ChapterThreeEnergyreleaseratetheoryEnergyreleaserateGisanotherimportantconceptinfracturemechanics.ThereisarelationshipbetweenGandK.Oneisknown.Anothercanalsobeknown.Sometime,itisdifficulttoobtainKfactor.However,itisperhapseasytocalculateG§3-1ConceptofenergyreleaserateBasicconceptsStrainenergyWorkdonebytheexternalforceischangedtothestrainenergystoredintheelasticdeformation.Thestrainenergycanbereleasedandtheelasticdeformationthendisappears.Strainenergydensityw:strainenergyperunitvolume.-,1w=Jg*ds*-—<5s,(forlinearelasticbody)0ijij2司司TotalstrainenergyU(Internalforcepotential):totalenergystoredinthevolumeVU-JwdVVExternalforcepotentialUp.thenegativevalueofvirtualworkdonebytheexternalforce.AssumethatthebodyforceisBandthesurfaceforceisTonthestressboundary.TheexternalforcepotentialUpisU=-(JBudV+JTudS)TOC\o"1-5"\h\zPiiQiiVS5TotalpotentialH:n-u+u-JwdV-(JBudV+JTudS)PVV11S115EnergyreleaserateGThebodyforceB,thesurfaceforceTonSandthedisplacementuonSaregiven.Assumethatthecracksizeischangedfromatoa+Aa.Accordingly,the

displacement,strain,stress,stainenergydensity,internalforcepotential,externalforcepotentialandtotalpotentialarealsochanged.Thetotalpotentialnischangedton+An.Anistheincrementcausedbythecrackgrowthka.Assumethattheplatethicknessist.AS=t-Aadenotesthesinglesurfaceareaincrement.TheenergyreleaserateGisdefinedasAn而G——lim———Aa_0ASdSIftheplatethicknesstisaconstant,theenergyreleaserateGistdaAn1dnG——tdaAa^0t"AaConstantforceandconstantdisplacementconditions(1)ConstantforceconditionAplatewithacrackisappliedbyaconstantforceFasshowninFig.3.1(b).TheexternalforcevirtualworkWandexternalforcepotentialUprespectively,areW=F5,Up=-W=-F5ThetotalstrainenergyU(internalforcepotential)isU=1F2Then,thetotalpotentialnisNow,W=2U,n=-U.TheenergyreleaserateGundertheconstantforceconditioncanbewrittenas

八anQU、G=_a、-(泌)f，forconstantF.Inthiscase,aU>0,thestrainenergyinthebodyinfactincreasesratherthanreleaseswiththecrackgrowth.Gcannotbecalledasthestrainenergyreleaserate.(2)ConstantdisplacementconditionAfteradisplacement5occurs,theplateisclamped.ThisistheconstantdisplacementconditionasshowninFig.3.1(c).Inthiscase,thereisW=0,n=U-W=U八an,aU、G=-~s--)5,forconstantdisplacement5.Itisseenthatonlyfortheconstantdisplacementcondition,Gcanbecalledasthestrainenergyreleaserate.SinceW=0,theenergyneededbythecrackgrowthcomesfromthereleaseofstrainenergystoredinthebody.Thatisthestrainenergystoredinthebodydecreaseswiththecrackgrowth.Foetheconstantforcecondition,theincrementofexternalworkisdW=Fd5inwhichapartisusedtoincreasestrainenergydUwhiletheotherpartisusedforcrackgrowth.However,thevaluesofGfortwocasesareequal.Constantforcecase:F2ac.Cisthecomplianceoftheplate.UF2ac.Cisthecomplianceoftheplate.ModeIcrack22asFConstantdisplacementcase:Q52U=-F5=—,2CaEa52152-1acp2ac=-(~as)5~~~dSQ52U=-F5=—,2CItisseenthatthevaluesofGfortwocasesareequal.§3-2RelationbetweenGandKTherelationshipbetweenGandKisoneofthemostimportantrelationsinfracturemechanics.betweenKandGcanbeestablished.Fig.3.2AsketchofcrackclosureAssumethatasegmentofcracklengthisclosure.Tothisend,adistributedstress%,isapplied.Thisdistributedstress%isthestressfieldinthevicinityinthecracktipfor0=0,r=x,i.e.K

b—i

yyv'2kxInaddition,fortheplanestraincase,thedisplacementproducedinthecrackclosureprocesscanbeknownfromtheasymptoticsolutionofdisplacementfieldinthevicinityofthecracktip.Thepointo’istakenastheorigifiainppin知aFox,thedisplacementalongy-axisis心)=eK蘭2(1-v)=411X22K."三Ei\丸Ei2兀Assumethattheplatethicknessis1.TheappliedforceFinthearea1-dxisAsshowninFig.3.2,thetotalcrackclosuredisplacementis8=2v.Sincetheactualworkdonebytheexternalforcebyinthelengthdxisequaltothestrainenergy,thereis

°1f1…C7U=萬(wàn)F8=2b-1-dx-2v=bv-dxThecrackclosurelengthisasothatthechangeofsystemstrainenergyisIaovdxThechangeofsystemstrainenergybyclosingtheunitcrackareais和1Ia/濟(jì)1JasvdxOntheotherhand,thecrackgrowthcaseissimilartothecrackclosurecase.Theenergyforcrackgrowthisequaltotheenergyforcrackclosure.Therefore,inthecaseofconstantdisplacement,thedissipativeenergyfortheunitareaincrementofcrackarea,i.e.G[,canalsobewrittenasG=一竺=1』七vdx

1daao'Theexpressionofsandvcanbeinsertedintotheaboveeq.togiveG=-jasvdx=-』，仁MzX21k；三dx1a0'a0x.：2kxEZ2兀=MK三j'Udx=里K2E1an0\xE1i.e.1—v2G=—e—K2,forplanestraincaseFortheplanestresscase,thereisG=—K2,forplanestresscase.iEModeIIcrackForthemodeIIcrack,therelativeslidedisplacementofcracksurfacesalongthex-axisbytheshearstresstis2u.TheenergyreleaserateG〃isG=—j氣udxTheshearstresst勺anddisplacementuarerespectivelyknownasK八t一,,ii,0=0,r=x4(1-v2),:a—x八u=K,0=0,r=a一xEiit2兀Wehave

GiiEl—v22n.a-xl—v2K2——Jaqdx=—e—K2,torplanestraincaseG=—K2,forplaneGiiEiiEiiModeIIIcrackForthemodeIIIcrack,therelativedisplacementbytheanti-planeshearstresstzyalongthez-axisis2w.TheenergyreleaserateGmcanalsobewrittenasG=1j與wdxmaozyTheanti-planeshearstresstandanti-planedisplacementwareknownast="ii,0=0,r=xzy<2kxw=w=4(1+v)kq,iii\2兀0=0,r=a一xWehaveGIII14GIII1441+T2K三dx=ao\.'2兀xE叭2兀1+vEK2IIIForthecomplexcrack,K產(chǎn)0,K^0,K日產(chǎn)0,thetotalenergyreleaserateis—1—v21+vG]=~e~(Ki+K2)+虧峰§3-3Bi-cantileverbeamproblemItisamodeIcrackproblem.1.Longcrackcase:l>>hWhenl>>h,thecracksegmentisequivalenttoacantileverbeam.c偵13,1…「.、25=2-3EI，I=12th3(rectangularsection),8Fl3d=——Eth31,1,4F213U=-Fd,Up=-Fd,n=U+Up=—-Fd=-—-—-八1dn12F212G=—=tdlEt2h3Fortheplanestressproblem,thereis1G=—K2EiThestressintensityfactorcanbeknownasK9舊FlK廣入3布2.Shortcrackcase:Inthiscase,thecracksegmentisequivalenttoadumpybeam(短粗梁).Thesheardeformationofthecracksegmentandthedeformationintheuncrackedsegmentmustbeconsidered.Thedisplacementdcanbedividedintothreeparts.d=d1+d2+d3

dlisproducedbybendandd2byshear.Itisknownthatd汶13,d汶12d3isthebodydisplacementinducedbythedeformationoftheuncrackedsegment.d=a'13+a'12+a'1123Anormalizedcompliancecoefficient入isintroduced.dtE1tE1d3isthebodydisplacementinducedbythedeformationoftheuncrackedsegment.d=a'13+a'12+a'1123Anormalizedcompliancecoefficient入isintroduced.dtE1tE1tE1人=一(tE)=一ah3(-)3+一ah2(-)2+一a'h(-)F1hF2hF3h1=a(—)3+a2+a3Thecoefficientsa,aandacanbedeterminedbytheoryorexperiment.IfthenormalizedcomplianceF111d=[a()3+a()2+a()]tE1h2h3hTotalpotentialofthesystemis入isknown,thedisplacementdcanbedetermined.n=-U=-2Fd=-2FL[ai(h)3+a2(1)2+aiTheenergyreleaserateistdl2Et211h2、h3Fortheplanestressproblem,thestressintensityfactorcanbedeterminedfrom1G=—K2Eithat1F1_111K=~2^—[叫(h)3+2a2(h”+氣(h)]2andl2Therearethreespecimens.Thecrackandl2Therearethreespecimens.Thecracklengthsarel,Fversusdcurvescanbeobtainedbytest.Fanddarethegeneralizedforceanddisplacement.Inlightof人=F(tE)=a(1)3+a(1)2+awecanhavethreeequationsEtF=a(h)3+a(11)2+a1EtF=a(?)3+a(§)2+a(，)2EtF=a(?)3+a(,)2+a3Thecoefficients;,andnowcanbeknown.Then,wecalculatetheenergyreleasereGteandthestressintensity.factor§3-4DeterminationofSIFbyexperimentThemethodinthelastsectioncanbeappliedtothegeneralcase.ConsideraModeIcrackproblem.Thecrackcanbeacentralorfedgeddrack.denotethegeneralizedforceanddisplacementrespectively.Forthedifferentcracksa,a,…，a,wecanobtainagroUpvfrsudstraightlinesbyexperiment.Foraconstantappliedffprwecanhavethedisplacementvdl,ueds,…，d.Thetotalpotentialcanbecalculatedfordifferentcracksizes.：Fd.pfora」i1,2,,nFinally,agroupofdc^ta.)(isobtained.The^ersuaicanbeplotted.Fig."

Fig."FortheModeIandplanestressproblem,inviewof—1g1K2C^—^K2tdaE1thestressintensityfactorK【is,dnwhere-——da1dn)tdaisthe,dnwhere-——daAnothertreatment:AssumethatthecomplianceX(a)ofthesystemdependsonthecracksizea.ThegeneralizeddisplacementqandforceFarerelatedby入asd-人(a)F11n--—Fd--—F2從a)TOC\o"1-5"\h\z22WhenF=constant,thederivativeisdn1d人(a)_F2da2daK-、E^-F^l1\tda\2tdaForaconstantappliedforceF,wecanhavethedisplacementvaluesd,d,…，dfromFig.3-8.Byusing人-d/F,agroupofdata(人,a)canbeobtained.The人iiiiversusacurvecanbedepicted.Byusingtheslopeatthepointa,wecancalculatethestressintensityfactorforthecracksizeaandtheappliedloadF.§3-5AninfiniteplatewithacentralcrackunderuniaxialtensionweresolveitbytheenergyThisproblemhasbeensolvedinchapterweresolveitbytheenergyapproach.Case(a)canbedividedintocase(b)pluscase(c).K(a)=K(b)+K(c)Case(b)isthesameasnocrackcase,K(b)=0,suchthatiK(a)=K(c)IICase(a)isequaltocase(c).1.Solutionforcase(c)Fig.3-12Fig.3-12Forthecase(c),assumethatthedisplacementdistributionontheuppercracksurfaceisanellipse.

匚)2+(三)2=1vaTherelationx=a-rissubstitutedtogive(V)2=1-(三)2=1-(a-r)2=1-a2-2ar+r2vaa2a2Whenthedistancerisverysmall,rTheexpressionofenergyreleaserateGInwhatfollows,theexpressionofenergyreleaserateGisTheexpressionofenergyreleaserateGInwhatfollows,theexpressionofenergyreleaserateGisderivedfromitsdefinition.Then,thedisplacementvcanbedeterminedandthesolutionfortheproblemcanbe,'2rv=v。、aIthasbeenknowninchaptertwothatthedisplacementasymptoticsolutionisv=W=sin0(k+1-2cos20)2.\2k22Ontheuppercracksurface,0=k,wehaveForplane3-vEstress,k=,r=,thestressintensityfactorcanbeobtained1+v2(1+v)thatand■，kK=Evi20*a廠1rkForplane■，kK=Evi20*aEi4a0Thedisplacementvremainsdetermined.

known.ThesystemtotalpotentialisH=-U=4[--fav^tdx]20H=-2btvLfa\：a2-x2dx=-—<jtav0a020Notethatvisrelatedtoa.—n=-—<5tav(a)dn—,da2-—bt(v+ada2_dn_dn_1dn_—G=-IS=一林=-E=4"0+aTheearlyresultis—Ev24a0AdifferentialequationcanbeobtainedthatdvvE—a+=v2,BernoulliequationThegeneralformoftheBernoulliequationis礦+p(x)y=q(x)ynThesolutioncanbefoundinthemathematichandbook.Theresultis2bav=0E+2ba2CCisanintegralconstant.Whena—s,thedisplacementvT3.ThisrequiresthatC=0.2bav=0EFinally,thesolutionisx2b。K[=gm,v=vi1-(—)2,%Itisidenticaltotheearlyresultinchaptertwo.§3-6ModeIcrackinthegeneralcasesolvedbyenergyapproach1.SymmetriccaseForthiscase,itisdifficultforthedirectanalyticalmethod.Thecase(b)isthesameasnocrackcase.K(a)=K(c)IICase(c)canalsobedividedintocase(1)andcase(2).Case(1)istheuniformappliedloadingcase.K(a)=K(c)=K11+K22Khasbeenknown.Theproblemistosolvethecase(2).TheenergyreleaserateGiswrittenasG=—K2=-L(K+K)2=-L(K2+2KK+K2)=G+G+—KKEiE1112E1111121212e1112G1=E%1Inaddition,theenergyreleaserateGcanbederivedfromthedefinition.n=-U=一4(上fapvtdx)=-2fa(p+p)(v+v)tdxTOC\o"1-5"\h\z2001212=-2fapvtdx-2fapvtdx-(2fapvtdx+2fapvtdx)=n+n-4fapvtdx22n12n2112n21VJV_equal,Bettitheorem〃dn1dn〃〃1da[、G=-==G+G(-4Japvtdx)dS2tda122tda021=G+G+2d(fapvdx)12da021Compa