:Characteristicofthediagramoftheaxialforce：Valueofsuddenchange=concentratedloadIfmeetingtheforcePtotheleft

，theincreaseoftheaxialforceNispositive；Ifmeetingtheforcetotheright

，theincreaseoftheaxialforceNisnegative.5kN8kN3kN+–3kN5kN8kN25PointsforAttention:截面不能刚好截在外力作用点处

求轴力时，外力不能沿作用线随意移动1.Onetoonecorrespondence2.Valueofsuddenchange=concentratedload3.Sameproportion设正Solution：qK

LxOExample2Lengthoftherodshowninthefigureis

L.Distributedforceq=kxisactedonit,directionoftheforceisshowninthefigure.Trytoplotthediagramofaxialforceoftherod.Lq(x)FN(x)xq(x)NxO–27Think:4、Conceptofstress1).Definition：Intensityoftheinternalforceduetotheexternalforces.292).Expressionofstress：①Averagestress：②Wholestress（sumstress）：

F

AM③Wholestressmaybedecomposedinto:p

M

a.Stressperpendiculartothesectioniscalled“normalstress”b.Stresslyinginthesectioniscalled“shearingstress”

33

§2–3Deformationandstrain(linear)strain

Shearstrainc’b’Angularstrain1、Internalforceonthecrosssection

§2–3InternalforceandstressesonthecrosssectionAccordingtothemethodofsection：Beforedeformation1).Hypothesisofplanesection：Hypothesisofplanesection：Crosssectionsremainplanesbeforeandafterdeformations.Deformationsoflongitudinalfibersarethesame.

abcdAfterloading35PPd´a´c´b´2、StressonthecrosssectionUniformdistributionofnormalstrainNoshearstrainFFFN=FFF即Normalstress—

：distributesuniformlyonthecrosssection.Hypothesisofplanesection

Straightrod、crosssectioniswithoutsuddenchange、thereisacertaindistancefromthesectiontothepointatwhichtheloadacts.3).Applicationconditions4).Saint-Venantprinciple：39P16Example3Acircularrodissubjectedtoatensileforce

P=25kN.Itsdiameterisd=14mm.Trytocalculatetheaxialforceandthestressonthecrosssectionoftherod.Solution：①Axialforce：FN

=P=25kN②Stress：PPkka

Aa:Areaoftheinclinedsection；Fa：Internalforceintheinclinedsection.

PkkaFaFromgeometricrelationSubstitutingitintotheaboveformulawegetWholestressintheinclinedsection：§2–5stressesontheobliquesection

Decomposition：pa=As

=90°，As

=0,90°，As

=0°，(Themaximumnormalstressexistsinthecrosssection)As

=±45°，(Themaximumshearingstressexistsintheinclinedsectionof45°)Wholestress:PPkkaPkkapatasaa：横截面外法线转至斜截面的外法线，逆时针转向为正，反之为负；：拉应力为正，压应力为负；：对所研究构件内一点产生顺时针力矩的剪应力为正，反之为负；目录正负规定:ɑExample4

Arod,whichthediameterd=1cmissubjectedtoatensileforceP=10kN.Determinethemaximumshearingstress,thenormalstressandshearingstressontheinclinedsectionofanangle30°aboutthecrosssection.Solution：Stressesintheinclinedsectionoftherodintensionorcompressioncanbedetermineddirectlybytheformula：Exercise:

AOC=2ACD，Determinethepositionsofthemaximalaxialforce,themaximalnormalstressandshearingstress.O3F4F2FBCD221133O3F4F2FBCDSolution：Calculatetheaxialforce221133O4FB22(compressiveforce)(tensileforce)(tensileforce)3、DiagramofaxialforceO3F4F2FBCD-diagram2211333F2F-F++-4、5、O3F4F2FBCD1133目录§2–6Mechanicalpropertiesofmaterialsinaxialtension1、Testingconditionsandinstruments

Mechanicalproperties：Somepropertiesaboutstrengthanddeformationshownbymaterialsundertheactionofexternalloads.1)、Testingconditions：normaltemperature(20℃)；staticload（loadedgradually)StandardspecimenL=5dorL=10dL:scaledistance2、Experimentalimplement：Universaltestingmachine；131Elasticdeformation:Plasticdeformation:Plasticmaterials:lowcarbonsteel

Brittlematerials:castiron2、Tensilediagramofthelowcarbonsteelspecimen(F--

Ldiagram)3、Stress-straincurveofthelowcarbonsteelspecimen(

--

diagram)Typicalpointsinthecurveofthelow-carbonsteel(1)Elasticrangeofthelow-carbonsteelintension(ob)135Stress-strainrelationshipintheelasticregion1、oa

–

Proportionalsegment:

p

–

Proportionallimit

2、ab

–

Curvedsegment:

e

–

Elasticlimitba(2)Yielding(flowing)rangeofthelowcarbonsteelintension

137

Yieldingrange:

s---YieldinglimitFailurestressofplasticmaterial:

s

。Single-crystalCu-A1specimenaftertensionSlidinglines(3)、Hardeningrangeofthelowcarbonsteel(ce)１、

ｂ--Strengthlimitce２、Unloadedlaw３、Coldhardeningreload(4)、Necking(crack)stage(bf)ofthelowcarbonsteelintension141Typicalpointsinthecurveofthelow-carbonsteel目录1、Percentelongation

2、Percentreductioninarea

3、relativityofbrittleness

andductility4、Plasticmaterialwithoutobviousyield0.2s0.2Nominalyieldstress:

0.2

isthefailurestressofthistypeofmaterial.5、Mechanicalpropertiesofthecastironintension

ｂL---Tensilestrengthlimitofcastiron(fail stress）143§2–7Mechanicalpropertiesofmaterialsinaxialcompression

Stress-straincurveofthelowcarbonsteelspecimen(

--

diagram)Comparedwiththetension:Thesame:Thedifference:

Stress-straincurveofthecastironspecimen(

--

diagram)

c

–

Compressivestrengthlimitofcastiron；

c

（4～6）

ｂ

1、Conceptsoffailure,safetyfactorandlimitstressn>12)、Permissiblestress(allowablestress)：3)、Limitstress(ultimatestress)：4)、Safetyfactor：§2–8Calculationoffailure,safetyfactorandstrength1)、Failure：UltimatestressPermissiblestressFailuretypesMaterialtypesPlasticmaterialYieldFractureBrittlematerialCriterionofthestrengthdesign：②Designthedimensionofthesection：Threekindsofcalculationofstrengthmaybedoneaccordingtothecriterionofstrength:①Checkthestrength：③Determinetheallowableload：

Example5Acircularrodissubjectedtoatensileforce

P=25kN.Itsdiameterisd=14mmanditsallowablestressis[

]=170MPa.Trytocheckthestrengthoftherod.Solution：①Axialforce：FN

=P

=25kN②Stress：③Checkthestrength：④Conclusion：Thestrengthoftherodsatisfiesrequest.Therodcanworknormally.Example6A

simplecraneisshowninthefigure.ACisarigidbeam，sumweightofthehoistandheavybodythatisliftedisP.Whatshouldbetheangle

sothattherodBDhastheminimumvolume？

Theallowablestressoftherod[

]isknown.Analysis：LhqPABCD

Thecross-sectionareaAoftherodBD：Solution：

Internalforce

FNBD((q

)oftherodBD：

TakeACasourstudyobjectasshowninthefigure.YAXAqFNBDLPABC55YAXAqNBDLPABC③DeterminetheminimumvalueofVBD

：67

12CBA1.5m2mF

Exercise:Thediameterofrod1isd=14mmanditsallowablestressis；,thesidelengthofrod2isa=100mm,anditsallowableStressis,(1)whenF=2000KN,trytocheckthestrengthoftherods

(2)CalculatetheallowableloadatBF1、Calculatetheaxialforce:

21

12CBA1.5m2mFB2、F=2000kN，checkthestrengthRod1：Rod2：Itissafe.3、Calculatetheallowableload:§2–9Deformationofaxialtensionandcompression

1)、Thewholelongitudinal deformationoftherod：

2)、Averagestain：1、Deformationandstrainoftherodintensionorcompression

abcdLCrosssectionSolution：Deformationmayexceedtherangeoflinearelasticity,thereforetheelasticlawisnotappliedhere.Calculationshouldbedoneinthefollowing:Example7Diameterofacopperwireisd=2mmanditslengthisL=500mm.Tensilecurveofcopperisshowninthefigure.Tomakeelongationofthecopperwireis30mmwhatistheforcePthatwemustact

？

Fromthetensilecurve：s(MPa)e(%)4)、Lateralstrain：3)、Lateraldeformationoftherod：PPd´a´c´b´L15)、Elasticlaw

6)、Possion

,sratio（orcoefficientofthelateraldeformation）

ElasticLaw=Hook’sLawTheElasticLawistheimportantfoundationofsolidmechanicssuchMechanicsofMaterials.Generallyitisconsideredtobe

proposedfirstlybytheEnglishscientistHook(1635-1703),SotheElasticLawisalsocalledHook’sLaw.orThatis792、Elasticlawoftherodintensionorcompression

1)、Caseofequalinternalforces

※“EA”iscalledtheaxialrigidityoftherodintensionorcompression.

PP2)、Caseofvariableinternalforcesandcrosssection

Wheninternalforcesinnsegmentsareconstant

Exercise:P1=30kN，P2=10kN，AAB=ABC=500mm2，ACD=200mm2，E=200GPa。ABCDP1P2100100100Trytocalculate:(1)theinternalforcesandstressesonthecrosssectionineachsegment.(2)ΔLSolution:①thediagramofaxialforce+

ABCDP1P210010010020kN10kNN②CalculatetheinternalforceonthecrosssectionCalculateΔL

C'1)、Howtoplottheenlargementsketchofthesmalldeformation

Accuratemethodtoplotdiagramofdeformation，thearclineasshowninthefigure；

Determinedeformation△LiofeachrodasshowninFigl.

Approximatemethodtoplotthediagramofdeformation;thetangentofthearclineshowninthefigure.3.EnlargementsketchofthesmalldeformationandthemethodtodeterminedisplacementsABCL1L2PC"2)、WritetherelationbetweenthedisplacementofpointBshowninFig.2and deformationsoftworods.Example8Supposethecrossbeam

ABCDisrigid.Asteelcablewiththecross-sectionarea76.36mm²isaroundapulleywithoutfriction.KnowingP=20kN，E=177GPa.Determinethestressofthesteelcableandtheuprightdisplacementofpoint

C.Solution：method1：Enlargementsketchmethodofthesmalldeformation.1)Determinetheinternalforceofthesteelcable:TakeABDasourstudyobject：2)StressandelongationofthesteelcablePABCDTTYAXA89800400400DCPAB60°60°CPAB60°60°800400400DAB60°60°DB'D'C3）Deformationisshowninthefigure.

UprightdisplacementofpointCis：§2–10Strainenergyofaxialtensionandcompression1、Elasticstrainenergy：Theworkdonebytheexternalforcesresultsintheincreaseofsomeenergyassociatedwiththedeformationoftherod.2、Calculationofthestrainenergyoftherodintensionandcompression：PLL

oLBPAInternalforceisconstantinasmallsegment

FN(x)dxx3、Strain－energydensityuoftherodintensionandcompression

Strainenergyperunitvolume.N(x)dxxdxN(x)N(x)95Solution：Method2：Energymethod：（Workofexternalforcesisequaltothestrainenergy）

（1）Determinetheinternalforceofthesteelcable：TakeABCDasourstudyobject：Example9Supposethecrossbeam

ABCDisrigid.Asteelcablewiththecross-sectionarea76.36mm²isaroundapulleywithoutfriction.KnowingP=20kN，E=177GPa.Determinethestressofthesteelcableandtheuprightdisplacementofpoint

C.800400400CPAB60°60°PABCDTTYAXA（2)Stressofthesteelcableis：（3)Displacementofpoint

Cis：800400400CPAB60°60°§2–11

Staticallyindeterminateproblems1)、Staticallyindeterminateproblems：1、StaticallyindeterminateproblemsandtheirtreatmentmethodsFP

DBACEFP

DBAC2、Methodtosolvestaticallyindeterminateproblems：EquilibriumequationsCompatibilityequationofdeformationPhysicalequationsExample10

Rods1，2and3areconnectedtogetherwithapinasshowninthefigure.Knowingthelengthofeachrodis：L1=L2、

L3=L；andtheareaofeachrodisA1=A2=A,A3；modulusofelasticityofeachrodis：E1=E2=E、E3.Externalforceisalongtheuprightdirection.Determinetheinternalforceofeachrod.CPABD213Solution：Equilibriumequations:PAFN2FN3FN1

Geometricequation—compatibilityequationofdeformation：

Physicalequation—elasticlaw：

Complementaryequation：detainingfromthegeometricequationandphysicalequation.

Solvingtheequilibriumequationsandcomplementaryequationweget:3、Methodandstepsforsolvingstaticallyindeterminateproblems：①Equilibriumequations；

②Geometricequations—compatibilityequationsofdeformation；

③Physicalequations—elasticlaws；

④Complementaryequations：gettingfromgeometricequationsandphysicalequations；

⑤Solvingthecombinedequationsincludingofequilibriumequationsandcomplementequations.Example11

Fouranglesofawoodenpolearereinforcedwithfourequalleganglesteelof40

40

4.Theallowablestressesofsteelandwoodarerespectively[

]1=160MPaand[

]2=12MPa，moduleofelasticityofthemareE1=200GPaandE2=10GPa.DeterminetheallowablepermissibleloadP.

Geometricequation

Physicalequationandcomplementaryequation：Solution：Equilibriumequations:PPy4N1N2PPy4N1N2

Solvingequilibriumequationsandthecomplementaryequationweget:

Determinethepermissibleloadofthestructure：

Method1:Sectionareaoftheanglesteelmaybeobtainedfromthetableofhot-rolledsteel:A1=3.086cm21)、Thereisnothetemperaturestressinstaticallydeterminatestructure.

2)、Thereisthetemperature

stressinstaticallyindeterminatestructure.1、Temperaturestress

Materialsanddimensionsofrod1and2areallthesameasshowninthefigure.IftemperatureofthestructurechangesfromT1to

T2,determinethetemperatureinternalforcesofeachrod.（linearthermalexpansioncoefficientofeachrod

i;△T=T2-T1)§2–12

Temperaturestressandassemblystress

GeometricequationSolution：Equilibriumequations:

Physicalequation：AFN2FN3FN1Complementaryequation：Solvingequilibriumequationsandthecomplementaryequationweget:

aaaaN1N2Example12

Theupperandlowerendsofaladder-likesteelshaftarefixedattemperatureT1=5℃asshowninthefigure.Areasoftheupperandlowersegmentsarerespectively

1=cm2and

2=cm2.WhenitstemperaturereachesT2=25℃,determinethetemperaturestressofeachrod.(Linearthermalexpansioncoefficient

=

12.5×10-61/0C；modulusofelasticity

E=200GPa)

Geometricequation：Solution：Equilibriumequation：125

Physicalequation:Solvingequilibriumequationsandthecomplementaryequationweget:

Complementaryequation:

、Temperaturestresses:

GeometricequationSolution：Equilibriumequations:

2、Assemblestresses—initialstresses1)、Thereisnotheassemblestressinstaticallydeterminatestructure.

2)、Thereistheassemblestressinstaticallyindeterminatestructure.

Thedimensionerrorofrod3is

asshowninthefigure.Determinetheassembleinternalforceofeachrod.ABC12DA13

Physicalequationandcomplementaryequation：Solvingequilibriumequationsandthecomplementaryequationweget:dA1N1N2N3AA11.Conceptofstressconcentration:§2–13StressConcentration2.Theoreticalstressconcentrationfactor§2–14Calculationofshearingandextrusion1、Characteristicsofloadsanddeformationofconnectingmembers:1)Connectingmember

Thestructurememberthatconnectsonemembertoanotheriscalledtheconnectingmember.

Suchas：bolts、rivets、keysetc.Theconnectingmemberissmall，butitplaystheroleofpassingloads.

Characteristic：Itcanpassgeneralloadsandcanbedismounted.

PPboltPPrivetCharacteristic:Itcanpassgeneralloads,butcannotbedismounted,forexample,thetrussinabridgeisconnectedbyit.nogapCharacteristic:Itcanpasstorques.mShaftKeyGearm2)Characteristicsofloadsanddeformation：nn（Resultant）（Resultant）PPUsearivetasanexample：①Characteristicofloads：

Therivetissubjectedtotwoequalandoppositeforces.Theactinglinesofthesetwoforcesareveryclose.②Characteristicofdeformation:

Twopartssubjectedtotwoequalandoppositeforcestendtoshiftoveroneanotheralongthejunctionplaneoftwoforces.SHEARnn（Resultant

）（ResultantPP③Shearingplane：

Theplanealongwhichtwopartsofthemembertendtoshiftoveroneanother.Suchasn–n

.④Internalforceonshearingplane：

Internalforce—ShearingforceFs

，itsactinglineisparalleltotheshearingplane.PnnFsShearingplaneSHEARnn(Resultant)(Resultant

）PP3)Threekindsofbreakageatjoint：

①Failureduetoshear

Snipalongtheshearingplaneoftherivet,

suchasalongsectionn–n.

②Breakageduetobearing

Failduetomutualbearingbetweentherivetandthesteelplateintheirconnectingplane.

PnnFsShearingplaneSHEARttdPPP112233P/4bPP③BreakageduetotensionThesteelplateisweakenedinthesectionin whichtherivetholesexistandstressintheweakenedsectionincreasessothatthesteelplateiseasilybrokenduetotensionattheconnectingposition.

2、PracticalcalculationofshearMethodofthepracticalcalculation：Accordingtopossibilityofbreakageofthemembersomeassumptionsbywhichbasiccharacteristicsubjectedtoforceactionscanbereflectedandcalculationscanbesimplifiedareused.Thencalculateitsnominalstress,determinethecorrespondingpermissiblestressinaccordancewiththeresultofdirecttest.Atlastdothestrengthcalculation.Applyingrange：volumeofthememberisnotlargeandrealstressisquitecomplex.Suchastheconnectingpiecesetc.Assumptionofpracticalcalculation：Assumethatshearingstressisdistributeduniformlyintheshearingplaneandequaltotheaverageshearingstress.SHEAR1)Shearingplane--As

：Shiftingplane.

Shearingforce--Fs：Internalforceontheshearingplane.2)Nominalshearingforce--

：3)Strengthconditionofshear：nn（Resultant

）(Resultant

）PPPnnFsShearingplaneWorkingstressshouldnotexceedthepermissiblestress.SHEAR,where3、Practicalcalculationsofbearing1)Bearingforce―Fbs

：Theresultantforceactingonthetouchingplane.Bearing：Thephenomenonthatthereispressureonthepartialareaofamember.Bearingforce：Theresultantforceactingonthebearingplane，designatedbyFbs.Assumption：Bearingstressesaredistributeduniformlyovertheeffective bearingplane.SHEAR2)Bearingarea：Areaoftheprojectionplaneofthetouchingplaneinthedirectionperpendicularto

Fbs

3)Strengthconditionofbearing：

WorkingbearingstressshouldnotexceedthepermissiblebearingstressBearingareaSHEAR4、ApplicationsSHEAR1)Checkthestrength2)Selectthecrosssectionarea3)DeterminetheexternalloadExample1

Awoodentenonjointisshowninthefigure.Knowingthatthequantitiesarea=b

=12cm，h=35cm，c=4.5cmand

P=40KN.Trytodeterminetheshearingstressandbearingstressforthejoint.Solution：：Freebodydiagramis showninthefigure：ShearingstressandbearingstressShearingforceisBearingforceisSHEARPPPPPPbachhSolution：

Freebodydiagramofthekeyisshown inthefigureExample2Agearandashaftareconnectedbyakey（b×h×L=20×12×100).Thetorquethatthekeycantransmitism=2KNm.Knowingthediameteroftheshaftisd=70mm，thepermissibleshearingstressandthepermissiblebearingstressofthekeyarerespectively[

]=60MPaand[

bs]=100MPa.Trytocheckthestrengthofthekey.

mbhLSHEARmdPAccordingtotheabovecalculation,strengthconditionsofthekeyaresatisfied.

Checkthestrengthofshearandbearing

bhLSHEARdmFsSolution：

Freebodydiagramofthekeyis showninthefigure[Example3]

Agearandashaftareconnectedbyakey（b=16mm,h=10mm).Thetorquethatthekeycantransmitism=1600Nm.Knowingthediameteroftheshaftisd=50mm，thepermissibleshearingstressandthepermissiblebearingstressofthekeyarerespectively[

]=80MPaand[

jy]=240MPa.Trytodesignthelengthofthekey.

bhLSHEARmmmdPbhL

Strengthconditionsoftheshearingstressandthebearingstress

AccordingtotheabovecalculationSHEARdmQSolution：

Freebodydiagram isshowninthefigure[Example4]

Arivetedtie-inactedbyforceP=110kNisshowninthefigure.Knowingthethicknessist=1cm，widthofitisb=8.5cm.Thepermissiblestressis[

]=160MPa.Diameteroftherivetisd=1.6cmandthepermissibleshearingstressis[

]=140MPa，thepermissiblebearingstressis[

jy]=320MPa.Trytocheckthestrengthoftheriveting.(Assumetheforceactedoneachrivetisequal.）bPPttdPPP112233P/4SHEAR

Thesections2—2and3—3ofthesteelplatearethecriticalsections.

ThestrengthconditionsofshearandbearingTherefore，thetie-inissafe.ttdPPP112233P/4SHEAR1、Internal-forceand

axial-forcediagramsof

therodintensionand

compression1)ExpressionofAxialforce?2)Methodtodetermineaxialforce?3)Positiveandnegativeofaxialforce?ExerciselessonsoftensionandcompressionandshearWhydoweplottheaxial-forcediagram?Whatshouldwepayattentionwhenwedoit?4)Axial-forcediagram：

ExpressedbythediagramofN=N(x)?PANASimplesketchBCPPNxP+Simplemethodtodetermineaxialforces:①Taketheleftpartofthesectionxastheobject,theaxialforceonthesectionxcanbecalculatedbyfollowingformula:

Where“

P()”and“

P()”expressthesumofleftdirectionforcesandthesumofrightdirectionforcesoftheleftpartofthesectionx.

②Taketherightpartofthesectionxastheobject,theaxialforceN(x)ofpointxcanbecalculatedbythefollowingformulate:Where“”and“”denotethesumofrightdirectionforcesandthesumofleftdirectionforcesoftherightpartofsectionx.SHEAR[Example1]Forces5P,8P,4PandPareactedatpointsA,B,CandDontherodrespectively，theirdirectionsareshowninthefigure.Trytoplottheaxial-forcediagramoftherod.ABCDO5P4PP8PFNx–3P5PP2PSHEAR⊕⊕⊕○Positiveandnegativeofstress?1)Stressonthecrosssection:

2、Stressoftherodintensionorcompression

Criticalsectionandmaximumworkingstress?

2)Stressontheinclinedsection

Saint-Venantprinciple?

Stressconcentrations?sFN(x)P

tasaxs0SHEAR3、Strengthdesigncriterion：1)Strengthdesigncriterion

①Checkstrength：②Designthecrosssectionarea：③Designtheload：SHEAR1)Elasticlawofthe