Dislocations-part-B

•structuredeterminesb,e.g.1/2110inFCC,1/2111inBCC.•italsodeterminestheglideplaneandeaseofdislocationmovement.How?1.StackingFaultsConsidercrystaltobeconstructedfromstackingatomicplanesontoeachother(ratherthanfromunitcells).EffectsofCrystalStructureonDislocationPropertiesGlideplanes:-havethelowestPeierlsstressdislocationcoretendsto'spread'-inmetals,theytendtobethemostwidelyspaced,i.e.mostdenselypacked-innon-metalsionicityandcovalentbondingalsohaveaninfluenceConsiderFCCmetalsonly-theirabilitytocontainstackingfaultsaffectsdislocationproperties.Perfectcrystal=planeshavespecificatomicarrangementandarestackedinaspecificsequence.Stackingfault=errorinthissequence{111}planesinFCCmetalshavea3-foldstackingsequence.....ABCABCABC.....andareseparatedby1/3111:Eachplanecontainsthree110andthree112directionse.g.(111)plane:Stablefaultscanoccurinthissequence:(a)Twinfault.....ABCABACBA.....[planeBisa'mirror']Thecrystalhasanadditionalenergyproportionaltotheareaofthefault.AAABBBCCC{111}planes111TheenergyperunitareaoffaultisthetwinboundaryenergyT≈10-100mJm-2.(c)Extrinsicfault[planeCinserted+normaldisplacement1/3111].....ABCACBCABC.....–similarfaultenergyto(b)AAABBBCCCperfectstackingBAABCCCAAfaultedcrystal(d)Intrinsicfault[sheardisplacement1/6112atfault]–1/6112takesAB,BC,CA:.....ABCACABCA.....–samefaultasin(b)above(b)Intrinsicfault[planeBremoved+normaldisplacement1/3111].....ABCACABCA.....-theenergyperunitareaisthestackingfaultenergy≈20-150mJm-2(≈2T)2.PartialDislocationsIfastackingfaultdoesnotrunallthewaythroughacrystal,itisboundedbyapartialdislocation,i.e.itsbisapartiallatticevector.(i)Shockleypartial•boundsaslip-typeintrinsicfault(d)abovewithb=1/6112•canbeedge,screwormixed•threepossibleb=1/6112foreach{111}planeT(ii)Frankpartial•hasb=1/3111andboundsfaults(b)or(c)abovei.e.vacancyorinterstitialloops:bb•boththelinedirectionandblieintheplaneofthefaultitisglissileintheplaneofthefault•bdoesnotlieinanyother{111}planescrewcannotcross-slip•dislocationenergyperunitlengthb2–inthiscase|b|2=b2=a2(1+1+4)/36=a2/6•bisperpendiculartothefaultplaneitispureedgeandsessile•b2=a2(1+1+1)/9=a2/3•b2:a2/6+a2/6a2/18largereductionindislocationenergyTfaultsstair-rod(iii)Stair-rodpartial•junctionlinewheretwointrinsicstackingfaultsondifferent{111}planesmeet1•bisthesumofthetwoindividualbse.g.on(111):1/6[11],1/6[11],1/6[11]andtheirreversesandon(11):1/6[12],1/6[12],1/6[211]andtheirreverses36possiblecombinations–themostfavourablegivesthesmallest|b|i.e.1/6[2]+1/6[12]=1/6[011]12121112•partialissessile,i.e.b=1/6110doesnotlieineitherofthe{111}planese.g.on(111):1/2[10]=1/6[11]+1/6[2]b2:a2/2a2/6+a2/6reductionindislocationenergy1211AperfectdislocationintheFCCmetalshasb=1/2110Inthestackingsequence.....ABCABC.....thistakesAA,BB,CCe.g.BBbyb1inthefigure:However,the'easy'routefromBtoBisviaCe.g.BCBby(b2+b3)VectorsBC,CA,AB,etc.are1/6112i.e.bofShockleypartialdislocationsThus,theperfectdislocationdissociatesintotwoShockleypartialdislocationsseparatedbyaribbonofintrinsicstackingfault:faultb2b3d3.DissociationofPerfectDislocationsTheanglebetweenb2andb3is60°andthetwopartialsrepeleachother.Theymoveapartuntiltherepulsiveforce,Frep,duetotheirelasticstressfieldsisbalancedbythestackingfaultenergy(=forceperunitlength).faultb2b3dFromsectionondislocationinteractions,fortwodislocationsofspacingd:Frep≈Gb2/2πdifbsareparallel=0ifbsareperpendiculartakeFrep≈Gb2cos(60°)/2πd=Gb2/4πd,withb2=a2/6forShockleypartialsequilibriumspacingd≈Ga2/24π,i.e.d/a≈Ga/24πe.g.Cu:G=40GPa,a=0.35nm,=40mJm-2d≈4.8a=1.6nmAg:(=20mJm-2)d≈7aAl:(=140mJm-2)d≈1adissociationisduetothestabilityofstackingfaultsonthe{111}planesoftheFCCmetalstheycanalsoformonthebasalplanesoftheHCPmetalsstablefaultsdonotoccurintheBCCmetalstheycanoccurinsomenon-metalsdissociationofdislocationsduetotheexistenceofstackingfaultsplaysakeyroleintheplasticbehaviourofcrystallinematerialstherefore,thedislocationcorespreadsalonga{111}slipplanelowPeierlsstresslowCRSSinFCCmetals(particularlyAu,Ag,Cu)ComputersimulationofperfectedgedislocationinCu4.ImplicationsforDeformationoftheFCCMetals•Dislocationsmoveeasilyinpuremetalslowyieldstress.•WhenShockleypartialdislocationsonintersecting{111}slipplanesmeet,theycanformalow-energystair-rodpartial(see2(iii)above).Its1/6110Burgersvectordoesnotlieineitherofthetwo{111}planesandsoitissessile.•ThetwodissociateddislocationsformaLomer-Cottrelllockwhichpreventsdislocationsfollowingbehindfrommovingandsotheypile-upbehindthelockonepossiblemechanismofwork-hardening.•Thepiled-updislocationscangetpasttheobstacleiftheycancross-slipontoanother{111}plane.HowcantheyiftheyaredissociatedandaShockleypartialcannotcross-slip?•Byconstriction–thetwopartialsofadissociateddislocationcometogetherandreformaperfectdislocationthatcancross-slipifinthescreworientation:constriction•Cross-sliprequireshighenoughRSSonthecross-slipplaneandisassistedby-highT-high-highs•Itcanthendissociateonthecross-slipplanee.g.on(111):1/6[2]+1/6[11]=1/2[10]on(11):1/2[10]