UNIT 2

Elements of applied physical metallurgy

UNIT 2aElements of mechanical metallurgy

The project of Unit 2

macroscopic object M

( )txF ,

system of(generalised)forces applied tothe object

Initial state

macroscopic objectM

state at time t

( )txF ,

The deformation of the object at time t can be descibed with the “displacement field”

( )txS ,

Our problem is to formalise the relationship between the input ( )txF ,

and the output ( )txS ,

mediated by the object M.

We can express this relationship with the operator α̂

Thus our project can be formalised as:

( ) ( ){ }txFtxS ,ˆ, α=

The operator α̂ can be specified with a model of the nature of M.

The typical hierarchy of models is represented by the following list:

1) Continuum, homogeneous, elastic (isotropic or anisotropic) (realm of continuum mechanics)2) Continuum, homogeneous, plastic (isotropic or anisotropic) (realm of continuum mechanics)3) Continuum, heterogenous elastic (realm of composite mechanics)4) Continuum, heterogenous plastic (realm of mechanical metallurgy)

- plasticity of perfect single crystals (heterogeneous owing to dislocations = line defects)- plasticity of single crystals with defects (heterogeneous owing to point & line defects)- plasticity of single-phase polycrystals (presence of grain boundaries & 1D and 2D defects)- plasticity of multi-phase polycrystals (different dimensions of grains of each phase)

5) Discontinuous (realm of fracture mechanics)

Of course: ( ) ( ){ } ( ) ( ) const : with ,ˆ, 1 ==⇒= ExεExσtxFtxS α

That, for a 1D problem, simplifies to: εσ E=( )εα E=2ˆCase (2) in 1D geometry is the usual uniaxial plastic flow curve:

Part 1: crystals and polycrystals

COMMENTS ON THE KEY POINTS OF THE FIRST COURSE IN METALLURGY

1.1) From the metallic bond according the jellium model to crystals- cation assembly and electron “jam”- Aufbau of an Alkaline (s electrons) metal (hinting at energy levels)- Lennard-Jones-type force for the dimer (see next slide) and extension to n-atom case- cohesion in metals

with the very crude (indeed) assumptions that: (i) the cores are fully screened, (ii) the electron density is low (n small and r1-r2>r-R) and (iii) the electron-to-core distance is ca. costant: r-R≅rs≅a/2 (with a internuclear distance),one can write:

Chapter 1 – (A very naïve account of the) Atomistic background

negative means that the Aufbautransformation is spontaneuos

(explanation on 2nd next slide)

r1

r1

r2

r2

pure electrostaticattraction

equilibrium positionelectrostaticrepulsion

r3

r3

r4

r4

trimer ⇒ 2D lattice tetramer ⇒ 3D lattice

Electrical energy:

electron-cationattractive (<0)

cation-cationrepulsive (>0)

electron-electronrepulsive (>0)

electron density: jam is a continuum again, jam is a continuum cations are discrete

n.b. 1: this is the energy keeping the metal together(Eel<0).n.b. 2: the shape of the cation assembly depends of Eel.

i.e.: the structure is determined by the separation of the ions.n.b. 3: elesticity is related to cohesive energy

(proof on next slides)

n.b. 4: Lennard-Jones potentials justify compressive behaviourn.b. 5: one can try and predict the UTS from Eel, but this yields a dramaticoverestimate ⇒ real metals fail in a different way (see below).n.b. 6: metal oxidation processes can be naturally incorporated into the jellium model

a

r

Elesticity is related to cohesive energy: (toy-)proof

Above, we have shown that the electronic energy of the metal crystal is:

If the crystal is strained in a lattice direction (called “axial”), leaving the coordinated onesunaffected, Eel energy becomes a function of the variable axial lattice spacing r.

In the usual way, one can define an electric force related Fel to the electric enegy Eel.

stress can be defined as:

the meaning of “F=-Fel” is that the stessing force balances the cohesion force

Now, we are interested in Young’s modulus EY, that can be defined as:

we then need dσ and dε. dσ can be derived by applying the 1st variation operator to the above expression for σ :

while, by definition:

eventually:

order-of-magnitudeestimate

Bravais’ lattices

Space tassellation with full coverage

Take-home message of analysis of mechanical properties in termsof perfect single crystal (without any kind of lattice defect):

Quantitative description of mechanical propertiesof crystalline materials requires

ingredients beyond perfect lattice geometry and cohesion energy

Project: provide a continuum descriptionincorporating parametrically information about defects

First approach to defect theory: the energetic scenario at a crystal surface

2.1 - POINT DEFECTS: self-defects & hetero-element defects

self

hetero

vacancy interstitial

interstitialsubstitutional

A better view of hetero-point defects(e.g. solid solutions)

Substitutional

Interstitial

Their location within the lattice can be understoodon the basis of an “Eel”-type of approach

e.g.:

Fe

Ni, Si

C, N, H, O, B

Thermal disorder brings about an equilibrium concentration of point self-defects.

Cu

“N” “V” “N” “S”

( ) [ ][ ]C

STK

SN

eq =

↔

Point defects (both self- and hetero-) can be viewed as:a “different” material in equilibrium with “normal” material

In an Arrhenius framework:

Eact: energy for a “jump” (go back to Eel)

As a “different” material, the defect can undergo “mass-transport”, typically by diffusion:

Complex defects, including both self- and hetero-point defects(aggragating in particular combinations)

Frenkel: one cation of a specific type (say A) is dislocated yielding an A-vacancy and an A-interstitial.

Case of a 2-cation (A and B) lattice

Schottky: coupled A-B vacancy.

B in B out

B generated (algebraically)

Knowing source (reaction) and mass-transport (flux) information in point defects,we can set up balance equations

St

Coutin +−=

∂∂ φφ

( )CSCDt

C +∇=∂∂ 2For: Fickian diffusion

& monomolecular reaction

The boundary between slipped and unslipped regions

Dislocation in a slip plane

Dislocation (ethymologically) line:atoms on half-plane α, displaced (slipped) 1 atomic distance in slip directionatoms on half-plane β, not yet slipped

Edge dislocation produced by slip in SC lattice.Dislocation lies along line AD, ⊥ to slip direction.

α β

Energetics: for a first approximation refer to “crystal of cubes”

2.2 - LINE DEFECTS (DISLOCATIONS): (a) GEOMETRY

First of two basic types

Screw dislocation produced by slip in SC lattice: slip has occurred overarea ABCD. Dislocation lies along line AD, // to slip direction.

Atomic arrangement around screw dislocation.

Second of two basic types

Macroscopic deformation of a cube produced by glide of:edge (above) and screw (below) dislocations.

n.b.: the same type of deformation is obtained by glide of both types of dislocations.

Two basic types (in fact: limiting cases)

Observing dislocations

Slip lines: (Cu)

200 µm

Etch pits on slip bands in brass Slip lines: (Cu)

20 µm

Hexagonal network of dislocations in AgCl by decoration technique

Network of dislocationsin cold-worked Al (TEM)

20 µm 2.5 µm

Elementary dislocation energetics (below, more details)

(a) Atom movements in the neighbourhoodof a dislocation in slip

(b) Movement of an edge dislocation

Energy changefrom unslipped to slipped state

Model: plastic deformation as transition from unslipped (higher-energy owingto accumulation of elastic energy) to slipped (lower-energy owing to releaseof elastic energy) state. Process opposed by energy barrier ∆E, located at the transformation Interface (that can be interpreted as the dislocation).

Stages in growth of slipped region

Assuming a sinusoidal potential profile (resulting from a L-J-type model, recall τmax estimate for perfect lattice):

↓⇒↑=∆

∆Eb

a

plane slipin atomsbetween distance

planes slipbetween distance

Shear stress required to move a dislocation:

−≈b

aexpτ

b

a

Dislocations can be characterised/quantified by Burgers vectors

i) Consider and atom-to-atom circuit around the slip feature.ii) Join each atom along the circuit with a vector:iii) The vector sum of such joining vectors is the Burgers vector:

Special cases:i) No slip:ii) Edge dislocation:iii) Screw dislocation:

edge screw

notable bearingon disclocationmovement throughlattice

Burgers vector notation

Direction indicated by direction indices and magnitude resulting from coefficient and the direction indices as illustrated below

Premise - direction indices : traslations parallel to the crystal axes yieldingthe desired direction

[a1 a2 a3 c]

exam

ples

ofhe

xago

nall

attic

e

e.g.

in a cubic crystal means that the Burger vector has components of2a/3, -a/3 and a/3 along the [100], [010] and [001] directions, respectively,with a lattice parameter and magnitude:

x

z

y

example in cubic lattice

Specific types of movement for edge and screw dislocations

EDGE DISLOCATION: glide in slip plane,but:under appropriate conditions it can move out of the slip plane onto a parallel planedirectly above or below it: climb .

diffusion of vacancyto edge dislocation

diffusion-controlled process

dislocation climbs up one lattice spacing

dislocation line

Burgers vector

SCREW DISLOCATION: glide in slip plane and cross slippingSince the line of a screw dislocation and b are //, b not define a specific planeas with the edge dislocation (where b is ⊥ to the dislocation line): Hence, to a screw dislocation, all directions around its axis look the same and it can glide on any plane as long as it moves // to its original orientation.

part of a screw dislocation line AB cross slippingfrom the primary slip plane PQ into the plane RS

Reactions between dislocations (see above “limiting cases”)

Couples of dislocations can combine and form a third dislocation.The Burgers vector of the product dislocation will be the vector sum of theBurger vectors of the reacting dislocations.

The reaction takes places if it is energatically favourable, i.e. if it lowers theenergy of the system: below we shall prove that:

Therefore the reaction is energetically favourable if:

The notion of disclocation reactions puts on a quantitative basis the interpretation ofedge and screw dislocations as limiting cases of actual dislocations that will exhibitmixed characteristics.

Since dislocations describe slip and slip is a property of a crystalline object, preferential slip planes and directions for dislocations will be present within a crystalthat is perfect apart from the presence of dislocations(the story will be different in the presence of hindrances to slip), depending on the specific crystal geometry.

Since (see above):

high-density planes will tend to be preferential slip planes and high-density directions, slip directions. The ensemble of the preferential slip planes (i.e. the numberof slip planes multiplied by the number of slip directions) for a particular crystal isdenominated “slip system” peculiar to that crystal structure.

−≈b

aexpτ density↑ ⇒ b↓ ⇒ τ↓

A practically important example of reaction between dislocations : “partial dislocations” in FCC crystals

In FCC crystals slip occurs on {111} planes ({111} denotes the ensemble of all(111)-type planes) in ⟨110⟩ directions (⟨110⟩ denotes the ensemble of all [110]-typedirections).

Consider the specific case of displacement:

A dislocation with this Burgers vectorcan dissociate into two partial dislocations,according to the reaction:

[ ] [ ] [ ] 22

222

35.01012

07.01216

1126

aa

aaa =<=+ dissocation reaction:

energetically favourable

Partial dislocations in FCC crystals lead to formation of “st acking faults”

If a single (a/6)⟨112⟩ partial dislocation passes through an FCC crystal, …

⟨112⟩… it leaves behind a region in which the sequence of stacking of the close-packed{111} planes does not correspond to the normal FCC lattice: In normal stacking, the third plane is overneither the first, nor the second one.A (a/6)⟨112⟩ partial dislocation changes the position of the third plane so that it ends updirectly over the first one: this is in fact the HCP stacking of {111} planes and, sincethe natural arrangement in FCC isdifferent, this fault yields a fault energy

Stacking of atomically compact planes and faulted structu res

Perfect FCC packing

Scheme of deformation by twinning

Digression: DEFORMATION BY TWINNING

Stacking fault in FCC Perfect HCP packing

Twin in FCC

Dissociation of a dislocation into two partial dislocations in an FCC crystal

Looking down ion (111) along [1 1 -1]:

Again: Partial dislocations in FCC crystals lead to “stacking fau lts”

Perfect dislocation linehaving the full slip vector b1.

screw dislocation edge dislocation

Dealing with defects in a continuum frameworkAlias: “energetics of dislocations”

Fundamentals of Continuum DescriptionBalance of Bulk and Surface Forces

A) No structure (no dependence on orientation of object) → hydrostatic equilibrium

bulk force (e.g. gravitational)

p∇r

grρ

( )xpr

gprr

ρ=∇ A

B) With structure (dependence on orientation) → equilibrium of deformation of solids

resultant of bulk forces

σ∇r

fr

( )norientatio surface;xpr

ifr

frr

=∇σ BIn general: a system of equations; structure ⇒ system

C) Just one unknown in balance equation → displacement ur

( ) ( )ufurrrr

=∇σSimplest option: linear elasticity

ii

i

x

u εdef

=∂∂ ( )( )

i

iijiijiii x

uEEu

∂∂== εεσ

( ) ( )iii

ii uFt

uuf

rr+

∂∂=

2

2

ρ

inertial static

C

D) Approximations: (i) elastostatic, (ii) without bulk forces

D

02

2

≡+∂∂= i

ii F

t

uf

rrρ

(i) (ii)

3 equations in6 unknowns

z

yzy

xzxyx

στσττσ

constitutive equations: links among strains due to material properties (ν)

{ }( )νσσ ;ˆklij O=

F

E) 6×6 PDE system & ways to solve it

{ }( )νσσ ;ˆklij O=

F) Ways to solve this PDE system: a convenient transformation of unknowns, the Beltrami tensor

E

ΦΦΦΦΦΦΦΦΦ

zzzyzx

yzyyyx

xzxyxx

such that:

G G) Ways to solve this PDE system: a special case of the Beltrami tensor, the Maxwell stress function

ΦΦ

Φ

zz

yy

xx

00

00

00

H H) Ways to solve this PDE system: a special case of the Maxwell stress function, the Airy stress function

( )

=Φ yxfzz ,00

000

000

Suitable for two-dimensional problems.

PREMISE: ELEMENTS OF ELASTICITY THEORY

The key problem of elasticity is:to calculate the deformation field (or, via the constitutive equations, the stress field)

prevailing in an (elastic) body, resulting from the application of a system of forces.

This is, of course, the key energetic problem with dislocations.

2.2 - LINE DEFECTS: (b) ENERGY

dislocation

compression zone

tension zone

Consider a column of water in hydrostatic equilibrium . All the forces on the water are in balance and the water is motionless. On any given drop of water, two forces are balanced. The first is gravity, which acts directly on each atom and molecule inside. The gravitational force per unit volume is ρg, where g is the gravitational acceleration. The second force is the sum of all the forces exerted on its surface by the surroundingwater. The force from below is greater than the force from above by just the amountneeded to balance gravity. The normal force per unit area is the pressure p. The average force per unit volume inside the droplet is the gradient of the pressure, so the force balance equation is:

i) 1st premise: Hydrostatic equilibrium

A

The momentum balance equations (in particular their static version) can be extendedto more general (than fluids) materials, including solids. For each surface with normal in direction i and force in direction j, there is a stress component σij. The nine components make up the Cauchy stress tensor σ, which includes both pressure and shear. The local equilibrium is expressed by:

ii) 2°Premise: equilibrium of deformable solids

where f is the body force and ∇⋅σ represent surface forces.

B

iii) The general problem of linear elasticity

C

C bis

In the case of interest only surfaceforces are present, thus Fx,Fy,Fz=0

iv) The linear elastostatic problem without body forces and with an isotropic homogeneous medium in stress formulation“Stress formulation” means that the surface tractions are prescribed everywhereon the surface boundary

a) Equation of motion

c) Constitutive equations

the Beltrami-Michell compatibility equations

Combining the strain-displacement equations and the constitutive equations, one gets:

D

n.b.:

The Beltrami-Michell compatibility equations in engineering notation

E

In the case of interest, the linear elasticity problem thus reduces to:

Equilibrium equations:

Compatibility equations(i.e. strain-displacementequations, combined withconstitutive equations):

E bis

It can be shown hat a complete solution to the equilibrium equations may be written as:

where Φmn is an arbitrary second-rank tensor field that is continuously differentiableat least four times, and is known as the Beltrami stress tensor.

The Maxwell stress functions are defined by assuming that the Beltrami stress tensoris restricted to be of the diagonal form:

whence:

F

G

The Airy stress function is a special case of the Maxwell stress functions, in which it is assumed that A=B=0 and C is a function of x and y only. This stress function can therefore be used only for two-dimensional problems.

The stress function C is customarily represented by φ

In polar coordinates the expressions are:

H

NOW WE HAVE ENOUGH BACKGROUND TO GO BACK TO DISLOCA TION PROBLEMSa) Energetics of a screw dislocation.

Consider the right-handed screw dislocation with the geometry depicted below,and imagine that the cylindrical region is a portion of an infinite continuum:

In cylindrical coordinates, the displacements are:

uz(θ=0)

uz(θ=2π)

Plugging the diplacement information into the elastostatic equations, we get:

Using the constitutive equations (with µ denoting the shear modulus) on getsthe stress field of a screw dislocation:

The elastic energy of a screw dislocation can be computed as follows:

the strain field of a screw dislocation.

dr

b

Elastic energy per unit length: ( ) ( )rdArUR

r

z

o

⋅= ∫ ϑσ2

1

r

b

( ) ( )

( )

222

ln88

22

1

2

1

br

Rb

r

drb

drbr

rdArU

o

R

r

R

r

z

R

r

z

o

o

o

≈⋅=⋅

=

⋅⋅

=⋅=

∫

∫

∫

πµ

πµ

σ

σ

ϑ

ϑ

( ) ( ) drb

rArb

rA ⋅≅⇒⋅≅

2d

2

Comments on the “core cutoff”

( ) ( ) ∫∫ ≈⋅=R

r

R

r

z

oor

drrdArU ϑσ

2

1

core cutoff(avoids core divergence)

ro~atomic scale (e.g. ro=b/r)

b) Energetics of an edge dislocation.

Consider the edge dislocation with the geometry depicted belowand again imagine that the cylindrical region is a portion of an infinite continuum:

The displacements are:

This is a plane-strain problem, that can be attacked with the Airy function formalism.

The appropriate Airy function is:

Plugging this Airy function into the equilibrium and compatibiilty equations:

( )

( )

( ) r

b

rr

r

b

r

r

b

rrr

r

rr

θνπ

µθϕσ

θνπ

µϕσ

θνπ

µθϕϕσ

θ

θθ

cos

12

1

sin

12

sin

12

11

2

2

2

2

2

⋅−

=

∂∂⋅

∂∂−=

⋅−

−=∂∂=

⋅−

−=∂∂⋅+

∂∂⋅=

Whence the axial stress can be deduced:

Hence, a hydrostatic stress exists around an edge dislocation:

This can be expressed more eicastically in cartesian coordinates:

That is plotted below:

Elastic energy per unit length: ( ) ( )rdArUR

r

z

o

⋅= ∫ ϑσ2

1

b

b·R/R=b

δδδδ=b·r/R

( ) ( ) drR

bLrAL

R

rbLrA ⋅≅⇒⋅=⋅≅ d δ

L

( ) ( ) ( )

( ) ( ) ( ) o

R

r

R

r

r

R

r

r

r

R

R

Lb

r

dr

R

Lb

drR

bLrrdArU

o

oo

ln14

0cos14

2

1

2

1

22

⋅−

=⋅=⋅−

=

⋅⋅=⋅=

∫

∫∫

νπµϑ

νπµ

σσ ϑϑ

for a crystallite exhibiting the shape of a right cylinder 2R=L:

( )2

2

ln12

br

RbU

o

≈⋅−

=νπ

µ

Mixed dislocations, compound Burgers vectorsand the notion of “dislocation loop”

Vector sum of all (local) bspure screw

pure edge

w: pure edge >0,y: pure edge <0x: pure right-handed screwz: pure left-handed screw

pure screw, can cross-slip

Cross-slip in a FCC crystaldouble cross-slip

during cross-sliponly screw component has moved

Dislocation loop lying in a slip plane

ds

dl

b

When an external force of sufficient magniture is appliedto a crystal, the dislocations move and produce slip.

Here we discuss the simple case of a dislocation linemoving in the direction of the Burgers vector under theinfluence of a uniform shear stress τ.

Let an element of dislocation line ds be displaced in thedirection of b, taken here normal to ds, by an amount dl:this displacement involves a fraction of the slip plane A, amounting to:

A

dlds ⋅

corresponding to a fractional slip of:

bA

dlds ⋅

⋅

2.2 - LINE DEFECTS: (c) FORCES ON DISLOCATIONS

Whence we can define a force F acting on the dislocation line per per unit length:

bdlds

dWF τ=

⋅=def

Since the force Fslip plane applied to the slip plane is: AF τ=plane slip

the work dW expressed in the fractional slip is:

( )dldsbbA

dldsAdW ⋅⋅=⋅

⋅⋅= ττ

Extension to case of general orientation of Burgers vector

( )σξ ,,bfFrrrr

=dl

ds

INTERACTIONS OF DISLOCATIONSThe force equations developed in the previous few slides allow to estimate interactionforces among dislocations. Here we shall develop an explicit expression for the simplecase of parallel disocations. More complex cases can be handled in a similar way, though at the expense of notably larger computational effort.

We condider the case of the following couple of // edge dislocations A and B:

The glide force on B, caused by the presence of A (i.e. the force experienced by B, located at a distance λ from A) in the stress field τA of A is:

Note that:

( ) BABA

BAB bbbb

bF ⋅≈⋅−

==λνπ

µτ12

For an arbitrary dislocation in a deformed crystal the determination of the energy ∆∆∆∆E/Lper unit length is a most formidable task. However, a rough approximation can be made.

Since like-sign dislocations repel and opposite-sign ones attract,a dislocation tends to be surrounded by opposite-sign ones at an average spacing λ.

But: a pair of dislocations of opposite sign are sources of equal and opposite stress fieldsthat tend to cancel one another over distances greater than λ.

As a result, the strain field generated by a dislocation tends to be limited to a radius R≈λ.

Moreover, λ can be estimated as:

Where ρ is the dislocation density (m/m3).

The line tension has units of energy per unit length (i.e. N) and it is the 1D analogueof 2D surface tension.

The line tension Γ produces a forcetending to straighten the dislocationline, that will remain curved only ifthere is a shear stress τ whichproduces a force F=τb on the dislocation resisting the line tension.

dislocation line

Fshear

We have proved that the force per unit lengthF acting on a dislocation b owing to the applicationof shear τ is:

bdlds

dWF τ=

⋅=

Therefore the outward force acting on the dislocation element ds is:

dsbdsF ⋅=⋅ τ

Deduction of the shear force term in balance

n.b.: [ ] NFdsF ==⋅ shear

LINE TENSION

dislocation line

Fline tension

The angle subtended by the dislocation element dsis dθ:

Deduction of the line tension term in balance

R

dsd =ϑ

The inward force on the dislocationline due to line tension is:

ϑϑd

dF ⋅Γ≅

Γ=2

sin2 tensionline

Balance of forces establishes the equilibrium:

R

dsddsbFF ⋅Γ=⋅Γ=⋅⇒= ϑτ tensionlineshear

whence: bbgbR

µµτ 7.0≈≈Γ=

taking Γ ≅ elastic energy of dislocation per unit length

On the one hand, we have proved above that a force Fon disl on a dislocationcan be written as:

In the case of small bow-out, the force on a dislocation segment of length λcan be expressed as (see Figure):

Dislocation motion, in practice, becomes hindered by a range of obstaclesand break-away from obstacles becomes rate-controlling.

The simplest way to consider the effect of obstacles, is that they pin dislocationsand cause their bow-out.

2.2 - LINE DEFECTS: (d) MOTION IN IMPERFECT CRYSTALS

On the other hand, in the line tension approximation, the force Fline acting at the pinning point is:

φ

But the force on the dislocation Fon disl is transmitted to the pinning pointsvia line tension forces Fpin acting on the pinning points, thus:

Fline

Fpin

( )2

cos2

cos2

cos2

cos 2linelinepin

φµφµφτφ ⋅=⋅⋅=⋅=⋅= bgbbgbFF

If there exists a critical stress σc for breakaway of the dislocation from thepinning point, for:

the bow-out is stable and the dislocation will be held pinned.

A SELECTION OF STRENGTHENING MECHANISMS

i) Solid solution strengthening (by alloying: single atoms of alloying element)

ii) Dispersion strengthening (by alloying: crystalline domains of alloying element)

iii) Work hardening (by cold working: increases dislocation density via “sources”,see below)

iv) Grain refining: multiplicity of grains, related to changes in orientation ofcrystal planes, higher energy required toactivate slip systems.Hall-Petch relationship:

D

baRs +=

Grain A

Grain

B

Grain boundary

Slip plane

INTERSECTION OF DISLOCATIONS

Since even well-annealed crystals contain many dislocations , it is quite common thata dislocation, moving along its slip plane, will intersect other dislocations crossingthat particular slip plane. Dislocation intersections play an important role in the strain-hardening process.

The intersection of two dislocations gives rise to a sharp break – a few atom spacingsin length – in the dislocation line, that can be of two types:a jog : a sharp break in the dislocation, moving it out of the slip plane;a kink : a sharp break in the dislocation line, which remains in the slip plane

Example 1: intersection of two ⊥⊥⊥⊥ edge dislocations with ⊥⊥⊥⊥ Burgers vectors

Dislocation XY moving on plane PXY cuts through dislocation AD moving on plane PAD.Their intersection produces the jog PP’ in dislocation AD.

Jog // to b1 has length |b1| and has Burgers vector b2,since it ∈ to the dislocation line APP’D .

The jog resulting from the intersection of two edge dislocations has an edge orientationand therefore it can glide with the rest of the jogged dislocation A(PP’)D .

Dislocation XY “slides”the half crystalcontaining dislocation AD

Example 2: intersection of two ⊥⊥⊥⊥ edge dislocations with // Burgers vectors

Both dislocations are jogged: PP’, QQ’.

Before intersection After intersection

b1

b2

b1

b2

b1

b2

Effect of blue on red:

|b2 |

half crystal slidby blue dislocation

half crystal unslid

b1

b2

b1

b2

Effect of red on blue:

half crystal slidby red dislocation

half crystal unslid

|b1|

b1

b2

Joint effects of“red on blue” & “blue on red”:

|b1||b2|

b1

b2

Both jogs have a screw orientation (// to b)and are called kinks .

Example 3: intersection of a screw dislocation with an edg e dislocation

Produces jogs with edge orientations in both edge and screw dislocations.

Example 4: intersection of two screw dislocations

Produces jogs with edge orientations in both screw dislocations.

SYNOPSIS OF NATURE OF “STEPS” FROM INTERSECTIONS OF ELEMENTARY DISLOCATIONS

edge screw

edge

screw

edge

edgeedge

⊥b: edge//b:

screw

Impact of dislocation intersection on plastic deformat ion

i) Jogs produced by intersection of two edge dislocations of either orientationare able to glide readily , because they lie in the slip planes of the original dislocations.The only difference is that instead of gliding along a single plane, they do so overa stepped surface .

Movement of jogon Screw dislocation:constrained to plain AA’BB’

The only way the screw dislocation can slip to a new position such as MNN’O taking its jog with it, is by climb .

ii) Jogs produced by intersection involving at least one screw dislocation all haveand edge orientation. Since an edge dislocation can glide freely only in a planecontaining its line and its Burgers vector, the only way the jog can move by slipis along the axis of the screw dislocation .

Motion of screw dislocations impeded by jogs .

Climb of jogged screw dislocations

Section of edge jog in a screw dislocation gliding on the plane of the slide in direction BD. Edge climb (alias: “non-conservative” movement) must remove or add matter.

generate interstitials

climb(A→B) + slip(B→C) ⇓

forms row of vacancies

climb(A→D) + slip(D→E) ⇓

forms row of interstitials

generate vacancies

vaca

ncy

inte

rstit

iala

tom

Under applied shear stress acting in the slip direction, the jogs will act as pinning points . The dislocation will thus tend to bow out between jogs.

Vacancy jogs and interstitial jogs tend to annichilate mutually during dilocation motion, ⇒ a net concentration of jogs of the the same type will tend to accumulate .Because of their mutual repulsion , they will tend to spread out along the dislocationline at approximately evenly spaced intervals.

At some critical shear , non-conserevative climb will occur and the dislocations will move, leaving behind a trail of vacancies or interstitials.

On the one hand, an estimate of the energy to form a vacancy or interstitial from a jog is:

( ) ( ) ( ) bGb ⋅≈⋅≈ 2defect oflength lengthunit per energy defect form energy to

On the other hand, the energy to move the jog forward by the length of a defect is:

( ) ( )lbbWdldsbdW ⋅⋅≈⇒⋅⋅=

⋅⋅≈ττ

jogsbetween spacingdefect oflength forceenergy

whence:

l

GblbGb ≈⇒≈ ττ 23

The shear stress required to generate a defect and move the dislocation by merelymechanical means (i.e. without thermal activation of climb):

2.2(e) – DISLOCATION SOURCES: the Frank-Read case

Suppose that the pinned dislocation is ca. “flat”, i.e. φ/2≅π/2.Since:

maxσσ >

If the stress is increased to σmax, the semi-circular equilibrium position is reached.

If the stress is increased further, beyond the equilibrium value:

the equilibrium conditions no longer holds and the dislocation keeps growingunder the force:

0) :(i.e. 2

cos max ==⇒⋅= φλµσφ

λµσ bgbg

bF τ=

but, if the pinning conditions still hold, i.e.

the dislocation will start spiralling around the pinning pointsand when the two lobes D and D’ touch one another,being equal and opposite dislocations, will annihiliate generating acomplete, unpinned loop, while the portion contrained by the pinningpoints can again outbow and repeat the process.

lbFrrr

⊥=τin the case:

D D’

Dislocation pile-ups and “continuous dislocations”

slip plane

⊥⊥⊥⊥shear: τ

Fother dislocations

Fobstacte

Dislocations frequently pile up on slip planesat obstacles such as grain boundaries andsecond phases.

yieldingon other sideof obstacle

crackingof obstacle

back stress

distribution ofdislocationsalong slip plane

GbL

n τ≈

giant dislocation: nb

continuous dislocation: b·dn