北京航空航天大学-材料力学课件-Chapt6

Chapter6DeflectionsofBeams梁的变形,1Introduction引言2Differentialequationofbeamdeflection梁变形微分方程3Integrationmethodtocalculatedeflections计算梁位移的积分法4Macaulaysmethodtocalculatedeflections计算梁位移的奇异函数法5Methodofsuperpositiontocalculatedeflections计算梁位移的叠加法6StaticallyIndeterminateBeams静不定梁7Stiffnessanalysisanddesignofbeams梁的刚度条件与合理设计,Purposeofthischapter:,1Stiffnessofbeam梁刚度,2Staticallyindeterminatebeam静不定梁,3ProblemsofStability稳定问题,Lastchapter:,1Internalforceofbeam,2Stressandstrengthanalysisofbeam,3,1Introduction引言,Deflectioncurve挠曲轴Deflectionandslope挠度与转角,deflectioncurve挠曲轴,Neglectingtheeffectofshearforces,theentiretransversesectionofthebeam,remainsplaneandnormaltothedeflectioncurve忽略剪力对弯曲变形的影响,因而横截面仍保持平面,并与挠曲轴正交,挠曲轴,Theaxisofabeamthatisbentintoanarc,iscalleddeflectioncurve(变弯后的梁轴称为挠曲轴),Underthecaseofsymmetricbending,thedeflectioncurveisintheplaneofsymmetry对称弯曲时，挠曲轴为位于纵向对称面的平面曲线,Deflectionandslope挠度与转角,挠曲轴,转角,挠度,Simplification:,(smalldeflection小变形),Deflection(挠度)verticaldisplacementofcentroidofcrosssection(横截面形心在垂直于梁轴方向的位移),挠曲轴方程,Slope(转角)angleofrotationofcrosssection(横截面绕中性轴的转角),转角方程,（Neglecttheeffectofshearforces忽略剪力影响）,（rad）,小变形的条件下，水平位移比挠度小得多，可以略而不计。,2Differentialequationofbeamdeflection梁变形微分方程,Differentialequationofdeflectioncurve挠曲轴微分方程ApproximateDifferentialequationofdeflectioncurve挠曲轴近似微分方程,Differentialequationofdeflectioncurve挠曲轴微分方程,（purebending纯弯）,（推广到非纯弯non-uniformbending）,wdeflection挠度smaxb时）,例4-2计算截面A的挠度,解：,（）,例4-3建立通用挠曲轴微分方程，写出位移边界条件,解：,5Methodofsuperpositiontocalculatedeflections计算梁位移的叠加法,Methodofsuperposition(I)叠加法(I)Methodofsuperposition(II)叠加法(II)(逐段变形效应叠加法)Examples例题,Methodofsuperposition(I)叠加法,Thedeflectionofthebeamcausedbyseveraldifferentloadsactingsimultaneouslycanbefoundbysuperimposingthedeflectionscausedbytheloadsactingseparately.当梁上同时作用几个载荷时的总位移，等于各载荷单独作用时在该截面引起的位移的代数和或矢量和,分解载荷,（Smalldeflection小变形,Withinlinearlyelasticrange比例极限内）,Theconditionsofmethodofsuperpositiontobevalid:Loadanddeformationkeepslinearrelationship叠加法适用范围：力与位移之间的线性关系(小变形，比例极限内),两类情况：叠加单个载荷引起的位移;叠加分段变形引起的位移,Methodofsuperposition(II)逐段变形效应叠加法,Resolve分解,Calculatethedeflectioncausedbythedeformationofeachsectionofthebeam分别计算各梁段的变形在需求位移处引起的位移,Summationofdisplacement位移之和（代数和或矢量和）,注意：分段计算时，除所研究梁段变形外，其余梁段均视为刚体,广义迭加法（GeneralizedSuperpositionMethod）,Known：EI=const常数,Tofind:wA,零弯矩，不变形,相当于悬臂梁,SolutionI:,Known：EI=const常数,Tofind:wA,刚化AB段,刚化BC段,SolutionII:,例题,例6-6q(x)=q0cos(px/2l)，利用叠加法求wB=?,解：,(）,(）,例6-8,解：,刚化BC段,刚化AB段,例求自由端位移d,挠曲轴与外力作用面不重合,一般情况,解：,利用对称性求简支梁与悬臂梁的变形关系,令l2l,P2P,习题6-14(b),仅考虑1段变形：,仅考虑2段变形：,求C点挠度与转角,对称,反对称,C点挠度为0,弯矩为0,例:求,组合梁的变形分析,用叠加法求AB梁上E处的挠度,简支梁中点挠度,刚化BC，CD为悬臂梁,刚化CD，BC看作悬臂梁,DeterminetheslopeanddisplacementofendDofthecantileveredbeamsubjectedtotwouniformlydistributedloads.,DisplacementatD:,where,where,SlopeatD:,TheslopeanddisplacementofpointCaredeterminedbymodelingthebeamasacantileveredbeamoflength2a.BetweenpointsCandDthebeamremainsstraight,TheslopeanddisplacementofpointBaredeterminedbymodelingthebeamasacantileveredbeamoflengtha.BetweenpointsBandDthebeamremainsstraight.,位移与变形的相依关系,比较二梁的受力、弯矩、变形与位移,位移除与变形有关外，还与约束有关;总体变形是微段变形累加的结果;有位移不一定有变形;有变形不一定处处有位移。,6StaticallyIndeterminateBeams静不定梁(超静定梁),Degreeofstaticalindeterminacyandredundantconfinement静不定度与多余约束Analyticalmethodofsimplystaticallyindeterminatebeams简单静不定梁分析方法Examples例题,Degreeofstaticalindeterminacyandredundantconfinement静不定度与多余约束,Redundantconfinement多余约束,Redundantreaction多余反力,Degreeofstaticalindeterminacy:Thenumberofreactionsinaccessofthenumberthatcanbefoundfromequationsofstaticequilibrium静不定度支反力（力偶）数有效平衡方程数,静不定度多余约束数,4-3=1度静不定,5-3=2度静不定,约束,多余约束：增加了未知力个数，同时增加对变形的限制与约束，前者使问题变为不可解，后者使问题变为可解。,Analyticalmethodofsimplystaticallyindeterminatebeams简单静不定梁分析方法,ChooseFbyasredundantreaction选Fby为多余力,compatibleequation变形协调条件,physicalequation物理方程,complementaryequation补充方程,staticequilibrium平衡方程,Onedegreeofstaticalindeterminacy一度静不定,Example,综合考虑三方面,Tofind:reactionsofthebeam求梁的支反力,判断梁的静不定度,用多余力代替多余约束的作用，得受力与原静不定梁相同的静定梁所谓相当系统,计算相当系统在多余约束处的位移，并根据变形协调条件建立补充方程,由补充方程确定多余力，由平衡方程求其余支反力,相当系统,通过相当系统计算内力、位移与应力等,求解依据综合考虑三方面,求解关键确定多余支反力,分析方法与步骤,相当系统,例6-9Tofind:Reactions,1.Analysis,小变形，水平反力忽略不计,2.Solution,Redundantreaction多余未知力:2,例题,Discussion:,(3)Initialstress,(2)Choicesofredundantreaction,(1)Whena=b,itssymmetricproblem:,OnlyoneofthecomplementaryequationisneededtofindMAorMB,例6-11直径为d的圆截面梁,支座B下沉d，smax=?,解：,例6-10悬臂梁AB，用短梁DG加固，试分析加固效果,解：1.静不定分析,2.加固效果分析,与Fa相比，减少50%,与相比，减少39.9%,例试求杆BC的轴力,解：,梁截面形心的轴向位移一般忽略不计,如计及梁截面形心的轴向位移，如何求解?,Tofindreactions,Tofind:deflectionofpointB(B点位移),C,习题6-24：已知弯曲刚度EI，梁单位长度重q，外伸长a,求拱起部分b。,工程问题：,力学分析与力学模型,求解,ABC段的力学分析：,位移的约束条件：,力：,(1)悬臂梁+B点多余约束,(2)简支梁:未知量b，多余约束,近似解法:忽略摩擦；刚性支撑？,7Stiffnessanalysisanddesignofbeams梁的刚度条件与合理设计,Stiffnessanalysisofbeams梁的刚度条件Designofbeamswithconsiderationofstiffness梁的合理刚度设计Examples例题,Stiffnessanalysisofbeams梁的刚度条件,最大位移控制,指定截面的位移控制,例如滑动轴承处,Designofbeamswithconsiderationofstiffness梁的合理刚度设计,Selectionofabeamcrosssection选择横截面形状,SelectionofMaterial材料的合理选择,使用较小的截面面积A，获得较大惯性矩I的截面形状，例如工字形与盒形