Pertemuan 5 Cacat Kristal6

7/21/2019 Pertemuan 5 Cacat Kristal6

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HAPTER 4:

HAPTER 4:

IMPERFECTIONS IN SOLIDSIMPERFECTIONS IN SOLIDS



Objectives

bjectives

Introduce the three basic types ofIntroduce the three basic types of

imperfections point defects! "ine defects #orimperfections point defects! "ine defects #or

dis"ocations$! and surfacedis"ocations$! and surface%area%area defects&defects&

E'p"ore the nature and effects of different typesE'p"ore the nature and effects of different typesof defects&of defects&



2

• Vacancy atoms• Interstitial atoms• Substitutional atoms

• Dislocations

• Grain Boundaries

Point defects

Line defects

Area/surface defects

TYPES OF

YPES OF

IMPERFE TIONSMPERFE TIONS



hapter Outlie

hapter Outlie

Point DefectsPoint Defects Other Point DefectsOther Point Defects

Dis"ocationsDis"ocations Obser(in) Dis"ocationsObser(in) Dis"ocations Si)nificance of Dis"ocationsSi)nificance of Dis"ocations Schmid*s La+Schmid*s La+

Inf"uence of Crysta" StructureInf"uence of Crysta" Structure Surface DefectsSurface Defects Importance of DefectsImportance of Defects



Point defectsPoint defects , Imperfections! such as (acancies! that are, Imperfections! such as (acancies! that are"ocated typica""y at one #in some cases a fe+$ sites in the"ocated typica""y at one #in some cases a fe+$ sites in thecrysta"&crysta"&

E'tended defectsE'tended defects , Defects that in(o"(e se(era" atoms%ions, Defects that in(o"(e se(era" atoms%ionsand thus occur o(er a finite (o"ume of the crysta""ine materia"and thus occur o(er a finite (o"ume of the crysta""ine materia"

#e&)&! dis"ocations! stac-in) fau"ts! etc&$&#e&)&! dis"ocations! stac-in) fau"ts! etc&$& .acancy.acancy , /n atom or an ion missin) from its re)u"ar, /n atom or an ion missin) from its re)u"ar

crysta""o)raphic site&crysta""o)raphic site&

Interstitia" defectInterstitia" defect , / point defect produced +hen an atom is, / point defect produced +hen an atom isp"aced into the crysta" at a site that is norma""y not a "atticep"aced into the crysta" at a site that is norma""y not a "attice

point&point& Substitutiona" defectSubstitutiona" defect , / point defect produced +hen an atom, / point defect produced +hen an atom

is remo(ed from a re)u"ar "attice point and rep"aced +ith ais remo(ed from a re)u"ar "attice point and rep"aced +ith adifferent atom! usua""y of a different si0e&different atom! usua""y of a different si0e&

Point Defects



3

• Vacancies:

!acant atomic sites in a structure"

Vacancydistortionof #lanes

• SelfInterstitials:$e%tra$ atoms #ositioned bet&een atomic sites"

selfinterstitialdistortion

of #lanes



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Figure 4.1 Point defects: (a) vacancy, (b) interstitial atom, (c) smallsubstitutional atom, (d) large substitutional atom, (e) Frenkel defect,(f) Schottky defect. ll of these defects disru!t the !erfectarrangement of the surrounding atoms.



!cetrati! !"

cetrati! !"

vacaciesacacies

/t room temperature /t room temperature!! the concentration of (acancies isthe concentration of (acancies is

sma""! but thesma""! but the concentration of (acancies increasesconcentration of (acancies increases

e'ponentia""y as +e increase the temperature! ase'ponentia""y as +e increase the temperature! as

sho+n by the fo""o+in) /rrhenius type beha(iorsho+n by the fo""o+in) /rrhenius type beha(ior



Calculate the concentration of vacancies in copper at roomtemperature (25oC). What temperature will be needed to heattreat copper such that the concentration of vacancies producedwill be 1000 times more than the euilibrium concentration of

vacancies at room temperature! "ssume that 20#000 cal arereuired to produce a mole of vacancies in copper.

$%ample &.1 '*+,-

+he lattice parameter of CC copper is 0./151 nm. +he basis

is 1# therefore# the number of copper atoms# or lattice points#per cm/ is

$%ample &.1+he $ffect of +emperature on

acanc3 Concentrations

322

38atoms/cmcopper10!"8

)cm10#1$1"3(

atoms/cell×=

×=

−n



$%ample &.1 '*+,- (Continued)

"t room temperature# + 4 25 26/ 4 278 9

38

3

22

/cm%acancies1081$"1

2&8' ' mol

cal1"&8!

mol

cal20000

ep "

cm

atoms10!"8

ep

×=

×−

−

×=

= RT nn Qν

ν

We could do this b3 heatin: the copper to a temperature atwhich this number of vacancies forms

C102 ))&8!"1/(00020ep()10!"8(

ep1081$"1

o22

11

=×−×=

=×=

T T

RT

Qnn

ν ν



Determine the number of vacancies needed for a ;CC ironcr3stal to have a densit3 of 6.86 :<cm/. +he lattice parameterof the iron is 2.8 10=8 cm.

$%ample &.2 '*+,-

+he e%pected theoretical densit3 of iron can be calculated fromthe lattice parameter and the atomic mass.

$%ample &.2acanc3 Concentrations in ,ron




et>s calculate the number of iron atoms and vacancies

that would be present in each unit cell for the reuireddensit3 of 6.86 :<cm/

r# there should be 2.00 ? 1.7761 4 0.0027 vacancies per

unit cell. +he number of vacancies per cm/ is



,nterstitialc3 = " point defect caused when a @@normal>>atom occupies an interstitial site in the cr3stal.

renAel defect = " pair of point defects produced whenan ion moves to create an interstitial site# leavin: behinda vacanc3.

'chottA3 defect = " point defect in ionicall3 bondedmaterials. ,n order to maintain a neutral char:e# astoichiometric number of cation and anion vacanciesmust form.

9rB:er=inA notation = " s3stem used to indicate pointdefects in materials. +he main bod3 of the notationindicates the t3pe of defect or the element involved.

'ection &.2ther Point Defects



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Figure 4." #hen a divalent cation re!laces a monovalentcation, a second monovalent cation must also be removed,creating a vacancy.



'

(&o outcomes if im#urity )B* added to +ost )A*:

• Solid solution of B in A )i"e", random dist" of #ointdefects*

• Solid solution of B in A #lus #articles of a ne& #+ase )usually for a lar-er amount of B*

.

Substitutional alloy)e"-", 0u in 1i*

Interstitial alloy

)e"-", 0 in e*

Second #+ase #articledierent com#ositionoften dierent structure"

POINT #EFE TS IN A$$OYS

OINT #EFE TS IN A$$OYS

http://alloy.ppt/

http://alloy.ppt/



45

De6nition: Amount of im#urity )B* and +ost )A*

in t+e system"

• 7ei-+t 8

(&o descri#tions:

• Atom 8

0B 9

mass of B

total mass% 4550

B 9

; atomsof B

total ; atoms

% 455

• 0on!ersion bet&een &t 8 and at8 in an AB alloy:

0B 90BAB

0AAA < 0BAB

% 455 0B 90B/AB

0A/AA <0B/AB • Basis for con!ersion:

mass of B 9moles of B % AB

atomic &ei-+t of B

mass of A 9moles of A % AA

atomic &ei-+t of A

OMPOSITION

OMPOSITION



Dislocation = " line imperfection in a cr3stalline material. 'crew dislocation = " dislocation produced b3 sAewin: a

cr3stal so that one atomic plane produces a spiral rampabout the dislocation.

$d:e dislocation = " dislocation introduced into thecr3stal b3 addin: an @@e%tra half plane>> of atoms. i%ed dislocation = " dislocation that contains partl3

ed:e components and partl3 screw components. 'lip = Deformation of a metallic material b3 the

movement of dislocations throu:h the cr3stal.

'ection &./ Dislocations



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Figure 4.4 the !erfect crystal (a) is cut and sheared one atoms!acing, (b) and (c). $he line along %hich shearing occurs is ascre% dislocation. &urgers vector b is re'uired to close a loo!of e'ual atom s!acings around the scre% dislocation.



Figure 4. $he !erfect crystal in (a) is cut and an etra!lane of atoms is inserted (b). $he bottom edge of theetra !lane is an edge dislocation (c). &urgersvector b is re'uired to close a loo! of e'ual atoms!acings around the edge dislocation. ( Adapted from

J.D. Verhoeven, Fundamentals of Physical *etallurgy,Wiley, 1975 .)



Figure 4.+ mied dislocation. $hescre% dislocation at the front face

of the crystal gradually changes toan edge dislocation at the side ofthe crystal. ( Adapted from W.T.Read, islocations in -rystals.M!ra"#$ill, 195%.)



Figure 4. Schematic of sli! line, sli! !lane, and sli!(&urgers) vector for (a) an edge dislocation and (b) for ascre% dislocation. ( Adapted from J.D. Verhoeven,Fundamentals of Physical *etallurgy , Wiley, 1975 .)



Figure 4./ (a) #hen a shear stress is a!!lied to the dislocation in (a),the atoms are dis!laced, causing the dislocation to move one &urgersvector in the sli! direction (b). -ontinued movement of the dislocationeventually creates a ste! (c), and the crystal is deformed. ( Adaptedfrom A.!. !&y , 0ssentials of *aterials Science, M!ra"#$ill, 197'.)(b) *otion of cater!illar is analogous to the motion of a dislocation.



Calculate the len:th of the ;ur:ers vector in copper.

$%ample &.8;ur:ers ector Calculation

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Figure 4.1

(a) &urgersvector for F--co!!er. (b)$he atomlocations on a(11) !lane ina &-- unit cell

(for eam!le4./ and 4.2,res!ectively)



+he len:th of the ;ur:ers vector# or the repeat distance# is

b 4 1<2(0.51125 nm) 4 0.255/ nm

$%ample &.8 '*+,-

Copper has an CC cr3stal structure. +he latticeparameter of copper (Cu) is 0./151 nm. +he close=pacAed directions# or the directions of the ;ur:ers vector#are of the form . +he repeat distance alon: thedirections is one=half the face dia:onal# since lattice pointsare located at corners and centers of faces i:ure&.10(a)E.

⟩⟨110 ⟩⟨110



+he planar densit3 of the (112) plane in ;CC iron is 7.7& 101& atoms<cm2. Calculate (1) the planar densit3 of the (110) planeand (2) the interplanar spacin:s for both the (112) and (110)planes. n which plane would slip normall3 occur!

$%ample &.7,dentification of Preferred 'lip Planes

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Figure 4.1(a) &urgersvector for F--co!!er. (b)$he atomlocations on a(11) !lane in

a &-- unit cell(for eam!le4./ and 4.2,res!ectively)



$%ample &.7 '*+,-

1. +he planar densit3 is

2. +he interplanar spacin:s are

+he planar densit3 and interplanar spacin: of the (110)plane are lar:er than those for the (112) planeF therefore#the (110) plane would be the preferred slip plane.



$tch pits = +in3 holes created at areas where dislocationsmeet the surface. +hese are used to e%amine thepresence and number densit3 of dislocations.

'lip line = " visible line produced at the surface of a

metallic material b3 the presence of several thousanddislocations. 'lip band = Collection of man3 slip lines# often easil3

visible.

'ection &.&bservin: Dislocations



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Figure 4.11 sketch illustrating dislocations, sli! !lanes,

and etch !it locations. ((o&re) Adapted from Physical*etallurgy Princi!les , Third *dition, by R.*. Reed#$ill andR. Abba+hian, p. 9, -i+. /#7 and /#0. opyriht 23199 4roo+6ole Thom+on earnin. Adapted by

permi++ion.3



Figure 4.13 !tical image of etch !its in siliconcarbide (Si-). $he etch !its corres!ond tointersection !oints of !ure edge dislocations %ith&urgers vector a5" and the dislocation linedirection along 617 (!er!endicular to theetched surface). 8ines of etch !its re!resent lo%angle grain boundaries (o&rte+y of Dr. Mare(o"ron+i, arneie Mellon 8niver+ity .)

⟩⟨ 2011



Figure 4.1" 0lectron !hotomicrogra!hs of dislocations inTi % Al : (a) islocation !ileu!s (3+,). (b) *icrogra!h at

1 sho%ing sli! lines and grain boundaries in 9. (c)Schematic of sli! bands develo!ment.

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Plastic deformation refers to irreversible deformation orchan:e in shape that occurs when the force or stressthat caused it is removed.

$lastic deformation = Deformation that is full3 recovered

when the stress causin: it is removed. Dislocation densit3 = +he total len:th of dislocation line

per cubic centimeter in a material.

'ection &.5'i:nificance of Dislocations



'ection &.'chmid>s aw

'chmid>s law =+he relationship between shear stress# theapplied stress# and the orientation of the slip s3stemGthat is#

Critical resolved shear stress = +he shear stress reuiredto cause a dislocation to move and cause slip.

φ λ σ τ coscos=



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Figure 4.14 (a) resolved shear stress τ is !roduced on a sli!

system. (ote: (; < =) does not have to be 2>.) (b)*ovement of dislocations on the sli! system deforms thematerial. (c) ?esolving the force.



"ppl3 the 'chmid>s law for a situation in which the sin:lecr3stal is at an orientation so that the slip plane isperpendicular to the applied tensile stress.

$%ample &.10Calculation of Hesolved 'hear 'tress

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Figure 14.1 #hen thesli! !lane is !er!endicularto the a!!lied stress σ , theangle λ is 2> and no shearstress is resolved.



$%ample &.10 '*+,-

'uppose the slip plane is perpendicular to theapplied stress σ# as in i:ure &.15. +hen# I 4 0o# λ 4

70o# cos λ 4 0# and therefore τ r 4 0. "s noted before# thean:les f and l can but do not alwa3s add up to 70o. $venif the applied stress s is enormous# no resolved shearstress develops alon: the slip direction and thedislocation cannot move. 'lip cannot occur if the slip

s3stem is oriented so that either λ or I is 70o.



We wish to produce a rod composed of a sin:le cr3stal ofpure aluminum# which has a critical resolved shear stress of1&8 psi. We would liAe to orient the rod in such a mannerthat# when an a%ial stress of 500 psi is applied# the roddeforms b3 slip in a &5o direction to the a%is of the rod andactuates a sensor that detects the overload. Desi:n the rod

and a method b3 which it mi:ht be produced.

$%ample &.11 '*+,-

Dislocations be:in to move when the resolved shear stressτ r euals the critical resolved shear stress# 1&8 psi. rom

'chmid>s law τ r 4 σ cos λ cos I= or

1&8 psi 4 (500 psi* cos λ cos I

$%ample &.11Desi:n of a 'in:le Cr3stal

Castin: Process




;ecause we wish slip to occur at a &5o an:le to the a%is ofthe rod# 4 &5o# and

+herefore# we must produce a rod that is oriented suchthat 4 &5o and I 4 5.2o. -ote that I and do notadd to 70o.

We mi:ht do this b3 a solidification process. We

would orient a seed cr3stal of solid aluminum at thebottom of a mold. iuid aluminum could be introducedinto the mold. +he liuid be:ins to solidif3 from thestartin: cr3stal and a sin:le cr3stal rod of the properorientation is produced.



Critical Hesolved 'hear 'tress -umber of 'lip '3stems Cross=slip = " chan:e in the slip s3stem of a

dislocation.

'ection &.6,nfluence of Cr3stal 'tructure



" sin:le cr3stal of ma:nesium (:)# which has a JCP cr3stalstructure# can be stretched into a ribbon=liAe shape four to si%times its ori:inal len:th. Jowever# polycrystalline : and othermetals with a JCP structure show limited ductilities. *se thevalues of critical resolved shear stress for metals with different

cr3stal structures and the nature of deformation inpol3cr3stalline materials to e%plain this observation.

$%ample &.12Ductilit3 of JCP etal 'in:le Cr3stals

and Pol3cr3stalline aterials



$%ample &.12 '*+,-

rom +able &=2# we note that for JCP metals such as :#the critical resolved shear stress is low (50?100 psi). We

also note that slip in JCP metals will occur readil3 on thebasal planeGthe primar3 slip plane. When a sin:le cr3stalis deformed# assumin: the basal plane is suitabl3oriented with applied stress# a ver3 lar:e deformation canoccur. +his e%plains wh3 sin:le cr3stal : can be

stretched into a ribbon four to si% times the ori:inal siKe.When we have a pol3cr3stalline :# thedeformation is not as simple. $ach cr3stal must deformsuch that the strain developed in an3 one cr3stal isaccommodated b3 its nei:hbors. ,n JCP metals# there are

no intersectin: slip s3stems# thus dislocations cannot:lide over from one slip plane in one cr3stal (:rain) ontoanother slip plane in a nei:hborin: cr3stal. "s a result#pol3cr3stalline JCP metals such as : show limitedductilit3.



'ection &.8'urface Defects

'urface defects = ,mperfections# such as :rainboundaries# that form a two=dimensional plane withinthe cr3stal.

Jall=Petch euation = +he relationship between 3ieldstren:th and :rain siKe in a metallic materialGthat is#

"'+ :rain siKe number (n) = " measure of the siKe of

the :rains in a cr3stalline material obtained b3 countin:the number of :rains per suare inch a ma:nification100.

'mall an:le :rain boundar3 = "n arra3 of dislocationscausin: a small misorientation of the cr3stal across thesurface of the imperfection.

2/1' 0−

= + d y σ σ



Figure 4.1+ (a) $he atoms near theboundaries of the three grains do not havean e'uilibrium s!acing or arrangement. (b)

@rains and grain boundaries in a stainlesssteel sam!le. (o&rte+y Dr. A. Deardo.)




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Figure 4.1 $he effect of grain siAe on the yieldstrength of steel at room tem!erature.



$%ample &.1/ '*+,- (Continued)

or the :rain siKe of 0.006 mm# the 3ield stress is &0.875 Pa 4 265.8 Pa. +herefore# a:ain usin: the Jall=Petch euation

'olvin: these two euations 9 4 18.&/ Pa=mm1<2# and: 0 4 55.5 Pa. -ow we have the Jall=Petch euation as

: y 4 55.5 18.&/ d=1<2

,f we want a 3ield stress of /0#000 psi or /0 .875 420.7 Pa# the :rain siKe will be 0.01&8 mm.



Figure 4.1/ *icrostructure of

!alladium ( 1). (-rom S*Bandbook, Vol. 9, Metalloraphyand Miro+tr&t&re 219053, A(M

;nternational, Material+ <ar, =$//>7%.)

$ l & 1&



'uppose we count 1 :rains per suare inch in aphotomicro:raph taAen at ma:nification 250. What is the"'+ :rain siKe number!

$%ample &.1& '*+,-

,f we count 1 :rains per suare inch at ma:nification 250#then at ma:nification 100 we must have

N 4 (250<100)2 (1) 4 100 :rains<in.2 4 2n=1

o: 100 4 (n ? 1) lo: 2

2 4 (n ? 1)(0./01)

n 4 6.&

$%ample &.1&Calculation of "'+ Lrain 'iKe -umber



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Figure 4.12 $he small anglegrain boundary is !roduced by

an array of dislocations,causing an angular mismatch ? bet%een lattices on either sideof the boundary.



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Figure 4.3 !!lication of a stress to the !erfect crystal (a) maycause a dis!lacement of the atoms, (b) causing the formation of at%in. ote that the crystal has deformed as a result of t%inning.




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Figure 4.3 (c) microgra!h of t%ins %ithin agrain of brass (3).



Figure 4.31 omains in ferroelectric bariumtitanate. (o&rte+y of Dr. Rodney Ro+eman,8niver+ity of ininnati .) Similar domainstructures occur in ferromagnetic andferrimagnetic materials.



$ffect on echanical Properties via Control of the 'lipProcess

'train Jardenin: 'olid='olution 'tren:thenin:

Lrain='iKe 'tren:thenin: $ffects on $lectrical# ptical# and a:netic Properties

'ection &.7,mportance of Defects

Pertemuan 5 Cacat Kristal6

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