Strong and Weak Forms for One-Dimensional Problems
Strong and Weak Forms for One-Dimensional Problems
GregDeamons,New Zealand,Professional
Published Date:03-08-2017
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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
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