弹性力学课件七

第三章平面问题的直角坐标解答3—1逆解法与半逆解法3—2矩形梁的纯弯曲3—3位移分量的求出3—4简支梁受均布荷载3—5楔形体受重力和液体压力SolutionofPlaneProblemsinRectangularCoordinatesInverseMethodandSemi-InverseMethodPureBendingofaRectangularBeamDeterminationofDisplacmentsBendingofaSimpleBeamUnderUniformLoadsTriangularGravityWall3—2矩形梁的纯弯曲xyh/2h/2MMh1例1.图示矩形截面梁，设板厚板长和板宽（平面应力情况），若板厚板长和宽（平面应变情况），取厚度为1，在两端受图示力偶M作用（单位宽度的力偶），体力不记,求梁内各点的应力分量,变形及位移分量.PureBendingofaRectangularBeamForconvenience,weconsideronlyaunitwidthofthebeam.LetthebendingmomentontheunitwidthbeM,whichhasadimensionof[force][length]/[length],orsimply[force].1、求应力分量a、取（x，y）=ay3无论a取何值，均能满足相容方程Thecompatibilityequationisalsosatisfiedforallvaluesoftheconstanta.应力分量：Thestresscomponentsareb、利用边界条件求系数axyh/2h/2MM上、下面左、右面自然满足自然满足M与应力形成一致时取+；反之，取-（1）（2）式（1）自然满足由式（2）应力分量：注意：梁截面惯性矩momentofinertia应力分量：上述结果与材料力学中的完全相同：梁的各纤维只受按直线分布的弯应力Eqscoincidecompletelywiththeelementarysolutiongiveninmechanicsofmaterials.注意：组成力偶的面力按直线规律分布时，上述结果完全精确，如果两端的面力按其他方式分布，上述结果有误差。Thesolutionisexactonlywhenthesurfaceforcesontheendsofthebeamconsistofnormaltractionsproportionaltoy.Ifthecouplesontheendsareappliedinanyothermanner,thissolutionisnolongerexact,becauseitwillnotpreciselysatisfytheboundaryconditionsontheends.按照圣维南原理，只在梁端附近有显著误差，在离开梁端较远处，误差可以不计2、求应变分量（平面应力问题）straincomponentsplanestressWhenthestresscomponentshavebeenfoundbysolutionintermsofstresses,wecanfindthedisplacementcomponentsbyphysicalandgeometricequations.Thiswillbeillustratedwiththeproblemofpurebendingasanexample.3、求位移分量将u、v代入下式：Substitutingu,vinto：移项得：左边是y的函数，右边是x的函数，两边要相等，只可能等于常数Itcanberewrittenas：Wenotethattheleftsideisafunctionofyalonewhiletherightsideisafunctionofxalone.

Afunctionofycanbeequaltoafunctionofxonlywhentheyarebothequaltoaconstant,say.上述两式积分后得：（1）（2）Byintegration,wehave：上述两式代入u、v的表达式得：Thus,thedisplacementcomponentscanbeexpressedas：(1)(2)From(1),wecanseethat,nomatterhowthebeamissupported,theangleofrotationofanyverticallineelementinthebeamisOnagivencrosssectionofthebeam,isconstant,sincexisconstant.Thisshowsthatalltheverticallineelementsinthecrosssectionhavethesameangleofrotationduetobendingofthebeam,andconsequentlyprovesthatthecrosssectionremainsplaneafterbending.From(2),weseethatallthelongitudinallines(fibres)ofthebeamwillhavethesamecurvatureafterbending.Thisisthebasicformulaforcomputationofbeamdeflectionsinmechanicsofmaterials.根据边界条件求积分常数将上述纯弯曲梁加上约束：xyh/2h/2MM（1）使之成为简支梁Ifthebeamissimplysupported：边界条件：TheconditionsofconstraintareTherewillbeneitherhorizontalnorverticaldisplacementsatthehingedsupport,andtherewillbenoverticaldisplacementeither,atthebarsupport.代入位移表达式得：最后得简支梁的位移分量：FromwhichwehaveInsertionofthesevaluesinto(1)and(2)yields梁轴线的挠曲线方程：xyh/2h/2MMThedeflectionofthebeamaxisis：Whichisthesameastheresultobtainedinmechanicsofmaterials.（2）若梁是悬臂梁Inthecaseofacantileverbeam边界条件：Theconditionsofconstraint：由此求得：悬臂梁的位移分量：Fromwhichwehave梁轴净线的优挠曲扶线方黄程：上述结宪论与材顿料力学欢中的相创同验正静材料荒力学训中的纲平面丢假定夺是否排成立采？（1）（2）The停de斥fle旷cti羞on冷of渠the妇be裳am煤axi民si筛s：Whi熄ch闯is聋als倍ot辽he愉sam喇ea敲st波he围res捏ult岗ob针tai尊ned常in犁me厕cha欢nic杜so竟fm杨ate纱ria乔ls.由（胶1）由（2宗）无论舍约束川情况右如何范，铅奸垂直兰线的怀转角醉都是，恭同一净截面盛，x为常欺数，莲所以洁为笔常数糟，即驰同一池平面阅转角株相同净，横担截面鹊保持远为平汉面无论约束情况如何，梁的纵向曲率为与材礼料力傲学求瓣梁的露挠度采时所属用的膝基本馆公式石一致对平面应变问题，