材料科学基础课后习题答案5

Problems-Chapter51.FIND:Calculatethestressonatensionedfiber.GIVEN:Thefiberdiameteris25micrometers.Theelongationalloadis25g.ASSUMPTIONS:Theengineeringstressisrequested.DATA:Accelerationduetogravityis9.8m/sec2.ANewtonisakg-m/sec2.APascalisaN/m2.AMPais106Pa.SOLUTION:Stressisforceperunitarea.Thecross-sectionalareaisR2=1963.5squaremicrometers.Theforceis25g(kg/1000g)(9.8m/sec2)=0.245N.Thus,thestressis=F/A=MPamumumN125.0101960245.0262COMMENTS:Youmustlearntodothesesortsofproblems,includingtheconversions.2.GIVEN:FCCCuwithao=0.362nmREQUIRED:A)LowestenergyBurgersvector,B)LengthintermsofradiusofCuatom,C)FamilyofplanesSOLUTION:WenotethattheBurgersvectoristheshortestvectorthatconnectscrystallographicallyequivalentpositions.Adiagramofthestructureisshownbelow:FCCstructurewith(111)shownWenotethatatomslyingalongfacediagonalstouchandarecrystallographicallyequivalent.Therefore,theshortestvectorconnectingequivalentpositionsis½facediagonal.Forexample,onesuchvectoris10]1[2aoasshownin(111).A.Thelengthofthisvectoris0.256nm=20.362=2a=4a+4ao2o2oB.Byinspection,thesizeofthevectoris2Cuatomradii.C.Slipoccursinthemostdenselypackedplanewhichisofthetype{111}.ThesearethesmoothestplanesandcontainthesmallestBurgersvector.Thismeansthatthedislocationsmoveeasilyandtheenergyislow.3.GIVEN:b=0.288nminAgREQUIRED:FindlatticeparameterSOLUTION:RecalltheAgisFCC.ForFCCstructurestheBurgersvectoris½afacediagonalasshown.Weseethat4.A.FCCstructureThe(111)planeisshowninaunitcellwithallatomsshown.Atomstouchalongfacediagonals.The(111)planeisthemostcloselypacked,andthevectorsshownconnectequivalentatomicposition.Thus10]1[21=betc.Theningeneral1102a=bB.ForNaC1Weseethattheshortestvectorconnectingequivalentpositionsis10]1[2aasshown.Thisdirectionliesinboththe{100}and{110}planesandbotharepossibleslipplanes.However{110}aretheplanesmostfrequentlyobservedastheslipplanes.Thisisbecauserepulsiveinterionicforcesareminimizedontheseplanesduringdislocationmotion.Thusweexpect1/2110Burgersvectorsand{110}slipplanes.5.GIVEN:Mocrystal0.272nm=bao=0.314nmREQUIRED:Determinethecrystalstructure.IfMowereFCC,then0.222nm=20.314=b__but|b|=0.272MoisnotFCC.0.407nm=a0.288x2=b2=a2a=2a2=boooodiagonal)cube(?2a3=bBCC,Fordiagonal)face(?2a=bFCC,Foroo____AssumingMoisBCC,then0.272nm.=0.314X23=b__ThustheBurgersvectorisconsistentwithMobeingBCC.6.FIND:Isthefracturesurfaceinionicsolidsroughorsmooth?SOLUTION:Cleavagessurfacesofionicmaterialsaregenerallysmooth.Onceacrackisstarted,iteasilypropagatesinastraightlineinaspecificcrystallographicdirectiononaspecificcrystallographicplane.Ceramicfracturesurfacesareroughwhenfailureproceedsthroughthenoncrystallineboundariesbetweensmallcrystals.7.GIVEN:BCCCrwith|b|=0.25nmREQUIRED:FindlatticeparameteraASSUME:1112a=bforBCCstructureSOLUTION:2a3=4a+4a+4a=b222__fromtheformulaforthemagnitudeofavector:8.GIVEN:Normalstressof123MPaappliedtoBCCFein[110]directionREQUIRED:Resolvedshearin[101]on(010)SOLUTION:Recallthattheresolvedshearstressisgivenby:=coscos(1)where=anglebetweenslipdirectionandtensileaxis;=anglebetweennormaltoslipplaneandtensileaxis0.289nm=30.5=0.25x32=b32=a__ThusMPa43.5=2121123=9.GIVEN:Stressin[123]directionofBCCcrystalREQUIRED:FindthestressneededtopromoteslipifcR=800psi.Theslipplaneis(11_0)andslipdirectionis[111].SOLUTION:Recall=coscos(1)=[123][111][123][111]=[123][111]cos=[123][11_0][123][11_0]=[123][11_0]cos45=,1oversqrt2=1=21=[110][010]=[110][010]30=,21=_0+0+1=22=[101][110]=[101][110][110][010]=[101],[110]=oocoscoscos____coscoscos____76=7322233=3722x3=3146=6=3+2+1=314=coscos10.Burgersvectorslieintheclosestpackeddirectionssincethedistancebetweenequivalentcrystallographicpositionsisshortestintheclose-packeddirections.ThismeansthattheenergyassociatedwiththedislocationwillbeminimumforsuchdislocationssincetheenergyisproportionaltothesquareoftheBurgersvector.11.Closepackedplanesareslipplanessincethesearethesmoothestplanes(onanatomiclevel)andwouldthenbeexpectedtohavethelowestcriticalresolvedshearstress.12.GIVEN:Dislocationlieson(11_1)paralleltointersectionof(11_1)and(111)withBurgersvectorparallelto[1_1_0].StructureisFCC.REQUIRED:A)Burgersvectorofdislocationand,B)Characterofdislocation.SOLUTION:A)SincethestructureisFCC,theBurgersvectorisparallelto110andhasmagnitude.2aForaBurgersvectorparallelto[1_1_0]thescalarmultipliermustbea/2.Thusb_=a/2[1_1_0].B)Wemustdeterminethelinedirectionofthedislocation.FromthediagramweseethattheBVandlinedirectionareat60owhichmeansthedislocationismixed.13.GIVEN:Dislocationreactionbelow:REQUIRED:Showitisvectoriallycorrectandenergeticallyproper.SOLUTION:[100]a=]111[2a+[111]2aThesumofthex,y&zcomponentsontheLHSmustbeequaltothecorrespondingcomponentontherighthandside.xcomponent(LHS)=xcomponent(RHS)on)(compressipsi4,572-=3147800-=214x67x800-==2141=1-=0+2-1=214=cRcoscoscoscosycomponent(LHS)=ycomponent(RHS)zcomponent(LHS)=zcomponent(RHS)Energy:Thereactionisenergeticallyfavorableif|b1|2+|b2|2|b3|3Thusthereactionisfavorablesinceaa43+a4322214.GIVEN:DislocationinFCCyesa