中国石油大学工程力学课件tm405mat.ppt

2020/9/25,Kylinsoft,MOM-5-1,,第五章弯曲变形Chapter5DeflectionsofBeams,海盗船长,2020/9/25,Kylinsoft,MOM-5-2,5.1Differentialequationsofthedeflectioncurve5.2Integrationofthebendingmomentequation5.3ofsuperposition5.4Staticallyindeterminatebeams5.5StrainenergyofbendingAppendixBDeflectionsandslopesofbeams,Contents,2020/9/25,Kylinsoft,MOM-5-3,,,5.1Differentialequationsofthedeflectioncurve,,Deflectioncurve挠曲线,,Deflection挠度,v，横截面形心沿垂直于梁轴线方向的位移，向上为正，向下为负。,Slope、Angleofrotation转角,横截面变形前后的夹角，逆时针为正，顺时针为负。,C,,,1.Terms术语,2020/9/25,Kylinsoft,MOM-5-4,C,,,Centerofcurvature曲率中心O,Radiusofcurvature曲率半径r,5.1Differentialequationsofthedeflectioncurve,Curvature曲率,或,2020/9/25,Kylinsoft,MOM-5-5,C,,,5.1Differentialequationsofthedeflectioncurve,,Forpurebending,2.Differentialequationsofthedeflectioncurve挠曲线微分方程,2020/9/25,Kylinsoft,MOM-5-6,横力弯曲时,剪力与弯矩都引起弯曲变形；但是，当梁的跨度远大于横截面高度时，剪力引起的弯曲变形与弯矩引起的弯曲变形相比可略去不计；不过这时弯矩M为截面位置x的函数。,C,,,5.1Differentialequationsofthedeflectioncurve,,Signconvention,,,,2020/9/25,Kylinsoft,MOM-5-7,挠度、转角、弯矩、剪力、载荷之间的微分关系关系）,C,,,5.1Differentialequationsofthedeflectioncurve,若EI为Constant,可得其它形式控制方程,3.*Alternativesofthegoverningequation控制方程的其它形式,2020/9/25,Kylinsoft,MOM-5-8,5.2Integrationofthebendingmomentequation,C,,,积分常数C、D的确定（需要两个方程）,若EI为Constant,1.Integrationofthebendingmomentequation弯矩方程的积分,2020/9/25,Kylinsoft,MOM-5-9,bContinuityconditions连续性条件,5.2Integrationofthebendingmomentequation,C,,,aBoundaryconditions边界条件,2020/9/25,Kylinsoft,MOM-5-10,5.2Integrationofthebendingmomentequation,C,,,,,2.*BoundaryconditionsContinued,2020/9/25,Kylinsoft,MOM-5-11,5.2Integrationofthebendingmomentequation,C,,,,,2020/9/25,Kylinsoft,MOM-5-12,[v]为许用挠度，[q]为许用转角，具体数值可查有关手册。,5.2Integrationofthebendingmomentequation,C,,,3.Criterionsofrigidity刚度条件,建筑钢梁的许可挠度,机械传动轴的许可转角,精密机床的许可转角,2020/9/25,Kylinsoft,MOM-5-13,4.Measurementsimprovingtherigidityofbeams提高梁刚度的措施改善结构形式，减小弯矩；增加支承，减小跨度；选择合理的截面形状，提高惯性矩，如工字形截面、空心截面等；选用合适的材料，增加弹性模量（效果并不显著）。,5.2Integrationofthebendingmomentequation,C,,,2020/9/25,Kylinsoft,MOM-5-14,5.2Integrationofthebendingmomentequation,C,,,Sample5.1积分法均布载荷简支梁挠度Sample5.2积分法集中载荷简支梁挠度Sample5.3积分法均布载荷悬臂梁挠度Sample5.4积分法集中载荷悬臂梁挠度Sample5.5积分法均布载荷静不定梁挠度1Sample5.6积分法均布载荷静不定梁挠度2,5.Samples例子,2020/9/25,Kylinsoft,MOM-5-15,Sample5.3积分法均布载荷悬臂梁挠度,挠度曲线微分方程,边界条件x0v0,弯矩方程,挠度曲线微分方程,D0,边界条件x0q0,,C0,2020/9/25,Kylinsoft,MOM-5-16,挠度曲线方程,转角方程,,最大挠度在梁自由端B,最大转角在梁自由端B,注意挠度曲线在固支端A点与轴线x是相切的,,,2020/9/25,Kylinsoft,MOM-5-17,Sample5.2a积分法集中载荷简支梁挠度,挠度曲线微分方程,弯矩方程,约束反力,0xa,axl,挠度曲线微分方程,0xa,axl,2020/9/25,Kylinsoft,MOM-5-18,挠度曲线微分方程,边界条件1x0v0,由1D10,边界条件2xlv0,0xa,axl,连续条件3,连续条件4,由3,由2,,由4,2020/9/25,Kylinsoft,MOM-5-19,梁端转角,挠度曲线方程,转角方程,AC0xa,CBaxl,,,,若ab则qmaxqBqA,,截面C的转角,2020/9/25,Kylinsoft,MOM-5-20,最大挠度应是在转角为零的截面,则最大挠度应是在AC段某处，而不是在C点,若ab则,令,因为qA0;qB0,x0的位置是比较接近梁的中点。,2020/9/25,Kylinsoft,MOM-5-21,Discussion1.Whenab0.5l,x0.5l.Vmaxoccursinthecenterofthebeam.,2.WhenPsupportB,i.e.b0,,2020/9/25,Kylinsoft,MOM-5-22,C,,,5.3ofsuperposition,AppendixBDeflectionsandslopesofbeamsP231-232Table1DeflectionsandslopesofsimplebeamsTable2DeflectionsandslopesofcantileverbeamsTable3Deflectionsandslopesofbeamswithanoverhang,1.查表求挠度和转角,2020/9/25,Kylinsoft,MOM-5-23,C,,5.3ofsuperposition,,2.叠加法的可行性证明,前提材料服从虎克定律、弯曲变形很小,,2020/9/25,Kylinsoft,MOM-5-24,C,,,5.3ofsuperposition,4.叠加法（二）,Sample5.7叠加法简支梁挠度Sample5.8叠加法悬臂梁挠度1Sample5.9叠加法悬臂梁挠度2,Sample5.10叠加法外伸梁挠度1Sample5.11叠加法外伸梁挠度2Sample5.12叠加法刚架,5.叠加法（三）,Sample5.13叠加法简支梁挠度3,3.叠加法（一）,2020/9/25,Kylinsoft,MOM-5-25,Sample5.7叠加法简支梁挠度,,,,C点挠度,A,B截面转角,2020/9/25,Kylinsoft,MOM-5-26,Solution1.DuetoPFromAppendixBtable1,2.Duetoq,Sample5.9叠加法悬臂梁挠度2,Findthemaximumdeflectionandmaximumrotationforthebeam.,2020/9/25,Kylinsoft,MOM-5-27,2020/9/25,Kylinsoft,MOM-5-28,FindyCandBforthebeamwithloadsshowninthefigure.,SolutionConsiderthedeflectionsforthetwosegmentsABABCandBCseparately.,Fromtable1,Sample5.10叠加法外伸梁挠度1,2020/9/25,Kylinsoft,MOM-5-29,Sample5.13叠加法悬臂梁挠度3,设在x处有一微小力dPqdx,,查表P232第8式令axPqdx,2020/9/25,Kylinsoft,MOM-5-30,C,,,5.4Staticallyindeterminatebeams,一个平面梁可以列三个平衡方程SX0SY0SM0,静定梁约束反力数目与平衡方程数目相等的梁。,静不定梁约束反力数目多于静力平衡方程数目的梁。,静不定次数约束反力数目与静力平衡方程数目之差。,,,,,,,,,一次静不定,一次静不定,二次静不定,三次静不定,1.Staticallyindeterminatebeams静不定梁,2020/9/25,Kylinsoft,MOM-5-31,将静定基与静不定梁相比较，在多余约束处找到变形协调条件；,qA0,C,,,5.4Staticallyindeterminatebeams,静定基ReleasedStructure指将静不定梁上的多余约束除去并把相应约束反力作为主动力所得到的“静定基本体系”。,vB0,代入力位移关系（Force-displacementrelations）得到求解静不定问题所需的相容方程（Compatibilityequations）。,和平衡方程（Equilibriumequations）联立求解出所有约束反力。,,,,,,,2.ofsolution求解方法,2020/9/25,Kylinsoft,MOM-5-32,Sample5.14静不定梁多跨度梁1Sample5.15静不定梁多跨度梁2Sample5.16静不定梁支撑悬臂梁Sample5.17静不定梁弹性支撑梁1Sample5.18静不定梁弹性支撑梁2,C,,,5.4Staticallyindeterminatebeams,3.Samples例子,2020/9/25,Kylinsoft,MOM-5-33,Sample5.14静不定梁多跨度梁1,2.变形协调条件vCvCR-vCq0,3.建立补充方程.,查表,1.建立静定基（作图）,2020/9/25,Kylinsoft,MOM-5-34,Sample5.16静不定梁支撑悬臂梁,2.找到变形协调条件,静定基,vB0,,,vBq,vBR,3.建立补充方程.,查表,vBvBR-vBq0,1.建立静定基（作图）,2020/9/25,Kylinsoft,MOM-5-35,对该系统求剪力方程和弯矩方程；作剪力图和弯矩图,2020/9/25,Kylinsoft,MOM-5-36,2.变形协调条件qAqAMqAq0,静定基,qA0,3.建立补充方程.,,,查表,1.建立静定基（作图）,,2020/9/25,Kylinsoft,MOM-5-37,1.建立静定基（作图）,2.找变形协调条件,3.建立补充方程,查表,补充方程,,DBvB,Sample5.17静不定梁弹性支撑梁1,2020/9/25,Kylinsoft,MOM-5-38,,C,,,5.5StrainEnergyofBending,1.StrainEnergyofBending弯曲应变能,,线弹性范围内纯弯曲梁,线弹性范围内横力弯曲梁内，剪力引起的应变能与弯矩引起的应变能相比可略去不计,2020/9/25,Kylinsoft,MOM-5-39,C,,,5.5StrainEnergyofBending,另外的推导方法,忽略剪力引起的应变能,应变能是载荷的二次函数，叠加原理不成立,2020/9/25,Kylinsoft,MOM-5-40,C,,,5.5StrainEnergyofBending,2.Samples例子,Sample5.19弯曲应变能,2020/9/25,Kylinsoft,MOM-5-41,C,,,AppendixBDeflectionsandslopesofbeams,说明,ndeflectionqangleofrotationEIconstant,2020/9/25,Kylinsoft,MOM-5-42,C,,Table1Deflectionsandslopesofsimplebeams,AppendixBDeflectionsandslopesofbeams,,2020/9/25,Kylinsoft,MOM-5-43,C,,Table2Deflectionsandslopesofcantileverbeams,AppendixBDeflectionsandslopesofbeams,,2020/9/25,Kylinsoft,MOM-5-44,C,,,Table3Deflectionsandslopesofbeamswithanoverhang,AppendixBDeflectionsandslopesofbeams,2020/9/25,Kylinsoft,MOM-5-45,位移与变形的相依关系,比较二梁的受力、弯矩、位移与变形,①位移除与变形有关外，还与约束有关;②总体变形是微段变形累加的结果;③有位移不一定有变形;④有变形不一定处处有位移。,几点重要结论,结论与讨论,2020/9/25,Kylinsoft,MOM-5-46,已知刚度为EI的简支梁的挠曲线方程为,据此推知的弯矩图有四种答案，试分析哪一种是正确的,结论与讨论,梁的连续光滑挠曲线,①由M的方向确定轴线的凹凸性;②由约束性质及连续光滑性确定挠曲线的大致形状及位置。,答案A,2020/9/25,Kylinsoft,MOM-5-47,结论与讨论,答案D,2020/9/25,Kylinsoft,MOM-5-48,结论与讨论,答案D,2020/9/25,Kylinsoft,MOM-5-49,结论与讨论,答案C,2020/9/25,Kylinsoft,MOM-5-50,正六边形截面的简支梁承载如图所示，若按a，b两种方式放置，则Aa有较高的抗弯刚度Bb有较高的抗弯刚度Ca、b的抗弯刚度相同Da、b的抗弯刚度相对大小要视具体材料性质而定,,其它,结论与讨论,答案C,2020/9/25,Kylinsoft,MOM-5-51,已知长度为l的等截面直梁的挠曲线方程为,据此推知x0和xl处梁的支撑情况为Ax0处为固定端，xl处为自由端Bx0处为自由端，xl处为固定端Cx0和xl处都为铰支座Dx0处为固定端，xl处为铰支座,结论与讨论,答案A,2020/9/25,Kylinsoft,MOM-5-52,Homework,P110-5.3c*、P111-5.8、P111-5.9、P112-5.13*,2020/9/25,Kylinsoft,MOM-5-53,,制作群,素材收集海盗船长多媒体制作海盗船长总编辑海盗船长审核海盗船长总策划海盗船长赞助海盗船长,本章结束,