工程材料科学与设计答案.docx

工程材料科学与设计答案.docx

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工程材料科学与设计答案

Problems-Chapter5

1.FIND:

Calculatethestressonatensionedfiber.

GIVEN:

Thefiberdiameteris25micrometers.Theelongationalloadis25g.

ASSUMPTIONS:

Theengineeringstressisrequested.

DATA:

Accelerationduetogravityis9.8m/sec2.ANewtonisakg-m/sec2.APascalisaN/m2.AMPais106Pa.

SOLUTION:

Stressisforceperunitarea.Thecross-sectionalareaisR2=1963.5squaremicrometers.Theforceis25g（kg/1000g）（9.8m/sec2）=0.245N.Thus,thestressis

=F/A=

COMMENTS:

Youmustlearntodothesesortsofproblems,includingtheconversions.

2.GIVEN:

FCCCuwithao=0.362nm

REQUIRED:

A）LowestenergyBurgersvector,B）LengthintermsofradiusofCuatom,C）Familyofplanes

SOLUTION:

WenotethattheBurgersvectoristheshortestvectorthatconnectscrystallographicallyequivalentpositions.Adiagramofthestructureisshownbelow:

FCCstructurewith（111）shown

Wenotethatatomslyingalongfacediagonalstouchandarecrystallographicallyequivalent.Therefore,theshortestvectorconnectingequivalentpositionsis½facediagonal.Forexample,onesuchvectoris

asshownin（111）.

A.Thelengthofthisvectoris

B.Byinspection,thesizeofthevectoris2Cuatomradii.

C.Slipoccursinthemostdenselypackedplanewhichisofthetype{111}.ThesearethesmoothestplanesandcontainthesmallestBurgersvector.Thismeansthatthedislocationsmoveeasilyandtheenergyislow.

3.GIVEN:

b=0.288nminAg

REQUIRED:

Findlatticeparameter

SOLUTION:

RecalltheAgisFCC.ForFCCstructurestheBurgersvectoris½afacediagonalasshown.Weseethat

4.A.FCCstructure

The（111）planeisshowninaunitcellwithallatomsshown.Atomstouchalongfacediagonals.The（111）planeisthemostcloselypacked,andthevectorsshownconnectequivalentatomicposition.Thus

etc.Theningeneral

B.ForNaC1

Weseethattheshortestvectorconnectingequivalentpositionsis

asshown.Thisdirectionliesinboththe{100}and{110}planesandbotharepossibleslipplanes.However{110}aretheplanesmostfrequentlyobservedastheslipplanes.Thisisbecauserepulsiveinterionicforcesareminimizedontheseplanesduringdislocationmotion.Thusweexpect1/2<110>Burgersvectorsand{110}slipplanes.

5.GIVEN:

Mocrystal

ao=0.314nm

REQUIRED:

Determinethecrystalstructure.

IfMowereFCC,then

but|b|=0.272MoisnotFCC.

AssumingMoisBCC,then

ThustheBurgersvectorisconsistentwithMobeingBCC.

6.FIND:

Isthefracturesurfaceinionicsolidsroughorsmooth?

SOLUTION:

Cleavagessurfacesofionicmaterialsaregenerallysmooth.Onceacrackisstarted,iteasilypropagatesinastraightlineinaspecificcrystallographicdirectiononaspecificcrystallographicplane.Ceramicfracturesurfacesareroughwhenfailureproceedsthroughthenoncrystallineboundariesbetweensmallcrystals.

7.GIVEN:

BCCCrwith|b|=0.25nm

REQUIRED:

Findlatticeparametera

ASSUME:

forBCCstructure

SOLUTION:

fromtheformulaforthemagnitudeofavector:

8.GIVEN:

Normalstressof123MPaappliedtoBCCFein[110]direction

REQUIRED:

Resolvedshearin[101]on（010）

SOLUTION:

Recallthattheresolvedshearstressisgivenby:

=coscos

（1）

where=anglebetweenslipdirectionandtensileaxis;=anglebetweennormaltoslipplaneandtensileaxis

Thus

9.GIVEN:

Stressin[123]directionofBCCcrystal

REQUIRED:

FindthestressneededtopromoteslipifcR=800psi.Theslipplaneis（1

0）andslipdirectionis[111].

SOLUTION:

Recall=coscos

（1）

=[123][111]

[123][111]=[123][111]cos

=[123][1

0]

[123][1

0]=[123][1

0]cos

10.Burgersvectorslieintheclosestpackeddirectionssincethedistancebetweenequivalentcrystallographicpositionsisshortestintheclose-packeddirections.ThismeansthattheenergyassociatedwiththedislocationwillbeminimumforsuchdislocationssincetheenergyisproportionaltothesquareoftheBurgersvector.

11.Closepackedplanesareslipplanessincethesearethesmoothestplanes（onanatomiclevel）andwouldthenbeexpectedtohavethelowestcriticalresolvedshearstress.

12.GIVEN:

Dislocationlieson（1

1）paralleltointersectionof（1

1）and（111）withBurgersvectorparallelto[

0].StructureisFCC.

REQUIRED:

A）Burgersvectorofdislocationand,B）Characterofdislocation.

SOLUTION:

A）SincethestructureisFCC,theBurgersvectorisparallelto<110>andhasmagnitude

ForaBurgersvectorparallelto[

0]thescalarmultipliermustbea/2.Thus

=a/2[

0].B）Wemustdeterminethelinedirectionofthedislocation.FromthediagramweseethattheBVandlinedirectionareat60

owhichmeansthedislocationismixed.

13.GIVEN:

Dislocationreactionbelow:

REQUIRED:

Showitisvectoriallycorrectandenergeticallyproper.

SOLUTION:

Thesumofthex,y&zcomponentsontheLHSmustbeequaltothecorrespondingcomponentontherighthandside.

xcomponent（LHS）=xcomponent（RHS）

ycomponent（LHS）=ycomponent（RHS）

zcomponent（LHS）=zcomponent（RHS）

Energy:

Thereactionisenergeticallyfavorableif|b1|2+|b2|2>|b3|3

Thusthereactionisfavorablesince

14.GIVEN:

DislocationinFCC

Parallelto[

01]i.e.

=[

01]

REQUIRED:

Characterandslipplane

SOLUTION:

Characterisfoundbyanglebetween

.Note

=-1+0+1=0.Thus

.Since

thedislocationispureedge.

Tofindtheslipplanewenotethatthecrossproduceof

givesavectorthatisnormaltotheplaneinwhich

lie.Thisvectorsoformedhasthesameindicesastheplanesincewehaveafundamentallycubicstructure.

Weseefromthediagramthatthesevectorslieon（010）.

Thus,wehavetheplane（0

0）whichisthesameasthe（010）plane.Thisdoesnotmovebyglidesinceplanesofthekind{100}arenotslipplanesfortheFCCstructure.

15.FCCmetalsaremoreductilethanBCCorHCPbecause:

1）thereisnoeasymechanismfornucleationofmicrocracksinFCCasthereisforBCCandHCP;2）thestressesforplasticdeformationarelowerinFCCduetothe（generally）smootherplanes.ThismeansthatthemicrocracksthatforminBCC&HCPwillhavehighstressestendingtomakethempropagate.

16.Forasimplecubicsystem,thelowestenergyBurgersvectorsareofthetype<001>sincethisistheshortestdistanceconnectingequivalentatomicpositions.ThismeansthattheenergyislowestsincethestrainenergyisproportionaltothesquareoftheBurgersvector.

17.GIVEN:

At.wt.0=16

At.wt.Mg=24.32SamestructureasNaCl

=3.65g/cm3

REQUIRED:

FindlengthofBurgersVectorinMgO

SOLUTION:

ThestructureofMgOisshownschematicallybelowalongwiththeshortestBurgersvector.Tosolvetheproblemwefirstnotethatwerequirethelatticeparameterao.Wecantakeasub-sectionoftheunitcell（cross-hatchedcube）whoseedgeis

unitslong.

Wecancalculatethetotalmassofthiscubeandthevolumeandcalculatethedensity.Sincethemassisknownandthedensityisknown,thevolumemaybecalculatedfromwhichaomaybeextracted.

Wenotethatthereare40=ionsand4Mg++ionslocatedatthecorners.However,anionatacornerissharedby8suchcubes.Thus,wehave½O=and½Mg++ionsinourcube.

Thus

18.GIVEN:

Criticalresolvedshearstress（0.34MPa）,slipsystem（111）[

10],andtensileaxis[101]

REQUIRED:

Appliedstressatwhichcrystalbeginstodeformandcrystalstructure.

SOLUTION:

（A）Thesituationisshownbelow

crss=coscos

=anglebetweentensileaxisandslipdirection

=anglebetweentensileaxisandnormaltoslipplane

=[111][101]=[101][

01]

[111][101]=[111][101]cos[101][110]=[101][110]cos

（B）:

Tohavea{111}<110>slipsystem,thematerialmusthaveanFCCstructure.

19.GIVEN:

crss=55.2MPa,（111）[

01]slipsystem,[112]tensileaxis

REQUIRED:

Findthehighestnormalstressthatcanbeappliedbeforedislocationmotioninthe[10

]direction.

SOLUTION:

Thesituationisshownbelow.Essentiallytheproblemreducestofindingthevalueofthetensilestresswhenthecriticalresolvedshearisreached.

crss=coscos

=[112][

01]=[112]{111}

B.WouldhaveexactlythesamestressforaBCCmetal（&wouldbeinterchanged）.

20.GIVEN:

atyield=3.5MPa;（111）[1

0]slipsystem[1

1]tensileaxis

REQUIRED:

Computecrss

SOLUTION:

crss=coscos

=[1

1][1

0]=[1

1][111]

21.ItemEdgeScrew

Lineardefect?

YesYes

ElasticDistortion?

YesYes

Glide?

YesYes

Climb?

YesNo

Cross-slip?

NoYes

BurgersVector（BV）toline//toline

Uniqueslipplane?

YesNo

Offset//toBV//toBV

Motion//toBVtoBV

22.GIVEN:

BCCmetalwithcrss=7MPa[001]tensileaxis.

REQUIRED:

（a）Slipsystemthatwillbeactivatedand（b）normalstressforplasticdeformation.

SOLUTION:

RecallthatforBCCmetalstheusualslipsystemis<111>{110}.Deformationoccursontheplaneanddirectionforwhichcoscosisamaximumsincethiswillhavethemaximumresolvedshearstress.Thesituationisshownbelow.

（Notethattheslipdirectionsareshownshortenedinthisview）

Possibleslipsystemsarelistedbelow:

sketch（also[1

1]on（011））（also[

11]on（0

1））（also[

1]on（101））

similartoplanesshowninsketch.Also[111]on（

01）

Weseebyinspectionthattheresolvedshearduetoatensileforcein[001]willallbethesame.Theresolvedshearonallother{110}<111>systemsiszero.

B.Tocomputethenormalstressattheonsetofplasticdeformationwewillconsider（011）[

1]

crss=coscos=7

=[001][

1];cos=[001][011]

Noteifweconsidered（101）[

1]wewouldhave

andwewould