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时间:2020-08-31
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1、Problems-Chapter51.FIND:Calculatethestressonatensionedfiber.GIVEN:Thefiberdiameteris25micrometers.Theelongationalloadis25g.ASSUMPTIONS:Theengineeringstressisrequested.DATA:Accelerationduetogravityis9.8m/sec2.ANewtonisakg-m/sec2.APascalisaN/m2.AMPais106Pa.SOLUTION:Stressisf
2、orceperunitarea.Thecross-sectionalareaisR2=1963.5squaremicrometers.Theforceis25g(kg/1000g)(9.8m/sec2)=0.245N.Thus,thestressis=F/A=COMMENTS:Youmustlearntodothesesortsofproblems,includingtheconversions.2.GIVEN:FCCCuwithao=0.362nmREQUIRED:A)LowestenergyBurgersvector,B)Lengthi
3、ntermsofradiusofCuatom,C)FamilyofplanesSOLUTION:WenotethattheBurgersvectoristheshortestvectorthatconnectscrystallographicallyequivalentpositions.Adiagramofthestructureisshownbelow:FCCstructurewith(111)shownWenotethatatomslyingalongfacediagonalstouchandarecrystallographical
4、lyequivalent.Therefore,theshortestvectorconnectingequivalentpositionsis½facediagonal.Forexample,onesuchvectorisasshownin(111).A.ThelengthofthisvectorisB.Byinspection,thesizeofthevectoris2Cuatomradii.C.Slipoccursinthemostdenselypackedplanewhichisofthetype{111}.Thesearethesm
5、oothestplanesandcontainthesmallestBurgersvector.Thismeansthatthedislocationsmoveeasilyandtheenergyislow.3.GIVEN:½b½=0.288nminAgREQUIRED:FindlatticeparameterSOLUTION:RecalltheAgisFCC.ForFCCstructurestheBurgersvectoris½afacediagonalasshown.Weseethat4.A.FCCstructureThe(111)pl
6、aneisshowninaunitcellwithallatomsshown.Atomstouchalongfacediagonals.The(111)planeisthemostcloselypacked,andthevectorsshownconnectequivalentatomicposition.Thusetc.TheningeneralB.ForNaC1Weseethattheshortestvectorconnectingequivalentpositionsisasshown.Thisdirectionliesinbotht
7、he{100}and{110}planesandbotharepossibleslipplanes.However{110}aretheplanesmostfrequentlyobservedastheslipplanes.Thisisbecauserepulsiveinterionicforcesareminimizedontheseplanesduringdislocationmotion.Thusweexpect1/2<110>Burgersvectorsand{110}slipplanes.5.GIVEN:Mocrystalao=0
8、.314nmREQUIRED:Determinethecrystalstructure.IfMowereFCC,thenbut
9、b
10、=0.272ÞMoisnotFCC.Assum
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