平面体系的机动分析PPT（双语71）

（GeometricConstructionAnalysisofPlaneSystems）第二二章章ChapterII平面面体体系系的的机机动动分分析析§2-1引言言Introduction结构构：：由由杆杆件件、、结结点点和和支支座座组组成成的的杆杆件件体体系系Structureconsistsofmembers,jointsandsupports.结构构必必须须是是在在不不考考虑虑材材料料变变形形的的条条件件下下能能保保持持几几何何形形状状和和位位置置不不变变的的杆杆件件体体系系。。Structuremustmaintainitsgeometricshapeandpositionswithoutconsiderationofthedeformationofmaterials.在不不考考虑虑材材料料变变形形的的条条件件下下，，杆杆件件体体系系可可分分为为如如下下两两种种类类型型Ifthedeformationofmaterialsisneglected,thenframedsystemscanbeclassifiedintotwocategories：几何何不不变变体体系系(geometricallystablesystem)几何何可可变变体体系系(geometricallyunstablesystem)几何何不不变变体体系系(geometricallystablesystem)在任任意意荷荷载载作作用用下下，，几几何何形形状状及及位位置置均均保持持不不变变的的体体系系。。（（不不考考虑虑材材料料的的变变形形））Undertheactionofanyloads,thesystemstillmaintainitsshapeandremainsitslocationifthedeformationsofthemembersareneglected.几何何可可变变体体系系(geometricallyunstablesystem)在一一般般荷荷载载作作用用下下，，几几何何形形状状及及位位置置将将发发生生改改变变的的体体系系。。（（不不考考虑虑材材料料的的变变形形））Undertheactionofanyloads,thesystemwillchangeitsshapeanditslocationifthedeformationsofthemembersareneglected.结构构机构构几何不变体系geometricallystablesystem几何可变体系geometricallyunstablesystem体系系组组成成分分析析的的目目的的ThepurposeofgeometricConstructionanalysis：1.判定定体体系系是是否否几几何何不变变toestimatewhetherornotasystemisgeometricallystable；2.研究究几几何何不不变变体体系系的的组组成成规规则则todiscussthegeometricconstructionrulesofstablesystems；3.区分分静静定定和和超超静静定定的的组组成成distinguishstaticallydeterminatestructuresandstaticallyindeterminatestructures。刚片(rigidbody)———平面刚体体。形状可任任意替换换maybereplacedbybodyofanyshape.杆件，几几何不变变部分均均可视为为刚片membersorstablepartsmaybelookedatasrigidbodies§2-1平面体系系的自由由度(degreesoffreedomofplanarsystem)自由度--确定物体体位置所所需要的的独立坐坐标数目目或体系运运动时可可独立改改变的几几何参数数数目Degreesoffreedomofasystemarethenumbersofindependentmovementsorcoordinateswhicharerequiredtolocatethesystemfully.xy平面内一一点forapointinplanen=2AxyBForplanerigidbody平面刚体体——n=3联系或约约束(linkorrestraint)一根链杆杆为一个个约束onelinkisequivalenttoonerestraint联系（约约束）--减少自由由度的装装置。linkorrestraint–devicesorconnectionsreducingthedegreesofasystem平面刚体——刚片n=31个单铰=2个联系onesimplejointequivalentto2restraints单铰联后后n=4xyαβ每一自由由刚片3个自由度度forecerybodyn=3两个自由由刚片共共有6个自由度度2bodieshave6degrees单铰simplejointxyBAC两刚片用用两链杆杆连接,两相交链链杆构成成一虚铰2rigidbodiesareconnectedby2linkswhichformonevirtualhingen=41连接n个刚片的的复铰=(n-1)个单铰Onemultiplejointconnectingnbarsisequivalentto(n-1)simplejointsn=5复铰等于多少少个单铰？A复刚结点点multiplerigidjoint=(n-1)simplerigidjoints连接n个杆的复复刚结点点等于多少少个单刚刚结点？？单刚结点点相当于于3个联系onerigidjoint=3restraintsn=3W=3m-(2h+b)m---刚片数thenumbersofrigidbodies（excludingfoundation不包括地地基）h---单铰数thenumbersofsimplejointsb---单链杆数数（含支支杆）thenumbersoflinks体系的计算自由度：：计算自由由度等于于刚片总总自由度度数减总总联系数数Thecomputeddegreesoffreedom=thetotalnumbersofdegreesoffreedomofrigidbodies–totalnumbersofrestraints铰结链杆杆体系---完全由两两端铰结结的杆件件所组成成的体系系linksystemconnectedbyhinges–systemofbarsconnectedbyhingesattheendsofthebars.铰结链杆杆体系的的计算自自由度Thecomputeddegreesoffreedom：W=2j-bj--结点数thenumbersofhinges;b--链杆数,含支座链链杆thenumbersoflinksincludingthelinksatthesupports例1：计算图图示体系系的自由由度DeterminethenumbersofdegreesoffreedomofthefollowingsystemGW=3×8-(2××10+4)=0ACCDBCEEFCFDFDGFG32311有几个个刚片？有几个个单铰铰？例2：计算算图示示体系系的自自由度度DeterminethenumbersofdegreesoffreedomofthefollowingsystemW=3×9-(2××12+3)=0332112按刚片片计算算9根杆,9个刚片片有几个个单铰铰？3根单链链杆另一种种解法法anothersolutionW=2×6-12=0按铰结结计算算6个铰结结点12根单链链杆W=0,体系是否一一定几何不不变呢呢？讨论W=3×9-(2××12+3)=0体系W等于多多少？可变吗吗？322113有几个个单铰铰？能够减减少体体系的的自由由度的的联系系称为为必要联联系Restraintswhichreducethedegreesoffreedomisnamedasnecessaryrestraints,otherwisetheyarecalledredundantrestraints.因为除除去图图中任任意一一根杆杆，体体系都都将有有一个个自由由度，，所以以图中中所有有的杆杆都是是必要的的联系系。Becausetheremovalofanybarinthesystemwillincreaseonedegreeoffreedom,thereforeallbarsarenecessaryrestraints除去联联系后后，体体系的的自由由度并并不改改变，，这类类联系系称为为多余联联系Restraints,removalofwhichdoesn’’tchangethedegreesoffreedom,isnamedasredundantrestraints.下部正正方形形中任任意一一根杆杆，除除去都都不增增加自自由度度，都都可看看作多余的的联系系。图中上上部四四根杆杆和三三根支支座杆杆都是是必要的的联系系。例3：计算图示体系的自由度W=3×9-(2××12+3)=0W=0,但布置不不当几何可可变。。上部有有多余联系系，下部缺缺少联系。。W=2×6-12=0W=2×6-13=-1<0例4：计算图示体系的自由度W<0,体系是否一一定几何不不变呢呢？上部具有多多余联系系W=3×10-(2××14+3)=-1<0计算自由度=体系真实的自由度？W=3×9-(2××12+3)=0W=2×6-12=0缺少联联系几何可可变W=3×8-(2××10+3)=1W=2×6-11=1W>0,缺少足足够联联系，，体系系几何何可变变Restraintsarenotenough,unstable。W=0,具备成成为几几何不不变体体系所所要求求的最最少联联系数数目hastheminimumnecessarynumbersofrestraintsforstablesystem。W<0，体系具具有多多余联联系hasredundantrestraints。W>0体系几何可变unstableW<0体系几何不变?小结结summary二刚片片规则则：two-rigid-bodyrule:两个刚刚片用用一个个铰和和一根根不通过过此铰铰的链杆杆相联联，组组成无多余余联系系的几几何不不变体体系。。2rigidbodies,connectedby1hingeand1linkthatdoesnotcrossthehinge,formaninternallystablesystemwithnoredundantrestraints.两刚片规则则two-rigid-bodyrule§2-3几何不变体体系的基本本组成规则则Geometricconstructionrulesofplanarstableframedsystems虚铰：联结两个刚刚片的两根根相交链杆杆的作用，，相当于在在其交点处处的一个单单铰，这种种铰称为虚虚铰（瞬铰铰）If2noncol-linearlinksconnecting2rigidbodiesintersectatapointoutsidethe2rigidbodies,thentheintersectionisreferredtoasavirtualorinstantaneoushinge。EF二刚片规则则：two-rigid-bodyrule:两个刚片用用三根不全平行也也不交于同同一点的链杆相联联，组成无无多余联系系的几何不不变体系2rigidbodies,connectedby3links,whicharenonparallelandnonconcurrentcrossthehinge,formaninternallystablesystemwithnoredundantrestraints.。三边在两边边之和大于于第三边时时,能唯一地组组成一个铰铰接三角形形——基本出发点点3bars,whenthesummationofthelengthsofany2barsisgreaterthanthelengthof3-done,canformuniquelyatriangular.铰结三角形形-几何不变形形triangularjoinedpairwisebyhingesisstable.三刚片规则则3-rigid-bodyrule：三个刚片用用不在同一直直线上的三个单单铰两两相相连，组成成无多余联联系的几何何不变体系系。3rigidbodiesjoinedpair-wisebyhinges,providedthatthe3hingesdon’tlieinthesamestraightline,formaninternallystablesystemwithnoredundantrestraints.例如三铰拱拱Threehingedarch大地、AC、BC为刚片;A、B、C为单铰无多余几何何不变二元体binarysystem---不在一直线线上的两根根链杆连结结一个新结结点的装置置。2non-collinearlinksconnectedbyahinge二元体规则则：Binarysystemrule:在一个体系系上增加或或拆除二元元体，不改改变原体系系的几何构构造性质Thegeometricconstructionpropertyofasystemwillnotchangeifabinarysystemisattachedtoordetachedfromthesystem。C减二元体简简化分析加二元体组组成结构structureformedbyAttachingofbinarysystems如何减二元元体？Howtoremovebinarysystems?IIIIIIOO是虚铰吗？有二元体吗？是什么体系？O不是有无多不变试分析图示示体系的几几何组成Analyzethegeometricconstruction.有虚铰吗Anyvirtualhige？有二元体吗吗？Anybinarysystem?是什么体系系？Whatsystem?无多余几何何不变stablewithoutanyredundantrestraints没有none有Yes三个分析规规则的统一一性瞬变体系--原为几何可可变，经微微小位移后后即转化为为几何不变变的体系。。instantaneouslyunstablesystem––initiallyunstable,butbecomesstableafterinfinitesimaldisplacement.ABCPC1§2-4瞬变体系Instantaneouslyunstablesystems微小位移后后不能继续续位移Cannotdisplaceafterinfinitesimaldisplacement.不能平衡cannotmaintainequilibrium瞬变体系的的其它几种种情况severalcasesforinstantaneouslyunstablesystem：常变体系movablesystem瞬变体系instantaneouslyunstablesystem§2-4分析示例Examples加、减二元元体removingorAttachingbinarysystems去支座后再再分析无多几何不不变stablewithoutanyredundantrestraints瞬变体系instantaneouslyunstablesystem加、减二元元体removingorattachingbinarysystems无多几何不不变stablewithoutanyredundantrestraints找虚铰findvirtualhinges无多几何不不变stablewithoutanyredundantrestraints行吗？它可变吗？找刚片、找虚虚铰无穷行吗？ⅠⅡⅢO13O12O23无多几何不不变stablewithoutanyredundantrestraints瞬变体系instantaneouslyunstablesystemDEFG找刚片无多几何不不变stablewithoutanyredundantrestraintsABCDEF找刚片内部可变性ABCDE可变吗？有多余吗？如何才能不不变？ABCDE加减二元体体removingorattachingbinarysystems几何组成分分析的一般般步骤分析一个体体系几何组组成时，应应注意刚体体形状可任任意改换。。1.先去除二元元体，简化化体系。2.然后后找找铰铰结结三三角角形形作作刚刚片片。。3.如果果得得到到的的刚刚片片数数多多，，那那么么尝尝试试用用三三刚刚片片规规则则，，否否则则用用二二刚刚片片规规则则。。Restraintsubstitutionshouldbeusedinanalysis.1.Atfirstweshouldremovebinarysystemstoreducethesystemtosimplifyanalysis.2.Thenfindhingedtriangularsasrigidbodies.3.Ifthenumbersofrigidbodiesisgreat,thenweshouldtrytouse3-bodyrule,otherwiseweshouldtrytouse2-bodyrule.(a)一铰铰无无穷穷远远情情况况Thecasesof1virtualhingesatinfinite§2-5三刚刚片片虚虚铰铰在在无无穷穷远远处处的的讨讨论论Thecasesofvirtualhingesatinfinite不平行相当于不全平行的3根连杆连接。几何不变体系stablewithoutanyredundantrestraints几何瞬变体系instantaneouslyunstablesystem

平行此体体系系等等同同于于两两刚刚片片I、II用三三个个平平行行的的不不等等长长的的链链杆杆1、2、O13O23连接四杆不全平行几何不变体系stablewithoutanyredundantrestraints(b)两铰无穷远情情况Thecasesof2virtualhingesatinfinite三个铰O12、O13和O23不共线四杆全平行几何瞬变体系instantaneouslyunstablesystem

由于下面2对不等长的杆杆微小转动后后汇交于有限限远处，而成为几何不不变几何常变体系mechanism平行等长四杆平行等长几何常变体系mechanism三铰无穷远如何?请大家自行分析!静定结构staticallydeterminatestructures：仅由静力平衡衡条件就可唯唯一确定全部部反力和内力力。allsupportreactionsandinternalforcescanbedeterminedbysolvingtheequilibriumequations超静定结构staticallyindeterminatestructures：仅由静力平衡衡条件不能唯唯一确定全部部反力和内力力。allsupportreactionsandinternalforcescannotbedeterminedbysolvingtheequilibriumequations：§2-7几何组成与静静定性的关系系静定结构staticallydeterminatestructures—无多余联系的的几何不变体体系stablesystemswithoutredundantrestraints超静定结构staticallyindeterminatestructures——有多余联系的的几何不变体体系stablesystemswithredundantrestraints静定结构staticallyindeterminatestructureFFBFAyFAx无多余联系几几何不变stablesystemswithoutredundantrestraints。如何求支座反反力Howtodeterminesupportreactions?FFBFAyFAxFC超静定结构staticallyindeterminatestructure有多余联系几几何不变stablesystemwithredundantrestraints。能否求全部反反力?Whetherallsupportreactionscanbedetermined?DEFG唯一吗unique？如何变静定Howtoobtainstaticallydeterminatesystem？体系systems几何不变体系系stable几何可变体系系unstable有多余联系withredundantrestraints无多余联系withoutredundantrestraints常变movable瞬变instantaneouslyunstablesystem可作为结构canserveasstructure静定结构staticallydeterminate超静定结构staticallyindeterminate不可作结构cannotserveasstructure小结Summary结论与讨论Conclusionsanddiscussions正确区分静定定、超静定，，正确判定超超静定结构的多余联系系数十分重要要Determiningcorrectlystaticallydeterminateandstaticallyindeterminatesyst