Direct Approach for Discrete Systems

Direct Approach for Discrete Systems
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GregDeamons,New Zealand,Professional
Published Date:03-08-2017
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2 Direct Approach for Discrete Systems Thefiniteelementmethod(FEM)consistsofthefollowingfivesteps: 1. Preprocessing:subdividingtheproblemdomainintofiniteelements. 2. Elementformulation:developmentofequationsforelements. 3. Assembly:obtainingtheequationsoftheentiresystemfromtheequationsofindividualelements. 4. Solvingtheequations. 5. Postprocessing: determining quantities ofinterest, such asstresses and strains, and obtainingvisua- lizationsoftheresponse. Step1,thesubdivisionoftheproblemdomainintofiniteelementsintoday’scomputeraidedengineering (CAE)environment,isperformedbyautomaticmeshgenerators.Fortrussproblems,suchastheoneshown inFigure2.1,eachtrussmemberisrepresentedbyafiniteelement.Step2,thedescriptionofthebehaviorof eachelement,generallyrequiresthedevelopmentofthepartialdifferentialequationsfortheproblemand itsweakform.Thiswillbethemainfocusofsubsequentchapters.However,insimplesituations,suchas systems of springs or trusses, it is possible to describe the behavior of an element directly, without consideringagoverningpartialdifferentialequationoritsweakform. Inthis chapter,we focuson step 3,how tocombine the equationsthat governindividual elements to obtaintheequationsofthesystem.Theelementequationsareexpressedinmatrixform.Priortothat,we developsomesimplefiniteelementmatricesforspringassemblagesandtrusses,step2.Wealsointroduce theproceduresforthepostprocessingofresults. 2.1 DESCRIBING THE BEHAVIOR OF A SINGLE BAR ELEMENT Atrussstructure,suchastheoneshowninFigure2.1,consistsofacollectionofslenderelements,often called bars. Bar elements are assumed to be sufficiently thin so that they have negligible resistance to torsion,bendingorshear,andconsequently,thebending,shearandtorsionalforcesareassumedtovanish. The only internal forces of consequence in such elements are axial internal forces, so their behavior is similartothatofsprings.SomeofthebarelementsinFigure2.1arealignedhorizontally,whereasothersare positionedatanarbitraryangleasshowninFigure2.2(b).Inthissection,weshowhowtorelatenodal e e internalforcesactingatthenodestothecorrespondingnodaldisplacements,whicharedenotedbyðF ;F Þ 1 2 e e andðu ;u Þ,respectively,forthebarinonedimensionasshowninFigure2.2(a).Intwodimensions,the 1 2 e e e e e e e e nodalforcesofanelementareðF ;F ;F ;F Þandthenodaldisplacementsareðu ;u ;u ;u Þ. 1x 1y 2x 2y 1x 1y 2x 2y A First Course in Finite Elements J. Fish and T. Belytschko 2007 John Wiley & Sons, Ltd ISBNs: 0 470 85275 5 (cased) 0 470 85276 3 (Pbk)12 DIRECT APPROACHFOR DISCRETESYSTEMS Figure 2.1 Abridgetruss. Notation.Throughoutthistextbook,thefollowingnotationisused.Elementnumbersaredenotedby superscripts.Nodenumbersaredenotedbysubscripts;whenthevariableisavectorwithcomponents,the componentisgivenafterthenodenumber.Whenthevariablehasanelementsuperscript,thenthenode numberisalocalnumber;otherwise,itisaglobalnodenumber.Thedistinctionbetweenlocalandglobal ð5Þ node numbers will be described later in this section. For instance, u is the y-component of the 2y displacementatnode2ofelement5.WewillstartbyconsideringahorizontallyalignedelementinSection 2.1.Two-dimensionalproblemswillbeconsideredinSection2.4. Consider abarelementpositioned alongthex-axisasshowninFigure2.2(a). Theshapeofthe cross section is quite arbitrary as shown in Figure 2.3. In this chapter, we assume that the bar is straight, its materialobeysHooke’slawandthatitcansupportonlyaxialloading,i.e.itdoesnottransmitbending,shear e e e ortorsion.Young’smodulusofelementeisdenotedbyE ,itscross-sectionalareabyA anditslengthbyl . Because of the assumptions on the forces in the element, the only nonzero internal force is an axial internalforce,whichiscollinearwiththeaxisalongthebar.Theinternalforceacrossanycrosssectionof e thebarisdenotedbyp .Theaxialstressisassumedtobeconstantinthecrosssectionandisgivenbythe internalforcedividedbythecross-sectionalarea: e p e  ¼ : ð2:1Þ e A Theaxialforceandthestressarepositiveintensionandnegativeincompression. Thefollowingequationsgovernthebehaviorofthebar: 1. Equilibriumoftheelement,i.e.thesumofthenodalinternalforcesactingontheelementisequalto zero: e e F þF ¼ 0: ð2:2Þ 1 2 ee Fu , 22 yy ee Fu , 22 x x ee Fu , 2 11 yy e ee e ee Fu , e Fu , 11 22 x 1 2 ee Fu , 11 xx 1 (a) (b) Figure 2.2 Variousconfigurationsofbarelements:(a)horizontallyalignedbarand(b)barelementpositionedatan arbitraryangleintwodimensions(seeSection2.4).DESCRIBING THE BEHAVIOR OF A SINGLE BAR ELEMENT 13 Figure 2.3 Examplesofcrosssectionsofabarelement. e 2. Theelasticstress–strainlaw,knownasHooke’slaw,whichstatesthatthestress isalinearfunctionof e thestraine : e e e  ¼ E e : ð2:3Þ 3. The deformation of the structure must be compatible, i.e. no gaps or overlaps can develop in the structureafterdeformation. Itisimportanttorecognizethedifferencebetweenthesignconventionfortheinternalaxialforce(and e thestress)andthatforthenodalinternalforces.Theinternalforcep ispositiveintensionandnegativein e compression,i.e.p ispositivewhenitpointsoutfromthesurfaceonwhichitisacting;thenodalinternal forces are positivewhen they point in the positive x-direction and are not associated with surfaces, see Figure2.4. WewillalsoneedadefinitionofstraininordertoapplyHooke’slaw.Theonlynonzerostrainistheaxial e e straine ,whichisdefinedastheratiooftheelongation totheoriginalelementlength: e  e e ¼ : ð2:4Þ e l Wewill now develop the element stiffness matrix, which relates the element internal nodal forces to e the element nodal displacements. The element internal force matrix is denoted by F and element e displacementmatrixbyd .Forthistwo-nodeelement,thesematricesaregivenby   e e F u e e 1 1 F ¼ ; d ¼ : e e F u 2 2 e l new e e F F 1 2 e e 1 2 p p e e u u 2 1 e l e e Figure 2.4 Elongationofanelementandfree-bodydiagrams,showingthepositivesenseofp andF . I14 DIRECT APPROACHFOR DISCRETESYSTEMS e TheelementstiffnessmatrixK thatrelatesthesematriceswillnowbedeveloped.Thematrixisderivedby applyingHooke’slaw,strain–displacementequationsandequilibrium: e e e e F ¼ p ¼ A  definition of stress ðEquation ð2:1ÞÞ 2 e e e 0 ¼ A E e Hookes law ðEquation ð2:3ÞÞ ð2:5Þ e  e e ¼ A E definition of strain ðEquation ð2:4ÞÞ: e ‘ Theelongationofanelementcanbeexpressedintermsofthenodaldisplacements(seeFigure2.4)by e e e  ¼ u u ; ð2:6Þ 2 1 e e e e e e e whichisobtainedasfollows:l ¼ l þu u ,sofrom ¼ l l ,(2.6)follows. new 2 1 new e e Notethatwhenu ¼ u ,whichisrigidbodytranslation,theelongationvanishes.Substituting(2.6)into 1 2 (2.5)gives e e e e F ¼ k ðu u Þ; ð2:7Þ 2 2 1 e wherek isgivenby e e A E e k ¼ : ð2:8Þ e l Fromequilibriumofthebarelement(2.2)and(2.7),itfollowsthat e e e e e F ¼F ¼ k ðu u Þ: ð2:9Þ 1 2 1 2 Equations(2.7)and(2.9)canbewritteninthematrixformas  e e e e F k k u 1 1 ¼ : ð2:10Þ e e e e F u k k 2 2 fflfflzfflffl fflfflfflfflfflfflfflfflfflfflzfflfflfflfflfflfflfflfflfflfflfflfflzfflffl e e e F K d Using the underscored definitions, we can write the relation between the nodal forces and nodal displacementsas   e e e e A E k k 1 1 e e e e F ¼ K d ; where K ¼ ¼ : ð2:11Þ e e e k k 11 l e Intheabove,K istheelementstiffnessmatrix.Wecanusethiselementstiffnessforanyconstantareabar elementinonedimension.ThisuniversalityofelementstiffnessmatricesisoneoftheattributesofFEM e thatleadstoitsversatility:foranybarelementwithconstantarea A inonedimension,Equation(2.11) givesthestiffnessmatrix.Wewilllaterdevelopelementmatricesthatapplytoanytriangularelementor quadrilateral element based on the weak solution of differential equations rather than on physical arguments. Equation(2.10)describestherelationshipbetweennodalforcesanddisplacementsforasingleelement, i.e. it describes the behavior of an element. Note that this is a linear relationship: The nodal forces are linearlyrelatedtothenodaldisplacements.Thislinearitystemsfromthelinearityofalltheingredientsthat describethiselement’sbehavior:Hooke’slaw,thelinearitybetweenaxialforceandstress,andthelinearity oftheexpressionforthestrain. e eT Animportantcharacteristicoftheelementstiffnessmatrixisthatitissymmetric,i.e.K ¼ K .EQUATIONSFOR A SYSTEM 15 (2) (2) E , A (a) (1) (1) E , A f f 3 2 x (1) (2) l l (b) f u f u , , r , u 3 3 2 2 1 1 (1) (2) 3 2 1 Figure 2.5 (a) Two-element bar structure and (b) the finite element model (element numbers are denoted in parenthesis). 2.2 EQUATIONS FOR A SYSTEM Theobjectiveofthissectionistodescribethedevelopmentoftheequationsforthecompletesystemfrom element stiffness matrices. Wewill introduce the scatter and assembly operations that are used for this purpose. These are used throughout the FEM in even the most complex problems, so mastering these proceduresisessentialtolearningtheFEM. Wewilldescribetheprocessofdevelopingtheseequationsbyanexample.Forthispurpose,considerthe two-barsystemshowninFigure2.5,whichalsogivesthematerialproperties,loadsandsupportconditions. At a support, the displacement is a givenvalue;wewill specify it later. Nodal displacements and nodal forcesarepositiveinthepositivex-direction. ThefirststepinapplyingtheFEMistodividethestructureintoelements.Theselectionandgenerationof ameshforfiniteelementmodelsisanextensivetopicthatwewilldiscussinsubsequentchapters.Inthecase ofadiscretestructuresuchasthis,itisnecessaryonlytoputnodeswhereverloadsareappliedandatpoints wherethesectionpropertiesormaterialpropertieschange,sothefiniteelementmeshconsistingoftwo elementsshowninFigure2.5(b)isadequate. The elements are numbered 1 and 2, and the nodes are numbered 1 to 3; neither the nodes nor the elements need to be numbered in a specific order in FEM. We will comment about node numbering in Section2.2.2.Ateachnode,eithertheexternalforcesorthenodaldisplacementsareknown,butnotboth; for example, at node 1 the displacement u ¼ u is prescribed, therefore the force to be subsequently 1 1 referredtoasreactionr isunknown.Atnodes2and3theexternalforcesf andf areknown,andtherefore 1 2 3 thedisplacementsu andu areunknown. 2 3 ForeachbarelementshowninFigure2.6,thenodalinternalforcesarerelatedtothenodaldisplacements bythestiffnessmatrixgiveninEquation(2.11). The stiffness equations of the elements, derived in Section 2.1.1, are repeated here for convenience (e¼ 1;2):  e e e e F u k k e e e 1 1 F ¼ K d or ¼ : ð2:12Þ e e e e F k k u 2 2 (1) (1) (1) (1) (2) (2) (2) (2) F u F u , F , u F , u , 1 1 2 2 2 2 1 1 Figure 2.6 Splittingthestructureinfigure2.5intotwoelements.16 DIRECT APPROACHFOR DISCRETESYSTEMS ff 3 r 2 1 x (a) 2 1 3 (1) (1) (2) (2) F F F F 1 2 1 2 (b) 1 1 2 2 ff r 3 2 1 (1) (1) (2) (2) F F F F 1 2 1 2 (c) Figure 2.7 Free-bodydiagramsofthenodesandelements(externalforcesareshownabovethenodesbutactinthe sameline):(a)completesystemwithglobalnodenumbers;(b)free-bodydiagramsofelementswithlocalnodenumbers and(c)free-bodydiagramsofnodes. The global system equations will be constructed by enforcing compatibility between the elements and nodalequilibriumconditions. Todevelopthesystemequations,wewillwritetheequilibriumequationsforthethreenodes.Forthis purpose,weconstructfree-bodydiagramsofthenodesasshowninFigure2.7(c).Notethattheforcesonthe elementsareequalandoppositetothecorrespondingforcesonthenodesbyNewton’sthirdlaw. 2 3 2 3 2 3 2 3 2 3 ð2Þ 0 F r 0 r 1 1 2 6 7 ð1Þ 6 7 6 7 6 7 6 7 6 ð2Þ7 F 4 5þ ¼4f 5¼4f 5þ4 0 5: ð2:13Þ 2 2 2 F 4 5 1 ð1Þ F f f 0 3 3 0 1 fflfflzfflffl fflfflzfflffl fflfflfflfflzfflfflfflffl fflfflfflfflzfflfflfflffl r ð1Þ f ð2Þ F F Eachrowoftheabovematrixequationisanequilibriumequationatanode.Ontheright-handsidearethe appliedexternalforcesandreactions, whicharearrangedinmatricesfandr,respectively.Thematrixf consists of the prescribed (known) external forces at the nodes, f and f ; the matrix r consists of the 2 3 unknownforceatnode1,denotedbyr . 1 Theaboveequationmaybesummarizedinwordsasfollows:Thesumoftheinternalelementforcesis equaltothatoftheexternalforcesandreactions.Thisdifferssomewhatfromthewell-knownequilibrium conditionthatthesumofforcesonanypointmustvanish.Thereasonforthedifferenceisthattheelement nodalforces,whicharetheforcesthatappearintheelementstiffnessmatrix,actontheelements.Theforces exertedbytheelementsonthenodesareequalandopposite. Noticethattheelementforcesarelabeledwithsubscripts1and2;thesearethelocalnodenumbers.The nodes of the mesh are the global node numbers. The local node numbers of a bar element are always numbered1,2inthepositivex-direction.Theglobalnodenumbersarearbitrary.Theglobalandlocalnode numbersforthisexampleareshowninFigure2.7(a)and(b),respectively. Wewillnowusetheelementstiffnessequationstoexpresstheelementinternalnodalforces(LHSin (2.13)),intermsoftheglobalnodaldisplacementsoftheelement. Forelement1,theglobalnodenumbersare2and3,andthestiffnessequation(2.12)gives ""  ð1Þ ð1Þ ð1Þ u F k k 3 1 ¼ : ð2:14Þ ð1Þ ð1Þ ð1Þ u k k F 2 2 Noticethatwehavereplacedthenodaldisplacementsbytheglobalnodaldisplacements.Thisenforces compatibilityasitensuresthatthedisplacementsofelementsatcommonnodesareidentical.EQUATIONSFOR A SYSTEM 17 Forelement2,theglobalnodenumbersare1and2,andthestiffnessequation(2.12)gives " "  ð2Þ ð2Þ ð2Þ F u k k 2 1 ¼ : ð2:15Þ ð2Þ ð2Þ ð2Þ k k u 1 F 2 Theaboveexpressionsfortheinternalnodalforcescannotbesubstituteddirectlyintotheleft-handside of(2.13)becausethematricesarenotofthesamesize.Therefore,weaugmenttheinternalforcesmatrices in (2.14) and (2.15) by adding zeros; we similarly augment the displacement matrices. The terms of theelementstiffnessmatricesin(2.14)and(2.15)arerearrangedintolargeraugmentedelementstiffness matricesandzerosareaddedwheretheseelementshavenoeffect.Theresultsare 2 3 2 32 3 0 00 0 u 1 ð1Þ ð1Þ ð1Þ ð1Þ ð1Þ 4 5 4 54 5 F ¼ 0 k k u or F ¼ K d: ð2:16Þ 2 2 ð1Þ ð1Þ ð1Þ 0 k k u F 3 1 fflfflfflfflzfflfflfflffl fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflzfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflzfflffl ð1Þ ð1Þ d F K Notethatwehaveaddedarowofzerosinrow1correspondingtotheforceatnode1,aselement1exertsno forceonnode1,andacolumnofzerosincolumn1,asthenodaldisplacementatnode1doesnotaffect element1directly.Similarly,anaugmentedequationforelement2is 2 3 2 32 3 ð2Þ ð2Þ ð2Þ u F k k 0 1 2 6 7 ð2Þ ð2Þ 6 76 7 ð2Þ ð2Þ 6 ð2Þ7 ¼ u or F ¼ K d: ð2:17Þ 4k k 054 25 F 4 5 1 00 0 u 3 0 fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflzfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflzfflffl fflfflfflfflzfflfflfflffl ð2Þ d ð2Þ K F Thematricesintheaboveequationsarenowofthesamesizeasin(2.13)andwecansubstitute(2.16)and (2.17)into(2.13)toobtain 2 32 3 2 32 3 2 3 2 3 ð2Þ ð2Þ 00 0 u u 0 r k k 0 1 1 1 ð1Þ ð1Þ ð2Þ ð2Þ 4 54 5 4 54 5 4 5 4 5 0 k k u þ u ¼ f þ 0 ; 2 k k 0 2 2 ð1Þ ð1Þ 0 k k u 00 0 u f 0 3 3 3 fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflzfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflzfflffl fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflzfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflzfflffl fflfflzfflffl fflfflzfflffl r ð1Þ ð2Þ d d f K K orinthematrixform ð1Þ ð2Þ ðK þK Þd¼ f þr: ð2:18Þ Theabovearetheassembledstiffnessequationsandthevariablewithintheparenthesesistheassembled stiffnessmatrix,whichinthiscaseisgivenby 2 3 ð2Þ ð2Þ 2 k k 0 X e ð2Þ ð1Þ ð2Þ ð1Þ 4 5 K¼ K ¼ k k þk k : ð2:19Þ ð1Þ ð1Þ e¼1 0 k k ThestiffnessmatrixKissingular,ascanreadilybeseenbycheckingthedeterminant.Toobtainasolvable system,theboundaryconditionsmustbeprescribed. Wewillnowsummarizewhatwehavedonetoobtaintheglobalstiffnessmatrix.First,wescatteredthe termsinanelementstiffnessintolargermatricesofthesameorderastheglobalstiffnessaccordingtothe18 DIRECT APPROACHFOR DISCRETESYSTEMS Table 2.1 Matrix scatter and add and direct assembly. Matrixscatterandadd Element1scatter,globalnodes3and2 2 3  00 0 ð1Þ ð1Þ ð1Þ k k ð1Þ ð1Þ ð1Þ 4 5 K ¼ ) K ¼ 0 k k ð1Þ ð1Þ k k ð1Þ ð1Þ 0 k k Element2scatter,globalnodes2and1 2 3 ð2Þ ð2Þ  k k 0 ð2Þ ð2Þ ð2Þ k k ð2Þ 4 ð2Þ ð2Þ 5 K ¼ ) K ¼ k k 0 ð2Þ ð2Þ k k 00 0 Addmatrices 2 3 ð2Þ ð2Þ 2 k k 0 X e 4 ð2Þ ð1Þ ð2Þ ð1Þ5 K¼ K ¼ k k þk k ð1Þ ð1Þ e¼1 0 k k Directassembly 2 3 " ð2Þ ð2Þ ½1 k k 0 ð1Þ ð1Þ k k ½3 ð1Þ 6 7 ð2Þ ð1Þ ð2Þ ð1Þ K ¼ K¼4k k þk k 5½2 ð1Þ ð1Þ ½2 k k ð1Þ ð1Þ ½3 0 k k ½3½2 ½1½2½3 " ð2Þ ð2Þ ½2 k k ð2Þ K ¼ ð2Þ ð2Þ k k ½1 ½2½1 globalnodenumbers.Then,weaddedtheseaugmentedstiffnessestoobtaintheglobalstiffnessmatrix. Thus, the process of obtaining the global stiffness matrix consists of matrix scatter and add. This is summarizedinTable2.1. Wecan bypassthe additionof zeros andassemble the matrix directly byjust addingthe terms in the elementstiffnessaccordingtotheirglobalnodenumbersasshowninTable2.1.Thisprocessiscalleddirect assembly.Theresultisequivalenttotheresultfromthematrixscatterandadd.Assemblingofthestiffness matrixincomputerprogramsisdonebydirectassembly,buttheconceptofmatrixscatterandaddisuseful inthatitexplainshowcompatibilityandequilibriumareenforcedatthegloballevel. 2.2.1 Equations for Assembly Wenextdeveloptheassemblyproceduresintermsofequations.Inthisapproach,compatibilitybetween elements is enforced by relating the element nodal displacements to the global displacementEQUATIONSFOR A SYSTEM 19 T matrixd¼½u u u  byequations.Theseequationsarewrittenasfollows: 1 2 3 2 3 2 3 " " u u   ð1Þ 1 ð2Þ 1 u 001 u 01 0 ð1Þ ð1Þ ð2Þ ð2Þ 1 1 4 5 4 5 d ¼ ¼ u ¼ L d; d ¼ ¼ u ¼ L d; 2 2 ð1Þ ð2Þ 010 10 0 u u 2 2 fflfflfflfflfflfflfflfflzfflfflfflfflfflfflfflffl fflfflfflfflfflfflfflfflzfflfflfflfflfflfflfflffl u u 3 3 ð1Þ ð2Þ L L ð2:20Þ oringeneral e e d ¼ L d: ð2:21Þ e ThematricesL arecalledthegathermatrices.Thenamegatheroriginatesfromthefactthatthesematrices gatherthenodaldisplacementsofeachelementfromtheglobalmatrix.Notethattheseequationsstatethat the element displacement at a node is the same as the corresponding global displacement, which is equivalenttoenforcingcompatibility. e ThematricesL areBooleanmatricesthatconsiststrictlyofonesandzeros.Theyplayanimportantrole indevelopingmatrixexpressionsrelatingelementtoglobalmatrices. Using(2.11),theelementequationscanbewrittenas e e e K L d¼ F : ð2:22Þ CompatibilityisautomaticallyenforcedbyEquation(2.20). Itcanbeobservedthatthefirsttermontheleft-handsideof(2.13)canbeexpressedas 2 3 2 3 " 0 00 ð1Þ F ð1Þ 6 7 6 7 1 ð1ÞT ð1Þ F ¼ 01 ¼ L F ; 4 5 4 5 2 ð1Þ ð1Þ F 2 10 F 1 whereasthesecondtermontheleft-handsideof(2.13)isequalto 2 3 2 3 ð2Þ " F 01 ð2Þ 2 6 7 F 6 7 1 ð2ÞT ð2Þ ð2Þ 6 7 ¼4105 ¼ L F : F 4 5 ð2Þ 1 F 2 00 0 e T NotethatðL Þ scattersthenodalforcesintotheglobalmatrix.Substitutingtheabovetwoequationsinto (2.13)gives 2 X eT e L F ¼ fþr: ð2:23Þ e¼1 Althoughwehaveshowntherelationbetweeninternal,externalforcesandreactionsforaspecificexample, (2.23)alwaysholds.ThegeneralrelationisderivedinSection2.5. InordertoeliminatetheunknowninternalelementforcesfromEquation(2.22),wepremultiply(2.22) T T e e by L and then add them together. Thus, premultiplying the element equations (2.22) by L yields eT e e eT e L K L d¼ L F ; e¼ 1;2:20 DIRECT APPROACHFOR DISCRETESYSTEMS Wenowdefinethesystemofequationsfortheentiresystem.Byaddingtheelementequations(e¼ 1;2), weget Kd¼ fþr; ð2:24Þ whereKiscalledtheglobalstiffnessmatrixandisgivenby n el X eT e e K¼ L K L ð2:25Þ e¼1 wheren isthenumberofelements;inthiscasen ¼ 2.Theabovegivestheassemblyprocedureinterms el el of an equation. It is equivalent to direct assembly and matrix scatter and add. Whenever this equation appears,itindicatesassemblyoftheelementmatricesintotheglobalmatrix(forgeneralmeshes,therange ofewillbe1ton ).Bycomparisonwith(2.19),wecanseethat el e eT e e K ¼ L K L : ð2:26Þ e eT e So the stiffness matrix scatter corresponds to pre- and postmultiplications of K by L and L , respectively. Substitutingtheexpressionsoftheelementstiffnessmatrices(2.12)into(2.24)andusing(2.25)gives theglobalequation 2 32 3 2 3 ð2Þ ð2Þ u r k k 0 1 1 ð2Þ ð1Þ ð2Þ ð1Þ 4 54 5 4 5 u ¼ f : ð2:27Þ k k þk k 2 2 ð1Þ ð1Þ 0 k k u f 3 3 Theabovesystemofthreeequationscanbesolvedforthethreeunknownsu ,u andr asdescribedinthe 2 3 1 nextsection. 2.2.2 Boundary Conditions and System Solution Wenowproceedwiththeprocessofsolvingtheglobalsystemofequations.Forthepurposeofdiscussion, ð2Þ weconsiderprescribeddisplacement u ¼ 4=k atnode1andexternalforcesf ¼4andf ¼ 10acting 1 2 3 atnodes2and3asshowninFigure2.8. Theglobalsystemofequations(2.27)isthen: 2 32 3 2 3 ð2Þ ð2Þ  u r k k 0 1 1 ð2Þ ð1Þ ð2Þ ð1Þ 4 54 5 4 5 k k þk k u ¼ 4 : ð2:28Þ 2 ð1Þ ð1Þ 0 k k u 10 3 Thereareseveralwaysofmodifyingtheaboveequationstoimposethedisplacementboundaryconditions. Inthefirstmethod,theglobalsystemispartitionedbasedonwhetherornotthedisplacementatthenodeis prescribed.WepartitionthesystemofequationsintoE-nodesandF-nodes.TheE-nodesarethosewhere thenodaldisplacementsareknown(Estandsforessential,themeaningofthiswillbecomeclearinlater chapters),whereasF-nodesarethosewherethedisplacementsareunknown;(orfree).ThesubscriptsEand 4 f f = 10 =−4 (1) (2) u = 3 2 1 (2) k (1) (2) r 1 1 3 2 kk Figure 2.8 Two-elementtrussstructurewithappliedexternalforcesandboundaryconditions.- EQUATIONSFOR A SYSTEM 21    d f E E Fintheglobaldisplacementmatrix,d¼ ,theglobalforcematrix,f ¼ andreactionmatrix, d f F F  r E r¼ denote the corresponding blocks; r ¼ 0 because there are no reactions on free nodes; the F r F externalforcesinthischaptercorrespondingtoE-nodesareassumedtovanish,f ¼0. E For convenience, when solving the equations either manually or by utilizing the MATLAB program (Chapter12),theE-nodesarenumberedfirst.Ingeneral,theoptimalnumberingisbasedoncomputational efficiencyconsiderations. Thesystemequation(2.28)isthenpartitionedasfollows: 2 32 3 2 3 ð2Þ ð2Þ    k k 0 u r 1 1  K K d r E EF E E 4 ð2Þ ð1Þ ð2Þ ð1Þ54 5 4 5 u ¼ 4 or ¼ ; ð2:29Þ k k þk k 2 T K K d f F F EF F ð1Þ ð1Þ u 10 0 k k 3 where " ð1Þ ð2Þ ð1Þ k þk k ð2Þ ð2Þ K ¼½k ; K ¼½k 0; K ¼ ; E EF F ð1Þ ð1Þ k k   4 u 2 ð2Þ  r ¼½r ; d ¼½u¼½4=k ; f ¼ ; d ¼ : E 1 E 1 F F 10 u 3 ð1Þ ð2Þ  Theunknownsintheabovesystemofequationsared andr ,whereasd ,f ,k andk areknown.Ifwe F E E F writethesecondrowofEquation(2.29),wehave T  K d þK d ¼ f : E F F F EF 1 IfwesubtractthefirsttermfrombothsidesoftheaboveequationandpremultiplybyK ,weobtain F 1 T  d ¼ K ðf K d Þ: ð2:30Þ F F E F EF This equation enables us to obtain the unknown nodal displacements. The partitioning approach also enablesustoobtainthereactionforce,r .Writingthefirstrowof(2.29)gives E  r ¼ K d þK d : ð2:31Þ E E E EF F Asd isknownfromEquation(2.30),wecanevaluatetheright-handsideoftheaboveequationtoobtainthe F reactionsr . E Forthetwo-barproblem,thesolutionoftheunknowndisplacementsbyEquation(2.30)using(2.29) gives   1 ð1Þ ð2Þ ð1Þ ð2Þ u k þk k 4 k 2 ð2Þ ¼  ½4=k  ; ð1Þ ð1Þ u 10 k k 0 3 whichyields  10 1 1 u ¼ ; u ¼ 10 þ : 2 3 ð2Þ ð1Þ ð2Þ k k k22 DIRECT APPROACHFOR DISCRETESYSTEMS ThereactionforceisfoundfromEquation(2.31)andisgivenby r ¼6: 1 ItcanbeshownthatK ispositivedefinite(seeProblem12.3inChapter12). F The second method for imposing the displacement boundary conditions is to replace the equations correspondingto prescribed displacements bytrivialequations thatset the nodaldisplacements totheir correctvalue,orinmanualcomputations,todropthemaltogether.Weputtheproductofthefirstcolumnof Kand u ontheright-handsideandreplacethefirstequationbyu ¼ u .Thisgives 1 1 1 2 32 3 2 3 10 0 u  u 1 1 ð1Þ ð2Þ ð1Þ ð2Þ 4 54 5 4 5 0 k þk k u ¼ 4ðk Þu : ð2:32Þ 2 1 ð1Þ ð1Þ 0 k k u 10ð0Þ u 3 1 Again, it can be seen that the above equations can be solved manually by just considering the last two equations. Thereactionscanthenbecomputedbyevaluatingtherowsofthetotalstiffnessequationsthatgivethe reactions.Fromrow1ofEquation(2.29),weobtain 2 3  u hi 1 ð2Þ ð2Þ 4 5 r ¼ k k 0 u ¼6: 1 2 u 3 The third method for imposing the boundary conditions is the penalty method. This is avery simple methodtoprogram,butshouldbeusedformatricesofmoderatesize(uptoabout10000unknowns)only becauseittendstodecreasetheconditioningoftheequations(seeSaad(1996)andGeorgeandLiu(1986)). In this method, the prescribed displacements are imposed by putting a very large number in the entry correspondingtotheprescribeddisplacement.Thus,fortheexamplewehavejustconsidered,wechange theequationsto 2 32 3 2 3 ð2Þ u b u b k 0 1 1 ð2Þ ð1Þ ð2Þ ð1Þ 4 54 5 4 5 k k þk k u ¼ 4 ; ð2:33Þ 2 ð1Þ ð1Þ u 10 0 k k 3 where b is a very large number. For example, in a computer with eight digits of precision, we make 7 b 10 averageðK Þ.Theothertermsinrow1andcolumn1thenbecomeirrelevantbecausetheyare ii much smaller than the first diagonal term, and the equations are almost identical to those of (2.32). Themethodcanphysicallybeexplainedinstressanalysisasconnectingaverystiffspringbetweennode 1andthesupport,whichisdisplacedby u .Thestiffspringthenforcesnode1tomovewiththesupport.The 1 penaltymethodismosteasilyunderstoodwhenu ¼ 0;thenitcorrespondstoastiffspringlinkedtothe 1 stationarysupportandthedisplacementofthenode1isverysmall.Thereactionscanbeevaluatedaswas doneforthepreviousmethod.WewillelaborateonthepenaltymethodinChapters3and5. Example2.1 ThreebarsarejoinedasshowninFigure2.9.Theleftandrightendsarebothconstrained,i.e.prescribed displacement is zero at both ends. There is a force of 5N acting on the middle node. The nodes are numberedstartingwiththenodeswheredisplacementsareprescribed.- EQUATIONSFOR A SYSTEM 23 f= 5 3 (1) k (3) k (2) k 2 3 1 Figure 2.9 Three-barexampleproblem. Theelementstiffnessmatricesare ½1½3 ½1½3 ½3½2    ð1Þ ð1Þ ð2Þ ð2Þ ð3Þ ð3Þ k k k k k k ð1Þ ½1 ð2Þ ½1 ð3Þ K ¼ K ¼ K ¼ ð1Þ ð1Þ ð2Þ ð2Þ ð3Þ ð3Þ ; k k ½3 k k k k ½3; wheretheglobalnumberscorrespondingtotheelementnodesareindicatedaboveeachcolumnandtothe rightofeachrow. Bydirectassembly,theglobalstiffnessmatrixis ½1½2½3 2 3 ð1Þ ð2Þ ð1Þ ð2Þ ½1 k þk 0 k k ð3Þ ð3Þ 4 5 K¼ 0 k k ½2 ð1Þ ð2Þ ð3Þ ð1Þ ð2Þ ð3Þ k k k k þk þk ½3 Thedisplacementandforcematricesforthesystemare 2 3 2 3 2 3 0 0 r 1 4 5 4 5 4 5 d¼ 0 ; f ¼ 0 ; r¼ r 2 u 5 0 3 Theglobalsystemofequationsisgivenby 2 32 3 2 3 ð1Þ ð2Þ ð1Þ ð2Þ 0 r k þk 0 k k 1 ð3Þ ð3Þ 4 54 5 4 5 0 ¼ r : 0 k k 2 ð1Þ ð2Þ ð3Þ ð1Þ ð2Þ ð3Þ k k k k þk þk u 5 3 Asthefirsttwodisplacementsareprescribed,wepartitionaftertworowsandcolumns 2 32 3 2 3 ð1Þ ð2Þ ð1Þ ð2Þ 0 r k þk 0 k k 1 6 76 7 6 7 ð3Þ ð3Þ 0 ¼ r 4 0 k k 54 5 4 25 ð1Þ ð2Þ ð3Þ ð1Þ ð2Þ ð3Þ k k k k þk þk u 5 3 or    K K d r E EF E E ¼ ; T K K f d F F F EF24 DIRECT APPROACHFOR DISCRETESYSTEMS where " " ð1Þ ð2Þ hi ð1Þ ð2Þ k þk 0 k k ð1Þ ð2Þ ð3Þ K ¼ K ¼ k þk þk K ¼ E F EF ð3Þ ð3Þ 0 k k   0 r 1  d ¼ d ¼½ u f ¼½ 5 r ¼ E F 3 F E 0 r 2 Thereducedsystemofequationsisgivenby ð1Þ ð2Þ ð3Þ ðk þk þk Þu ¼ 5; 3 whichyields 5 u ¼ : 3 ð1Þ ð2Þ ð3Þ k þk þk 1 2.3 APPLICATIONS TO OTHER LINEAR SYSTEMS The methods described for one-dimensional bars can also be used directly for other networks. For the methodstobeapplicable,thesystemsmustbecharacterizedby 1. abalanceorconservationlawfortheflux; 2. alinearlawrelatingthefluxtothepotential; 3. acontinuouspotential(i.e.acompatiblepotential). Twoexamplesare described inthe following:steady-state electricalflowinacircuitandfluidflowina hydraulicpipingsystem. Inanelectricalsystem,thepotentialisthevoltageandthefluxisthecurrent.Anelementofacircuitis showninFigure2.10.ByOhm’slaw,thecurrentfromnode1tonode2isgivenby e e e e e 2 1 i ¼ ; ð2:34Þ 2 e R e e e wheree ande arethevoltages(potentials)atthenodesandR istheresistanceofthewire.Thisisthelinear 2 1 flux–potentiallaw.Bythelawofchargeconservation,ifthecurrentisinsteadystate, e e i þi ¼ 0; ð2:35Þ 1 2 whichisthesecondoftheaboveconditionsontheelementlevel.Writing(2.34)and(2.35)inmatrixform, wehave   e e i 1 1 1 e 1 1 ¼ : ð2:36Þ e e e i 11 e R 2 2 fflzffl fflfflfflfflfflfflfflfflfflfflfflzfflfflfflfflfflfflfflfflfflfflfflfflfflzfflffl e e e f K d Thecontinuityofthevoltageatthenodesisenforcedby e e d ¼ L d: ð2:37Þ 1 Recommended for Science and Engineering Track.APPLICATIONS TO OTHER LINEARSYSTEMS 25 e e e e e P e e e P e e 1 2 Q 1 2 Q i i 1 2 1 2 e 2 1 e R 1 Figure 2.10 Aresistanceelementforacircuitandahydraulicelementforapipingnetwork;thenodalfluxispositive whenitexitsthedomainoftheelement. Currentbalanceatthenodesgives n el X eT e L F ¼ f þr ð2:38Þ e¼1 DetailscanbeseeninExample2.2. Thesystemequationcanthenbeobtainedbyenforcingtheconditionthatthesumofthecurrentsatany nodeisequaltoanyexternalsourcesofcurrents.Theprocessisidenticaltowhatwedidforbarelements. n el X eT e fþr¼ L F by Equation ð2:38Þ e¼1 n el X eT e e ¼ L K d by Equation ð2:36Þ e¼1 n el X eT e e ¼ L K L d by Equation ð2:37Þ: e¼1 fflfflfflfflfflfflfflfflfflfflzfflfflfflfflfflfflfflfflfflffl K Asindicatedbytheunderscore,theassembledsystemmatrixisgivenby n el X eT e e K¼ L K L : ð2:39Þ e¼1 Thissystemisobtainedbyasequenceofscatterandaddoperations,whichcorrespondstodirectassembly. For a piping system, a similar procedure can be developed if the flow rate is linearly related to the pressure drop between two points. A network model is constructed as shown in Figure 2.11. Nodes are needed only where two pipes join or where the fluid is withdrawn or added. In each element, the nodal e e e outflowrateQ atnodeisproportionaltothenodalpressuredropðP P Þ(seeFigure2.10),so I 2 1 e e e e Q ¼  ðP P Þ; ð2:40Þ 2 2 1 e where dependsonthecross-sectionalareaofthepipe,theviscosityofthefluidandtheelementlength. Linearlawsofthistypeapplyoveralargerangeofflows. Conservationoffluidinanelementisexpressedby e e Q þQ ¼ 0: ð2:41Þ 1 2 Thesystemequationsarethenobtainedbywritingtheequationfortheconservationoffluidatnodesand usingthecontinuityofthepressurefield.TheprocessisidenticaltothatusedinobtainingEquation(2.39). Thisisleftasanexercise,althoughitwillbecomeapparentintheexample. Thesimilarityofthesedifferentsystemsissurprisingandcanprovideadeeperunderstandingoflinear systems.Allofthesesystemspossessapotentialandaconservationlaw.Inthemechanicalbar,thepotential isnotasobvious:itisthedisplacement.Thedisplacementhasallofthepropertiesofapotential:itmustbe continuous(compatible)anditschangedeterminestheflux,whichinthiscaseisthestress.26 DIRECT APPROACHFOR DISCRETESYSTEMS Example2.2 SetupthediscreteequationsforthesystemsshowninFigure2.11andsolvethem.Thethreesystemsin Figure2.11allhavethesamebasictopology,i.e.thesamerelationshipbetweennodesandelements.We firstassemblethesystemmatrixbyscatterandadd.Thenthespecificequationsaresetupbyenforcing e 1 e constantsonthefluxorpotential.Weusek ¼ ¼  todenotetheelementcoefficientsforthethree e R differentsystems. Thescatteroperationsontheelementsthengivethefollowing(IandJgivetheglobalnodenumbersof theelement): Element1,I ¼ 1,J ¼ 4: 2 3 ð1Þ ð1Þ k 00 k  6 7 ð1Þ 1 1 000 0 ð1Þ ð1Þ 6 7 K ¼ k ) K ¼ : 4 5 11 000 0 ð1Þ ð1Þ k 00 k Element2,I ¼ 4,J ¼ 2: 2 3 00 00  ð2Þ ð2Þ 6 7 ð2Þ 1 1 0 k 0 k ð2Þ ð2Þ 6 7 K ¼ k ) K ¼ : 4 5 11 00 00 ð2Þ ð2Þ 0 k 0 k Element3,I ¼ 1,J ¼ 3: 2 3 ð3Þ ð3Þ k 0 k 0  6 7 1 1 ð3Þ 000 0 ð3Þ ð3Þ 6 7 K ¼ k ) K ¼ : ð3Þ ð3Þ 4 5 11 k 0 k 0 000 0 (2) k (1) k (1) 2 uk =10 / (4) (5) 2 k k 4 1 (3) k 3 p = 0 1 e = 0 1 1 4 (3) (1) 1 (1) (3) κ κ R R (4) κ 3 3 (4) R 4 (2) R (2) κ (5) (5) κ R 2 2 (1) (1) p =10 /κ eR =10 2 2 Figure 2.11 Example2.2:mechanical,electricalandhydraulicsystemswithanidenticalnetworkstructure.- TWO-DIMENSIONAL TRUSSSYSTEMS 27 Element4,I ¼ 4,J ¼ 3: 2 3 00 0 0  6 7 1 1 ð4Þ 00 0 0 ð4Þ ð4Þ 6 7 K ¼ k ) K ¼ : ð4Þ ð4Þ 4 5 11 00 k k ð4Þ ð4Þ 00 k k Element5,I ¼ 3,J ¼ 2: 2 3 00 00  ð5Þ ð5Þ 6 7 ð5Þ 1 1 0 k k 0 ð5Þ ð5Þ 6 7 K ¼ k ) K ¼ : ð5Þ ð5Þ 4 5 11 0 k k 0 00 00 Assembledsystemmatrix: 2 3 ð1Þ ð3Þ ð3Þ ð1Þ k þk 0 k k 5 X 6 ð2Þ ð5Þ ð5Þ ð2Þ 7 e 0 k þk k k 6 7 K¼ K ¼ : 4 ð3Þ ð5Þ ð3Þ ð4Þ ð5Þ ð4Þ 5 k k k þk þk k e¼1 ð1Þ ð2Þ ð4Þ ð1Þ ð2Þ ð4Þ k k k k þk þk Equationsforthemechanicalsystem: 2 3 2 3 ð1Þ ð3Þ ð3Þ ð1Þ r k þk 0 k k 1 " 6 7  ð2Þ ð5Þ ð5Þ ð2Þ 6 7  6 7 d r r 0 k þk k k E 2 E 6 7 6-7 ¼ ¼6 7 6 7 ð3Þ ð5Þ ð3Þ ð4Þ ð5Þ ð4Þ 4 5 k k k þk þk k d f 0 4 5 F F ð1Þ ð2Þ ð4Þ ð1Þ ð2Þ ð4Þ k k k k þk þk 0 wherethesolutionmatrixformechanical,pipingandelectricalsystemsis       u p e u p e 0 3 3 3 1 1 1  d ¼ ¼ ¼ ; d ¼ ¼ ¼ ¼ F E ð1Þ u p e  u p e 10=k 4 4 4 2 2 2 Partitioningaboveaftertworowsandcolumnsgives    ð3Þ ð4Þ ð5Þ ð4Þ ð5Þ 10 k þk þk k 0 k d ¼  F ð4Þ ð1Þ ð2Þ ð4Þ ð2Þ ð1Þ k k þk þk 0 k k e Lettingk ¼ 1fore¼ 1to5andsolvingabovegives     u p e 5 3 3 3 d ¼ ¼ ¼ ¼ F u p e 5 4 4 4 2 2.4 TWO-DIMENSIONAL TRUSS SYSTEMS Trussstructures,suchastheoneshowninFigure2.1,consistofbarelementspositionedatarbitraryangles in space joined by pin-like joints that cannot transmit moments. In order to analyze such general truss 2 Recommended for Structural Mechanics Track.28 DIRECT APPROACHFOR DISCRETESYSTEMS ee uF , 22 yy e x ee Fu , 22 yy ee ee uF , Fu , 22 xx 22 xx e 2 ee k e 2 Fu , 11 yy ee k uF , 11 yy e e x x y ee e uF , ee 11 xx y Fu , 1 11 xx 1 e x x 0e 0e Figure 2.12 Atwo-dimensionaltrusselementinthelocalcoordinatesystemx ,y . 1 1 structures,itisnecessarytodevelopanelementstiffnessmatrixforabarelementalignedarbitrarilyintwo-or three-dimensionalspace.Wewillfirstconsiderthetwo-dimensionalcasewherethebarelementsareinthexy- plane as shown in Figure 2.2(b). Trusses differ from networks such as electrical systems in that the nodal displacementsinmultidimensionalproblemsarevectors.Theunknownsofthesystemarethenthecomponents ofthevector,sothenumberofunknownspernodeis2and3intwoandthreedimensions,respectively. Webeginbydevelopingtheelementstiffnessmatrixforabarelementintwodimensions.Agenericbar elementisshowninFigure2.12,alongwiththenodaldisplacementsandnodalforces.Ateachnode,the nodalforcehastwocomponents;similarly,ascanbeseenfromFigure2.12,eachnodaldisplacementhas twocomponents,sotheelementforceanddisplacementmatricesare,respectively, T T e e e e e e e e e e F ¼½F F F F  and d ¼½d d d d  : 1x 1y 2x 2y 1x 1y 2x 2y e e ToobtainageneralrelationbetweentheelementinternalforcesF anddisplacementsd ,westartwiththe 0e 0e 0e stiffnessequationsinthelocalelementcoordinatesystemðx ; y Þ;asshowninFigure2.12,x isaligned e alongtheaxialdirectionofthebarelementandispositivefromnode1tonode2.Theangle isdefinedas positiveinthecounterclockwisesense. 0e 0e Intheelementcoordinatesystemðx ; y Þ,theelementstiffnessgivenbyEquation(2.10)applies,so   e e 0e 0e k k u F 1x 1x ¼ : 0e 0e e e k k u F 2x 2x 0e 0e The above equation can be expanded by adding the equations F ¼ F ¼ 0. These nodal force 1y 2y components perpendicular to the axis of the element can be set to zero because we have assumed that theelementissoslimthattheshearforcesarenegligible. The nodal forces in the element are independent of the normal displacements in small displacement theory.Thisisbecausetheelongationisaquadraticfunctionofthenodaldisplacementsnormaltothebar. As the nodal displacements are assumed to be small, the effect of the normal displacements on the elongation is therefore of second order, and hence the effects of these displacement components on the stressandstraincanbeneglected.Sothestiffnessmatrixintheelementcoordinatesystemisgivenby 2 3 2 32 3 0e 0e F u 10 10 1x 1x 6 7 6 76 7 0e 0e F u 6 7 60 00076 7 1y 1y e 6 7 6 76 7 ¼ k ; 6 0e7 6 76 0e7 F u 1 010 4 5 4 54 5 2x 2x 0e 0e F 0 000 u 2y 2y fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflzfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl fflfflfflzfflfflffl fflfflfflzfflfflffl 0e 0e 0e K F dTWO-DIMENSIONAL TRUSSSYSTEMS 29 orintermsoftheunderscorednomenclature e e e 0 0 0 F ¼ K d : ð2:42Þ 0e Itiseasytoseethatfortheabovestiffnessmatrix,they -componentsoftheforcesatthetwonodesalways 0e vanishandthatthey -componentsofthedisplacementshavenoeffectonthenodalforces;thestiffness matrixin(2.42)issimplythematrix(2.11)embeddedinamatrixofzeros.Inotherwords,wehavesimply scattered theaxial barstiffness into alargermatrix; this is valid whenthe elementcoordinate system is alignedwiththeaxisoftheelement. TherelationbetweenthedisplacementcomponentsinthetwocoordinatesystemsshowninFigure2.12 atthenodes(I ¼ 1;2)isobtainedbymeansoftherelationforvectortransformations: 0e e e e e u ¼ u cos  þu sin  Ix Ix Iy 0e e e e e u ¼u sin  þu cos  Iy Ix Iy Theseequationscanbewritteninthematrixformasfollows: e 0 e e d ¼ R d ; ð2:43Þ where 2 3 2 3 e e e u cos sin 00 1x 6 e 7 6 e e 7 u sin cos 00 6 1y7 6 7 e e d ¼ ; R ¼ : 6 7 6 7 e e e 4u 5 4 0 0 cos sin 5 2x e e e u 00 sin cos 2y e R is the rotation matrix. The above combines the vector transformation at the two nodes. As these transformationsareindependentofeachother,blocksofthematrixrelatingdifferentnodesarezero,e.g.the upper right 22 block is zero as the element components of the nodal displacement at node 1 are independentofthedisplacementatnode2. T T e e e e e NotethatR isanorthogonalmatrix:itsinverseisequaltoitstranspose,i.e.ðR Þ R ¼ R ðR Þ ¼ Ior e 1 eT ðR Þ ¼ R : ð2:44Þ T e PremultiplyingEquation(2.43)byðR Þ ,weobtain e eT 0 eT e e e R d ¼ R R d ¼ d ; wherethesecondequalityfollowsfromtheorthogonalityrelation(2.44).Thecomponentsoftheelement forcematricesarerelatedbythesamecomponenttransformationrule: 0e e e e eT 0e ðaÞ F ¼ R F ; ðbÞ F ¼ R F : ð2:45Þ e e WearenowinapositiontodeterminetherelationbetweenF andd .Startingwith(2.45b), e eT 0e F ¼ R F by Equation ð2:45bÞ eT 0e 0e ¼ R K d by Equation ð2:42Þ eT 0e e e ¼ R K R d by Equation ð2:43Þ fflfflfflfflfflfflzfflfflfflfflfflffl e K30 DIRECT APPROACHFOR DISCRETESYSTEMS Asindicatedabove,theunderscoredtermistheelementstiffnessintheglobalcoordinatesystem: e e eT 0 e K ¼ R K R : ð2:46Þ e e e AnexplicitexpressionforK isobtainedbysubstitutingthematrixexpressionsforK andR intoEquation (2.46),whichgives 2 3 2 e e e 2 e e e cos  cos sin cos  cos sin e e 2 e e e 2 e 6 7 cos sin sin  cos  sin  sin  e e 6 7 K ¼ k : ð2:47Þ 2 e e e 2 e e e 4 5 cos  cos sin  cos  cos sin 2 2 e e e e e e cos sin sin  cos sin sin  e ItcanbeseenthatK isasymmetricmatrix. 3 2.5 TRANSFORMATION LAW Inthefollowing,wedevelopamoregeneralmethodfortransformationofstiffnessmatricesbymeansof energyconcepts.Bytransformationherewemeaneitherarotationfromonecoordinatesystemtoanother orascatteroperationfromanelementtotheglobalcoordinatesystem.Wewilldenotesuchatransformation e e matrixbyT .ThematrixT transformstheelementdisplacementmatrixfromacoordinatesystemwhere e e   thestiffnessrelationK isknowntoanothercoordinatesystemwherethestiffnessmatrixK isunknown. Westartwith e e e e e e b  b b b ðaÞ d ¼ T d ; ðbÞ F ¼ K d : ð2:48Þ e 0 e e e e In the case of rotation from one coordinate system to another (Section 2.4), d ¼ R d ,so T ¼ R , e e 0 e e e e e e e e b  b d ¼ d andd ¼ d ;inthecaseofscatteroperation(Section2.2),d ¼ L d,soT ¼ L ,d ¼ d and e e e   b d¼ d .Inthefollowing,wewilldescribehowtorelateF toF andhowtoestablishthestiffnessrelation e e e    F ¼ K d . e e b b Let F betheinternalelementforcematrixandd beanarbitraryinfinitesimalelementdisplacement matrix.Thenodalinternalforcesmustbechosensothattheworkdonebytheinternalforces,denotedby W ,isgivenby int eT e b b W ¼ d F : ð2:49Þ int e e b b Notethat d hastobeinfinitesimalsothattheinternalforcematrixF remainsconstantastheelement deforms. For example, for the two-node element in one dimension, the work done by element e is e e e e b b W ¼ b u F þb u F . int 1 1 2 2 Wenowshowthatif(2.48)holdsthen e eT e e  b K ¼ T K T : ð2:50Þ Wefirstshowthatif(2.48)holdsthen e eT e  b F ¼ T F ð2:51Þ e e b b Thekeyconceptthatmakesthisproofpossibleisthattheinternalworkexpressedintermsof d andF e e   mustequaltotheinternalworkexpressedintermsofd andF,so eT e eT e b b   W ¼ d F ¼ d F : ð2:52Þ int 3 Optional for all tracks.

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