Plasticity ppt

8/10/2019 Plasticity ppt

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PLASTIC DEFORMATION

Mechanisms of Plastic Deformation

The Uniaxial Tension Test

Mechanisms of Plastic Deformation

Mechanical MetallurgyGeorge E Dieter

McGraw-Hill Book Company, London (1988)


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Plasticdeformation

Mechanisms / Methods by which a can Material canFAIL

Fracture

FatigueCreep Chemical /

Electro-chemical

degradation

Physical

degradation

Wear

Erosion

Microstructuralchanges

Phase transformations

Twinning

Grain growth

Elastic deformation

Particle coarsening

If failure is considered as change in desired performance*- which could involve changes inpropertiesand/orshape; then failure can occur by many mechanisms as below.

* Beyond a certain limit

Corrosion

Oxidation

Slip

Twinning


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Slip(Dislocation

motion)

Plastic Deformation in Crystalline Materials

Twinning Phase Transformation Creep Mechanisms

Grain boundary sliding

Vacancy diffusion

Dislocation climb

+ Other Mechanisms

Note:Plastic deformation in amorphous materials occur by other mechanisms including flow (~viscous fluid) and shearbanding

Plastic deformation in the broadest sense meanspermanent deformationin the absence

of external constraints (forces, displacements) (i.e. external constraints are removed).

Plastic deformation of crystalline materials takes place by mechanisms which are very

different from that for amorphous materials (glasses). The current chapter will focus on plastic

deformation of crystalline materials. Glasses deform by shear banding etc. below the glass transition temperature(Tg) and by flow above Tg.

Though plasticity by slipis the most important mechanism of plastic deformation, there

are other mechanisms as well. Many of these mechanisms may act inconjunction/parallel to give rise to the observed plastic deformation.

Grain rotation


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Tension/Compression

Bending

Shear

Torsion

Common types of deformation

Tension Compression

Shear TorsionDeformed configuration

Bending

Note: modes of deformation in other contexts will be defined in the topic on plasticity

Tension / Compression

Torsion

Modes

of

Deformation Shear

Bending

Review


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One of the simplest test which can performed to evaluate the mechanical properties of amaterial is the Uniaxial Tension Test.

This is typically performed on a cylindrical specimen with a standard gauge length. (Atconstant temperature and strain rate).

The test involves pulling a material with increasing load (force) and noting the elongation(displacement) of the specimen.

Data acquired from such a test can be plotted as: (i) load-stroke (raw data), (ii) engineeringstress- engineering strain, (iii) true stress- true strain. (next slide).

It is convenient to use Engineering Stress (s)and Engineering Strain (e) as defined belowas we can divide the load and change in length by constant quantities (A0and L0).Subscripts 0 refer to initial values and i to instantaneous values.

But there are problems with the use of s and e (as outlined in the coming slides) andhence we define True Stress ()and True Strain () (wherein we use instantaneous valuesof length and area).

Though this is simple test to conduct and a wealth of information about the mechanicalbehaviour of a material can be obtained (Modulus of elasticity, ductilityetc.) However,it must be cautioned that this data should be used with caution under other states of stress.

The Uniaxial Tension Test (UTT)

0A

Ps

0L

Le

0 initial

i instantaneousSubscript

Note: quantities obtained by performing an

Uniaxial Tension Test are valid only underuniaxial state of stress


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The Tensile Stress-Strain Curve

Stroke L

oad

e

s

Gauge Length L0

Possible axes

Tensile specimen

Initial cross sectional area A0

Important Note

We shall assume cylindrical specimens (unless otherwise stated)


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Problem with engineering Stress (s) and Strain (e)!!

Consider the following sequence of deformations:

L0

2L0

L0

e12= 1

e23=

e13= 0

1

2

3

[e12+ e23] =

It is clear that from stage 1 3 there is no strain

But the decomposition of the process into 1 2 & 2 3 gives a net strain of

Clearly there is a problem with the use (definition) of Engineering strain

Hence, a quantity known as True Strain is preferred (along with True Stress) as

defined in the next slide.


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True Stress () and Strain ()

iA

P

0

ln

0L

L

L

dLL

L

Ai instantaneous area

0A

Ps

0L

Le

The definitions of true stress and true strain are based on instantaneous values of area (Ai)and length (Li).


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Same sequence of deformations considered before:

L0

2L0

L0

12= Ln(2)

23= Ln(2)

13= 0

1

2

3

[12+ 23] = 0

With true strain things turn out the way they should!


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Schematic s-e and -curves

Information gained from the test:

(i) Youngs modulus(ii) Yield stress (or proof stress)(iii) Ultimate Tensile Stress (UTS)(iv) Fracture stress

UTS- Ultimate Tensile StrengthSubscripts:

y- yield, F,f- fracture,u- uniform (for strain)/ultimate (for stress)

Points and regions of the curves are explained in the next slide

These are simplified schematics which are close to thecurves obtained for some metallic materials like Al, Cuetc. (polycrystalline materials at room temperature).

Many materials (e.g. steel) may have curves which arequalitatively very different from these schematics.

Most ceramics are brittle with very little plastic

deformation. Even these diagrams are not to scale as the strain at

yield is ~0.001(eelastic~103)

[E is measured in GPa and yin MPa thus givingthis small strains]the linear portion is practically vertical and stuck tothe Y-axis (when efractureand eelasticis drawn to the samescale).

Schematics: not to scale

Note the increasing stress requiredfor continued plastic deformation

Neck


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O unloaded specimen

OYElasticLinear Region in the plot (macroscopiclinear elastic region)

Ymacroscopic yield point (there are many measures of yielding as discussed later)

Occurs due to collective motion of many dislocations. YFElastic + Plastic regime

If specimen is unloaded from any point in this region, it will unload parallel to OY and theelastic strain would be recovered. Actually, more strain will be recovered than unloading from Y(and hence in some sense in the region YF the sample is more elastic than in the elastic regionOY).

In this region the material strain hardensflow stress increases with strain.This region can further be split into YN and NF as below.

YNStable region with uniform deformation along the gauge length

NInstability in tension Onset of neckingTrue condition of uniaxiality broken onset of triaxial state of stress (loading remains uniaxialbut the state of stress in the cylindrical specimen is not).

NF most of the deformation is localized at the neckSpecimen in a triaxial state of stress

FFracture of specimen(many polycrystalline materials like Al show cup and cone fracture)

Sequence of events during the tension test

Notes:

In the -plot there is no distinct point N and there is no drop in load (as instantaneous area has been taken into account in the definition of

) in the elastic + plastic regime (YF) The stress is monotonically increasing in the region YF true indicator of strain hardening


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Comparison between true strain and engineering strain

True strain () 0.01 0.10 0.20 0.50 1.0 2.0 3.0 4.0

Engineering strain (e) 0.01 0.105 0.22 0.65 1.72 6.39 19.09 53.6

e)( 1ln

e)s( 1

Comparison between Engineering and True quantities:

Note that for strains of about 0.4, true and engineering strains can be assumed to be equal. At large strains thedeviations between the values are large.

In engineering stress since we are dividing by a constant number A0(and there is a localreduction in area around the neck)

Engineering and true values are related by the equations as below.

At low strains (in the uniaxial tension test) either of the values work fine. As we shall see that during the tension test localized plastic deformation occurs after some

strain (called necking). This leads to inhomogeneity in the stress across the length of thesample and under such circumstances true stress should be used.

iA

P

0

ln

0 L

L

L

dLL

L

0

ln 1 1 ln(1+e)L

L

0

0 0 0

1 1 (1 )i i

i

A L LPs s s e

A A L L

00 0 i i

0

From volume constancy A L =A L i

i

A L

A L

Valid tillnecking starts


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Yielding can be defined in many contexts.Truly speaking (microscopically) it is point at which dislocationsleave the crystal (grain) and cause microscopic plastic deformation

(of unit b) this is best determined from microstrain (~106)experiments on single crystals. However, in practical terms it isdetermined from the stress-strain plot (by say an offset as describedbelow).

True elastic limit(microscopically and macroscopically elasticwhere in there is not even microscopic yielding)

~106[OAportion of the curve] Microscopically plastic but macroscopically elastic

[AYportion of the curve]

Proportional limitthe point at which there is a deviation from thestraight line elastic regime

Offset Yield Strength(proof stress) A curve is drawn parallel tothe elastic line at a given strain like 0.2% (= 0.002) to determine theyield strength.

Where does Yielding start?

In some materials (e.g. pure annealed Cu, gray cast iron etc.) the linear portion of the curvemay be limited and yield strength may arbitrarily determined as the stress at some given

strain (say 0.005).

Microscopic

elastic

Macroscopic

elastic


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Important Note

yis yield stress in an uniaxial tension test and should not be used in other states of stress(other criteria of yield should be used for a generalized state of stress).

Tresca and von Mices criterion are the two most popular ones.


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Slip is competing with other processes which can lead to failure.

In simple terms a ductile materialis one which yields before failure (i.e. y< f).

Ductility depends on the state of stress used during deformation.

We can obtain an measure of the ductility of a material from the uniaxial tension testasfollows (by putting together the fractured parts to make the measurement):Strain at fracture(ef) (usually expressed as %), (often called elongation, although it is a dimensionless quantity)Reduction in area at fracture (q) (usually expressed as %)

q is a better measure of ductilityas it does not depend on the gauge length (L0); while, efdepends on L0. Elongation/strain to necking (uniform elongation) can also be used to avoidthe complication arising from necking.Also, q is a more structure sensitive ductility parameter.

Sometimes it is easier to visualize elongation as a measure of ductility rather than areduction in area. For this the calculation has be based on a very short gauge length in thenecked region called Zero-gauge-length elongation (e0). e0 can be calculated from qusing constancy of volume in plastic deformation (AL=A0L0).

What is meant by ductility?

0

0(%) 100

f

f

L L

e L

0

0(%) 100

fA A

q A

Note: this is ductility inUniaxial Tension Test

0

0

(%) 100uuL L

eL

0

0

1

1

AL

L A q

0 00

0

11 1

1 1

L L A qe

L A q q

0 1

q

e q


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We had seen two measures of ductility:Strain at fracture(ef)Reduction in area at fracture (q)

We had also seen that these can be related mathematically as:

However, it should be noted that they represent different aspects of material behaviour.

For reasonable gauge lengths, eis dominated by uniform elongation prior to necking and thus isdependent on the strain hardening capacity of the material (more the strain hardening, more will be thee). Main contribution to q(area based calculation) comes from the necking process (which is moregeometry dependent).

Hence, reduction in area is not truly a material property and has geometry dependence.

Comparison between reduction in area versus strain at fracture

0

0

f

f

L Le

L

0

0

fA Aq

A

01

qe

q


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What happens after necking?

Following factors come in to picture due to necking: Till necking the deformation is ~uniform along the whole gauge length. Till necking points on the -plot lie to the left and higher than the s-e plot (as below).

After the onset of necking deformation is localized around the neck region. Formulae used for conversion of e to and s to cannot be used after the onset of

necking. Triaxial state of stress develops and uniaxiality condition assumed during the testbreaks

down. Necking can be considered as an instability in tension.

Hence, quantities calculated after the onset of necking (like fracture stress,F) has to becorrected for: (i) triaxial state of stress, (ii) correct cross sectional area.

e)( 1lne)s( 1

Fractured surfacesFractured surfaces

Neck


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Beyond necking

True values beyond necking

Calculation of true strain beyond necking: True strain values beyond necking can be obtained by using the concept of zero-gauge-

length elongation (e0). This involves measurement of instantaneous area.

e)( 1ln0

1

qe

q

0

0

iA AqA

0 0ln 1 ) ( e 01

ln 1 ln1 1

q

q q

Note: Further complications arise at strains close to fracture as microvoid nucleation & growthtake place and hence all formulations based on continuum approach (e.g. volumeconstancy) etc. are not valid anymore.


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True values beyond necking

Calculation of true Stress beyond necking: Neck acts like a diffuse notch. This produces a triaxial state of stress (radial and transverse

components of stress exist in addition to the longitudinal component)this raises thevalue of the longitudinal stress required to case plastic flow.

Using certain assumptions (as below) some correction to the state of stress can be made*(given that the state of stress is triaxial, such a correction should be viewed appropriately).

Assumptions used in the correction*: neck is circular (radius = R), von Mises criterioncan be used for yielding, strains are constant across the neck.

The corrected uniaxial stress (uniaxial) is calculated from the stress from the experiment

(exp=Load/Alocal), using the formula as below.

* P.W. Bridgman, Trans. Am. Soc. Met. 32(1944) 553.

exp

21 ln 1

2

uniaxialR a

A R

The Correction

Cotd..


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Unloading the specimen during the tension test

If the specimen is unloaded beyond the yield point (say point Xin figure below), elasticstrain is recovered (while plastic strain is not). The unloading path is XM.

The strain recovered ( ) is more than that recovered if the specimen was to be

unloaded from Y ( ) i.e. in this sense the material is more elasticin the plasticregion (in the presence of work hardening), than in the elastic region!

If the specimen is reloaded it will follow the reverse path and yielding will start at ~X.Because of strain hardening* the yield strength of the specimen is higher.

Xelastic

Y

elastic

* Strain Hardeningis also called work hardeningThe material becomes harder with plastic deformation(on tensile loading the present case)

We will see later that strain hardening is usually causedby multiplication of dislocations.

If one is given a material which is at point M, then the

Yield stress of such a material would be ~X(i.e. as wedont know the prior history of the specimen loading,

we would call ~Xas yield stress of the specimen).

The blue part of the curve is also called the flow stress

T


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Variables in plastic deformation T,,, In the tension experiment just described the temperature (T) is usually kept constant and

the sample is pulled (between two crossheads of a UTM) at a constant velocity. Thecrosshead velocity can be converted to strain rate ( ) using the gauge length (L0) of the

specimen. At low temperatures (below the recrystallization temperature of the material, T < 0.4Tm)

the material hardens on plastic deformation (YFin the -plot known as workhardeningor strain hardening). The net strain is an important parameter under suchcircumstances and the material becomes a partial store of the deformation energy provided.The energy is essentially stored in the form of dislocations and point defects.

If deformation is carried out at high temperatures (above the recrystallization temperature;wherein, new strain free grains are continuously forming as the deformation proceeds),strain rate becomes the important parameter instead of net strain.

Range of strain rates obtained in various experiments

TestRange of strain rate

(/s)

Creep tests 108to 105

Quasi-static tension tests (in an UTM) 105to 101

Dynamic tension tests 101to 102

High strain rate tests using impact bars 102to 104

Explosive loading using projectiles (shock tests) 104to 108

0/v L


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In the -plot the plastic stress and strain are usually expressed by the expression givenbelow. Where, n is the strain hardening exponent and K is the strength coefficient.

,

n

plastic plasticT

K

Usually expressed as (for plastic)

For an experiment done inshear on single crystalsthe equivalent region to OYcan furtherbe subdivided into:True elastic strain (microscopic) till the true elastic limit (E)Onset of microscopic plastic deformation above a stress of A.For Mo a comparison of these quantities is as follows: E= 0.5 MPa, A= 5 MPa and

0(macroscopic yield stress in shear) = 50 MPa.

Deviations from this behaviour often observed (e.g. in Austenitic stainless steel) at lowstrains (~103) and/or at high strains (~1.0). Other forms of the power law equation are alsoconsidered in literature (e.g. ).

An ideal plastic material (without strain hardening) would begin to neck right at the onsetof yielding. At low temperatures (below recrystallization temperature- less than about

0.5Tm) strain hardening is very important to obtain good ductility.This can be understoodfrom the analysis of the results of the uniaxial tension test. During tensile deformationinstability in the form of necking localizes deformation to a small region (which nowexperiences a triaxial state of stress). In the presence of strain hardening the neck portion(which has been strained more) hardens and the deformation is spread to other regions,thus increasing the ductility obtained.

n

y plasticK


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Process parameters (characterized by parameters inside the sample)Mode of deformation, Sample dimensions, Stress, Strain, Strain Rate,

Temperature etc. Material parameters

Crystal structure, Composition, Grain size, dislocation density, etc.

Variables/parameters in mechanical testing

T


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,

n

T

K

Variables in plastic deformation T,,,

When true strain is less than 1, the smaller value of n dominates over a larger value of n

K strength coefficientn strain / work hardening coefficient

Cu and brass (n ~ 0.5) can be given large plastic strain moreeasily as compared to steels with n ~ 0.15

Material n K (MPa)

Annealed Cu 0.54 320

Annealed Brass (70/30) 0.49 900Annealed 0.5% C steel 0.26 530

0.6% carbon steelQuenched and Tempered (540C)

0.10 1570

n and K for selected materials ,

ln

ln Tn


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,

m

TA

A a constantm index of strain rate sensitivity

If m = 0 stress is independent of strain rate (stress-straincurve would be same for all strain rates)

m ~ 0.2 for common metalsIf m (0.4, 0.9) the material may exhibit superplastic behaviourm = 1 material behaves like a viscous liquid (Newtonian flow)

The effect of strain rate is compared by performing tests to a

constant strain

At high temperatures (above recrystallization temperature) where strain rate is theimportant parameter instead of strain, a power law equation can be written as belowbetween stress and strain rate.

,

ln

ln Tm

Thermal softening coefficient () is also defined as below:

ln

lnT


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Funda Check What is the important of m and n

We have seen that below recrystallization temperature n is the important parameter.

Above recrystallization temperature it is m which is important.

We have also noted that it is necking which limits the ductility in uniaxial tension.

Necking implies that there is locally more deformation (strain) and the strain rate is alsohigher locally.

Hence, if the locally deformed material becomes harder (stronger) then the deformation

will spread to other regions along the gauge length and we will obtain more ductility.

Hence having a higher value of n or m is beneficial for obtaining good ductility.

Pl i d f i b li


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As we noted in the beginning of the chapter plastic deformation can occur by manymechanisms SLIP is the most important of them. At low temperatures (especially inBCC metals) twinning may also be come important.

At the fundamental level plastic deformation (in crystalline materials) by slip involves themotion of dislocations on the slip planefinally leaving the crystal/grain*(creating astep of Burgers vector).

Slip is caused by shear stresses(at the level of the slip plane). Hence, a purely hydrostaticstate of stress cannot cause slip (or twinning for that matter).

A slip system consists of a slip direction lying on a slip plane.

Under any given external loading conditions, slip will be initiated on a particular slipsystem if the Resolved Shear Stress(RSS)** exceeds a critical value [the Critical ResolvedShear Stress(CRSS)].

For slip to occur in polycrystalline materials, 5 independent slip systemsare required.Hence, materials which are ductile in single crystalline form, may not be ductile in

polycrystalline form. CCP crystals (Cu, Al, Au) have excellent ductility. At higher temperatures more slip systems may become active and hence polycrystalline

materials which are brittle at low temperature, may become ductile at high temperature.

Plastic deformation by slip Click here to see overview of mechanisms/modes of plastic deformation

* Leaving the crystal part is important** To be defined soon

Sli t


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Crystal Slip plane(s) Slip direction Number of slip systems

FCC {111} 12

HCP (0001) 3

BCCNot close packed

{110}, {112}, {123} [111] 12

NaClIonic

{110}{111} not a slip plane

6

C

Diamond cubic {111} 12

Slip systems

In CCP, HCP materials the slip system consists of a close packed direction on a close packed plane.

Just the existence of a slip system does not guarantee slip slip is competing against other processeslike twinning and fracture. If the stress to cause slip is very high (i.e. CRSS is very high), then fracture

may occur before slip (like in brittle ceramics).


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Crystal Slip plane(s) Slip direction Slip systems

TiO2

Rutile {101}

CaF2, UO2, ThO2Fluorite

{001}

CsClB2

{110}

NaCl, LiF, MgORock Salt

{110} 6

C, Ge, SiDiamond cubic

{111} 12

MgAl2O4Spinel

{111}

Al2O3Hexagonal

(0001)

More examples of slip systems


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Crystal Slip plane(s) Slip direction (b) Slip systems

Cu (FCC)

Fm 3m {111} (a/2)< 1 10> 4 x 3 = 12

W (BCC)

Im 3m

{011}

{112}

{123}

(a/2)

6 x 2 = 12

12 x 1 = 12

24 x 1 = 24

Mg (HCP)

P63/mmc

{0001}

{10 10}

{10 11}

(a/3)

1 x 3 = 3

3 x 1 = 3

6 x 1 = 6

Al2O3

R 3c

{0001}

{10 10}(a/3)

1 x 3 = 3

3 x 1 = 3

NaCl

Fm 3m

{110}

{001}(a/2)< 1 10>

6 x 1 = 6

6 x 1 = 6

CsCl

Pm 3m{110} a 6 x 1 = 6

Polyethylene

Pnam

(100)

{110}

(010)

c

1 x 1 = 1

2 x 1 = 2

1 x 1 = 1

More examples of slip systems

Critical Resolved Shear Stress (CRSS) Wh i h i b P i l d CRSS?
http://localhost/var/www/apps/books/e-book/peierls_stress.ppthttp://localhost/var/www/apps/books/e-book/peierls_stress.ppt


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As we saw plastic deformation byslipis due to shear stresses. Even if we apply an tensile force on the specimen the shear stress resolved onto the slip

plane is responsible for slip.

When the Resolved Shear Stress (RSS) reaches a critical value Critical Resolved ShearStress (CRSS) plastic deformation starts (The actual Schmids law)

Slip planenormal

Slip direction

A

DArea

ForceStress

1

CosA

CosF

/

RSS Cos Cos

A

Cos

AA'

A

F

Schmid factor

Critical Resolved Shear Stress (CRSS)

Note externally only tensile

forces are being applied

What is the connection between Peierls stress and CRSS?
http://localhost/var/www/apps/books/e-book/peierls_stress.ppthttp://localhost/var/www/apps/books/e-book/peierls_stress.ppt


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Schmids law

CRSSy

Cos Cos

RSS CRSS Slip is initiated when

Yield strength of a single crystal

CRSSis a material parameter, which is determined from experiments

How does the motion of dislocations lead to a macroscopic shape change?


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How does the motion of dislocations lead to a macroscopic shape change?(From microscopic slip to macroscopic deformation a first feel!)

Net shape change

When one bents a rod of aluminium to a new shape, it involves processes occurring atvarious lengthscales and understanding these is an arduous task.

However, at the fundamental level slip is at the heart of the whole process. To understand how slip can lead to shape change?; we consider a square crystal

deformed to a rhombus (as Below).


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b

Dislocationformed bypushing ina plane

Step formedwhen dislocationleaves the crystal

Now visualize dislocations being punched in on successive planesmoving and finallyleaving the crystal


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Net shape change

This sequence of events finally leads to deformed shape which can be approximated to a

rhombus

Stress to move a dislocation: Peierls Nabarro (PN) stress


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Stress to move a dislocation:PeierlsNabarro (PN) stress

We have seen that there is a critical stress required to move a dislocation.

At the fundamental level the motion of a dislocation involves the rearrangement of bondsrequires application of shear stress on the slip plane.

When sufficient stress is applied the dislocation moves from one metastable energyminimum to another.

The original model is due to Peierls & Nabarro (formula as below) and the sufficient stress

which needs to be applied is called Peierls-Nabarro stress (PN stress) or simply Peierls stress.

Width of the dislocation is considered as a basis for the ease of motion of a dislocation in themodel which is a function of the bonding in the material.

Click here to know more about Peierls Stress

bw

PN eG

2

G shear modulus of the crystalw width of the dislocation !!!b |b|

How to increase the strength?
http://localhost/var/www/apps/conversion/tmp/scratch_4/peierls_stress.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_4/peierls_stress.ppt


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How to increase the strength?

We have seen that slip of dislocations weakens the crystal. Hence we have two strategies tostrengthen the crystal/material:

completely remove dislocationsdifficult, but dislocation free whiskers have been

produced (however, this is not a good strategy as dislocations can nucleate during loading)increase resistance to the motion of dislocations orput impediments to the motion ofdislocationsthis can be done in many ways as listed below.

Solid solution strengthening(by adding interstitial and substitutional alloying elements).

Cold Workincrease point defect and dislocation density

(Cold work increases Yield stress but decreases the % elongation, i.e. ductility). Decrease in grain sizegrain boundaries provide an impediment ot the motion of

dislocations (Hall-Petch hardening).

Precipitation/dispersion hardeningintroduce precipitates or inclusions in the path ofdislocations which impede the motion of dislocations.

Forest dislocation

Strengthening mechanisms

Solid solutionPrecipitate

&

Dispersoid

Grain boundary


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Classification of the obstacles to motion of a dislocation Based of if the obstacle can be


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Internal resisting stress (i)

Long range obstacles (G)

Short range obstacles (T)

Athermal f (T, strain rate)These arise from long range internal stress fields

Sources:Stress fields of other dislocationsIncoherent precipitates

Note: Ghas some temperature

dependence as G decreases with T

Thermal = f (T, strain rate)Short range ~ 10 atomic diametersT can help dislocations overcome these obstaclesSources:

Peierls-Nabarro stressStress fields of coherent precipitates & solute atoms

Classification of the obstacles to motion of a dislocation overcome by thermal activation


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Fe

W

18-8 SS

Cu

Ni

Al2O3

Si

150

300

450

0.2 0.4 0.60.0

Y

ieldStress(M

Pa)

T/Tm

Very high temperaturesneeded for thermal activationto have any effect

RT is like HT and P-Nstress is easily overcome

Fe-BCC

W-BCC

Cu-FCC

Ni-FCC

Metallic

Ionic

Covalent

What causes Strain hardening? multiplication of dislocations


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X

g p

Why increase in dislocation density ?

Why strain hardening ?

If dislocations were to leave the surfaceof the crystal by slip / glide then thedislocation density should decrease

on plastic deformation but observation is contrary to this

)1010(~

)1010(~

1412

ndislocatio

96

ndislocatio

materialStrongermaterialAnnealedworkCold

Strain hardening

This implies some sources of dislocation

multiplication / creation should exist

Dislocation sources


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Dislocation sources

Solidification from the melt

Grain boundary ledges and step emit dislocations

Aggregation and collapse of vacancies to form a disc or prismatic loop (FCC Frankpartials)

High stressesHeterogeneous nucleation at second phase particlesDuring phase transformation

Frank-Read source

Orowan bowing mechanism

It is difficult to obtain crystals without dislocations (under special conditions whiskers havebeen grown without dislocations).

Dislocation can arise by/form:

Solidification (errors in the formation of a perfect crystal lattice)Plastic deformation (nucleation and multiplication)Irradiation

Some specific sources/methods of formation/multiplication of dislocations include


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Formation of the Lomer-Cottrell barrier


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)111(

]101[2

1

)111(

]011[2

1

)100(

]110[2

1

Formation of the Lomer-Cottrell barrier

1 1 1


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)100(

]110[2

1

)111(

]011[2

1

)111(

]101[2

1

+

Lomer-Cottrell barrier

The redandgreendislocations attract each other and move towards their line of intersectionThey react as above to reduce their energy and produce the bluedislocation

The bluedislocation lies on the (100)plane which is not a close packed plane hence immobile acts like a barrier to the motion of other dislocations

]110[

Impediments (barriers) to dislocation motion


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ped e ts (ba e s) to d s ocat o ot o

Grain boundary

Immobile (sessile) dislocations

Lomer-Cottrell lockFrank partial dislocation

Twin boundary

Precipitates and inclusions

Dislocations get piled up at a barrier and produce a back stress

Stress to move a dislocation & dislocation density


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y

A0

0 base stress to move a dislocation in the crystalin the absence of other dislocations

Dislocation density A A constant

as (cold work) (i.e. strain hardening)

Empirical relation

(MN / m2) ( m/ m3) 0(MN / m2) A (N/m)

1.5 1010 0.5 10

100 1014 0.5 10

COLD WORK strength ductility

Grain size and strength Hall-Petch Relation


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g

d

kiy

y Yield stress [N/m2]

i Stress to move a dislocation in single crystal [N/m2]

k Locking parameter [N/m3/2]

(measure of the relative hardening contribution of grain boundaries)

d Grain diameter [m]

Grain size


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Grain size

101

10)2(

645

n

d d Grain diameter in meters

n ASTM grain size number

Effect of solute atoms on strengthening


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Effect of solute atoms on strengthening

Solid solutions offer greater resistance to dislocation motion than pure crystals(even solutewith a lower strength gives strengthening!)

The stress fields around solute atoms interact with the stress fields around the dislocation toleading to an increase in the stress required for the motion of a dislocation

The actual interaction between a dislocation and a solute is much more complex

The factors playing an important role are:Size of the solutemore the size difference, more the hardening effect

Elastic modulus of the solute (higher the elastic modulus of the solute greater thestrengthening effect)e/a ratio of the solute

A curved dislocation line configuration is required for the solute atoms to provide hindranceto dislocation motion

As shown in the plots in the next slide, increased solute concentration leads to an increasedhardening. However, this fact has to be weighed in the backdrop of solubility of the solute.Carbon in BCC Fe has very little solubility (~0.08 wt.%) and hence the approach of puresolid solution strengthening to harden a material can have a limited scope.

Size effect For the same size difference the


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Solute Concentration (Atom %)

y(MPa)

50

100

150

10 20 30 40

200

0

Solute strengthening of Cu crystalby solutes of different sizes

Matrix: Cu (r = 1.28 )

Be (1.12)

Sn (1.51)

Zn (1.31)

(Values in parenthesis are atomic radius values in )

Size effectSize difference

Size effect depends on:

Concentration of the solute (c) cy

~

ffsmaller atom gives a greaterstrengthening effect

By i (lattice friction)


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y

Often produce Yield Point Phenomenon

Solute atoms level of - curve

X

By i(lattice friction)

Relative strengthening effect of Interstitial and Substitutional atoms


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Interstitial

Solute atoms

Substitutional

3Gsolute

Relative strengthening effect / unit concentration

Gsolute/ 10 fielddistortionSpherical

fielddistortionsphericalNon

Interstitial solute atoms have a non-spherical distortion field and can elastically interactwith both edge and screw dislocations. Hence they give a higher hardening effect (per unitconcentration) as compared to substitutional atoms which have (approximately) a

spherical distortion field.


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Elastic Interstitial Edge and screw dl.


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Long range(T insensitive)

Solute-dislocation interaction

Short range

(T sensitive)

Modulus

Long range order

Elastic

Substitutional edge

Electrical

Short range order

Stacking fault

Mechanisms of interaction ofdislocations with solute atoms

The hardening effect of precipitates


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Glide through the precipitate

Dislocation

Get pinned by the precipitate

If the precipitate is coherent with the matrix

Precipitates may be coherent, semi-coherent or incoherent. Coherent (& semi-coherent)precipitates are associated with coherency stresses.

Dislocations cannot glide through incoherent precipitates.

Inclusions behave similar to incoherent precipitates in this regard (precipitates are part ofthe system, whilst inclusions are external to the alloy system).

A pinned dislocation (at a precipitate) has to either climb over it (which becomesfavourable at high temperatures) or has to bow around it.

A complete list of factors giving rise to hardening due to precipitates/inclusions will be considered later

Dislocation Glide through the precipitate


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Only if slip plane is continuous from the matrix through the precipitate precipitate is coherent with the matrix.

Stress to move the dislocation through the precipitate is ~ that to move it in thematrix (though it is usually higher as precipitates can be intermetallic compounds).

Usually during precipitation the precipitate is coherent only when it is small andbecomes incoherent on growth.

Glide of the dislocation causes a displacement of the upper part of the precipitatew.r.t the lower part by b ~ cutting of the precipitate.

IncoherentcoherentPartiallyCoherent

GrowthGrowth LargeSmall

Schematic views

d di l i lid h h h i i


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b

Precipitate particleb

edge dislocation glide through a coherent precipitate

If the particle is sheared, then how does the hardening effect come about?


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Hardening effect Part of the dislocation line segment (inside theprecipitate)could face a higher PN stress

Increase in surface area due to particle shearing

We have seen that as the dislocation glides through the precipitate it is sheared.

If the precipitate is sheared, then how does it offer any resistance to the motion of the dislocation? I.e.how can this lead to a hardening effect?

The hardening effect due to a precipitate comes about due to many factors (many of which are systemspecific). The important ones are listed in the tree below.

Pinning effect of inclusions Orowan bowing mechanism


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Dislocations can bow around widely separated inclusions. In this process they leavedislocation loops around the inclusions, thus leading to an increase in dislocation density.This is known as the orowan bowing mechanism as shown in the figure below. (This is in somesense similar to the Frank-Read mechanism).

The next dislocation arriving (similar to the first one), feels a repulsion from thedislocation loop and hence the stress required to drive further dislocations increases.Additionally, the effective separation distance (through which the dislocation has to bow)reduces from d to d1.

Precipitate Hardening effect (Complete List)


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The hardening effect of precipitates can arise in many ways as below:

Lattice Resistance: the dislocation may face an increased lattice friction stress in theprecipitate.

Chemical Strengthening: arises from additional interface created on shearing

Stacking-fault Strengthening: due to difference between stacking-fault energy betweenparticle and matrix when these are both FCC or HCP (when dislocations are split intopartials)

Modulus Hardening: due to difference in elastic moduli of the matrix and particle

Coherency Strengthening: due to elastic coherency strains surrounding the particle

Order Strengthening: due to additional work required to create an APB in case ofdislocations passing through precipitates which have an ordered lattice

Strain rate effects


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We had noted that strain rate can vary by orders of magnitude depending on deformationprocess (Creep: 108to Explosions: 105).

Strain rate effects become significant (on properties like flow stress) only when strain rate

is varied by orders of magnitude (i.e. small changes in flow stress do not affect the value of the properties much). Strain rate can be related to dislocation velocity by the equation below.

dd vb vd velocity of the dislocations

d density of mobile/glissiledislocations

b |b|

In a UTT why does the plot not continue along OY (straight line)?Funda Check


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When stress is increased beyond the yield stress the mechanism of deformation changes.

Till Y in the s-e plot, bond elongation (elastic deformation) gives rise to the strain. After Y, the shear stress resulting from the applied tensile force, tends to move

dislocations (and cause slip)rather than stretch bonds as this will happen at lowerstresses as compared to bond stretching (beyond Y).

If there are not dislocations (e.g. in a whisker) (and for now we ignore other mechanism ofdeformation), the material will continue to load along the straight line OYtilldislocations nucleate in the crystal.

Funda Check


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