PLASTIC DEFORMATIONeacharya.inflibnet.ac.in/data-server/eacharya... · Creep Mechanisms. Grain...

72
PLASTIC DEFORMATION PLASTIC DEFORMATION Modes of Deformation The Uniaxial Tension Test Mechanisms underlying Plastic Deformation Strengthening mechanisms Mechanical Metallurgy George E Dieter McGraw-Hill Book Company, London (1988)

Transcript of PLASTIC DEFORMATIONeacharya.inflibnet.ac.in/data-server/eacharya... · Creep Mechanisms. Grain...

Page 1: PLASTIC DEFORMATIONeacharya.inflibnet.ac.in/data-server/eacharya... · Creep Mechanisms. Grain boundary sliding. Vacancy diffusion. Dislocation climb + Other Mechanisms. Note: Plastic

PLASTIC DEFORMATIONPLASTIC DEFORMATION

Modes of Deformation

The Uniaxial

Tension Test

Mechanisms underlying Plastic Deformation

Strengthening mechanisms

Mechanical MetallurgyGeorge E Dieter

McGraw-Hill Book Company, London (1988)

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An Al rod when bent through a large angle does not come back to its original shape.

A steel piece is easier to deform when heated (as compared to when it is cold).

Let us start with some observations…

Click here to know about all the mechanisms by which materials fail Click here to know about all the mechanisms by which materials fail

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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 shear banding

Plastic deformation in the broadest sense means permanent deformation

in 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 slip

is the most important mechanism of plastic deformation, there are other mechanisms as well. Many of these mechanisms may act in conjunction/parallel to give rise to the observed plastic deformation.

Grain rotation

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A body can be deformed using many modes:

Tension/Compression

Bending

Shear

TorsionIt is important to note that these are macroscopically defined with respect to a body of given geometry (even in tensile loading

inclined planes will be subjected to shear stress)

Common types of deformation

Tension Compression

ShearTorsion

Deformed configuration

Bending

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

Tension / Compression

Torsion

Modesof

Deformation Shear

Bending

Review

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Mode I

Mode III

Modes

of

DeformationMode II

In addition to the modes of deformation considered before the following modes can be defined w.r.t

fracture.

Fracture can be cause by the propagation of a pre-existing crack (e.g. the notches shown in the figures below) or by the nucleation of a crack during deformation followed by its propagation.

In fracture the elastic energy stored in the material is used for the creation of new surfaces (when the crack nucleates/propagates)

Peak ahead

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The following aspects need to be understood to comprehend plasticity*:

External process parameters (Temperature, strain rate etc.)

Macroscopic and Microscopic aspects of plasticity

Continuum and Discrete views of plasticity

Plasticity in single crystals

Plasticity in polycrystals

Plasticity in multiphase materials

Plasticity in nanomaterials

Path to understanding plasticity

* Some of these aspects will be covered in the current chapter

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One of the simplest test which can performed to evaluate the mechanical properties of a material is the Uniaxial

Tension Test. The force/load applied is uniaxial.

This is typically performed on a cylindrical specimen with a standard ‘gauge length’. (At constant temperature and strain rate). Other types of specimens are also used.

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) engineering stress-

engineering strain, (iii) true stress-

true strain. (next slide).

It is convenient to use Engineering Stress (s)

and Engineering Strain (e) as defined below as we can divide the load and change in length by constant quantities (A0

and 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) and hence we define True Stress ()

and True Strain () (wherein we use instantaneous values of length and area).

Though this is simple test to conduct and a wealth of information about the mechanical behaviour of a material can be obtained (Modulus of elasticity, ductility

etc.) However, it must be cautioned that this data should be used with caution under other states of stress.

The Uniaxial

Tension Test (UTT)

0APs

0LLe

0 → initial

i → instantaneousSubscript Note: quantities obtained by performing an

Uniaxial

Tension Test are valid only under uniaxial

state of stress

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The Tensile Stress-Strain Curve

Stroke →

Load

e →

s →

Gauge Length → L0Possible axes

Tensile specimen

Initial cross sectional area → A0

Important Note

We shall assume cylindrical specimens (unless otherwise stated)

Note that L0

is NOT the length of the specimen, but the gauge length

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

Consider the following sequence of deformations:

L0

2L0

L0

e1→2

= 1

e2→3

=

½

e1→3

= 0

1

2

3

[e1→2

+ e2→3

] = ½

It is clear that from stage 1 → 3 there is no strainBut the decomposition of the process into 1 → 2 & 2 → 3 gives a net strain of ½► Clearly there is a problem with the use (definition) of Engineering strain (for large

strains as in the example above).► 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 ()

iAP

0

ln0

LL

LdLL

L

0A

Ps 0LLe

The definitions of true stress and true strain are based on instantaneous values of area (Ai

) and length (Li

) and not on the original measures (as for engineering stress and strain).

Ai

→ instantaneous area

Same sequence of deformations considered before:

L0

2L0

L0

1→2

= Ln(2)

2→3

=

Ln(2)

1→3

= 0

1

2

3

[

1→2

+

2→3

] = 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)

Young’s 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 the curves obtained for some metallic materials like Al, Cu etc. (polycrystalline materials at room temperature).

Many materials (e.g. steel) may have curves which are qualitatively 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

~10–3)

[E is measured in GPa

and y

in MPa

thus giving this small strains] the linear portion is practically vertical and stuck to the Y-axis (when efracture

and eelastic

is drawn to the same scale).

Schematics: not to scale

Note the increasing stress required for continued plastic deformation

Neck

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O unloaded specimen

OY

Elastic

Linear Region in the plot (macroscopic

linear elastic region)

Y macroscopic yield point (there are many measures of yielding as

discussed later)

Occurs due to collective motion of many dislocations finally leaving the grain boundary or crystal surface.

YF

Elastic + Plastic regime

If specimen is unloaded from any point in this region, it will unload parallel to OY

and the elastic 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 region OY).

In this region the material strain hardens

flow stress increases with strain.

This region can further be split into YN

and NF as below.

YN

Stable region with uniform deformation along the gauge length

N

Plastic Instability

in tension Onset of necking True condition of uniaxiality

broken onset of triaxial

state of stress (loading remains uniaxial

but the state of stress in the cylindrical specimen is not).

NF most of the deformation is localized at the neck

Specimen in a triaxial

state of stress

F

Fracture 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 strainTrue strain () 0.01 0.10 0.20 0.50 1.0 2.0 3.0 4.0Engineering strain (e) 0.01 0.105 0.22 0.65 1.72 6.39 19.09 53.6

e)(ε 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 the deviations between the values are large.

In engineering stress since we are dividing by a constant number

A0

(and there is a local reduction 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 the sample and under such circumstances true stress should be used.

iAP

0

ln0

LL

LdLL

L

0

ln 1 1 ln(1+e)LL

0

0 0 0

1 1 (1 )i i

i

A L LP s s s eA A L L

00 0 i i

0

From volume constancy A L =A L i

i

A LA L

Valid till necking starts

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Yielding can be defined in many contexts.

Truly speaking (microscopically) it is point at which dislocations leave the crystal (grain) and cause microscopic plastic deformation (of unit ‘b’) this is best determined from microstrain

(~10–6

) experiments on single crystals. However, in practical terms it is determined from the stress-strain plot (by say an offset as described below).

True elastic limit

(microscopically and macroscopically elastic

where in there is not even microscopic yielding) ~10–6

[OA

portion of the curve]

Microscopically plastic but macroscopically elastic

[AY

portion of the curve]

Proportional limit

the point at which there is a deviation from the straight line ‘elastic’

regime

Offset Yield Strength

(proof stress) A curve is drawn parallel to the elastic line at a given strain like 0.2% (= 0.002) to determine the yield strength.

Where does Yielding start?

In some materials (e.g. pure annealed Cu, gray cast iron etc.) the linear portion of the curve may be limited and yield strength may arbitrarily determined as the stress at some given strain (say 0.005).

Microscopicelastic

Macroscopicelastic

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Important Note

y

is yield stress in an uniaxial

tension test (i.e. plastic deformation will start after crossing yield stress

only under uniaxial

tensile loading) 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.

Important Note

Yielding begins when a stress equal to y

(yield stress) is applied; however to cause further plastic deformation increased stress has to be applied i.e. the material hardens with plastic deformation → known as work hardening/strain hardening.

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

In simple terms a ductile material

is 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 test

as follows (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 ductility

as it does not depend on the gauge length (L0

); while, ‘ef

depends on L0

. Elongation/strain to necking (uniform elongation) can also be used to avoid the 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 a reduction in area. For this the calculation has be based on a very short gauge length in the necked region called Zero-gauge-length elongation (e0

). ‘e0

can be calculated from ‘q’

using constancy of volume in plastic deformation (AL=A0

L0

).

What is meant by ductility?

0

0

(%) 100ff

L Le

L

0

0

(%) 100fA Aq

A

Note: this is ductility in Uniaxial

Tension Test

0

0

(%) 100uu

L LeL

0

0

11

ALL A q

0 00

0

11 11 1

L L A qeL A q q

0 1

qeq

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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, ‘e’

is dominated by uniform elongation prior to necking and thus is

dependent on the strain hardening capacity of the material (more

the strain hardening, more will be the ‘e’). Main contribution to ‘q’

(area based calculation) comes from the necking process (which is more geometry 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

ff

L Le

L

0

0

fA Aq

A

0 1qe

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 test breaks down.

Necking can be considered as an instability in tension.

Hence, quantities calculated after the onset of necking (like fracture stress, F

) has to be corrected for: (i) triaxial

state of stress, (ii) correct cross sectional area.

e)(ε 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)(ε 1ln0 1

qeq

0

0

iA AqA

0 0ln 1 )ε ( e 01ln 1 ln

1 1qε

q q

Note: Further complications arise at strains close to fracture as microvoid

nucleation & growth take place and hence all formulations based on continuum approach (e.g. volume constancy) 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 the value 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’

criterion can 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 12

uniaxial R aA R

The Correction

Cotd..

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0 1ln ln1f

f f

AA q

Calculation of fracture stress/strain:

To calculate true fracture stress (F

) we need the area at fracture (which is often not readily available and often the data reported for fracture stress could be in error).

Further, this fracture stress has to be corrected for triaxiality.

True strain at fracture can be calculated from the areas as below.

Fracture stress and fracture strain

The tensile specimen fails by ‘cup & cone’

fracture wherein outer regions fail by shear and inner regions in tension.

Fracture strain (ef

) is often used as a measure of ductility.

ff

f

PA

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The UTT

test as described was done on a ductile material, then why did it fracture? (don’t brittle materials fracture?)

Funda

Check

A ductile material is one whose fracture stress is above its yield stress: y

< f

.

Two factors contribute to this fracture:

(i) the triaxial

state of stress introduced by the necking (like at a crack tip).

(ii) the progressive work hardening, which makes y

> f

(i.e. the yield stress becomes more than the fracture stress).

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Unloading the specimen during the tension test

If the specimen is unloaded beyond the yield point (say point ‘X’

in figure below), elastic strain 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 elastic’

in the plastic region (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

Yelastic

* Strain Hardening

is also called work hardening

The material becomes harder with plastic deformation (on tensile loading the present case)

We will see later that strain hardening is usually caused by 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 we don’t know the ‘prior history’

of the specimen loading, we would call ~X

as yield stress of the specimen).

The blue part of the curve is also called the ‘flow stress’.

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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. The crosshead 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 (YF

in the -

plot known as work hardening

or strain hardening). The net strain is an important parameter under such circumstances 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

Test Range of strain rate (/s)

Creep tests 10–8

to 10–5

Quasi-static tension tests (in an UTM) 10–5

to 10–1

Dynamic tension tests 10–1

to 102

High strain rate tests using impact bars 102

to 104

Explosive loading using projectiles (shock tests) 104

to 108

0/v L

Page 25: PLASTIC DEFORMATIONeacharya.inflibnet.ac.in/data-server/eacharya... · Creep Mechanisms. Grain boundary sliding. Vacancy diffusion. Dislocation climb + Other Mechanisms. Note: Plastic

In the -

plot the plastic stress and strain are usually expressed by the

expression given below. Where, ‘n’

is the strain hardening exponent and ‘K’

is the strength coefficient.

,

nplastic plastic T

K

Usually expressed as (for plastic

)

For an experiment done in shear on single crystals

the equivalent region to OY

can further be 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 low strains (~10–3) and/or at high strains (~1.0). Other forms of the power law equation are also considered in literature (e.g. ).

An ideal plastic material (without strain hardening) would begin

to neck right at the onset of yielding. At low temperatures (below recrystallization temperature-

less than about 0.5Tm

) strain hardening is very important to obtain good ductility.

This can be understood from the analysis of the results of the uniaxial

tension test. During tensile deformation instability in the form of necking localizes deformation to a small region (which now experiences 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.

ny plasticK

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There are variables, which we have to ‘worry about’, when we do mechanical testing (for the test and to interpret the results):

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

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,

n

TK

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 coefficient

n

→ strain / work hardening coefficient

Cu and brass (n ~ 0.5) can be given large plastic strain (before

fracture) as compared to steels with n ~ 0.15.

Material n K (MPa)

Annealed Cu 0.54 320

Annealed Brass (70/30) 0.49 900

Annealed 0.5% C steel 0.26 530

0.6% carbon steel Quenched and Tempered (540C) 0.10 1570

‘n’

and ‘K’

for selected materials,

lnln T

n

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,

m

TC

C → a constant

m → index of strain rate sensitivity

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

m ~ 0.2 for common metals

If m (0.4, 0.9) the material may exhibit superplastic behaviour

m = 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 the important parameter instead of strain, a power law equation can be written as below between stress and strain rate.

,

lnln T

m

Thermal softening coefficient () is also defined as below:

lnlnT

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In some materials (due to structural condition or high temperature) necking is prevented by strain rate hardening.

Further aspects regarding strain rate sensitivity

,

m

TC

m PC

A

1/ 1/1m mPC A

From the definition of true strain

1 1dL dAL dt A dt

1/ 1/1 1m mdA P

A dt C A

1/

(1 ) /

1m

m m

dA Pdt C A

If m < 1→ smaller the cross-sectional area, the more rapidly the area is reduced.

If m = 1

→ material behaves like a Newtonian viscous liquid → dA/dt

is independent of A.

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1/

(1 ) /

1m

m m

dA Pdt C A

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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 also higher 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.

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

At the fundamental level plastic deformation (in crystalline materials) by slip involves the motion of dislocations on the slip plane finally leaving the crystal/grain*

(creating a step of Burgers vector).

Slip is caused by shear stresses

(at the level of the slip plane). Hence, a purely hydrostatic state 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 slip system if the Resolved Shear Stress

(RSS)** exceeds a critical value [the Critical Resolved Shear Stress

(CRSS)].

For slip to occur in polycrystalline materials, 5 independent slip systems

are 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

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

FCC {111} ½<110> 12

HCP (0001) <1120> 3

BCC

Not close packed {110}, {112}, {123}

½[111] 12

NaCl

Ionic

{110}

{111} not a slip plane

½<110> 6

C

Diamond cubic {111} ½<110> 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 processes like 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

TiO2Rutile {101} <101>

CaF2

, UO2

, ThO2Fluorite {001} <110>

CsCl

B2 {110} <001>

NaCl, LiF, MgO

Rock Salt {110} <110> 6

C, Ge, Si

Diamond cubic {111} <110> 12

MgAl2

O4Spinel {111} <110>

Al2

O3Hexagonal (0001) <1120>

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)<11 1>6 x 2 = 12

12 x 1 = 1224 x 1 = 24

Mg (HCP)P63

/mmc

{0001}{10 10}{10 11}

(a/3)<1120>1 x 3 = 33 x 1 = 36 x 1 = 6

Al2

O3

R 3c{0001}{10 10}

(a/3)<1120>1 x 3 = 33 x 1 = 3

NaClFm 3m

{110}{001}

(a/2)< 1 10>6 x 1 = 66 x 1 = 6

CsClPm 3m

{110} a<001> 6 x 1 = 6

PolyethylenePnam

(100){110}(010)

c<001>1 x 1 = 12 x 1 = 21 x 1 = 1

More examples of slip systems

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As we saw plastic deformation by slip

is 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 Shear Stress (CRSS) → plastic deformation starts (The actual Schmid’s

law)

Slip plane normal

Slip direction

A’

DAreaForceStress

1

CosACosF/

RSS Cos Cos

A

CosAA'

AF

Schmid

factor

Critical Resolved Shear Stress (CRSS)

Note externally only tensile forces are being applied

What is the connection between Peierls stress and CRSS?

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Schmid’s

law

CRSSy Cos Cos

RSS CRSS Slip is initiated when

Yield strength of a single crystal

CRSS

is a material parameter, which is determined from experiments

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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 at various 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

Dislocation formed by

pushing in

a plane

Step formed when dislocation

leaves the crystal

Now visualize dislocations being punched in on successive planes

moving and finally leaving 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

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Stress to move a dislocation: Peierls –

Nabarro

(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 bonds

requires application of shear stress on the slip plane.

When ‘sufficient stress’

is applied the dislocation moves from one metastable energy minimum 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 the model 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 crystal w → width of the dislocation !!! b → |b|

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How to increase the strength?

We have seen that slip of dislocations weakens the crystal. Hence we have two strategies to strengthen the crystal/material: completely remove dislocations

difficult, 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 or put impediments to the motion of dislocations this can be done in many ways as listed below.

Solid solution strengthening

(by adding interstitial and substitutional alloying elements).

Cold Work

increase point defect and dislocation density

(Cold work increases Yield stress but decreases the % elongation, i.e. ductility).

Decrease in grain size

grain boundaries provide an impediment ot

the motion of dislocations (Hall-Petch

hardening).

Precipitation/dispersion hardening

introduce precipitates or inclusions in the path of dislocations which impede the motion of dislocations.

Forest dislocation

Strengthening mechanisms

Solid solutionPrecipitate

& Dispersoid

Grain boundary

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Applied shear stress vs

internal opposition

Applied shear stress () Internal resisting stress (i

)

PN

stress is the ‘bare minimum’

stress required for plastic deformation. Usually there will be other sources of opposition/impediment to the motion of dislocations in the material. Some of these include :

Stress fields

of other dislocations

Stress fields

from coherent/semicoherent precipitates

Stress fields

from low angle grain boundaries

Grain boundaries

Effect of solute atoms and vacancies

Stacking Faults

Twin boundaries

Phonon drag

etc.

Some of these barriers (the short range obstacles) can be overcome by thermal activation (while other cannot be-

the long range obstacles).

These factors lead to the strengthening of the material.

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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 dislocations

►Incoherent precipitates

Note: G has some temperature dependence as G decreases with T

Thermal = f (T, strain rate) Short range ~ 10 atomic diameters T can help dislocations overcome these obstacles

Sources:

►Peierls-Nabarro

stress

►Stress fields of coherent precipitates & solute atoms

Classification of the obstacles to motion of a dislocation Based of if the obstacle can be overcome by thermal activation

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Effect of Temperature

Equilibrium positions of a dislocation

Q

Motion of a dislocation can be assisted by thermal energy.

However, motion of dislocations by pure thermal activation is random.

A dislocation can be thermally activated to cross the potential barrier ‘Q’

to the neighbouring metastable position.

Strain rate can be related to the temperature (T) and ‘Q’

as in the equation below.

This thermal activation reduces the Yield stress (or flow stress).

Materials which are brittle at room temperature may also become ductile at high temperatures.

QkTAe

rateStraindtd

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FeW

18-8 SS

Cu

Ni

Al2

O3

Si

150

300

450

0.2 0.4 0.60.0

Yie

ld S

tress

(MPa

) →

T/Tm

Very high temperatures

needed for thermal activation

to have any effect

RT is like HT and P-N

stress is easily overcome

Fe-BCCW-BCC

Cu-FCC

Ni-FCC

MetallicIonic

Covalent

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X

Strain hardening →

What causes Strain hardening? → multiplication of dislocations

Why increase in dislocation density ?

Why strain hardening ?

If dislocations were to leave the surface of the crystal by slip / glide then the

dislocation density should decrease on plastic deformation →

but observation is contrary to this

)1010(~

)1010(~

1412ndislocatio

96

ndislocatio

materialStrongermaterialAnnealed workCold

Strain hardening

This implies some sources of dislocation multiplication / creation should exist

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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 Frank partials)

High stresses

► Heterogeneous nucleation at second phase particles

► During phase transformation

Frank-Read source

Orowan

bowing mechanism

It is difficult to obtain crystals without dislocations (under special conditions whiskers have been 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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Strain hardening

We had noted that stress to cause further plastic deformation (flow stress) increases with strain strain hardening. This happens at

Dislocations moving in non-parallel slip planes can intersect with each other → results in an increase in stress required to cause further plastic deformation

Strain Hardening / work hardening

One such mechanism by which the dislocation is immobilized is the Lomer-Cottrell barrier.

Dynamic recovery

In single crystal experiments the rate of strain hardening decreases with further strain after reaching a certain stress level

At this stress level screw dislocations are activated for cross-slip

The RSS on the new slip plane should be enough for glide

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)111(

]1 0 1[21

)111(

]0 1 1[21

)100(

]1 1 0[21

Formation of the Lomer-Cottrell barrier

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)100(

]1 1 0[21

)111(

]0 1 1[21

)111(

]1 0 1[21

+

Lomer-Cottrell barrier → The red

and green

dislocations attract each other and move towards their line of intersection They react as above to reduce their energy and produce the blue

dislocation

The blue

dislocation lies on the (100)

plane which is not a close packed plane

→ hence immobile → acts like a barrier to the motion of other dislocations

]1 1 0[

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Impediments (barriers) to dislocation motion

Grain boundary

Immobile (sessile) dislocations

► Lomer-Cottrell lock

► Frank partial dislocation

Twin boundary

Precipitates and inclusions

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

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Stress to move a dislocation & dislocation density

A 0

0

→ base stress to move a dislocation in the crystal in 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 10100 1014 0.5 10

COLD WORK ► ↑ strength

► ↓ ductility

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Grain size and strength

dk

iy 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]

Hall-Petch

Relation

Hall-Petch

constantsMaterial I

[MPa] k [MPa

m1/2]

Cu 25 0.11

Ti 80 0.40

Mild steel 70 0.74

Ni3

Al 300 1.70

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Photo courtesy: Dr. Eswar Prasad (Johns Hopkins University, 2013)Mg alloy (AZ31, 3% Al , 1% Zn)

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Grain size

101 10)2(645 nd

d → Grain diameter in meters n

→ ASTM grain size number

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Effect of solute atoms on strengthening

Solid solutions offer greater resistance to dislocation motion than pure crystals

(even solute with a lower strength gives strengthening!)

The stress fields around solute atoms interact with the stress fields around the dislocation to leading 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 solute more the size difference, more the hardening effect

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

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

As shown in the plots in the next slide, increased solute concentration leads to an increased hardening. 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 pure solid solution strengthening to harden a material can have a limited scope.

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Solute Concentration (Atom %) →

y(M

Pa)→

50

100

150

10 20 30 40

200

0

Solute strengthening of Cu crystal by solutes of different sizes

Matrix: Cu (r = 1.28 Å)

Be (1.12)

Si (1.18)

Sn

(1.51)

Ni (1.25)

Zn (1.31)

Al (1.43)

(Values in parenthesis are atomic radius values in Å)

Size effectSize difference

Size effect depends on:Concentration of the solute (c) cy

~

For the same size difference the

smaller atom gives a greater

strengthening effect

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y

Often produce Yield Point Phenomenon

Solute atoms ↑

level of

-

curve

X

By ↑

i

(lattice friction)

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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 interact with both edge and screw dislocations. Hence they give a higher hardening effect (per unit concentration) as compared to substitutional atoms which have (approximately) a spherical distortion field.

Relative strengthening effect of Interstitial and Substitutional

atoms

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Long range

(T insensitive)

Solute-dislocation interaction

Short range

(T sensitive)

Modulus

Long range order

ElasticSubstitutional → edge

Interstitial → Edge and screw dl.

Electrical

Short range order

Stacking fault

Mechanisms of interaction of dislocations with solute atoms

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The hardening effect of precipitates

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 of the system, whilst inclusions are external to the alloy system).

A pinned dislocation (at a precipitate) has to either climb over

it (which becomes favourable 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

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Dislocation Glide through the precipitate

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 the matrix (though it is usually higher as precipitates can be intermetallic compounds).

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

Glide of the dislocation causes a displacement of the upper part

of the precipitate w.r.t

the lower part by b

→ ~ cutting of the precipitate.

IncoherentcoherentPartially CoherentGrowthGrowth LargeSmall

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b

Precipitate particleb

Schematic views edge dislocation glide through a coherent precipitate

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Hardening effectPart of the dislocation line segment (inside the precipitate)

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 system specific). The important ones are listed in the tree below.

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

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Pinning effect of inclusions

Dislocations can bow around widely separated inclusions. In this

process they leave dislocation 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 ‘some sense’

similar to the Frank-Read mechanism).

The next dislocation arriving (similar to the first one), feels a repulsion from the dislocation 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

’.

Orowan

bowing mechanism

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Precipitate Hardening effect

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

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

Chemical Strengthening: arises from additional interface created on shearing

Stacking-fault Strengthening: due to difference between stacking-fault energy between particle and matrix when these are both FCC or HCP

(when dislocations are split into partials)

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 of dislocations passing through precipitates which have an ordered lattice

(Complete List)

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We had noted that strain rate can vary by orders of magnitude depending on deformation process (Creep: 10–8

to Explosions: 10–5).

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.

Strain rate effects

dd vb vd

→ velocity of the dislocations d

→ density of mobile/glissile

dislocations b → |b|

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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 lower stresses as compared to bond stretching (beyond ‘Y’).

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

till dislocations nucleate in the crystal.

In a UTT

why does the plot not continue along OY

(straight line)?Funda

Check

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