Strong and Weak Forms for One-Dimensional Problems

Strong and Weak Forms for One-Dimensional Problems
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Published Date:03-08-2017
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3 Strong and Weak Forms for One-Dimensional Problems Inthischapter,thestrongandweakformsforseveralone-dimensionalphysicalproblemsaredeveloped. Thestrongformconsistsofthegoverningequationsandtheboundaryconditionsforaphysicalsystem.The governingequationsareusuallypartialdifferentialequations,butintheone-dimensionalcasetheybecome ordinarydifferentialequations.Theweakformisanintegralformoftheseequations,whichisneededto formulatethefiniteelementmethod. Insomenumericalmethodsforsolvingpartialdifferentialequations,thepartialdifferentialequations canbediscretizeddirectly(i.e.writtenaslinearalgebraicequationssuitableforcomputersolution).For example,inthefinitedifferencemethod,onecandirectlywritethediscretelinearalgebraicequationsfrom thepartialdifferentialequations.However,thisisnotpossibleinthefiniteelementmethod. AroadmapforthedevelopmentofthefiniteelementmethodisshowninFigure3.1.Ascanbeseenfrom theroadmap,therearethreedistinctingredientsthatarecombinedtoarriveatthediscreteequations(also calledthesystemequations;forstressanalysistheyarecalledstiffnessequations),whicharethensolvedby acomputer.Theseingredientsare 1. thestrongform,whichconsistsofthegoverningequationsforthemodelandtheboundaryconditions (thesearealsoneededforanyothermethod); 2. theweakform; 3. theapproximationfunctions. The approximation functions are combined with the weak form to obtain the discrete finite element equations. Thus,thepathfromforthegoverningdifferentialequationsissubstantiallymoreinvolvedthanthatfor finitedifferencemethods.Inthefinitedifferencemethod,thereisnoneedforaweakform;thestrongformis directlyconvertedtoasetofdiscreteequations.Theneedforaweakformmakesthefiniteelementmethod more challenging intellectually. A number of subtle points, such as the difference between various boundaryconditions,mustbelearnedforintelligentuseofthemethod.Inreturnforthisaddedcomplexity, however,finiteelementmethodscanmuchmorereadilydealwiththecomplicatedshapesthatneedtobe analyzedinengineeringdesign. Todemonstratethebasicstepsinformulatingthestrongandweakforms,wewillconsideraxiallyloaded elasticbarsandheatconductionproblemsinonedimension.Thestrongformsfortheseproblemswillbe developedalongwiththeboundaryconditions.Thenwewilldevelopweakformsfortheseproblemsand showthattheyareequivalenttothestrongforms.Wewillalsoexaminevariousdegreesofcontinuity,or smoothness,whichwillplayanimportantroleindevelopingfiniteelementmethods. 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)42 STRONG AND WEAK FORMS FOR ONE-DIMENSIONAL PROBLEMS Strong form Weak form (Chapter 3) (Chapter 3) Discrete equations (Chapter 5) Approximation of functions (Chapter 4) Figure 3.1 Roadmapforthedevelopmentofthefiniteelementmethod. Theweakform is themost intellectually challenging part in thedevelopment offiniteelements, so a student may encounter some difficulties in understanding this concept; it is probably different from anything else that he has seen before in engineering analysis. However, an understanding of these proceduresandtheimplicationsofsolvingaweakformarecrucialtounderstandingthecharacteroffinite element solutions. Furthermore, the procedures are actually quite simple and repetitive, so once it is understoodforonestrongform,theprocedurescanreadilybeappliedtootherstrongforms. 3.1 THE STRONG FORM IN ONE-DIMENSIONAL PROBLEMS 3.1.1 The Strong Form for an Axially Loaded Elastic Bar ConsiderthestaticresponseofanelasticbarofvariablecrosssectionsuchasshowninFigure3.2.Thisisan example of a problem in linear stress analysis or linear elasticity, where we seek to find the stress distributionsðxÞinthebar.Thestresswillresultsfromthedeformationofthebody,whichischaracterized bythedisplacementsofpointsinthebody,uðxÞ.ThedisplacementresultsinastraindenotedbyeðxÞ;strain is a dimensionless variable. As shown in Figure 3.2, the bar is subjected to a body force or distributed loading bðxÞ. The body force could be due to gravity (if the bar were placed vertically instead of horizontallyasshown),amagneticforceorathermalstress;intheone-dimensionalcase,wewillconsider bodyforceperunitlength,sotheunitsofbðxÞareforce/length.Inaddition,loadscanbeprescribedatthe  endsofthebar,wherethedisplacementisnotprescribed;theseloadsarecalledtractionsanddenotedbyt. Theseloadsareinunitsofforceperarea,andwhenmultipliedbythearea,givetheappliedforce. ∆ x ∆ x p p(x) () xx + ∆ b(x+ ) 2 ux() ux() + ∆ x A(x) b(x) t x x = l x = 0 Figure 3.2 Aone-dimensionalstressanalysis(elasticity)problem.THE STRONG FORM IN ONE-DIMENSIONAL PROBLEMS 43 Thebarmustsatisfythefollowingconditions: 1. Itmustbeinequilibrium. 2. Itmustsatisfytheelasticstress–strainlaw,knownasHooke’slaw:sðxÞ¼ EðxÞeðxÞ. 3. Thedisplacementfieldmustbecompatible. 4. Itmustsatisfythestrain–displacementequation. ThedifferentialequationforthebarisobtainedfromequilibriumofinternalforcepðxÞandexternalforce bðxÞ acting on the body in the axial (along the x-axis) direction. Consider equilibrium of a segment of thebaralongthex-axis,asshowninFigure3.2.Summingtheforcesinthex-directiongives  x pðxÞþbxþ xþpðxþxÞ¼ 0: 2 Rearrangingthetermsintheaboveanddividingbyx,weobtain  pðxþxÞpðxÞ x þbxþ ¼ 0: x 2 Ifwetakethelimitoftheaboveequationasx 0,thefirsttermisthederivativedp=dxandthesecond termbecomesbðxÞ.Therefore,theabovecanbewrittenas dpðxÞ þbðxÞ¼ 0: ð3:1Þ dx Thisistheequilibriumequationexpressedintermsoftheinternalforcep.Thestressisdefinedastheforce dividedbythecross-sectionalarea: pðxÞ sðxÞ¼ ; so pðxÞ¼ AðxÞsðxÞ: ð3:2Þ AðxÞ Thestrain–displacement(orkinematical)equationisobtainedbyapplyingtheengineeringdefinitionof strainthatweusedinChapter2foraninfinitesimalsegmentofthebar.Theelongationofthesegmentis givenbyuðxþxÞuðxÞandtheoriginallengthisx;therefore,thestrainisgivenby elongation uðxþxÞuðxÞ eðxÞ¼ ¼ : originallength x Takingthelimitoftheaboveasx 0,werecognizethattherightright-handsideisthederivativeofuðxÞ. Therefore,thestrain–displacementequationis du eðxÞ¼ : ð3:3Þ dx Thestress–strainlawforalinearelasticmaterialisHooke’slaw,whichwealreadysawinChapter2: sðxÞ¼ EðxÞeðxÞ; ð3:4Þ whereEisYoung’smodulus.44 STRONG AND WEAK FORMS FOR ONE-DIMENSIONAL PROBLEMS Substituting(3.3)into(3.4)andtheresultinto(3.1)yields  d du AE þb¼ 0; 0 x l: ð3:5Þ dx dx Theaboveisasecond-orderordinarydifferential equation.In theaboveequation, u(x)isthedependent variable, which is the unknown function, and x is the independent variable. In (3.5) and thereafter the dependenceoffunctionsonxwillbeoftenomitted.Thedifferentialequation(3.5)isaspecificformofthe equilibrium equation (3.1). Equation (3.1) applies to both linear and nonlinear materials whereas (3.5) assumes linearity in the definition of the strain (3.3) and the stress–strain law (3.4). Compatibility is satisfied by requiring the displacement to be continuous. More will be said later about the degree of smoothness,orcontinuity,whichisrequired. Tosolvetheabovedifferentialequation,weneedtoprescribeboundaryconditionsatthetwoendsofthe bar.Forthepurposeofillustration,wewillconsiderthefollowingspecificboundaryconditions:atx¼ l,  the displacement, uðx¼ lÞ, is prescribed; at x¼ 0, the force per unit area, or traction, denoted by t,is prescribed.Theseconditionsarewrittenas  du pð0Þ  sð0Þ¼ E ¼ t; dx Að0Þ ð3:6Þ x¼0 uðlÞ¼ u: Notethatthesuperposedbarsdesignatedenoteaprescribedboundaryvalueintheaboveandthroughout thisbook.  Thetractionthasthesameunitsasstress(force/area),butitssignispositivewhenitactsinthepositive x-directionregardlessofwhichfaceitisactingon,whereasthestressispositiveintensionandnegativein compression, so that on a negative face a positive stress corresponds to a negative traction; this will be clarifiedinSection3.5.Notethateithertheloadorthedisplacementcanbespecifiedataboundarypoint, butnotboth. Thegoverningdifferentialequation(3.5)alongwiththeboundaryconditions(3.6)iscalledthestrong formoftheproblem.Tosummarize,thestrongformconsistsofthegoverningequationandtheboundary conditions,whichforthisexampleare  d du ðaÞ AE þb¼0on0 x l; dx dx  du ð3:7Þ  ðbÞ sðx¼ 0Þ¼ E ¼t; dx x¼0 ðcÞ uðx¼ lÞ¼ u:  Itshouldbenotedthatt, uandbaregiven.Theyarethedatathatdescribetheproblem.Theunknownisthe displacementuðxÞ. 1 3.1.2 The Strong Form for Heat Conduction in One Dimension Heat flow occurs when there is a temperature difference within a body or between the body and its surroundingmedium.Heatistransferredintheformofconduction,convectionandthermalradiation.The heatflowthroughthewallofaheatedroominthewinterisanexampleofconduction.Ontheotherhand,in convectiveheattransfer,theenergytransfertothebodydependsonthetemperaturedifferencebetweenthe surface of the body and the surrounding medium. In this Section, we will focus on heat conduction. A discussioninvolvingconvectionisgiveninSection3.5. 1 Reccommended for Science and Engineering Track.THE STRONG FORM IN ONE-DIMENSIONAL PROBLEMS 45 ∆ x Ax() + ∆ x qx()A() x qx() + ∆ x A() x + ∆ x Furring strips Ax() Concrete blocks ∆ x s (x+ ) 2 Siding Building paper Siding Building Furring Concrete x paper strips blocks l Figure 3.3 Aone-dimensionalheatconductionproblem. ConsideracrosssectionofawallofthicknesslasshowninFigure3.3.Ourobjectiveistodeterminethe temperaturedistribution.LetAðxÞbetheareanormaltothedirectionofheatflowandletsðxÞbetheheat generatedperunitthicknessofthewall,denotedbyl.Thisisoftencalledaheatsource.Acommonexample ofaheatsourceistheheatgeneratedinanelectricwireduetoresistance.Intheone-dimensionalcase,the rateofheatgenerationismeasuredinunitsofenergypertime;inSIunits,theunitsofenergyarejoules(J) 1 perunitlength(meters,m)andtime(seconds,s).Recallthattheunitofpoweriswatts(1 W¼1Js ).A heatsourcesðxÞisconsideredpositivewhenheatisgenerated,i.e.addedtothesystem,andnegativewhen heatiswithdrawnfromthesystem.Heatflux,denotedbyqðxÞ,isdefinedasatherateofheatflowacrossa 2 surface.Itsunitsareheatrateperunitarea;inSIunits,W m .Itispositivewhenheatflowsinthepositive x-direction.Wewillconsiderasteady-stateproblem,i.e.asystemthatisnotchangingwithtime. To establish the differential equation that governs the system, we consider energy balance (or con- servationofenergy)inacontrolvolumeofthewall.Energybalancerequiresthattherateofheatenergy (qA)thatisgeneratedinthecontrolvolumemustequaltheheatenergyleavingthecontrolvolume,asthe temperature,andhencetheenergyinthecontrolvolume,isconstantinasteady-stateproblem.Theheat energyleavingthecontrolvolumeisthedifferencebetweentheflowinatontheleft-handside,qA,andthe flowoutontheright-handside,qðxþxÞAðxþxÞ.Thus,energybalanceforthecontrolvolumecanbe writtenas sðxþx=2ÞxþqðxÞAðxÞqðxþxÞAðxþxÞ¼ 0: fflfflfflfflfflfflfflfflfflfflfflzfflfflfflfflfflfflfflfflfflfflffl fflfflfflfflfflzfflfflfflfflffl fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflzfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl heat generated heat flow in heat flow out Notethattheheatfluxesaremultipliedbytheareatoobtainatheheatrate,whereasthesourcesismultiplied bythelengthofthesegment.Rearrangingtermsintheaboveanddividingbyx,weobtain qðxþxÞAðxþxÞqðxÞAðxÞ ¼ sðxþx=2Þ: x Ifwetakethelimitoftheaboveequationasx 0,thefirsttermcoincideswiththederivativedðqAÞ=dx andthesecondtermreducestosðxÞ.Therefore,theabovecanbewrittenas dðqAÞ ¼ s: ð3:8Þ dx46 STRONG AND WEAK FORMS FOR ONE-DIMENSIONAL PROBLEMS Theconstitutiveequationforheatflow,whichrelatestheheatfluxtothetemperature,isknownasFourier’s lawandisgivenby dT q¼k ; ð3:9Þ dx where T is the temperature and k is the thermal conductivity (which must be positive); in SI units, the 1o 1 dimensions of thermal conductivity are Wm C . A negative sign appears in (3.9) because the heat flows from high (hot) to low temperature (cold), i.e. opposite to the direction of the gradient of the temperaturefield. Inserting(3.9)into(3.8)yields  d dT Ak þs¼ 0; 0 x l: ð3:10Þ dx dx WhenAkisconstant,weobtain 2 d T Ak þs¼ 0; 0 x l: ð3:11Þ 2 dx Atthetwoendsoftheproblemdomain,eitherthefluxorthetemperaturemustbeprescribed;thesearethe boundaryconditions.WeconsiderthespecificboundaryconditionsoftheprescribedtemperatureTatx¼ l andprescribedfluxqat x¼ 0.Theprescribedfluxqispositiveifheat(energy)flowsoutofthebar,i.e. qðx¼ 0Þ¼q.Thestrongformfortheheatconductionproblemisthengivenby  d dT Ak þs¼0on0 x l; dx dx ð3:12Þ dT q¼ k ¼ q on x¼ 0; dx T ¼ T on x¼ l: 2 3.1.3 Diffusion in One Dimension Diffusionisaprocesswhereamaterialistransportedbyatomicmotion.Thus,intheabsenceofthemotion of a fluid, materials in the fluid are diffused throughout the fluid by atomic motion. Examples are the diffusionofperfumeintoaroomwhenaheavilyperfumedpersonwalksin,thediffusionofcontaminantsin alakeandthediffusionofsaltintoaglassofwater(thewaterwillgetsaltybydiffusionevenintheabsence offluidmotion). Diffusion also occurs in solids. One of the simplest forms of diffusion in solids occurs when two materialscomeincontactwitheachother.Therearetwobasicmechanismsfordiffusioninsolids:vacancy diffusionandinterstitialdiffusion.Vacancydiffusionoccursprimarilywhenthediffusingatomsareofa similar size. A diffusing atom requires a vacancy in the other solid for it to move. Interstitial diffusion, schematicallydepictedinFigure3.4,occurswhenadiffusingatomissmallenoughtomovebetweenthe atomsintheothersolid.Thistypeofdiffusionrequiresnovacancydefects. 3 Letcbetheconcentrationofdiffusingatomswiththedimensionofatoms m .Thefluxofatoms,qðxÞ 2 1 (atoms m s ),ispositiveinthedirectionfromhighertolowerconcentration.Therelationshipbetween fluxandconcentrationisknownasFick’sfirstlaw,whichisgivenas dc q¼k ; dx 2 Recommended for Science and Engineering Track.THE WEAK FORM IN ONE DIMENSION 47 Lattice atoms x Diffusing atoms qx()A() x qx() + ∆ + x A() x ∆ x Figure 3.4 Interstitialdiffusioninanatomiclattice. 2 1 where k is the diffusion coefficient, m s . The balance equation for steady-state diffusion can be developedfromFigure3.4bythesameproceduresthatweusedtoderivetheheatconductionequationby imposingconservationofeachspeciesofatomsandFick’slaw.Theequationsareidenticalinstructureto thesteady-stateheatconductionequationanddifferonlyintheconstantsandvariables:  d dc Ak ¼0on0 x l: dx dx 3.2 THE WEAK FORM IN ONE DIMENSION Todevelopthefiniteelementequations,thepartialdifferential equationsmustberestatedinanintegral form called the weak form. A weak form of the differential equations is equivalent to the governing equation and boundary conditions, i.e. thestrong form. In many disciplines, theweakform has specific names;forexample,itiscalledtheprincipleofvirtualworkinstressanalysis. To show how weak forms are developed, we first consider the strong form of the stress analysis problemgivenin(3.7).Westartbymultiplyingthegoverningequation(3.7a)andthetractionboundary condition(3.7b)byanarbitraryfunctionwðxÞandintegratingoverthedomainsonwhichtheyhold:forthe governingequation,thepertinentdomainistheinterval½0;l,whereasforthetractionboundarycondition, itisthecross-sectionalareaatx¼ 0(nointegralisneededbecausethisconditiononlyholdsonlyatapoint, butwedomultiplybytheareaA).Theresultingtwoequationsare l Z  d du ðaÞ w AE þb dx¼ 0 8w; dx dx 0 ð3:13Þ   du ðbÞ wA E þ t ¼ 0 8w: dx x¼0 ThefunctionwðxÞiscalledtheweightfunction;inmoremathematicaltreatments,itisalsocalledthetest function.Intheabove,8wdenotesthatwðxÞisanarbitraryfunction,i.e.(3.13)hastoholdforallfunctions48 STRONG AND WEAK FORMS FOR ONE-DIMENSIONAL PROBLEMS wðxÞ.Thearbitrarinessoftheweightfunctioniscrucial,asotherwiseaweakformisnotequivalenttothe strongform(seeSection3.7).Theweightfunctioncanbethoughtofasanenforcer:whateveritmultipliesis enforcedtobezerobyitsarbitrariness. Youmighthavenoticedthatwedidnotenforcetheboundaryconditiononthedisplacementin(3.13)by theweightfunction.ItwillbeseenthatitiseasytoconstructtrialorcandidatesolutionsuðxÞthatsatisfythis displacementboundarycondition,sowewillassumethatallcandidatesolutionsofEquation(3.13)satisfy thisboundarycondition.Similarly,youwillshortlyseethatitisconvenienttohaveallweightfunctions satisfy wðlÞ¼ 0: ð3:14Þ Soweimposethisrestrictiononthesetofweightfunctions. As you will see, in solving a weak form, a set of admissible solutions uðxÞ that satisfy certain conditionsisconsidered.Thesesolutionsarecalledtrialsolutions.Theyarealsocalledcandidatesolutions. Onecoulduse(3.13)todevelopafiniteelementmethod,butbecauseofthesecondderivativeofuðxÞ in the expression, very smooth trial solutions would be needed; such smooth trial solutions would be difficult toconstruct in more thanone dimension. Furthermore, the resulting stiffnessmatrix wouldnot be symmetric, because the first integral is not symmetric in wðxÞ and uðxÞ: For this reason, we will transform (3.13) into a form containing only first derivatives. This will lead to a symmetric stiffness matrix, allow us to use less smooth solutions and will simplify the treatment of the traction boundary condition. Forconvenience,werewrite(3.13a)intheequivalentform: l l Z  Z d du w AE dxþ wbdx¼ 0 8w: ð3:15Þ dx dx 0 0 Toobtainaweakforminwhichonlyfirstderivativesappear,wefirstrecalltherulefortakingthederivative ofaproduct: d df dw df d dw ðwfÞ¼ w þf ) w ¼ ðwfÞf : dx dx dx dx dx dx Integratingtheaboveequationontherightoverthedomain0,l,weobtain l l l Z Z Z df d dw w dx¼ ðwfÞdx f dx: dx dx dx 0 0 0 Thefundamentaltheoremofcalculusstatesthattheintegralofa derivativeofafunctionisthefunction itself.Thistheoremenablesustoreplacethefirstintegralontheright-handsidebyasetofboundaryvalues andrewritetheequationas l l l Z Z Z df dw dw l w dx¼ðwfÞj  f dxðwfÞ ðwfÞ  f dx: ð3:16Þ 0 x¼l x¼0 dx dx dx 0 0 0 Theaboveformulaisknownasintegrationbyparts.Wewillfindthatintegrationbypartsisusefulwhenever werelatestrongformstoweakforms.THE WEAK FORM IN ONE DIMENSION 49 Toapplytheintegrationbypartsformulato(3.15),letf ¼ AEðdu=dxÞ.Then(3.16)canbewrittenas l l   Z Z l  d du du dw du  w AE dx¼ wAE  AE dx: ð3:17Þ  dx dx dx dx dx 0 0 0 Using(3.17),(3.15)canbewrittenasfollows:  0 1 l  l l  Z Z  B C du dw du B C wAE  AE dxþ wbdx¼ 0 8w with wðlÞ¼ 0: ð3:18Þ A dx dx dx  fflzffl 0 0  s 0 Wenotethatbythestress–strainlawandstrain–displacementequations,theunderscoredboundarytermis thestresss(asshown),sotheabovecanberewrittenas l l Z Z dw du ðwAsÞ ðwAsÞ  AE dxþ wbdx¼ 0 8w with wðlÞ¼ 0: x¼l x¼0 dx dx 0 0 The first term in the above vanishes because of (3.14): this is why it is convenient to construct weight functionsthatvanishonprescribeddisplacementboundaries.Thoughthetermlooksquiteinsignificant,it wouldleadtolossofsymmetryinthefinalequations. From(3.13b),wecanseethatthesecondtermequalsðwAtÞ ,sotheaboveequationbecomes x¼0 l l Z Z dw du AE dx¼ðwAtÞ þ wbdx 8w with wðlÞ¼ 0: ð3:19Þ x¼0 dx dx 0 0 Let us recapitulate what we have done. We have multiplied the governing equation and traction boundary by an arbitrary, smooth weight function and integrated the products over the domains where theyhold.Wehaveaddedtheexpressionsandtransformedtheintegralsothatthederivativesareoflower order. Wenowcometothecruxofthisdevelopment:Westatethatthetrialsolutionthatsatisfiestheabovefor allsmoothwðxÞwithwðlÞ¼ 0isthesolution.Sothesolutionisobtainedasfollows: Find uðxÞ among the smooth functions that satisfy uðlÞ¼ u such that l l Z Z ð3:20Þ dw du AE dx¼ðwAtÞ þ wbdx 8w with wðlÞ¼ 0: x¼0 dx dx 0 0 Theaboveiscalledtheweakform.Thenameoriginatesfromthefactthatsolutionstotheweakformneed not be as smooth as solutions of the strong form, i.e. theyhaveweaker continuity requirements. This is explainedlater. Understandinghowasolutiontoadifferentialequationcanbeobtainedbythisratherabstractstatement, andwhyitisausefulsolution,isnoteasy.Ittakesmoststudentsconsiderablethoughtandexperienceto comprehendtheprocess.Tofacilitatethis,wewillgivetwoexamplesinwhichasolutionisobtainedtoa specificproblem.50 STRONG AND WEAK FORMS FOR ONE-DIMENSIONAL PROBLEMS Wewillshowinthenextsectionthattheweakform(3.20)isequivalenttotheequilibriumequation(3.7a) andtractionboundarycondition(3.7b).Inotherwords,thetrialsolutionthatsatisfies(3.20)isthesolution ofthestrongform.TheproofofthisstatementinSection3.4isacrucialstepinthetheoryoffiniteelements. Ingettingto(3.19),wehavegonethroughasetofmathematicalstepsthatarecorrect,butwehavenobasis forsayingthatthesolutiontotheweakformisasolutionofthestrongformunlesswecanshowthat(3.20) implies(3.7). It is important to remember that the trial solutions uðxÞ must satisfy the displacement boundary conditions (3.7c). Satisfying the displacement boundary condition is essential for the trial solutions, so theseboundaryconditionsareoftencalledessentialboundaryconditions.WewillseeinSection3.4thatthe tractionboundaryconditionsemanatenaturallyfromtheweakform(3.20),sotrialsolutionsneednotbe constructedtosatisfythetractionboundaryconditions.Therefore,theseboundaryconditionsarecalled naturalboundaryconditions.Additionalsmoothnessrequirementsonthetrialsolutionswillbediscussed inSections3.3and3.9. A trial solution that is smooth and satisfies the essential boundary conditions is called admissible. Similarly,aweightfunctionthatissmoothandvanishesonessentialboundariesisadmissible.Whenweak formsareusedtosolveaproblem,thetrialsolutionsandweightfunctionsmustbeadmissible. Notethatin(3.20),theintegralissymmetricinwandu.Thiswillleadtoasymmetricstiffnessmatrix. Furthermore, the highest order derivative that appears in the integral is of first order: this will have importantramificationsontheconstructionoffiniteelementmethods. 3.3 CONTINUITY Although we have now developed the weak form, we still have not specified how smooth the weight functions and trial solutions must be. Before examining this topic, we will examine the concept of n smoothness, i.e. continuity. A function is called a C function if its derivatives of order j for 0 j n 0 1 1 existandarecontinuousfunctionsintheentiredomain.WewillbeconcernedmainlywithC ; C andC 0 functions. Examples of these are illustrated in Figure 3.5. As can be seen, a C function is piecewise continuouslydifferentiable,i.e.itsfirstderivativeiscontinuousexceptatselectedpoints.Thederivativeof 0 1 0 1 aC functionisaC function.Soforexample,ifthedisplacementisaC function,thestrainisaC 0 1 0 function.Similarly,ifatemperaturefieldisaC function,thefluxisaC functioniftheconductivityisC . n n1 Ingeneral,thederivativeofaC functionisC . 0 1 1 ThedegreeofsmoothnessofC ; C andC functionscanberememberedbysomesimplemnemonic 1 0 devices.AscanbeseenfromFigure3.5,aC functioncanhavebothkinksandjumps.AC functionhas 1 no jumps, i.e. discontinuities, but it has kinks. A C function has no kinks or jumps. Thus, there is a progressionofsmoothnessasthesuperscriptincreasesthatissummarizedinTable3.1.Intheliterature, jumpsinthefunctionareoftencalledstrongdiscontinuities,whereaskinksarecalledweakdiscontinuities. ItisworthmentioningthatCADdatabasesforsmoothsurfacesusuallyemployfunctionsthatareatleast 1 C ;themostcommonaresplinefunctions.Otherwise,thesurfacewouldpossesskinksstemmingfromthe 1 function description, e.g. in a car therewould be kinks in the sheet metal wherever C continuity is not 0 observed.WewillseethatfiniteelementsusuallyemployC functions. f(x) 1 C 0 C Jumps Kinks –1 C x 1 0 1 Figure 3.5 ExamplesofC ,C andC functions.THE EQUIVALENCE BETWEEN THE WEAK AND STRONG FORMS 51 Table 3.1 Smoothness of functions. Smoothness Kinks Jumps Comments 1 C Yes Yes Piecewisecontinuous 0 C Yes No Piecewisecontinuouslydifferentiable 1 C No No Continuouslydifferentiable 3.4 THE EQUIVALENCE BETWEEN THE WEAK AND STRONG FORMS In the previous section, we constructed the weak form from the strong form. To show the equivalence betweenthetwo,wewillnowshowtheconverse:theweakformimpliesthestrongform.Thiswillinsure thatwhenwesolvetheweakform,thenwehaveasolutiontothestrongform. Theproofthattheweakformimpliesthestrongformcanbeobtainedbysimplyreversingthestepsby whichweobtainedtheweakform.Soinsteadofusingintegrationbypartstoeliminatethesecondderivative ofuðxÞ,wereversetheformulatoobtainanintegralwithahigherderivativeandaboundaryterm.Forthis purpose,interchangethetermsin(3.17),whichgives l l  Z  Z  l  dw du du d du  AE dx¼ wAE  w AE dx:  dx dx dx dx dx 0 0 0 Substituting the above into(3.20) and placing theintegraltermsonthe left-hand side and theboundary termsontheright-handsidegives l   Z d du w AE þb dxþwAðtþsÞ ¼ 0 8w with wðlÞ¼ 0: ð3:21Þ x¼0 dx dx 0 ThekeytomakingtheproofpossibleisthearbitrarinessofwðxÞ.Itcanbeassumedtobeanythingweneedin ordertoprovetheequivalence.OurselectionofwðxÞisguidedbyhavingseenthisproofbefore–Whatwe willdoisnotimmediatelyobvious,butyouwillseeitworksFirst,welet   d du w¼ cðxÞ AE þb ; ð3:22Þ dx dx where cðxÞ is smooth, cðxÞ 0on0 x l and cðxÞ vanishes on the boundaries. An example of a function satisfying the above requirements is cðxÞ¼ xðlxÞ. Because of how cðxÞ is constructed, it follows that wðlÞ¼ 0, so the requirement that w¼ 0 on the prescribed displacement boundary, i.e. the essentialboundary,ismet. Inserting(3.22)into(3.21)yields l Z  2 d du c AE þb dx¼ 0: ð3:23Þ dx dx 0 The boundary term vanishes becausewe have constructed theweight function so that wð0Þ¼ 0. As the integrandin(3.23)istheproductofapositivefunctionandthesquareofafunction,itmustbepositiveat52 STRONG AND WEAK FORMS FOR ONE-DIMENSIONAL PROBLEMS everypointintheproblemdomain.Sotheonlywaytheequalityin(3.23)ismetisiftheintegrandiszeroat everypointHence,itfollowsthat  d du AE þb¼ 0; 0 x l; ð3:24Þ dx dx whichispreciselythedifferentialequationinthestrongform,(3.7a). From(3.24)itfollowsthattheintegralin(3.21)vanishes,soweareleftwith ðwAðtþsÞÞ ¼ 0 8w with wðlÞ¼ 0: ð3:25Þ x¼0 Astheweightfunctionisarbitrary,weselectitsuchthatwð0Þ¼ 1andwðlÞ¼ 0.Itisveryeasytoconstruct suchafunction,forexample,ðlxÞ=lisasuitableweightfunction;anysmoothfunctionthatyoucandraw ontheinterval0,lthatvanishesatx¼ lisalsosuitable. Asthecross-sectionalareaA(0)6¼0andwð0Þ¼ 6 0,itfollowsthat s¼t at x¼ 0; ð3:26Þ whichisthenatural(prescribedtraction)boundarycondition,Equation(3.7b). Thelastremainingequationofthestrongform,thedisplacementboundarycondition(3.7c),issatisfied byalltrialsolutionsbyconstruction,i.e.ascanbeseenfrom(3.20)werequiredthatuðlÞ¼ u.Therefore, wecanconcludethatthetrialsolutionthatsatisfiestheweakformsatisfiesthestrongform. Another waytoprovetheequivalencetothestrongform startingfrom(3.20)thatismoreinstructive aboutthecharacteroftheequivalenceisasfollows.Wefirstlet  d du rðxÞ¼ AE þb for 0 x l dx dx and r ¼ Að0Þsð0Þþt: 0 ThevariablerðxÞiscalledtheresidual;rðxÞistheerrorinEquation(3.7a)andr istheerrorinthetraction 0 boundarycondition(3.7b). Notethat when rðxÞ¼ 0,theequilibriumequation (3.7a)ismet exactly and whenr ¼ 0thetractionboundarycondition(3.7b)ismetexactly. 0 Equation(3.20)canthenbewrittenas l Z wðxÞrðxÞdxþwð0Þr ¼ 0 8w with wðlÞ¼ 0: ð3:27Þ 0 0 We now prove that rðxÞ¼ 0 by contradiction. Assume that at some point 0 a l, rðaÞ¼ 6 0. Then assumingrðxÞissmooth,itmustbenonzeroinasmallneighborhoodofx¼ aasshowninFigure3.6(a).We havecompletelatitudeintheconstructionofwðxÞasitisanarbitrarysmoothfunction.Soweconstructitas showninFigure3.6(b).Equation(3.27)thenbecomes l Z 1 wðxÞrðxÞdxþwð0Þr  rðaÞ¼ 6 0: 0 2 0THE EQUIVALENCE BETWEEN THE WEAK AND STRONG FORMS 53 rx rx () () • (a) x a a x wx () wx () 1 1 (b) a a x x δ δ wr wr (c) a a x x Figure 3.6 Illustrationoftheequivalencebetweentheweakandstrongforms:(a)anexampleoftheresidualfunction; (b)choiceoftheweightfunctionand(c)productofresidualandweightfunctions.Ontheleft,theprocedureisshownfora 1  C function;ontherightforaC function. Theaboveimpliesthat(3.27)isviolated,sobycontradictionrðaÞcannotbenonzero.Thiscanberepeated atanyotherpointintheopeninterval0 x l,soitfollowsthatrðxÞ¼ 0for0 x l,i.e.thegoverning equation(3.27)ismet.Wenowlet wð0Þ¼ 1;astheintegralvanishesbecauserðxÞ¼ 0for0 x l,it followsfrom(3.27)thatr ¼ 0andhencethetractionboundaryconditionisalsomet. 0 We can see from the above why we have said that multiplying the equation, or to be more precise the residual, by the weight function enforces the equation: because of the arbitrariness of the weight function, anything it multiplies must vanish. The proofs of the equivalence of the strong and weak forms hinge critically on the weak form holding for any smooth function. In the first proof (Equations (3.7)–(3.20)),weselectedaspecialarbitraryweightfunction(basedonforesightastohowtheproofwould evolve) that has to be smooth, whereas in the second proof, we used the arbitrariness and smoothness directly.TheweightfunctioninFigure3.6(b)maynotappearparticularlysmooth,butitisassmoothaswe needforthisproof. Example3.1 Developtheweakformforthestrongform:  d du ðaÞ AE þ10Ax¼ 0; 0 x 2; dx Ex 4 ðbÞ u  uð0Þ¼ 10 ; ð3:28Þ x¼0  du ðcÞ s ¼ E ¼ 10: x¼2 dx x¼2 Equation(3.28c)isaconditiononthederivativeofuðxÞ,soitisanaturalboundarycondition;(3.28b)isa conditiononuðxÞ,soitisanessentialboundarycondition.Therefore,astheweightfunctionmustvanish ontheessentialboundaries,weconsiderallsmoothweightfunctionswðxÞsuchthatwð0Þ¼ 0.Thetrial 4 solutionsuðxÞmustsatisfytheessentialboundaryconditionuð0Þ¼ 10 .54 STRONG AND WEAK FORMS FOR ONE-DIMENSIONAL PROBLEMS Westartbymultiplyingthegoverningequationandthenaturalboundaryconditionoverthedomains wheretheyholdbyanarbitraryweightfunction: 2    Z d du ðaÞ w AE þ10Ax dx¼ 0 8wðxÞ; dx dx ð3:29Þ 0 du ðbÞðwAðE 10ÞÞ ¼ 0 8wð2Þ: x¼2 dx Nextweintegratethefirstequationintheabovebyparts,exactlyaswedidingoingfrom(3.13a)to(3.17): 2 2    Z Z x¼2  d du du dw du  w AE dx¼ wAE  AE dx: ð3:30Þ  dx dx dx dx dx x¼0 0 0 Wehaveconstructedtheweightfunctionssothatwð0Þ¼ 0;therefore,thefirsttermontheRHSofthe abovevanishesatx¼ 0.Substituting(3.30)into(3.29a)gives 2 2 Z Z  dwdu du  AE dxþ 10wAxdxþ wAE ¼ 0 8wðxÞ with wð0Þ¼ 0: ð3:31Þ dx dx dx x¼2 0 0 Substituting(3.29b)intothelasttermof(3.31)gives(afterachangeofsign) 2 2 Z Z dwdu AE dx 10wAxdx10ðwAÞ ¼ 0 8wðxÞ with wð0Þ¼ 0: ð3:32Þ x¼2 dx dx 0 0 4 Thus,theweakformisasfollows:finduðxÞsuchthatforallsmoothuðxÞwith uð0Þ¼ 10 ,suchthat (3.32)holdsforallsmoothwðxÞwithwð0Þ¼ 0. Example3.2 Developtheweakformforthestrongform: 2 d u ¼0on1 x 3; 2 dx  ð3:33Þ du ¼ 2; uð3Þ¼ 1: dx x¼1 Theconditionsontheweightfunctionandtrialsolutioncanbeinferredfromtheboundaryconditions. The boundary point x¼ 1 is a natural boundary as the derivative is prescribed there, whereas the boundaryx¼ 3isanessentialboundaryasthesolutionitselfisprescribed.Therefore,werequirethat wð3Þ¼ 0andthatthetrialsolutionsatisfiestheessentialboundaryconditionuð3Þ¼ 1. Next we multiply the governing equation by the weight function and integrate over the problem domain;similarly,wemultiplythenaturalboundaryconditionbytheweightfunction,whichyields 3 Z 2 d u ðaÞ w dx¼ 0; 2 dx ð3:34Þ 1   du ðbÞ w 2 ¼ 0: dx x¼1THE EQUIVALENCE BETWEEN THE WEAK AND STRONG FORMS 55 Integrationbypartsoftheintegrandin(3.34a)gives 3 3   Z Z 2 d u du du dwdu w dx¼ w  w  dx: ð3:35Þ 2 dx dx dx dx dx x¼3 x¼1 1 1 Aswð3Þ¼ 0,thefirsttermontheRHSintheabovevanishes.Substituting(3.35)into(3.34a)gives 3  Z dwdu du  dx w ¼ 0: ð3:36Þ dx dx dx x¼1 1 Adding(3.34b)to(3.36)gives 3 Z dwdu dxþ2wð1Þ¼ 0: ð3:37Þ dx dx 1 Sotheweakformis:findasmoothfunctionuðxÞwithuð3Þ¼ 1forwhich(3.37)holdsforallsmoothwðxÞ withwð3Þ¼ 0. Toshowthattheweakformimpliesthestrongform,wereversetheprecedingsteps.Integrationby partsofthefirsttermin(3.37)gives 3 3  Z Z 3 2  dwdu du d u  dx¼ w  w dx: ð3:38Þ  2 dx dx dx dx 1 1 1 Nextwesubstitute(3.38)into(3.37),giving 3   Z 2 du du d u w  w  w dxþ2wð1Þ¼ 0: ð3:39Þ 2 dx dx dx x¼3 x¼1 1 Sinceontheessentialboundary,theweightfunctionvanishes,i.e.wð3Þ¼ 0,thefirsttermintheabove dropsout.Collectingtermsandchangingsignsgive 3 Z   2 d u du w dxþ w 2 ¼ 0: ð3:40Þ 2 dx dx x¼1 1 WenowusethesameargumentsasEquations(3.22)–(3.26).AswðxÞisarbitrary,let 2 d uðxÞ w¼ cðxÞ ; 2 dx where 8 0; x¼ 1; cðxÞ¼ 0; 1 x 3; : 0; x¼ 3:56 STRONG AND WEAK FORMS FOR ONE-DIMENSIONAL PROBLEMS Then(3.40)becomes 3  Z 2 2 d u cðxÞ dx¼ 0: 2 dx 1 As the integrand is positive in the interval½1;3, it follows that the only way that the integrand can vanishisif 2 d uðxÞ ¼ 0 for 1 x 3; 2 dx whichisthedifferentialequationinthestrongform(3.33). NowletwðxÞbeasmoothfunctionthatvanishesatx¼ 3butequalsoneatx¼ 1.Youcandrawan infinitenumberofsuchfunctions:anycurvebetweenthosepointswiththespecifiedendvalueswilldo. Aswealreadyknowthattheintegralin(3.40)vanishes,weareleftwith    du du w 2 ¼ 0 ) 2 ¼ 0; dx dx x¼1 x¼1 sothenaturalboundaryconditionissatisfied.Astheessentialboundaryconditionissatisfiedbyalltrial solutions,wecanthenconcludethatthesolutionoftheweakformisthesolutiontothestrongform. Example3.3 ObtainasolutiontotheweakforminExample3.1byusingtrialsolutionsandweightfunctionsofthe form uðxÞ¼ a þa x; 0 1 wðxÞ¼ b þb x; 0 1 where a and a are unknown parameters and b and b are arbitrary parameters. Assume that A is 0 1 0 1 5 constantandE¼ 10 .Tobeadmissibletheweightfunctionmustvanishatx¼ 0,sob ¼ 0.Forthetrial 0 4 4 solutiontobeadmissible,itmustsatisfytheessentialboundaryconditionuð0Þ¼ 10 ,soa ¼ 10 . 0 From this simplification, it follows that only one unknown parameter and one arbitrary parameter remain,and duðxÞ 4 uðxÞ¼ 10 þa x; ¼ a ; 1 1 dx ð3:41Þ dw wðxÞ¼ b x; ¼ b : 1 1 dx Substitutingtheaboveintotheweakform(3.32)yields 2 2 Z Z b a Edx b x10dxðb x10Þ ¼ 0: 1 1 1 1 x¼2 0 0 Evaluatingtheintegralsandfactoringoutb gives 1 b ð2a E2020Þ¼ 0: 1 1THE EQUIVALENCE BETWEEN THE WEAK AND STRONG FORMS 57 As the above must hold for all b , it follows that the term in the parentheses must vanish, so 1 4 a ¼ 20=E¼ 210 .Substitutingthisresultinto(3.41)givestheweaksolution,whichweindicate 1 lin 4 lin bysuperscript‘lin’ asit isobtainedfrom lineartrialsolutions: u ¼ 10 ð1þ2xÞands ¼ 20(the stress-strainlawmustbeusedtoobtainthestresses).TheresultsareshowninFigure3.7andcomparedto theexactsolutiongivenby ex 4 3 ex 2 u ðxÞ¼ 10 ð1þ3xx =6Þ; s ðxÞ¼ 10ð3x =2Þ: Observethateventhisverysimplelinearapproximationforatrialsolutiongivesareasonablyaccurate result,butitisnotexact.Wewillseethesamelackofexactnessinfiniteelementsolutions. Repeattheabovewithquadratictrialsolutionsandweightfunctions 2 2 uðxÞ¼ a þa xþa x ; wðxÞ¼ b þb xþb x : 0 1 2 0 1 2 4 Asbefore,becauseoftheconditionsontheessentialboundaries,a ¼ 10 andb ¼ 0.Substitutingthe 0 0 abovefieldswiththegivenvaluesofa andb intotheweakformgives 0 0 2 2 Z Z 2 2 ðb þ2b xÞðEða þ2a xÞÞdx ðb xþb x Þ10dxððb xþb x Þ 10Þ ¼ 0: 1 2 1 2 1 2 1 2 x¼2 0 0 Integrating,factoringoutb ,b andrearrangingthetermsgives 1 2   32a 200 2 b ½Eð2a þ4aÞ40þb 4a þ E ¼ 0: 1 1 2 2 1 3 3 Astheabovemustholdforarbitraryweightfunctions,itmustholdforarbitraryb andb .Therefore,the 1 2 coefficientsofb andb mustvanish(recallthescalarproducttheorem),whichgivesthefollowinglinear 1 2 algebraicequationina anda : 1 2 2 3 2 3 " 24 40 a 1 4 5 4 5 E ¼ : 32 200 4 a 2 3 3 60 30 quad 55 28 u (x) ex σ (x) ex 50 26 u (x) 45 24 quad σ (x) 40 22 35 20 30 lin 18 lin u (x) σ (x) 25 16 20 14 15 12 10 10 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 (b) (a) Figure 3.7 Comparisonof linear(lin) andquadratic(quad) approximations to theexact solutionof (a)displace- mentsand(b)stresses. 5 Displacements (×10 ) Stresses58 STRONG AND WEAK FORMS FOR ONE-DIMENSIONAL PROBLEMS 4 4 Thesolutionisa ¼ 310 anda ¼0:510 .Theresultingdisplacementsandstressesare 1 2 quad 4 2 quad u ¼ 10 ð1þ3x0:5x Þ; s ¼ 10ð3xÞ: TheweaksolutionisshowninFigure3.7,fromwhichyoucanseethatthetwo-parameter,quadratictrial solutionmatchestheexactsolutionmorecloselythantheone-parameterlineartrialsolution. 3.5 ONE-DIMENSIONAL STRESS ANALYSIS WITH ARBITRARY BOUNDARY CONDITIONS 3.5.1 Strong Form for One-Dimensional Stress Analysis Wewillnowconsideramoregeneralsituation,whereinsteadofspecifyingastressboundarycondition at x¼ 0 and a displacement boundary condition at x¼ l, displacement and stress boundary conditions can be prescribed at either end. For this purpose, we will need a more general notation for the boundaries. Theboundaryoftheone-dimensionaldomain,whichconsistsoftwoendpoints,isdenotedby.The portionoftheboundarywherethedisplacementsareprescribedisdenotedby ;theboundarywherethe u tractionisprescribedisdenotedby.Inthisgeneralnotation,both and canbeemptysets(nopoints), t u t onepointortwopoints.Thetractionanddisplacementbothcannotbeprescribedatthesameboundary point.Physically, this can be seen to be impossible by considering a bar such as that in Figure 3.2. If we could prescribe both the displacement and the force on the right-hand side, this would mean that the deformation of the bar is independent of the applied force. It would also mean that the material properties have noeffect on the force–displacementbehaviorofthebar.Obviously,thisisphysically unrealistic, so any boundary point is either a prescribedtractionoraprescribed displacement boundary. We write this as  \ ¼ 0. We will see from subsequent examples that this can be t u generalized to other systems: Natural boundary conditions and essential boundary conditions cannot be applied at the same boundary points. We will often call boundaries with essential boundary conditions essential boundaries; similarly, boundaries with natural boundary conditions will be called natural boundaries. We can then say that a boundarycannotbebothanaturalandanessentialboundary.Italsofollowsfromthetheoryofboundary valueproblemsthatonetypeofboundaryconditionisneededateachboundarypoint,i.e.wecannothave anyboundaryatwhichneitheranessentialnoranaturalboundaryconditionisapplied.Thus,anyboundary iseitheranessentialboundaryoranaturalboundaryandtheirunionistheentireboundary.Mathematically, thiscanbewrittenas  ¼. t u Tosummarizetheabove,atanyboundary,eitherthefunctionoritsderivativemustbespecified,butwe cannot specify both at the same boundary. So any boundary must be an essential boundary or a natural boundary,butitcannotbeboth.Theseconditionsareveryimportantandcanbemathematicallyexpressed bythetwoconditionsthatwehavestatedabove:   ¼;  \ ¼ 0: ð3:42Þ t u t u Thetwoboundariesaresaidtobecomplementary:theessentialboundaryplusitscomplement,thenatural boundary,constitutethetotalboundary,andviceversa. Usingtheabovenotation,wesummarizethestrongformforone-dimensionalstressanalysis(3.7)in Box3.1.ONE-DIMENSIONAL STRESS ANALYSIS WITH ARBITRARY BOUNDARY CONDITIONS 59 Box3.1.Strongformfor1Dstressanalysis  d du AE þb¼ 0; 0 x l; dx dx du ð3:43Þ sn¼ En ¼ t on  ; t dx u¼ u on  : u Intheabove,wehaveaddedaunitnormaltothebodyanddenoteditbyn;ascanbeseenfromFigure3.2, n¼1at x¼ 0andn¼þ1at x¼ l.Thistrickenablesustowritetheboundaryconditionintermsofthe tractionsappliedateitherend.Forexample,whenapositiveforceperunitareaisappliedattheleft-hand endofthebarinFigure3.2,thestressatthatendisnegative,i.e.compressive,andsn¼s¼ t.Atany right-handboundarypoint,n¼þ1andsosn¼ s¼ t. 3.5.2 Weak Form for One-Dimensional Stress Analysis Inthissection,wewilldeveloptheweakformforone-dimensionalstressanalysis(3.43),witharbitrary boundary conditions. We first rewrite the formula for integration by parts in the notation introduced in Section3.2: Z Z Z df dw dw w dx¼ðwfnÞj  f dx¼ðwfnÞj þðwfnÞj  f dx: ð3:44Þ    u t dx dx dx    Intheabove,thesubscriptontheintegralindicatesthattheintegralisevaluatedovertheone-dimensional problemdomain,i.e.thenotationindicatesanylimitsofintegration,suchas½0;l,½a;b.Thesubscript indicatesthattheprecedingquantityisevaluatedatallboundarypoints,whereasthesubscripts and u t indicatethattheprecedingquantitiesareevaluatedontheprescribeddisplacementandtractionboundaries, respectively. The second equality follows from the complementarity of the traction and displacement boundaries:Since,asindicatedby(3.42),thetotalboundaryisthesumofthetractionanddisplacement boundaries,theboundarytermcanbeexpressedasthesumofthetractionanddisplacementboundaries. Theweightfunctionsareconstructedsothatw¼0on ,andthetrialsolutionsareconstructedsothat u u¼ uon . u Wemultiplythefirsttwoequationsinthestrongform(3.43)bytheweightfunctionandintegrateoverthe domainsoverwhichtheyhold:thedomainforthedifferentialequationandthedomain forthetraction t boundarycondition.Thisgives Z  d du ðaÞ w AE þb dx¼ 0 8w; dx dx ð3:45Þ  ðbÞðwAðtsnÞÞj ¼ 0 8w:  t Denotingf ¼ AEðdu=dxÞandusingintegrationbyparts(3.44)ofthefirsttermin(3.45a)andcombining with(3.45b)yields Z Z dw du ðwAsnÞj þðwAtÞj  AE dxþ wbdx¼ 0 8w with w¼0on : ð3:46Þ u   u t dx dx  60 STRONG AND WEAK FORMS FOR ONE-DIMENSIONAL PROBLEMS Theboundarytermon vanishesbecausewj ¼ 0.Theweakformthenbecomes u  u Z Z dw du AE dx¼ðwAtÞj þ wbdx 8w with w¼0on : u  t dx dx   Atthispoint,weintroducesomenewnotation,sowewillnotneedtokeeprepeatingthephrase‘uðxÞissmooth enoughandsatisfiestheessentialboundarycondition’.Forthispurpose,wewilldenotethesetofallfunctions 1 1 0 1 0 thataresmoothenoughbyH .H functionsareC continuous.Mathematically,thisisexpressedasH  C . 0 1 However,notallC functionsaresuitabletrialsolutions.WewillfurtherelaborateonthisinSection3.9;H is aspaceoffunctionswithsquareintegrablederivatives. WedenotethesetofallfunctionsthatareadmissibletrialsolutionsbyU,where   1  U ¼ uðxÞ uðxÞ2 H ; u¼ u on : ð3:47Þ u AnyfunctioninthesetUhastosatisfyallconditionsthatfollowtheverticalbar.Thus,theabovedenotesthe set of all functions that are smooth enough (the first condition after the bar) and satisfy the essential boundary condition (the condition after the comma). Thus, we can indicate that a function uðxÞ is an admissibletrialsolutionbystatingthatuðxÞisinthesetU,oruðxÞ2 U. Wewillsimilarlydenotethesetofalladmissibleweightfunctionsby   1  U ¼ wðxÞ wðxÞ2 H ; w¼0on : ð3:48Þ 0 u Notice that this set of functions is identical to U, except that the weight functions must vanish on the essentialboundaries.ThisspaceisdistinguishedfromU bythesubscriptnought. 1 Suchsetsoffunctionsareoftencalledfunctionspaces,orjustspaces.ThefunctionspaceH containsan infinitenumberoffunctions.Therefore,itiscalledaninfinite-dimensionalset.Foradiscussionofvarious spaces,thereadermaywishtoconsultCiarlet(1978),OdenandReddy(1978)andHughes(1987). Withthesedefinitions,wecanwritetheweakform((3.45),(3.47)and(3.48))asinBox3.2. Box3.2.Weakformfor1Dstressanalysis FinduðxÞ2 U suchthat Z Z dw du AE dx¼ðwAtÞj þ wbdx 8w2 U : ð3:49Þ 0  t dx dx   NotethatthefunctionswðxÞanduðxÞappearsymmetricallyinthefirstintegralin(3.49),whereastheydo notin(3.45a).In(3.49),boththetrialsolutionsandweightfunctionsappearasfirstderivatives,whereasin thefirstintegralin(3.45a),theweightfunctionsappeardirectlyandthetrialsolutionappearsasasecond derivative. It will be seen that consequently (3.49) leads to a symmetric stiffness matrix and a set of symmetriclinearalgebraicequations,whereas(3.45a)doesnot. 3.6 ONE-DIMENSIONAL HEAT CONDUCTION WITH ARBITRARY 3 BOUNDARY CONDITIONS 3.6.1 Strong Form for Heat Conduction in One Dimension with Arbitrary Boundary Conditions Followingthesame procedureasinSection3.5.1,theportion oftheboundarywherethetemperature is prescribed,i.e.theessentialboundary,isdenotedby andtheboundarywherethefluxisprescribedis T 3 Recommended for Science and Engineering Track.

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