材料力学课件 1

Example1-1Theforceswithmagnitudes5P、8P、4PandPactrespectivelyatpointsA、B、C、Doftherod.Theirdirectionsareshowninthefigure.Trytoplotthediagramoftheaxialforceoftherod.[例1-1]图示杆的A、B、C、D点分别作用着大小为5P、8P、4P、P的力，方向如图，试画出杆的轴力图。ABCDPAPBPCPDOABCDPAPBPCPDN11Solution：DeterminetheinternalforceN1insegmentOA.Takethefreebodyasshowninthefigure.解：求OA段内力N1：设置截面如图ABCDPAPBPCPDOABCDPAPBPCPDN12Similarly，wegettheinternalforcesinsegmentAB、BC、CD.Theyarerespectively：

同理，求得AB、BC、CD段内力分别为：

N2=–3P

N3=5PN4=PBCDPBPCPDN2CDPCPDN3DPDN43Thediagramoftheaxialforceisshownintherightfigure.轴力图如右图Nx2P3P5PP++–Characteristicofthediagramoftheaxialforce：Valueofsuddenchange=concentratedload轴力图的特点：突变值=集中载荷

4Simplemethodtoplotthediagramofaxialforce：Fromthelefttotheright:轴力(图)的简便求法：自左向右:IfmeetingtheforcePtotheleft，theincreaseoftheaxialforceNispositive；遇到向左的P，轴力N

增量为正；5kN8kN3kN+–3kN5kN8kNIfmeetingtheforcetotheright，theincreaseoftheaxialforceNisnegative.遇到向右的P，轴力N增量为负。5Example1-2LengthoftherodshowninthefigureisL.Distributedforceq=kxisactedonit,directionoftheforceisshowninthefigure.Trytoplotthediagramofaxialforceofthetherod.[例1-2]图示杆长为L，受分布力q(x)=kx作用，方向如图，试画出杆的轴力图。Lq(x)ox6Solution：Thefreeendoftherodistheoriginofthecoordinateandcoordinatextotherightispositive.Takethesegmentoflengthxontheleftofpointx,itsinternalforceis解：x坐标向右为正，坐标原点在自由端。取左侧x段为对象，内力N(x)为：N(x)xq(x)NxO–K

LxOq7Example1-3AcircularrodissubjectedtoatensileforceP=25kN.Itsdiameterisd=14mmanditsallowablestressis[]=170MPa.Trytocheckthestrengthoftherod.[例1-3]已知一圆杆受拉力P=25kN，直径d=14mm，许用应力[]=170MPa，试校核此杆是否满足强度要求。8Solution：解：(2)Stress：应力：(3)Checkthestrength：强度校核：(4)Conclusion：Thestrengthoftherodsatisfiesrequest.Therodcanworknormally.结论：此杆满足强度要求，能够正常工作。(1)Axialforce：N=P=25kN轴力：N=P=25kN9

Example1-4Athree-pinhouseframeonwhichaverticaluniformload,withtheindensityintensityisq=4.2kN/misappliedisshowninthefigure.Diameterofthesteeltensilerodintheframeisd=16mmanditsallowablestressis[]=170MPa.Trytocheckthestrengthoftherod.[例1-4]已知三铰屋架如图，承受竖向均布载荷，载荷的分布集度为：q=4.2kN/m，屋架中的钢拉杆直径d=16mm，许用应力[]=170MPa。试校核钢拉杆的强度。Tiebar4.2m8.5m10(1)Determinethereactionsfirstaccordingtotheglobalequilibrium整体平衡求支反力Solution：解：Tiebar8.5m4.2mRARBHA11(3)Stress

应力：(2)Determinetheaxialforceaccordingtothepartialequilibrium：

局部平衡求轴力：RAHARCHCN12(4)Strengthcheckandconclusion：

强度校核与结论：Thisrodsatisfiestherequestofstrength.Itissafe.此杆满足强度要求，是安全的。RAHARCHCN13hCxLqPABDExample1-5Asimplecraneisshowninthefigure.ACisarigidbeam，sumweightofthehoistandheavybodythatisliftedisP.Whatshouldbetheanglesothattherod

BDhastheminimumweight？Theallowablestressoftherod[]isknown.[例1-5]简易起重机构如图，AC为刚性梁，吊车与吊起重物总重为P，为使BD杆最轻，角应为何值？已知BD杆的许用应力为[]。14Analysis：分析：CxLhqPABDSolution：解：(1)Internalforce

N((q)oftherodBD：TakeACasourstudyobjectasshowninthefigure.

BD杆内力N(q)：

取AC为研究对象，如图15(2)Thecross-sectionareaAoftherodBD：

BD杆横截面面积A：LNBDYAXAqxPABC16LYAXAqNBDxPABC(3)DeterminetheminimumvalueofVBD

：

求VBD的最小值：174.Stressesintheinclinedsectionoftherodintensionorcompression

拉(压)杆斜截面上的应力PPkka

PkkaPaAssumeastraightrodissubjectedtoatensileforceP.Determinethestressintheinclinedsectionk-k.

设有一等直杆受拉力P作用。求：斜截面k-k上的应力。18Aa:Areaoftheinclinedsection；

Aa：斜截面面积；PPkkaPkkaPaPa=PSolution:Adoptthemethodofsection.解：采用截面法Accordingtotheequilibriumequation：由平衡方程：

Then则Pa：Internalforceintheinclinedsection.Pa：斜截面上内力。19

PPkkaPkkaPaFromgeometricrelation由几何关系：Substitutingitintotheaboveformulaweget：代入上式，得：Wholestressintheinclinedsection：斜截面上全应力：20Decomposition：分解：pa=Itindicatesthechangeofstressesindifferentsectionsthroughapoint.反映：通过构件上一点不同截面上应力变化情况。Wholestressintheinclinedsection：斜截面上全应力：PPkkaPkkapatasaa21As=90°，As=0,90°，As=0°，(Themaximumnormalstressexistsinthecrosssection)(横截面上存在最大正应力)PPkkaPkkapatasaaAs=±45°，(Themaximumshearingstressexistsintheinclinedsectionof45°)当=0°时，当=90°时，当=±45°时，(45°斜截面上剪应力达到最大)当=0,90°时，22Example1-6Arod,whichthediameterd=1cmissubjectedtoatensileforceP=10kN.Determinethemaximumshearingstress,thenormalstressandshearingstressintheinclinedsectionofanangle30°aboutthecrosssection.[例1-6]直径为d=1cm杆受拉力P=10kN的作用，试求最大剪应力，并求与横截面夹角30°的斜截面上的正应力和剪应力。23Solution：Stressesintheinclinedsectionoftherodintensionorcompressioncanbedetermineddirectlybytheformula：解：拉压杆斜截面上的应力，直接由公式求之：24Example1-7

Atensilerodasshowninthefigureismadefromtwopartsgluedmutuallytogetheralong

mn.Itissubjectedtotheactionofforce

P.Assumethattheallowablenormalstressis[]=100MPaandallowableshearingstressis[]=50MPafortheadhesive.AreaofcrosssectionoftherodisA=4cm².Ifstrengthoftherodiscontrolledbytheadhesivewhatistheangle(:between00~600)togetthelargesttensileforce?[例1-7]图示拉杆沿mn由两部分胶合而成,受力P，设胶合面的许用拉应力为[]=100MPa；许用剪应力为[]=50MPa，并设杆的强度由胶合面控制,杆的横截面积为A=4cm²，试问:为使杆承受最大拉力,角值应为多大?(规定:在0~60度之间)。25Combine(1)、(2)andget：联立(1)、(2)得：Solution：解：PPmnaPa600300B0026Thecurvesofformula(1)and、(2)areshownintheTig.(2).ObviouslythestrengthoftherodontheleftofpointBiscontrolledbythenormalstress，thatontherightofpointBiscontrolledbytheshearingstress.Asa=60°，fromformula(2)wecanget(1)、(2)式的曲线如图(2)，显然，B点左侧由正应力控制杆的强度，B点右侧由剪应力控制杆的强度，当a=60°时，由(2)式得Pa600300B100Discussion：As讨论：若27Solution:

Atthepointofintersectionofcurves(1)and(2)：解:(1)、(2)曲线交点处：Pa600300B100281.Howtoplottheenlargementsketchofthesmalldeformation1.怎样画小变形放大图？2)Accuratemethodtoplotdiagramofdeformation，thearclineasshowninthefigure；

变形图严格画法，图中弧线；1)Determinedeformation△LiofeachrodasshowninFigl.

求各杆的变形量△Li，如图1；3)Approximatemethodtoplotthediagramofdeformation;thetangentofthearclineshowninthefigure.

变形图近似画法，图中弧之切线。Example1-8Enlargementsketchofthesmalldeformationtheandmethodtodeterminedisplacements[例1-8]小变形放大图与位移的求法。C'ABCL1L2PC"292.WritetherelationbetweenthedisplacementofpointBshowninFig.2anddeformationsoftworods.写出图2中B点位移与两杆变形间的关系ABCL1L2B'PFig.2图230Solution：ThediagramofdeformationisshownintheFig.2.PointBmovestopointB'，Fromthediagramofdisplacementwemayknow：解：变形图如图2，B点位移至B'点，由图知：ABCL1L2B'PFig.2图231800400400DCPAB60°60°Example1-9

SupposethecrossbeamABCDisrigid.Asteelcablewiththecross-sectionarea76.36mm²isaroundapulleywithoutfriction.KnowingP=20kN，E=177GPa.DeterminethestressofthesteelcableandtheuprightdisplacementofpointC.[例1-9]设横梁ABCD为刚梁，横截面面积为76.36mm²的钢索绕过无摩擦的定滑轮。设P=20kN，试求钢索内的应力和C点的垂直位移。设钢索的E=177GPa。321)Determinetheinternalforceofthesteelcable:TakeABDasourstudyobject：求钢索内力：以ABCD为研究对象2)Stressandelongationofthesteelcable钢索的应力和伸长分别为：

PABCDTTYAXA800400400DCPAB60°60°Solution：method1Enlargementsketchmethodofthesmalleformation.

解：方法1

：小变形放大图法：333）Deformationisshowninthefigure.

UprightdisplacementofpointCis：变形图如左图,C点的垂直位移为DCPAB60°60°800400400AB60°60°DB'D'C拉压AxialTensionandCompression34800400400CPAB60°60°PABCDTTYAXAExample1-9SupposethecrossbeamABCDisrigid.Asteelcablewiththecross-sectionarea76.36mm²isaroundapulleywithoutfriction.KnowingP=20kN，E=177GPa.DeterminethestressofthesteelcableandtheuprightdisplacementofpointC.[例1-9]设横梁ABCD为刚梁，横截面面积为76.36mm²的钢索绕过无摩擦的定滑轮。设P=20kN，试求钢索内的应力和C点的垂直位移。设钢索的E=177GPa。35800400400CPAB60°60°PABCDTTYAXASolution：Method2：Energymethod:(Workofexternalforcesisequaltothestrainenergy)解：方法2：能量法：（外力功等于变形能）1)Determinetheinternalforceofthesteelcable：TakeABCDasourstudyobject：求钢索内力：以ABCD为研究对象：362)Stressofthesteelcableis：钢索的应力为：3)Displacementofpoint

Cis：

C点位移为：Energymethod：Themethodbywhichtheproblemsrelativetotheelasticdeformationofthestructuremembersaresoloedaccordingtotheconceptofstrainenergyiscalledtheenergymethod.

能量法：利用应变能的概念解决与结构物或构件的弹性变形有关的问题，这种方法称为能量法。800400400CPAB60°60°37CPABD123PAN1N3N2Example1-10Rods1，2and3areconnectedtogetherwithapinasshowninthefigure.Knowingthelengthofeachrodis：L1=L2、L3=L；andtheareaofeachrodisA1=A2=A,A3；modulusofelasticityofeachrodis：E1=E2=E、E3.Externalforceisalongtheuprightdirection.Determinetheinternalforceofeachrod.[例1-10]设1、2、3三杆用铰链连接如图，已知：各杆长为：L1=L2、L3=L；各杆面积为A1=A2=A、A3；各杆弹性模量为：E1=E2=E、E3。外力沿铅垂方向，求各杆的内力。38

Equilibriumequations:

平衡方程:CPABD123PAN1N3N2Solution：

解：39Geometricequation—compatibilityequationofdeformation：

几何方程——变形协调方程：Physicalequation—elasticlaw：

物理方程——弹性定律：CABD123A140Solvingtheequilibriumequationsandcomplementaryequationweget:

解由平衡方程和补充方程组成的方程组，得:CABD123A1Complementaryequation：detainingfromthegeometricequationandphysicalequation.

补充方程：由几何方程和物理方程得。41P1mPExample1-11Fouranglesofawoodenpolearereinforcedwithfourequalleganglesteelof40404.Theallowablestressesofsteelandwoodarerespectively[]1=160MPaand[]2=12MPa，moduleofelasticityofthemareE1=200GPaandE2=10GPa.RespectivelyDeterminetheallowablepermissibleloadP.[例1-11]木制短柱的四角用四个40404的等边角钢加固，角钢和木材的许用应力分别为[]1=160MPa和[]2=12MPa，弹性模量分别为E1=200GPa和E2=10GPa；求许可载荷P。42Geometricequation

几何方程Physicalequationandcomplementaryequation：

物理方程及补充方程：Equilibriumequations:

平衡方程:PPy4N1N2Solution：解：43PPy4N1N2Solvingequilibriumequationsandthecomplementaryequationweget:

解平衡方程和补充方程，得:Determinethepermissibleloadofthestructure：

Method1:

求结构的许可载荷：方法1:44A1=3.086cm2PPy4N1N2Sectionareaoftheanglesteelmaybeobtainedfromthetableofhot-rolledsteel:角钢截面面积由型钢表查得:45Inthecaseof△1=△2，theanglesteelwillreachthelimitstatefirst,thatisthemaximumloadisdeterminedbytheanglesteel.

所以在△1=△2的前提下，角钢将先达到极限状态，即角钢决定最大载荷。Determinethepermissibleloadofthestructure：

求结构的许可载荷：

Method2:方法2:46

aaaaN1N2Example1-12

Theupperandlowerendsofaladder-likesteelshaftarefixedattemperatureT1=5℃asshowninthefigure.Areasoftheupperandlowersegmentsarerespectively1=cm2and

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

=

12.5×10-61/0C；modulusofelasticity

E=200GPa)

[例12]如图，阶梯钢杆的上下两端在T1=5℃时被固定,杆的上下两段的面积分别=cm2，

=cm2，当温度升至T2=25℃时,求各杆的温度应力。(线膨胀系数=；弹性模量E=200GPa12.5×１０－６/℃

)47

aaaaN1N2Geometricequation：

几何方程：

Equilibriumequation：

平衡方程：Physicalequation:物理方程Solution：解：48Solvingequilibriumequationsandthecomplementaryequationweget:

解平衡方程和补充方程，得:Complementaryequation:

补充方程Temperaturestresses:

温度应力49Example1-13Diameterofacopperwireisd=2mmanditslengthisL=500mm.Tensilecurveofcopperisshowninthefigure.Tomakeelongationofthecopperwireis30mmwhetistheforcePthatwemustact

？[例1-13]铜丝直径d=2mm，长L=500mm，材料的拉伸曲线如图所示。如欲使铜丝的伸长量为30mm，则大约需加多大的力P？

s(MPa)e(%)50Solution：Deformationmayexceedtherangeoflinearelasticity,thereforetheelasticlawisnotappliedhere.Calculationshouldbedoneinthefollowing:解：变形量可能已超出了“线弹性”范围，故，不可再应用“弹性定律”。应如下计算：Fromthetensilecurve：由拉伸图知：s(MPa)e(%)511.KnowingtheelasticmodulusofsteelisE＝200GPa，theelasticmodulusofAluminumisE＝71GPa.Trytocompare:①whichmaterialproducesalargerstrainwhentheyaresubjectedtothesamestress？②whichmaterialcorrespondstoalargestresswhentheyhavethesamestrain？

钢的弹性模量E＝200GPa，铝的弹性模量E＝71GPa。试比较在同一应力作用下，那种材料的应变大？在产生同一应变的情况下，那种材料的应力大？Chapter1Exercises第一章练习题522.Forthedifferentmembersmadeofasamematerial,dotheyhavethe