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CHAPTER 10:

Dislocations and StrengtheningMechanisms


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CHAPTER 7

1. Dislocations and Its types

2. How plastic deformation occurs by dislocation

3. Define slip system and example

.

5. How grain structure impede dislocation motion

6. Describe and explain solid solution strengthening

7. Recrystallization and Recovery


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SCREW DISLOCATION

Formed by meansof shear Distortion

Upper region isshifted one atomic

Linear, its followsspiral or helical

path arounddislocation line


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DISLOCATION MOVEMENT


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DISLOCATION MOVEMENT


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DISLOCATION INTERACTION Strain field from

one dislocation can

affect a neighboringdislocation

Two like

repel each other Unlike dislocations

attract and

annihilate eachother


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SLIP SYSTEM

Depends on crystal structure and Motion ofdislocation, where atomic distortion is minimum.


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PlaneaofArea

PlaneaoncenteredAtomsofNumberPD

...

......

=

Determines the dislocation plane ( closely

VectorDirectionofLength

vectordirectiononcenteredatomsofNumberLD

...

......

=

packed)

Determines the dislocation direction


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z

a

a

Slip Planes FCC

x

y


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zSlip Direction (111)

[ 1 0 1 ][ 0 1 1 ]

x

y

[ 1 0 1 ]

[ 1 1 0]

[ 1 1 0 ]

[ 0 1 1 ]


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z

a

a

Slip Planes FCC

x

y


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z

a

a

Slip Planes FCC

x

y


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z

a

a

Slip Planes FCC

x

z


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z

a

a

Slip Planes FCC

x

z


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z

a

a

Slip Planes FCC

x

z


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z

a

a

C

B

Slip Planes BCC

x

y

A

D


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z

a

Slip direction BCC

x

y

A

D


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z

a

a

C

B

Slip Planes BCC

x

y

A

D


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z

a

a

C

B

Slip Planes BCC

x

y

A

D


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z

a

a

C

B

Slip Planes BCC

x

y

A

D


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z

a

a

C

B

Slip Planes BCC

x

y

A

D


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z

a

a

C

B

Slip Planes BCC

x

y

A

D


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SLIP SYSTEM: EXAMPLES

FCC (Al, Cu, Ni, Ag, Au) Close packed planes: {111}, e.g., ADF

Close packed directions: , e.g., AD, DF, AF

Slip system: {111} (12 independent slip systems)

BCC (Fe, W, Mo): {110} (12 independent slipsystems)


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SLIP IN SINGLE CRYSTALS

Even though the applied Stress may be puretensile or compressive, shear components exist

at all parallel or perpendicular to stress directionResolved Shear Stresses

Magnitudes depends notonly on the applied stressand also slip plane,

direction


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Schmids Law


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Condition for dislocation motion: R > CRSS

Crystal orientation can make it easy or hard to move disl.

R= cos cos

CRITICAL RESOLVED SHEAR

STRESS (CRSS)

R = 0

=90

R = /2=45=45

R = 0

=90

Maximum possible R = /2; thus y = 2CRSS


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Zinc CrystalSLIP LINES

1.Slips occurs along no of

Equivalent and most favorableoriented planes and directionsat various positions

2. Slip deformation forms assmall steps on the surfaceof crystal and parallel.

3. Large no of Dislocationalong the same slip plane.

Slip plane

Slip plane


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Examples for solving Problem

Consider a single crystal of BCC iron oriented suchthat a tensile stress is applied along a [010] direction.Calculate shear stress along a (110) plane and in [1 1

1] direction when stress applied

x

y

a

a

A


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Slip planes & directions (,,,,) change from one crystal toanother.

R will vary from one crystalto another.

The crystal with the largest

DISL. MOTION IN POLYCRYSTALS

R y e s rs .

Other (less favorablyoriented) crystals yield later.

Polycrystalline materials

generally stronger than singlecrystals, due to geometricconstraints & the requirementof larger stresses for yielding 300 m


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Grains beforedeformation

Elongated grains afterdeformation


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MECHANISM OF STRENGTHENING

The ability of a metal to deform plastically depends

on the ability of dislocations to move.

Hardness and strength are related to the ease with

which plastic deformation can be made to occur.

of the dislocation.

RESTRICTING OR HINDERING A DISLOCATION

MOTION RENDERS A MATERIAL HARDER AND

STRONGER


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Grain boundaries are

barriers to slip.

Barrier "strength"

slip planeB

STRATEGIES FOR STRENGTHENING:

1: REDUCE GRAIN SIZE

7

ncreases w

misorientation.

Atomic disorder within

grain boundary results in

discontinuity of slip planes

grainboun

dar

grain Agr

a


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REDUCE GRAIN SIZE

Smaller grain size: more barriers to slip.

Not valid for very large grain as well as extremely

fine grained material.

Grain size reduction improves strength as well as

.

Small angle grain boundaries not effective in

interfering with slip process.

Twin boundaries effectivelyblock slip and improve strength.


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70wt%Cu-30wt%Zn brass alloy

yield = o + kyd1/2

Data:

GRAIN SIZE STRENGTHENING:

AN EXAMPLE

Hall-petch Eqn

8

Adapted from Fig. 7.13,

Callister 6e.

(Fig. 7.13 is adapted

from H. Suzuki, "The

Relation Between the

Structure and

Mechanical Properties

of Metals", Vol. II,

National Physical

Laboratory Symposium

No. 15, 1963, p. 524.)

[grain size (mm)]-0.5

yield(MPa)

50

100

150

200

04 8 12 16

10-1 10-2 5x10-3

,

1

ky

0

0.75mm

STRENGTHENING STRATEGY


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Smaller substitutional

STRENGTHENING STRATEGY

2: SOLID SOLUTIONS

Interstitial

11

,

atomic structure

Substitutional Replaces host atom byadding impurity


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LOCAL STRAIN FIELDS

When metal palstically deformed someamount of energy stores as strain energy

Compressive and tensile around dislocation Stress & strain fields decrease with radial

distance from dislocation line


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Impurity atoms distort the lattice & generate stress

on .

Stress can produce a barrier to dislocation motion.

Smaller substitutional

impurity

Larger substitutional

impurity

2: SOLID SOLUTIONS

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Impurity generates local shear at

A and B that opposes disl motion

to the right.

Impurity generates local shear at

C and D that opposes disl motion

to the right.

C

D

A

B


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Smaller impurity atom above dislocation line

Larger impurity atom below dislocation line


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Tensile strength & yield strength increase w/wt% Ni.

Adapted from Fig.7.14 (a) and (b),

Callister 6e.

th(MPa)180

lestrength(MPa)

300

400

EX: SOLID SOLUTION

STRENGTHENING IN COPPER

12

Empirical relation:

Alloying increases y.y ~ C

1/2

Yieldstr

en

wt. %Ni, (Concentration C)

600 10 20 30 40 50

Tensi

wt. %Ni, (Concentration C)

2000 10 20 30 40 50


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STRENGTHENING STRATEGY 3


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STRENGTHENING STRATEGY 3:

COLD WORK (%CW) Distance of separation between dislocation

decreases.

During movement of dislocation lines interaction

with other defects occur.

16

%CW =

Ao AdAo

x100

ovemen n ere , s ress requ re or p as c

deformation increased.

Referred as work hardening, strain hardening

or cold working measured as A/Ao


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ss

Yield strength ( ) increases. Tensile strength (TS) increases.

Ductility (%EL or %AR) decreases.

yIMPACT OF COLD WORK

Stre

%coldwork Strain

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