Theoretical shear strength of a solid

Slip in a perfect, defect free lattice

a

b

x1

τ

τ

maxτ

b/2

2sinmx

bπτ τ= shear stress

displacement relation

xG Ga

τ γ= = Elastic shear stress-strain relation

For small x/b2

mx

bπτ τ=

2

2 2

m

m

x xGa b

b G Ga

πτ

τπ π

=

= ≈ This is 100 – 1000 times larger than the measuredshear strength of solids.

Dislocation Mechanics

Defects in Solids

Perfect Crystal Vacancy Self - interstitial

Point Defects; these are thermodynamic in the sense that T and ΔGf define the concentration.

exp vv

GxRT∆ ≅ −

For Tmp ~ 103 K , xv ~ 10 -5

4010~ix −

Line defects - Dislocations

FF’

Crystal analogue

Movement of an “edge” dislocation

edge dislocation

Take the same path around the dislocation in the imperfect crystal. If the path encircles a dislocation (the area encircled by the path should be large enough so that it avoids the heavily distorted area around the supposed dislocation), then it will no longer close. In this case, the Burgers vector points from the start of the path to the end of the path (hence "start-finish").

ξ⊗

edge dislocation

The “strength” of an edgeDef: Burgers vector; b

In a perfect crystal trace out a right-handed path around the line. In the perfect crystal this path should be constructed so that it closes. "Right-handed path" means you point the thumb on your right hand in the line direction , then the path is traced following the sense of your curved fingers.

SF/RHconvention

Points in direction of extra half plane

b

×ξ

Useful rule:

⊗ξ

b

Continuum analogue of an edge dislocation

b

A cut is made along the cylinder axisand the surfaces are displaced by b.

bξ

plane-strain displacement ofan edge dislocation.

screw dislocation

ξ

b

Right-handed screw

Left-handed-handed screwˆb bξ• = −

ˆb bξ• =

Continuum analogue of an screw dislocation

⊗

Right-handed screw dislocation along the axis of a cylinder of radius R and length L. A Burgers circuitoperation shows the dislocation has a positive b.

mixed dislocation

( )

( ) ( )

ˆ ˆ

ˆ ˆˆ ˆ

ˆˆ

ˆ

screw edge

screw

edge

b b b

b b

b b e e

beb

ξ ξ

ξ ξ

ξξ

= +

= •

= × • ×

×=

×

where and is a unit vector normal to the slip plane

321 bbb

+=

“node”

3b

2b

1b

2ξ

3ξ

1ξ

Conservation of b

Dislocation must end at a free surface, dislocation node (intersection with other dislocations), grain boundary, or other type of defect (but not a point defect, for instance). A dislocation can not end in the middle of a perfect region of crystal.

Formal def. of b

ξ

dl

ub dll

∂=

∂∫

Reversing the sense of thedislocation line by reversingthe direction of causes

to reverse direction.ξ

b

Stress fields of straight dislocations in isotropic solids

Right-handed screw dislocation along the axis of a cylinder of radius R and length L. A Burgers circuitoperation shows the dislocation has a positive b.

Screw dislocationThe dislocation is “made” by making a cutdefined by and sliding the bottomof the cut surface over the top by an amountb in the +z direction. The displacement is discontinuous as the cut surface and is

0; 0y x= >

y η= −

y η= +( ) ( )

00

lim , ,z z zx

u x u x bη

η η→>

− − =

The diagram also shows that the displacement uz increases uniformly with θ to give the discontinuity;

( ) 1, tan2 2z

b yu r bx

θθπ π

−= =

Note that the displacement is only a functionof x and y. Using the definition of strains in termsof displacements;

12

jiij

j i

uux x

ε ∂∂

= + ∂ ∂

we can immediately obtain the strains and then usingthe linear elastic isotropic constitutive equations obtainthe stresses.

Rectangular coordinates Cylindrical polar coordinates

Edge dislocation

An edge dislocation with indicated plain strain displacements.

The displacement field and strains of theedge dislocation can not be determined byinspection as was the case for the screw.Instead, we must solve the bi-harmonicequation. It turns out the appropriate Airystress function takes the form,

( ) ( )2 2ln4 1edge

by x yµϕπ ν

= − +−

Using the relation between the stress functionand the stresses we obtain;

Rectangular coordinates

Cylindrical polar coordinates

Mixed dislocation

The stress components for edge dislocations include , , ,xx yy zz xyσ σ σ σ

while those for a screw dislocation are ,xz yzσ σ

Since a mixed dislocation has both screw and edge character it will include all the stresscomponents from both edge and screw. The only issue is that for the edge componentof the mixed dislocation in place of b in the stress field equations for an edge dislocation you must use only the edge component of b, bsinθ. Likewise, for the screw componentof the mixed dislocation in place of b in the stress field equations for a screw dislocation you must use only the screw component of b, bcosθ.

θ coscos

sin sinsin sin

scos c inos

xx xy

xy yy

zz

xz

yz

xz

m

yz

θσθ

θσ θσθσ θσ

θθ θσσ

σ σσ

=

Strain energy and self- energy of straight dislocations in isotropic media

Screw dislocation

It’s easiest to work this out for the stress filed in cylindrical polar coordinates since we have only one stress

2zbrθ

µσ

π=

2 2

0 2o

ij ij ij ij

Rz

ij ijr

dW d dV d dAL

dW d dA d rdrL

πθ

σ ε σ ε

σσ ε θ

µ

= =

= = ∫ ∫

The strain energy is

The strain energy per unit length of dislocation line is

2

ln4 o

dW b RL r

µπ

=

ro is called the “core” region of the dislocation and corresponds to a region around thedislocation where the displacements are so large that linear elasticity breaks down.

The size of this core region depends on crystal structure. For example dislocations infcc solids have a bit larger core regions than that in bcc solids. Generally one can write

, where α is 0.5 – 2 for metals, ~ 2 for an ionic solid such as NaCl and ~1.2 forcovalent crystals.

/or b α=

The self energy of a dislocation includes the elastic energy per unit length plus a termrelated to the core energy of the dislocation. For screw dislocations this additional term is

2

core energy ,4bµπ

≅

so, we can write the self energy of the dislocation line per unit length as

2

_ ln 14self screw

o

b Rur

µπ

= +

Edge dislocation

The strain energy is calculated for the edge dislocation the same way as it wasfor the screw except we have more terms!

( )22

2 2 2

0

1 22 2

o

Rxy

ij ij xx yy xx yy zzr

dW d dA d rdrL E

π σσ ε θ σ σ νσ σ σ

µ

= = + + − −

∫ ∫

( )2

ln .4 1 o

dW b RL r

µπ ν

=−

Substituting for the stresses and integrating, we obtain

This result is identical in form to that obtained for the screw except for the factor of1−ν. The core-cutoff size is the same as that discussed for the screw.

The core energy of the edge is ( )2

4 1bµ

π ν≅

−so the self energy of the edge dislocation

can be written as,

( )2

_ ln 14 1self edge

o

b Rur

µπ ν

= + −

Mixed dislocation

2

ln ,4 o

dW b RL r

µπκ

=

The strain energy of a mixed dislocation can be written as

where is1κ −

( )2

21 coscos1

θθ

κ ν= +

−

In a similar fashion the self energy of a mixed dislocation can be written as

2

_ ln 14self mixed

o

b Rur

µπκ

= +

Forces on dislocations

We can expect a gradient force to act on a dislocation if the mechanical energyof the system is a function of the position of the dislocation:

To a dislocation we can expect that it is immaterial whether the stresses actingon it are internal or external in origin.

We want to find a relation between an external or internal stress and it’s actionon the dislocation.

F U= −∇

An interesting and often not appreciated result is that since the strain energy of a body is independent of the position of the dislocation, the presence of the dislocation does not alter the elastic constants of the material.

causes an elastic displacement and the strain energyin the body is just the work done by the force,

( )1 1 1 2 3, ,u u x x x=

1T

Consider an unstrained body with no dislocation present. Apply and external forceper unit area over the surface of the body.

1T

1T

S ds

1T

S ds

( )1 1 11/ 2S

U T u dS= •∫

2T

Now form a dislocation by cutting and applying to the cut faces a force causing 2.u

( )2 2 21/ 2S

U T u dA= •∫

Also, there is work done by as is being applied. The total strain energy is 1T

2u

( ) ( )1 2 1 2 1 2S A

U U U T u dS T u dA= + + • + •∫ ∫

2T

A

Now we can do exactly the same thing in reverse. First form the dislocationby applying resulting in and then apply 2T

2u1.T

Alternatively we could remove and then in the manner of a thermodynamic cycle.

1T

2T

Since does not act on ( )2 2 1, 0 ; 0S

S T on S T u dS= • =∫

2T

Also, is single valued whereas is equal and opposite on the two sides of thecut surface; the work gained on one face of the cut is exactly balanced by the work lost on the other face,

1u 2T

( )2 1 0A

T u dA• =∫

( ) ( )1 2 2 1 2 1S A

U U U T u dS T u dA= + + • + •∫ ∫

( ) ( )1 2 1 2 1 2S A

U U U T u dS T u dA= + + • + •∫ ∫

So, therefore is independent of the position of the dislocation.1 2U U U= +

This means that the elastic constants are insensitive to the presence orlocation of stationary dislocations. *This is for an “infinite” solid as thepresence or effect of surfaces is not taken into account.

The “force” on the dislocation will be due entirely to the external work that isdone when it moves,

( )1 2A

W T u dA= − •∫

( ) ( )

( ) ( )

1 2 1 2

1 2 1 2

0S A

A S

T u dS T u dA

T u dA T u dS

• + • =

− • = •

∫ ∫

∫ ∫

Also, this means that the work done by over S and A as the dislocation is formed must sum to zero.

1T

and 1 21W WT u

A l x∂ ∂

= − • ≡∂ ∂

This may be considered as the force per unit length acting on a dislocation line

1 21 WF T u bl x

σ∂= − = • =

∂

where is the stress in the slip plane in the slip direction.σ

Edge dislocation S Screw dislocation

glide

S

only one glide plane(mixed dislocations also)

any crystallographically allowable glide plane

climb

S

edge & mixed dislocations

screw dislocations don’t climb (unless they have a small edge component)

. and both containing plane aon motion :motion) ive(conservat Glide

b

ξ

. direction in motion :motion) veconservati-(non Climb

b

×±ξ

coren dislocatio fromaway or toed transportbemust matter :motion veconservati-non

Dislocation Climb

Figure showing edge dislocation climb

Jogged edge dislocation indicating annihilation and removal of atomic planes.

Climb: important in high temperature creep (helps in dislocation annihilation and circumventing obstructions)

dislocation emitting or absorbing vacancies?

dislocation emitting or absorbing vacancies?

dislocations are important sinks and sources for vacancies

adsorbingvacancies

emittingvacancies

Mixed & edge dislocations have only one glide plane and can’t cross slip

S

another toplane glide one from change :slip-cross

Cross slip is important as a means of dislocations 1) rearranging themselves, 2) getting around obstacles and 2) annihilating (dynamic recovery)

+S

-S

Screw dislocations can cross-slip

b

rδξ

Consider a mixed dislocation of line direction ξ and Burgers vector b. A stress σ acts on the crystal.

Question: What is the force (per unit length) acting on the dislocation?

We know that, in principle, if the dislocation moves by a displacement δr, then the work done on the crystal will be δW.

rFW

δδ ⋅=

Force on a Dislocation: Peach Koehler Equation

glide component

climb component

Force on a Dislocation: Peach Koehler Equation

Let us calculate δW. To do so we need to move the dislocation by an amount δr. This is the way we move it: We first make a cut of length L(the length of the dislocation) and width δr. The area of this cut we denote |δA|, where the vector δA points perpendicular to the plane of the cut:

( )rLA

δξδ ×=

b

rδξ

LA

δ

Peach Koehler Equation (continued)Note: While making the cut we don’t want the faces of the cut to move in the presence of the applied stress. In order to prevent that from happening, we have to apply tractions to the surfaces of the cut that are equal and opposite to the applied stress. Next, while the tractions are applied, we shift the two faces with respect to each other by an amount b. In order to do this it might be necessary to take material away from the cut and place it on the surface of the crystal (or vice versa) so that the dislocation can climb. (This will be necessary if δr lies outside the glide plane).

While we are moving the two surfaces, two things are happening:1) The external stress is doing work on the crystal,2) The dislocation is effectively moving by an amount δr to the new location.

b

rδξ

LA

δ

Peach Koehler Equation (continued)The amount of work done is given by:

forcentdisplacemeAbW δσδ ⋅⋅=

This can be rearranged to give:

( )rLA

δξδ ×=

( )[ ]ξσδδ ˆ×⋅⋅= brLW

The force per unit length acting on the dislocation is therefore:

( ) ξσ ˆ×⋅= bf Peach-Koehler

Equation

with

The component of force per unit length acting in the direction of climb is:

The glide force (the most important component low temps) is:

where τrss is the resolved shear stress, that is, the component of stress acting on the glide plane and in the direction of the Burgers vector. The glide force acts perpendicular to the line of the dislocation in such a way as to expand the slipped area.

( )[ ] ( )ξ

ξξσˆ

ˆˆ

×

×⋅×⋅=

bbbfcl

bfff rssclgl τ=−=

Climb and glide forces

Examples

x

y

z

Mixed dislocation with a tangent vector in the negativez-direction and a Burgers vector

ξ ( ) ξσ ˆ×⋅= bf

ˆˆ ˆx y zb b i b j b k= + +

Generally• •

but the are equalif T is symmetric

xx x xy y xz zx xxx xy xz

yx yy

T

yz y

v v

y yx x yy y yz z

zx zy zz z z zx x zy y zz z

T

b b bb Gb G b b b Gb G b b b

σ σ σσ σ σσ σ σ σ σ σσ σ σ σ σ σ

≠

+ + = = + + = + +

ˆ ˆy xG G i G jξ× = − +

force for glideforce for climb

y

z

x

Dislocation with a tangent vector in the +x direction and a Burgers vector

ξ

Examples

Apply stressesAll other stresses are zero.

b

Examples

y

z

ξ

b

x

Note that if we had chosen a coordinate system like thisthen

x

ξ

b

z

y

x

y

z

Examples

x

y

z

rA B

We want to calculate the force (per unit length) that dislocation A exerts on dislocation B.

z

Since the stress field assumes that the dislocation is at the origin of the coordinate system,we place A there and note that the tangent vector is parallel to the z-direction. Then the PKEquation takes the form ( ) ˆ

A on B AB B A Bf f b σ ξ= = ×

ˆ

ˆˆ ˆA B x

A B

b b b i

kξ ξ

= =

= =

( )0

0 0 00 0 0 0

ˆ ˆ00 0 1

Bxx xy xx x x

Bx xy yy xy x y

zz

x y y x

b Gb b G

i j kG G G i G j

σ σ σσ σ σ

σ

= =

= −

( )( )( )

( )( )( ) ( )

2 2

22 2

2 2

22 2

0

12 1

32 1

2 1

xx

xxy

A

y x yb

x y

x x ybry

b

x

µσπ ν

µσ µν νπ π

=+

=− +

−=

−−=

+

Since

( ) ( )21 1ˆ ˆ

2 1 2 1

A Bx x x

AB yb b bf G i i

r rµ µπ ν π ν

= = =− −

0; 0xx yGσ = =

Examples

x

y

zr

A

B

We want to calculate the force (per unit length) that dislocation A exerts on dislocation B.

z

ˆ

ˆˆ ˆA B x

A B

b b b i

kξ ξ

= =

= =

The x-y position of dislocation B iscossin

x ry r

θθ

==

𝜃𝜃

( )( )( ) ( )

( )

( )( )( ) ( )

( )

2 2

22 2

2 2

2

2

2

2

2 2

sin 3cos sin2 1

cos cos sin21 1

32 1

2

Ax

Ax

xx

xy

br

y x yb

x y

x x yry

bb

x

θ θ θµπ ν

θ θ θµπ ν

µσπ ν

µσπ ν

+=

− +

−=

−

+

−

−+

−

=

( ) ( ) ( ) ( )2

2 21 1ˆ ˆ ˆcos cos sin sin 2cos 12 1 2 1

A Bx x

AB yb b bf G i i j

r rµ µ θ θ θ θ θπ ν π ν

= = = − − + − −

( )0

ˆ ˆ0 0 0 ; 00 0 0 0 0 0 1

Bxx xy xx x x

Bx xy yy xy x y x y y x

zz

b G i j kb b G G G G i G j

σ σ σσ σ σ

σ

= = = −

Suppose that there is a super saturation or deficit of vacancies: The crystal has N vacancies instead of N0, the equilibrium number for a given temperature and pressure. In this case the excess free energy per vacancy is

=

0

lnNNTkG Bv∆

A dislocation which is at least partly edge can climb to create or destroy vacancies, moving the crystal back toward equilibrium and thereby lowering the free energy of the crystal. Suppose the dislocation absorbs δN vacancies by climbing a distance δx,

Osmotic or Chemical Force

y per vacancvolumetheisofcomponent edgetheiswhere

v

e bbΩ

e

v

b L xN

δδ =

Ω

Osmotic force (continued)The change in free energy of the crystal is

( ) xNNTkLbNGG Bvev δΩδ∆δ

=×=

0

ln/

The osmotic force, per unit length, is

==

0

ln1NNTkb

NG

xLf

v

Beos Ωδ

δδ

v

v

vv

Line Tension of a dislocation

τ

dF dL bL

τ= ×

T

0

2

2 cos 2 sin

2

2

yF b dl T

dl rdbr T

Tbr

GbT

Gbr

Gbl

θ

τ θ θ

θτ

τ

τ

τ

= −

==

=

≈

=

=

∑ ∫

r

The line tension T can be approximated as

θ

θ θ

The maximum stress occurs at r=l/2

ll

Dislocation Sources in Crystals

The Frank-Read source

Gbl

τ =

Dislocation point defect interactions

The elastic interaction energy between a point defect and a dislocationresults from the interaction of the volume change associated with the defectand the hydrostatic stress field of the dislocation. The interaction is in the form of a p∆V work term:

The volume change associated with a point defect is

3 *4 ; o

oo

r rV r

rπ ε ε

−∆ = =

The mean stress of an edge dislocation is

( )( )

1 sin3 1m

Gbr

ν θσ

π ν+

=−

Dislocation point defect interactions

The interaction energy is

( )( )

( )( )

331 sin 4 1 sin4 sin

3 1 3 1o

o

Gb GbrU r A

r r rν θ ν ε θπ ε θπ ν ν+ +

= = =− −

For r ~ b the interaction energy is maximum at 3 / 2θ π=

θ

r

positive misfit

negative misfit

Strain caused by a glide dislocationmoving through a crystal

x

t

h

l

When a glide dislocation movesa distance x on its slip plane in acrystal of length l, height h, andthickness t the plastic shear strainthat results is

θ

tan xt btl h

γ θ = =

When there are N dislocationsparallel to the t direction and theyall move an average distance λ

the total strain in the crystal is .b Nlhλγ = Defining /N lh = Λ as the dislocation

density,bγ λ= Λ

Image forces on dislocations

A free surface or interface separating materials with “different” elastic constantswill exert a “force” on a dislocation. This is a gradient force in the sense that the dislocation can lower its energy by moving either toward or away from theinterface.

y

z xr

A

B

A screw dislocation A is near a free surface. B is the “image” of A and is ofopposite sign placed symmetrically wrt A so that the free surface boundaryconditions are satisfied; 0.ij jnσ =

A Bb b= −

Image forces on dislocations

The stress field of the screw dislocation is

( )

( )

2 2

2 2

2

2

xz

yz

Gb yx y

Gb xx y

σπ

σπ

−=

+

=+

at y = 0,

01

2

xz

yzGb

r

σ

σπ

=

=

Application of the PK equation for the force that B exerts on A yields

0xzσ =

For x = r at the free surface

0; i.e., 0.A Byz yzσ σ += =

2

4GbF

rπ= −

Image forces on dislocations

This is the equivalent of a gradient force. Consider the elastic energyper unit length of a screw dislocation;

2

2

ln4

.4

o

Gb rur

u GbFr r

π

π

=

∂= − = −

∂

The gradient force is given by