Module #13Module #13

Elastic Properties of DislocationsStacking faults

READING LISTDIETER: Ch. 5, Pages 160-168

Ch. 4, Pages 205-222 in Meyers & Chawla, 1st ed.

IntroductionIntroduction• Dislocations are line defects that distort the perfect crystal lattice.

• This lattice distortion produces an elastic stress field inside the crystal.

► The magnitude of the elastic stress field can be estimated using elasticity theory. The next viewgraph provides a general explanation of why.

► We can in turn use this information to determine the energy of the dislocation, the force it exerts on other dislocations, and its energy of interaction with other defects.

• Interactions between the stress fields (i.e., the distorted regions) around dislocations (and those of other defects) ultimately determine the mechanical properties of the lattice.

Strain energy of a dislocationStrain energy of a dislocation• The energy of a dislocation comes from the distortion associated with

the displacement of atoms from their equilibrium positions.

A few lattice spacings from the core, say at a distance ro, we can model things using elasticity theory as the displacements are very small. Hooke’s law applies here. ro is called the cutoff radius. It typically has a value near b.

d

b

Slip plane

Displacement u

B

A

• Near the center of the dislocation (i.e., along the dislocation line), the displacements are too large to be calculated with elasticity theory. Hooke’s law does not apply here. This region is called the dislocation core.

Tensile

Compressive

x

y

Strain EnergyStrain Energy• Strain increases the internal energy of a body. The

elastic strain energy per unit volume is given by:

• Thus, for an element of volume dV, the elastic strain energy can be expressed as:

• We can utilize this expression to estimate the elastic strain energy for a dislocation by incorporating the components of stress and strain that surround the dislocation.

, , , ,

Elastic strain energy 1Volume 2 ij ij

i x y z j x y z

e

elastic, , , ,

12 ij ij

i x y z j x y z

dE dV e

• A crystal containing a dislocation is not at its lowest energy state. There is a strain energy term (Etotal) that must be added to the lattice energy of the crystal.

• The distortions caused by the presence of edge and screw dislocations are simulated on cylindrical diagrams to the right. The diagrams depict each type of dislocation as a distorted cylinder.

• Using these diagrams, we can calculate the strain energy on the basis of elasticity theory. This is detailed in Hull and Bacon [1] and in Weertman and Weertman [2].

[1] D. Hull and D.J. Bacon, Introduction to Dislocations, 4th ed. (Butterworth-Heinemann, Oxford, 2001), pp.62-72.[2] J. Weertman and J.R. Weertman, Elementary Dislocation Theory(Oxford University Press, Oxford, 1992), pp. 32-40,

L

[Hull and Bacon ]

L

[Hull and Bacon ]

Screw DislocationScrew Dislocation

• Elastic stresses around a screw dislocation:

• All shear components parallel to the dislocation line.

2 2

2 2

sin2 2

cos2 2

0

xz zx

yz zy

xx yy zz xy yx

Gb y Gbrx y

Gb x Gbrx y

[Hull and Bacon ]

Edge DislocationEdge Dislocation• Elastic stresses around an edge dislocation:

• Stress field has dilatational and shear components.

2 2

22 2

2 2

22 2

2 2

22 2

2 2

32 1

2 1

2 1

10

xx

yy

xy yx

zz xx yy

xz zx yz zy

y x yGbx y

y x yGbx y

x x yGbx y

Gb yx y

[Hull and Bacon ]

Schematic representation of the stress field about an edge

dislocation

Stress fields around an edge dislocationStress fields around an edge dislocation

Compression

Tension

+Shear-Shear

y = -xy = x

θ x

y

• Etotal can be divided into two parts:

Etotal = Ecore + Eelastic (strain energy).

• The core contribution (Ecore) is difficult to calculate. It is estimated to have a value of ≈0.5 eV/plane threaded by a dislocation.

• From elasticity theory, the elastic strain energy per unit length of dislocation is given by:

2 2 22

2 2 22

2 2 2 2

ln4 4 4

ln 1.54 (1 ) 4 (1 ) (1 )

sin cos4 (1 ) 4

o o

o

R R

r ro

R

ro

Gb dr Gb dx Gb RE E Gbr x r

Gb dx Gb R GbE E Gbx r

Gb GbE

elastic

SCREW SCREWelastic elastic

EDGE EDGEelastic elastic

MIXED lno

Rr

• For all of the different types of dislocations, the value of Eelastic depends on ro and the crystal radius R through a slowly varying logarithmic term. This is illustrated in the figure on the next page.

2

2

0.5 1.0

E Gb

bOR

E

elastic

elastic

where

( ) ( ) ( )E screw E edge E mixed elastic elastic elastic

• The value of Eelastic depends weakly on ro and the crystal radius R through a slowly varying logarithmic term outside of the core region.

• We can approximate Eelastic for any type of dislocation as:

Elastic strain energy per unit length of dislocation

[Hull & Bacon]

• One of the important consequences of this relationship is that it allowsfor determination of whether or not it is energetically feasible for two dislocations to react/combine to form another.

• This is generally known as Frank’s Rule.

2

0.5 1.0E Gb

elastic

where

2 2 21 2 3

2 2 21 2 3

2 2 21 2 3

b b b

b b b

b b b

Yes if

No if

No net energy change if

IMPORTANTElastic strain energy Gb2b1 b2 b3

+ =b1 b2

b3

Φ

b1 + b2 = b3

Dislocations of unit strengthDislocations of unit strength

• The Burgers vector of a crystal must also connect one equilibrium lattice position to another.

• Crystal structure determines Burgers vector.

►If Burgers vector = one lattice spacing then the dislocation is of unit strength.

Dissociation of dislocationsDissociation of dislocations

• Because of energy considerations it is possible for somedislocations with strengths greater than or equal to unity to dissociate (split) into shorter segments.

• This is occurs in some close-packed (i.e., FCC and HCP)crystals such as where equilibrium positions are not the edges of the unit cell.

• Can have serious implications on plastic deformation.

1 2 3

2 2 21 2 3

2 2 21 2 3

b b b

b b b

b b b

Will the reaction occur?

Yes if

No if

IMPORTANTElastic strain energy Gb2

• In this example, the separation into partial dislocations is energetically favorable. There is a decrease in strain energy.

• The separation produces a stacking fault between the partials.

Dissociation of a unit dislocation in an FCC crystalDissociation of a unit dislocation in an FCC crystal

2

2 2 21 2 3 1 2 3

2 2

Reaction will occ

2 1 1 1126 6

6 6

ur if

1012

2

o o o

o oo

b b b b b b

a

a

a a

a a

2101

2 2o oa a

b

6211

6 6o oa a

b

6112

6 6o oa a

b

A BC

A

B

C

Forces on dislocationsForces on dislocations• When the stresses applied to a crystal are large enough,

dislocations will move resulting in plastic deformation. Deformation will occur via slip or climb.

• The applied load induces work on the crystal during dislocation motion.

• In turn, the dislocation responds to the applied stress as though it were experiencing a force. The force is given by:

We will consider the glide/slip force now and the climb force later.

Work req'd. to move FDistance moved

Glide force on a dislocationGlide force on a dislocation• If we consider a glissile dislocation () under an applied

shear stress (). The dislocation has a Burgers vector b.

• When a segment of the dislocation line dlmoves forward a distance ds, the regions of the crystal above and below the slip plane are displaced relative to each other by a distance that is equivalent to the Burgers vector, b.

• The average shear displacement of the crystal due to glide of segment dl is thus:

• where A is the area of the entire slip plane.

Displacement dsof an element dlin its slip plane used to determine the glide force.

b

ds

dl

ddxA

ds b

• The external force ( f ) acting uniformly on this small area (due to ) is:

• The work done when the small element dl glides forward is therefore:

• Since force = work divided by the distance over which it is applied (ds), the glide force per unit length on a dislocation (dl), FL, is:

• This force, which is the result of externally applied stress, acts normal to the dislocation line along its length. This is the Peach-Koehler equation. It is often given as:

Forces on dislocations (2)Forces on dislocations (2)

is the shear stress in the slip plane resolved in the direction of b.

ds d ddW fdx A d dAA

sb b b

force displacement

f A

Ldsf dW dW

d d dbF

A

Ld b dF b F A more advanced treatment is provided in the following five pages

General Peach-Koehler equation• Most dislocations are mixed. Mixed dislocations are oriented such that the tangent

to the dislocation line (i.e., the sense vector , t, or l) is neither parallel or perpendicular to the Burgers vector. In this case:

• In 3D space, all will have components parallel to the x,y,z axes. We also need to account for components of stress that are parallel to the x,y,z axes.

• If we let g = σijb, then:

where FL is the force per unit length of dislocation. This is essentially F/L for a straight dislocation where L is the length of the dislocation line.

This general form of the Peach-Koehler equation is used to calculate the magnitudes of the forces on and the forces between dislocations.

and x x xx y xy z xz

y x yx y yy z yz L x y z

z x zx y zy z zz x y z

g b b b i j kg b b b F g g g gg b b b

t̂

x y zb b i b j b k

B*

Forces exerted on a straight Forces exerted on a straight screwscrew dislocationdislocation• Elastic stresses around a screw dislocation:

• All shear components acting parallelto the dislocation line.

• Thus:

2 2

2 2

sin2 2

cos2 2

0

xz zx

yz zy

xx yy zz xy yx

Gb y Gbrx y

Gb x Gbrx y

0 0 0 00 0 0 0

0 0 0

xz xz x xz

screw yz yz y yz

zx zy xz yz z

g bg bg

and

x

y

z

Rr

B

B

b

0b

[001]b

B*

,b

• Taking the cross product of (b) and the line sense (ξ) we get:

This tells us that only τxz, and τyz can exert a force on this dislocation and that the force acts perpendicular to the dislocation line along its length.

• Consider a straight screw dislocation as illustrated below.

General Peach-Koehler equation – cont’d

0 0 1

L xz yz zz yz xz x y

i j kF b b b b b i b j F F

• For this screw dislocation: b = [001] and ξ = [001]. Therefore,

001

xx xy xz

yx yy yz

zx zy zz

xz yz zz

b b

b

$

x

y

z

$

yzxz

NOTE: This also proves that two shear stresses act upon this screw dislocation.

B*

• Elastic stresses around an edge dislocation:

• Thus:

2 2

22 2

2 2

22 2

2 2

22 2

2 2

32 1

2 1

2 1

10

xx

yy

xy yx

zz xx yy

xz zx yz zy

y x yGbx y

y x yGbx y

x x yGbx y

Gb yx y

Forces exerted on a straight Forces exerted on a straight edgeedge dislocationdislocation

0 00 0

0 0 0 0 0

xx xy xx xy x xx

edge yx yy xy yy y yz

zz zz z

g bg bg

and

Rr

A

A

x

y

z

b

0b

[100]b

B*

• Taking the cross product of (b) and the line sense (ξ) we get:

This tells us that only τxy and σxx can exert a force on this dislocation and that the force acts perpendicular to the dislocation line along its length. Fx is the glide force (+y direction). while Fy is the climb force (in –y direction).

• Consider a straight edge dislocation as illustrated below.

General Peach-Koehler equation – cont’d

0 0 1

L xx xy xz xy xx x y

i j kF b b b b b i b j F F

x

y

z

xy

b

• For this edge dislocation: b = [100] and ξ = [001]. Therefore,

100

xx xy xz

yx yy yz

zx zy zz

xx yx zx xx xy xz

b b

b b

xx

xyx y

NOTE: This also proves that one normal stress and one shear stress act upon this screw dislocation.

B*

Bending of Dislocations Bending of Dislocations -- 11• In addition to the Peach-Koehler force,

dislocations develop a line tension.

• Line tension develops because the strain energy of a dislocation is proportional to its length.

• Any increase in length, increases the strain energy of the dislocation in proportion to its length.

Bending of Dislocations Bending of Dislocations -- 22►Therefore, dislocations bend and/or always try to

straighten out to reduce their lengths.

• Straight dislocations are shorter and thus have lower strain energies than curved dislocations.

• The line tension, Γ, which is the increase in energy per unit increase in dislocation length can be expressed as:

2Gb

• Consider the curved dislocation illustrated below.

• A specific shear stress, τo, is required to overcome Γ and maintain the radius of curvature, R.

• τo is maximum when dθ = 90°

Line tension on dislocations (1)Line tension on dislocations (1)

sin d

A

R

R

0

Γ

Γ

b

dl2 ddR

od

Γ d

Line tension on dislocations (2)Line tension on dislocations (2)• The outward force (fout) due to applied stress τo is:

• The outward force is opposed by a line tension force acting along OA. It is given by:

• Curvature will be maintained if:

2out o of b d Rd

)

2 sin( )2

tensionf dd

(for small values of

2 2tension out

o

o

f fd bRd

GbbR R

[Peach-Koehler equation (FL = σb)]

A

R

R

0

Γ

b

dl2 ddR

od

• This equation provides an adequate approximation for most dislocations (it does, however, make some generally invalid assumptions).

• We will use a version of this when we discuss strengthening via the immobilization of dislocations.

• See Hirth and Lothe, Friedel, or another text on dislocations for more detailed treatment.

oGbR

Line tension on dislocations (3)Line tension on dislocations (3)

[Stress required to bend to radius R]

There is a significance to the concept of forces on dislocations.

Dislocations will always try to adopt configurations that will reduce their

total elastic strain energies.

Interactions/forces between dislocationsInteractions/forces between dislocations

• What is the force on dislocation 2 due to the presence of dislocation 1? The dislocations have parallel Burgers vectors.

• Parallel edge dislocations:

1 2

2

1 2 1 2 2 2 2

2 2 2 2

2

[001]; , [100]

0 1 0 00 0 0 0

0 0 0 1 0 1

00 0 1

a

x xx xy xx

y xy yy xy

z zz

xx xy xy xx

x xy

b b

F bF F b b b

F

i j kb b b i b j

F b

2nd y xxF b

[glide] [climb]

1

2

(x,y) Fx

Fy

o x

y

r

θ

Edge dislocations

B*

Forces between dislocations (4)Forces between dislocations (4)

2 21 2

2 2 2 2

2 21 2

2 2 2

2 2 21 2

1 2

( )2 (1 ) ( )

(3 )2 (1 ) ( )

cos cos sin sin 1 cos2 (1 )

cos cos2 sin 22 (1 )

x xy

y xx

Gb b x x yF bx y

Gb b y x yF bx y

Gb bF i jr

Gb b ir

Glide force:

Climb force:

General equation:

1 2

cos2

2 (1 ) x y

j

Gb b F Fr

Edge dislocations

1

2

(x,y) Fx

Fy

o x

y

r

θ

B*

0y 2y 4y 6y 8y 10y

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3R

epul

sive

Forc

e be

twee

n ed

ge d

islo

catio

ns, F

x

Distance, x

Attra

ctiv

e

Forces between dislocations (5)Forces between dislocations (5)

The implications of this distribution of stresses is that dislocations will assume different configurations depending upon their type, sign, and orientation.

1

2(x,y) Fx

Fy

o1

2(x,y) Fx

Fy

o

1

2(x,y) Fx

Fy

o1

2(x,y) Fx

Fy

o

Like

Unlike

Parallel Burgers vectors

Opposite Burgers vectors

Edge dislocations

1

2

(x,y) Fx

Fy

o$

$r

x

y

• What is the force on dislocation 2 due to the presence of dislocation 1? The dislocations have parallel Burgers vectors.

• Parallel screw dislocations:

1 2

2

1 2 1 2 2 2 2

2 2 2 2

2

[001]; , [100]

0 0 1 0 00 0 0 0 0

0 0 1 0 1

00 0 1

and

x xz xz

y yz yz

z xz yz

xz yz yz xz

x yz

b b

F bF F b b b

F

i j kb b b i b j

F b

2 y xzF b

Interactions/forces between dislocationsInteractions/forces between dislocationsScrew dislocations

B*

1 2 1 22 2 2

1 2 1 22 2 2

1 2

cos2 ( ) 2

sin2 ( ) 2

cos sin2

x yz

y yz

Gb b x Gb bF bx y r

Gb b y Gb bF bx y r

Gb bF i jr

Glide force:

Climb force:

General equation:

Forces between dislocations (6)Forces between dislocations (6)Screw dislocations

1

2

(x,y) Fx

Fy

o$

$r

x

y

B*

Climb ForceClimb Force• The component Fy is a mechanical climb

force per unit length.

• It results from the normal stress of dislocation #1 attempting to push the extra half plane of dislocation #2 fromthe crystal.

►This only occurs if point defects can be emitted from or absorbed by the core of dislocation #2.

►Creation and annihilation of point defects induces chemical forces due to changes in defect concentration.

exp / /

exp /

vf

o

o

c E F b kT

c F bkT

kT

c

where atomic volume, Boltzman's const.,

temperature,equilibrium vacancy conc.

Chemical Force

2 2 2

2 2 2

(3 )2 (1 ) ( )y

Gb y x yFx y

Mechanical Force

lnyo

bkT cfc

1

2(x,y) Fx

Fy

o1

2(x,y) Fx

Fy

o

Climb ForceClimb Force

• For negative climb which involves vacancy emission (i.e., F < 0), the sign of the chemical potential changes so that c > co. This causes f to build up to balance F in equilibrium.

• In the presence of a supersaturation of vacancies (c/co), the dislocation will climb up under the chemical force f until compensated by the mechanical force F (external stresses or line tension).

Figure 4.15 Mechanical and chemical forces for climb of an edge dislocation. Vacancies (shown as squares) have a local concentration c in comparison with the equilibrium concentration co in a dislocation-free crystal. From D. Hull and D.J. Bacon, Introduction to Dislocations, 4th Ed., (Butterworth-Heinemann, Oxford, 2001).

f

c > co

c > co

FF

c < co

c < co

f

Mechanical force F

Chemical force f

B*

Implications of forces between dislocations (1)Implications of forces between dislocations (1)

• Two like edge dislocations (i.e., both have parallel Burgers vectors) lying on the same slip plane will repel each other.

• We can approximate this from the equations developed for elastic strain energy.

Slip plane

b b

2 2 2(2 )E Gb Gb G b elastic

Note the increased elastic strain energy

Slip plane

b

b

Implications of forces between dislocations (2)Implications of forces between dislocations (2)

• Two unlike edge dislocations (i.e., both have opposite Burgers vectors) lying on the same slip plane will attracteach other and annihilate out.

2 2 0E Gb Gb elastic

Note the lack of elastic strain energy

Implications of forces between dislocations (3)Implications of forces between dislocations (3)

Slip plane 1

b

b

Slip plane 2

• Two unlike edge dislocations (i.e., both have opposite Burgers vectors) lying parallel slip planes separated by a few atomic spacings will attract each other and annihilateout leaving vacancies.

This results in unique configurations of dislocations.

Low-angle boundaries

b

D

Stable positions for two edge dislocations

45° 45°90°

Same sign Opposite sign

Figure 2Numerical simulation of the total stress field and arrangement of dislocations in a solid.

[M.-C. Miguel, A. Vespignani, S. Zapperi, J. Weiss and J.-R. Grasso, Nature v. 410, pp. 667-671(5 April 2001).]

The elastic stress fields around dislocations will interact either to repel or annihilate each other and to minimize elastic strain energy and

obtain an equilibrium configuration.

Implications of forces between dislocations Implications of forces between dislocations (7)(7)

Dislocation dipoles and work hardening

• Consider two dislocation sources (S) operating on parallel slip planes

• After application of a stress, the two circled segments will interact positively leading to a stable (45°) dislocation configuration for the two unlike segments. Once in this orientation, the two segments can no longer move past each other. The dislocations become locked. Slip is inhibited.

• This leads to hardening due to a reduction in dislocation mobility and a reduction in mobile dislocation density.

S

S

S

S

(Before)

(After)

• Image forces– Free surfaces offer no stress in opposition to the

displacements caused by an approaching dislocation.– Strain energy of the crystal decreases as a dislocation

approaches a free surface. This pulls the dislocation towards the free surface (i.e., it “pulls the dislocation out of the crystal.

– Another image and appropriate mathematics are detailed on the next 2 pages.

Forces between Dislocations and Free SurfacesForces between Dislocations and Free Surfaces

b

b

F

l

lF

CrystalFree surface

Image dislocation

Dislocation in crystal

ll ll

[Adapted from Argon, p. 37 (with modification)]

2 22 2

2 22 2

2

2 2

2 2

0

4

zx zx

zy zy

x zx

Gb y Gb yx l y x l y

x l yGb Gb yx l y x l y

x l yGbF b

l

and

at , , the force on the screw dislocation is:

2 22 2

2 22 22 2

3 2 4

32 2

2

2 1 2 1

2 6

2 1

0

4 1

yx yx

x yx

Gb x l x l y Gb x l x l y

x l y x l y

Gbl x l x l x x l y y

x l y

x l y

GbF bl

at , , the force on the edge dislocation is:

Screw Screw DislocationDislocation

Edge Edge DislocationDislocation