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    Lecture 09Lecture 09

    Dislocations & Strengthening MechanismsDislocations & Strengthening Mechanisms

    Chapter 7 - 1

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    ISSUES TO ADDRESS...ISSUES TO ADDRESS... Why are dislocations observed primarily in metals Why are dislocations observed primarily in metals

    and alloys?and alloys?

    How are stren th and dislocation motion related? How are stren th and dislocation motion related?

    How do we increase strength? How do we increase strength?

    How can heating change strength and other properties? How can heating change strength and other properties?

    Chapter 7 -

    2

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    Mechanical Behavior & StructureMechanical Behavior & Structure

    Controlled by

    Chemical bond

    Controlled by Processing

    via defects, crystal structure

    Chapter 7 -

    3

    Chemistry an m crostr cture

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    Dislocations & Materials ClassesDislocations & Materials Classes

    Metals: Disl. motion easier.

    -non-directional bonding ++++++++

    -close-packed directionsfor slip. electron cloud ion cores++++++++

    Covalent Ceramics

    (Si, diamond): Motion hard.-

    Ionic Ceramics (NaCl):

    Motion hard.-need to avoid ++ and - -

    nei hbors.

    +++

    + + + +

    - - -----

    - - -

    Chapter 7 - 4

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    Dislocation MotionDislocation Motion

    Dislocations & plastic deformation

    Cubic & hexagonal metals - plastic deformation by

    plastic shear or slip where one plane of atoms slides

    over adjacent plane by defect motion (dislocations).

    Chapter 7 - 5

    If dislocations don't move, deformation doesn't occur!

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    Dislocation Motion (Contd)Dislocation Motion (Contd)

    Caterpillar

    b

    Chapter 7 - 6

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    Dislocation MotionDislocation Motion

    Dislocation moves along slip plane in slip direction

    perpendicular to dislocation line

    Slip direction same direction as Burgers vector

    Edge dislocation

    Screw dislocation

    Chapter 7 - 7

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    Comparison of Dislocation MotionComparison of Dislocation Motion

    EdgeEdge vsvs Screw DislocationsScrew Dislocations

    Chapter 7 - 8

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    Deformation MechanismsDeformation Mechanisms

    Slip System

    Slip plane - plane allowing easiest slippage

    Wide interplanar spacings - highest planar densities

    Slip direction - direction of movement - Highest lineardensities

    FCC Slip occurs on {111} planes (close-packed) in

    directions (close-packed) => total of 12 slip systems in FCC

    Chapter 7 - 9

    in BCC & HCP other slip systems occur

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    Deformation MechanismsDeformation Mechanisms

    Common Slip SystemsCommon Slip Systems

    Chapter 7 - 10

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    Slip Systems in Common MetalsSlip Systems in Common Metals

    After Dieter,

    (1990)

    Chapter 7 - 11

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    Stress and Dislocation MotionStress and Dislocation Motion

    Crystals slip due to a resolved shear stress, tR.

    Applied tension can produce such a stress.

    Resolved shear

    stress: tR =Fs /As

    Relation between

    s and tR

    Applied tensile

    stress: = F/As

    F

    normal, ns

    AS

    tR R S S

    Fcosl

    A/cosf

    A

    tR

    lFS

    nS

    AS

    A

    SchmidsSchmids LawLaw

    Chapter 7 - 12

    coscosR

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    Critical Resolved Shear StressCritical Resolved Shear Stress

    Condition for dislocation motion: CRSSR

    it easy or hard to move dislocation 10-4 GPa to 10-2 GPa

    typically

    coscoss s s

    tR = 0 tR = s /2 tR = 0

    = =f=45 =

    Chapter 7 - 13maximum at

    =

    = 45

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    Single Crystal SlipSingle Crystal Slip

    Single

    crystal

    w re

    Chapter 7 - 14

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    Ex: Deformation of single crystalEx: Deformation of single crystal

    =60

    a) Will the single crystal yield?

    b) If not, what stress is needed?

    cos cos=35

    crss = ps

    6500 psi

    (6500 psi) (cos35)(cos60)

    (6500 psi) (0.41)

    2662 psi crss 3000 psi

    = 6500 psi

    Chapter 7 - 15

    So the applied stress of 6500 psi will not cause the crystal

    to yield.

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    Ex 1:Ex 1: Deformation of single crystalDeformation of single crystal

    What stress is necessary (i.e., what is the yield stress, y)?

    )41.0(coscospsi3000crss yy

    psi7325psi3000

    crss

    .

    So for deformation to occur the applied stress must be

    greater than or equal to the yield stress

    Chapter 7 - 16

    y

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    Ex 2: Deformation of Single crystalEx 2: Deformation of Single crystal

    Chapter 7 - 17After Dieter, Mechanical Metallurgy (1990)

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    Slip Motion inSlip Motion in PolycrystalsPolycrystals

    Stronger - grain boundaries

    pin deformations

    Slip planes & directionschange from one

    crys a o ano er.

    R will vary from one .

    The crystal with the

    .

    Other (less favorably

    Chapter 7 - 18

    yield later.300 mm

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    Anisotropy iny

    Can be induced by rolling a polycrystalline metal

    - before rollin - after rollin

    235 mm

    - isotropic - anisotropic

    since grains areapprox. spherical

    & randomly

    since rolling affects grainorientation and shape.

    Chapter 7 - 19

    oriented.

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    Anisotropy in DeformationAnisotropy in Deformation

    1. Cylinder of

    Tantalum

    machined

    2. Fire cylinder

    at a target.

    3. Deformed

    cylinder

    side view

    from a

    rolled plate:

    ire

    ction

    rolling

    The noncircular end view shows

    endview

    platethickness

    direction

    Chapter 7 - 20

    anisotropic deformation of rolled material.

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    Deformation byDeformation by MechlMechl TwinningTwinning

    Twins also re-orient slip planes and contribute to dislocation slip indirectly

    Twinning is an alternate (plastic)

    deformation mechanism observed when the

    strain rate is ver hi h or at low T dislocation sli is su ressed

    Chapter 7 -

    It is a strain-relief mechanism. Very common in BCC and HCP metals

    when slip is restricted!

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    MechlMechl Twinning vs.Twinning vs. Dislocation SlipDislocation Slip

    Homogeneous

    (Twin Band)

    Chapter 7 -

    After

    Dieter.

    Mechl Metallurgy

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    Strategies for Strengthening:Strategies for Strengthening:

    .. e uce ra n zee uce ra n ze

    Grain boundaries arebarriers to slip.

    " "

    increases with

    increasing angle ofm sor entat on.

    Smaller grain size (d):

    more barriers to slip.

    21/

    Chapter 7 -

    yoy e

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    Strategies for Strengthening:Strategies for Strengthening:

    . e uce ra n ze on. e uce ra n ze on

    Very difficult

    to synthesize polycrystalline

    materials with d

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    Strategies for Strengthening:Strategies for Strengthening:

    Primary mechanism

    ..

    for strengthening by grain

    boundaries

    Chapter 7 - 25

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    StrategiesStrategies for Strengthening:for Strengthening:

    Im urit atoms distort the lattice & enerate stress.

    . o. o o u onso u ons

    Stress can produce a barrier to dislocation motion. Smaller substitutional impurity Larger substitutional impurity

    A C

    B D

    Impurity generates local stress at A

    and B that o oses dislocation motion

    Impurity generates local stress at C

    and D that o oses dislocation motion

    Chapter 7 - 26

    to the right. to the right.

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    Stress Concentration atStress Concentration at DislocationsDislocationsWhat makes a dislocation move & interact?What makes a dislocation move & interact?

    e stress e roun t e s ocat one stress e roun t e s ocat oninteracts with the applied stress (viainteracts with the applied stress (via RR))

    and the Strain field of structuraland the Strain field of structural

    im er ections such as rain boundariesim er ections such as rain boundaries

    Chapter 7 -

    and solute atomsand solute atoms

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    Strengthening by AlloyingStrengthening by Alloying

    Small impurities tend to concentrate at dislocations

    Chapter 7 - 28

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    Strengthening byStrengthening by alloying (Contd)alloying (Contd)

    Large impurities concentrate at dislocations on low

    density side

    Chapter 7 -

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    Ex: SolidEx: Solid Solution StrengtheningSolution Strengthening in Copperin Copper

    Tensile strength & yield strength increase with wt% Ni.

    Pa) 400

    a)

    180

    rength(M

    300

    en

    gth(M120

    Tensiles

    2000 10 20 30 40 50

    Yieldstr

    600 10 20 30 40 50

    Empirical relation:21 /y C~

    . , . ,

    Chapter 7 -

    Alloying increases sy and TS.

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    DislocationDislocation Sources: FrankSources: Frank--Read SourcesRead Sources

    Chapter 7 - 31

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    StrategiesStrategies for Strengthening:for Strengthening:

    Hard precipitates are difficult to shear.

    . rec p a on. rec p a on reng en ngreng en ng

    Ex: Ceramics in metals (SiC in Iron or Aluminum).

    Large shear stress neededto move dislocation toward

    precipitate

    prec p a e an s ear .Side View

    Unsli ed art of sli lane advances butprecipi tates act as pinning sites with

    S.

    op ew

    Sspacing

    Result: ~1

    Slipped part of slip plane

    The smaller the S, the larger the

    dislocation line needs to bow out,

    Chapter 7 -

    the more surface energy needs to

    be spend!

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    ApplicationApplication:: Precipitation StrengtheningPrecipitation Strengthening

    Internal wing structure on Boeing 767

    Aluminum is strengthened with precipitates formedy a oy ng.

    Black s ots

    are the ppts.

    Chapter 7 - 33

    1.5mm

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    StrategiesStrategies for Strengthening:for Strengthening:

    . o. o oror

    Room temperature deformation. Common forming operations change the cross sectional area:

    -Forging force -Rolling

    Ao Ad

    die

    blank

    roll

    Ao

    Ad

    force-Drawing

    tensileAddie

    -Extrusion

    container die holder

    Ao

    forceo

    dieram billet

    container dieAdextrusion

    Chapter 7 -

    100x%

    oA

    CW

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    Dislocations During Cold WorkDislocations During Cold Work

    Ti alloy after cold working:

    Dislocations entangle with oneanother during cold work.

    Dislocation motion becomes

    more difficult.

    Strain hardening due to

    -

    interactions

    cutting through a forest ofcutting through a forest of

    Chapter 7 -

    s ocat onss ocat ons. mm

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    Dislocations During Cold WorkDislocations During Cold Work

    Edge and screw

    dislocation interaction

    leads to strain hardening

    After

    Dieter.

    Mechl Metallur

    Chapter 7 - 36

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    Result of Cold WorkResult of Cold Work

    Dislocation density =total dislocation length

    unit volume

    Carefully grown single crystal ca. 103 mm-2

    Deforming sample increases density 109-1010 mm-2

    eat treatment re uces ens ty - mm-

    Yield stress increases as rdincreases:

    large hardening

    y1

    small hardeningy0

    Chapter 7 -

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    Effects of Stress at DislocationsEffects of Stress at Dislocations

    Chapter 7 -

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    Cold Work AnalysisCold Work Analysis What is the tensile strength &

    ductility after cold working?22

    Cold

    Work

    Copper

    %6.35100x%2

    o

    do

    rCW

    Do =15.2mm Dd=12.2mm

    700

    yield strength

    (MPa)

    tensile strength

    (MPa)800

    ductility

    (%EL)60

    300

    500

    Cu300MPa

    400

    600

    20

    40

    Cu340MPa

    % Cold Work

    100200 40 60

    % Cold Work

    200Cu

    0 20 40 60% Cold Work20 40 600

    07%

    Chapter 7 -

    y = a = a %EL= 7%

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    -- Behavior vs. TemperatureBehavior vs. Temperature

    Results for

    polycrystalline iron:-200C

    800

    600

    MPa)

    -

    25C

    400

    200Stres

    s(

    and TS decrease with increasing test temperature.

    0

    Strain0 0.1 0.2 0.3 0.4 0.5

    %ELincreases with increasing test temperature.

    Why? Vacancies

    hel dislocations 2. vacancies

    3 . disl. glides past obstacle

    move past obstacles. replaceatoms on thedisl. half

    plane 1. disl. trapped

    obstacle

    Chapter 7 -

    y o s ac e

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    Effect of Heating After %Effect of Heating After %CWCW

    1 hour treatment at Tannealdecreases TS& increases %EL.

    Annealing temperature (C)200100 300 400 500 600 700

    th(MPa

    (%

    EL)tensile strength

    500

    50

    stagesstages to discussto discuss......

    ilestren

    ductility

    400

    40

    30

    te

    n ductility

    30020

    Chapter 7 -

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    RecoveryRecovery

    Annihilation reduces dislocation density.

    Scenario 1 extra half-plane

    Results from

    diffusion

    annihilate

    and form

    a perfect

    atomsdiffuse

    atomic

    plane.extra half-plane

    of atoms

    of tension

    cenar o

    tR3. Climbed disl. can nowmove on new slip plane

    4. opposite dislocationsmeet and annihilate

    .

    vacancy diffusionallowing disl. to climb1. dislocation blocked; Obstacle dislocation

    Chapter 7 -

    cant move to the right

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    RecrystallizationRecrystallization

    New grains are formed that:

    - Have a small dislocation density

    - re sma

    - Consume cold-worked grains.

    . .

    33% cold New crystals

    Chapter 7 -

    worked

    brass

    nucleate after

    3 sec. at 580C.

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    FurtherFurther RecrystallizationRecrystallization

    AllAll coldcold--worked grains are consumed.worked grains are consumed.

    0.6 mm0.6 mm

    After 4 After 8

    Chapter 7 -

    secon s secon s

    G i G hG i G h

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    Grain GrowthGrain Growth

    At longer times, larger grains consume smaller ones. Why? Grain boundary area (and therefore energy) is reduced.

    0.6 mm 0.6 mm

    After 8 s,

    580C

    After 15 min,

    580C

    Empirical Relation:

    Elapsed time

    Coefficient dependent

    on material and T.

    Grain diam.

    Eexponent typ. ~ 2

    Chapter 7 -

    o at time t.Ostwald Ripening

    G i G thG i G th

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    Grain GrowthGrain Growth

    TR = Recrystallization

    TR

    Cold work state, is

    A metastable state.

    The ener stored Due to cold work

    is the driving for

    Recrystallized

    grain are strain

    Chapter 7 -

    free.

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    RecrystallizationRecrystallization Temperature,Temperature, TTRR

    TR = recrystallization temperature = point of

    . m R . - . m

    2. Due to diffusion annealing time TR = f(t)

    shorter annealing time => higher T3. Higher %CW=> lower TR strain hardening

    4. Pure metals lower TR due to dislocation movements

    Easier to move in pure metals => lower TR

    Chapter 7 -

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    Recrystallization Temperature,Recrystallization Temperature, TTRR

    Chapter 7 - 49

    Stages ofStages of RecrystallizationRecrystallization

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    Stages ofStages of RecrystallizationRecrystallization

    33% CW Brass 3 s 580 oC Initial Recryst.

    Chapter 7 -

    Stages ofStages of RecrystallizationRecrystallization

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    Stages ofStages of RecrystallizationRecrystallization

    4 s 580 oC Recryst. Contd 8 s 580 oC Recryst. Complete

    Chapter 7 - 51

    Stages ofStages of RecrystallizationRecrystallization

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    Stages ofStages of RecrystallizationRecrystallization

    15 min 580 oC Grain Growth 10 min 700 oC Grain Growth

    Chapter 7 - 52

    C ld W kC ld W k C l l tiC l l ti

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    Cold WorkCold Work CalculationsCalculations

    A cylindrical rod of brass originally 0.40 in (10.2 mm)

    in diameter is to be cold worked by drawing. The

    circular cross section will be maintained during

    deformation. A cold-worked tensile strength in excess

    ,

    %EL are desired. Further more, the final diameter

    mus e . n . mm . xp a n ow s may e

    accomplished.

    Chapter 7 -

    C ld W kC ld W k C l l ti S l tiC l l ti S l ti

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    Cold WorkCold Work Calculations SolutionCalculations Solution

    If we directly draw to the final diameterIf we directly draw to the final diameter

    Brass

    Cold

    Do = 0.40 in

    Work

    D = 0.30 in

    1001100x% xAAA

    CW ffo

    %843100x300

    1100x4

    1

    2

    2

    2

    ..Df

    oo

    Chapter 7 -

    .o

    ColdworkColdwork Calc Solution: ContCalc Solution: Cont

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    ColdworkColdwork Calc Solution: Cont.Calc Solution: Cont.

    540420

    6

    For %CW= 43.8%

    y = 420 MPa TS = 540 MPa > 380 MPa

    Chapter 7 -

    This doesnt satisfy criteria what can we do?

    ColdworkColdwork Calc Solution: Cont.Calc Solution: Cont.

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    ColdworkColdwork Calc Solution: Cont.Calc Solution: Cont.

    380 15

    12 27

    ForFor %%ELEL < 15< 15

    oror aa

    < 27 %< 27 %CWCW

    Chapter 7 - 56

    OurOur working range is limited to %working range is limited to %CWCW= 12= 12--2727

    ColdworkColdwork CalcCalc SolnSoln:: RecrystallizationRecrystallization

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    ColdworkColdwork CalcCalc SolnSoln:: RecrystallizationRecrystallization

    Cold draw-anneal-cold draw again

    For objective we need a cold work of %CW 12-27

    Well use %CW= 20

    Diameter after first cold draw (before 2nd cold draw)?

    mus e ca cu a e as o ows:

    %11001%

    2

    2

    2

    2

    2

    2 CWDxD

    CW ff

    0202

    50

    2 %1

    .

    f CWD

    50

    202 .

    fDD

    02

    1001

    2050.

    Chapter 7 - 57

    100021 ..f

    n erme a e ame er =

    Cold Work Calculations:Cold Work Calculations: SolutionSolution

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    Cold Work Calculations:Cold Work Calculations: SolutionSolution

    Summary:

    = =

    %CW1 1 0.3350.4

    2

    x100 30

    2. Anneal above D02 =Df1

    =

    =

    . . .

    2010030

    1%

    2

    2

    x.

    CWMPa340y

    Therefore, meets all requirements

    .

    24% EL

    Chapter 7 -

    Rate ofRate of RecrystallizationRecrystallization

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    Rate ofRate of RecrystallizationRecrystallization

    kT

    ERtR logloglog 0 start

    50%

    T

    BCtlog RT1 finish

    t/R 1:note

    Hot work above TR

    Cold work below TR

    log t

    ma er gra ns

    stronger at low temperature

    Chapter 7 - 59

    SummarySummary

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    SummarySummary

    Dislocations are observed primarily in metals

    an a oys.

    Strength is increased by making dislocationmotion difficult.

    Particular ways to increase strength are to:

    --decrease grain size

    -- --precipitate strengthening

    --cold work

    eat ng annea ng can re uce s ocat on ens ty

    and increase grain size. This decreases the strength.

    Chapter 7 - 60

    HomeworkHomework

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    HomeworkHomework

    Problems: 7.9, 7.12, 7.13, 7.15, 7.17, 7.19

    7.27, 7.29, 7.31, 7.32, 7.36, 7.41,

    7.D3, 7.D4.

    Due date:

    .

    Chapter 7 - 61

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    Advanced Topics in DislocationAdvanced Topics in Dislocation

    PhenomenaPhenomena

    Chapter 7 - 62

    Slip in a Perfect LatticeSlip in a Perfect Lattice

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    Slip in a Perfect LatticeSlip in a Perfect Lattice

    First hint as to why therehould be cr stalline de ects

    involved!!!

    Chapter 7 - 63

    Slip in a Perfect LatticeSlip in a Perfect Lattice

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    Slip in a Perfect LatticeSlip in a Perfect Lattice

    Chapter 7 - 64

    Slip in a Perfect LatticeSlip in a Perfect Lattice

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    Slip in a Perfect LatticeSlip in a Perfect Lattice

    Chapter 7 - 65

    Atomic movements near dislocation inAtomic movements near dislocation in

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    slip (plastic deformation)slip (plastic deformation)

    Chapter 7 - 66

    Why ceramics do not exhibitWhy ceramics do not exhibit plastic deformationplastic deformation

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    yy pp

    at ambient Temperatures?at ambient Temperatures?

    Chapter 7 - 67

    Why ceramics do not exhibitWhy ceramics do not exhibit plastic deformationplastic deformation

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    Chapter 7 - 68

    Why ceramics do not exhibitWhy ceramics do not exhibit plastic deformationplastic deformation

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    yy pp

    at ambient Temperatures?at ambient Temperatures?

    Chapter 7 - 69

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    Chapter 7 - 70

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    Stages of DislocationStages of Dislocation slip in FCC Metalsslip in FCC Metals

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    Chapter 7 - 72

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    Chapter 7 - 73

    the slide

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    Forces on DislocationsForces on Dislocations

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    Chapter 7 - 75

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    DislocationDislocation SourcesSources

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    Chapter 7 -77