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12

Are Dislocations Unimportant?

CHAPTER PREVIEW

We will begin our discussion with a reminder of the basic properties of single dislocations and conclude by considering combinations of dislocations, which will lead nicely into the next three chapters, which consider different types of interfaces.

Most students understand dislocations best by thinking about schematic diagrams and highresolution transmission electron microscopy (TEM) images. Dislocations are line defects, but like all crystal defects, they are actually volume defects; i.e., we should think of them as tubes, or pipes, whose properties change across the tube radius and that generally do not have cylindrical symmetry.

We are not trying to cover everything about dislocations here; rather we will review the basic features of dislocations and then introduce the complexities of ceramics. Remember the important experimental observation that led to the “invention” of dislocations: the stress required to deform a metal single crystal is at least 103 times smaller than the theoretical value.

Two vectors define the fundamental properties of any dislocation:

The line direction

The Burgers vector

The glide plane of the dislocation is the plane that contains both vectors. To summarize:

Geometry

Burgers vector and line direction define the glide plane.

Displacement

When a dislocation is present, atoms are displaced from their positions in the

 

perfect crystal; the material is strained so there must be a stress.

Movement

Dislocations move and interact (they even intersect).

Reacting

We generate dislocations, they multiply and combine (by intersecting).

Why consider dislocations in ceramics? Conventional wisdom says that dislocations are not nearly as important in the mechanical deformation of ceramics as they are for metals. The reason is that dislocations in ceramics do not move as easily as those in metals and they are usually not as numerous. So we should be asking why the last sentence is true. Dislocations in ceramics are extremely important because of what they do not do: they do not glide easily. Si devices would not work for long and ceramics, in general, would not be brittle if dislocations could glide easily. Understanding dislocations also helps us understand other more complex defects, how they interact with point defects, and how they can cause planar defects. Dislocations become very important when we use thin crystalline ceramic films, particularly when grown on crystalline substrates.

What is special about dislocations in ceramics?

Complex and large unit cells are the norm rather than the exception.

Charge—if you insert an extra half plane to make an edge dislocation, you must consider the charge.

Directional bonds—if you break a bond, does it reform?

For the student, dislocations are a great test of whether the structure of a crystal is understood. They give a fine probe of what is happening in the material. Be aware though that much of

C H A P T E R P R E V I E W ...........................................................................................................................................................

201

our understanding of dislocations comes from metallurgy, particularly from studies on Cu alloys. We often compare our dislocations to these, since those in metals are quite well understood, but ceramics do introduce new complications. A detailed understanding of dislocations and planar defects in ceramics is not as advanced as that of point defects.

12.1 A QUICK REVIEW OF DISLOCATIONS

Line defects in a crystalline material are known as dislocations (unless they are disclinations, which will be ignored because they are much more difficult and not nearly as important in ceramics). In contrast to point defects, dislocations never exist in thermodynamic equilibrium because they have formation energies of 1 eV (or more) per atom along the line and there is no significant balancing entropy contribution as there is for point defects. They are almost always present in crystals because of how the crystal grew or because it was deformed. Therefore dislocations usually form due to nonequilibrium conditions, such as thermal and mechanical processing, or for thin films and single crystals, during growth. There are two special types of dislocation.

Edge dislocations

Screw dislocations

12.1a. The Burgers vector is perpendicular to the dislocation line, u. The Burgers vector will be opposite if the extra half plane is below the glide plane.

Screw dislocation: Successive atomic planes are connected to form the surface of a helix (or screw) around the dislocation line as shown in Figure 12.2 where the dislocation is perpendicular to the planes: like the core of a spiral, parking ramp. The Burgers vector is parallel to the dislocation line and can point up or down for a leftor right-handed screw.

Glide plane: This is the plane containing both the dislocation line and the Burgers vector.

1

2

 

 

All other dislocations are referred to as “mixed.”

 

 

 

Defining the Burgers Vector and the

 

 

 

Glide Plane

 

 

 

The Burgers vector is defined by constructing a closed

 

S

 

circuit around the dislocation line. We first draw a circuit

 

 

 

 

3

around the dislocation in a clockwise (right-handed screw)

 

F

direction from the start (S) to the finish (F) as shown in

 

 

 

Figure 12.1a. We then transfer this circuit to a perfect

 

 

 

crystal as shown in Figure 12.1b. If there is a dislocation,

 

 

 

this second loop will not close on itself. We then define

 

1

2

the vector FS, which is required to close the loop in the

 

 

 

perfect crystal, as the Burgers vector. This method of

 

 

 

defining the Burgers vector, b, is known as the FS/RH

 

 

 

perfect-crystal convention. (The vector is named after

 

 

 

J.M. Burgers so there is no apostrophe.) The Burgers

 

 

 

vector is then defined with respect to the perfect crystal;

 

 

 

you would not want to define it in the imperfect crystal!

 

 

 

It is important that you are consistent in using this

 

 

 

convention. Some other texts, and even some of the

 

 

 

classic papers, use a convention that produces the opposite

 

S

 

sign for the Burgers vector. For example, they might use

 

 

 

 

 

an anticlockwise circuit, set b = SF, or draw the circuit in

t2

b

F

the perfect crystal first. You must use a convention

 

consistently.

 

 

 

Edge dislocation: An extra half-plane of atoms is

 

t1

 

FIGURE 12.1 The Burgers circuit in the imperfect and perfect

inserted above the glide plane as illustrated in Figure

lattices for an edge dislocation in a simple-cubic crystal.

202 ................................................................................................................................

A R E D I S L O C A T I O N S U N I M P O R T A N T ?

Circuit B

u2

b2

u3

b3

0

Circuit A

u1

b1

 

FIGURE 12.3 Diagram used for proving Frank’s rule for conserva-

FIGURE 12.2 Schematic of a screw dislocation.

tion of b.

 

The glide plane is not

 

 

SIGN OF b

 

same,

i.e.,

b1. It

is

more

usually an actual plane of

 

 

 

usual to define the Burgers

 

The sign of the dislocation is important: it determines

atoms, rather it is a plane

 

circuits by making a clock-

 

the direction of motion when you apply a stress.

between

two

planes

of

 

wise

circuit

around

each

 

 

 

 

 

 

 

 

atoms. (Where actually is

 

 

Edge: b points to the left or right: this determines

dislocation line looking out-

the dislocation line?) Dis-

 

ward from the nodal point.

 

 

whether the extra half-plane is above or below the

locations

are

 

often

more

 

 

This reverses the line sense

 

 

 

glide plane.

 

 

 

 

 

complicated

 

than

sug-

 

 

 

 

 

 

 

(and

 

hence

the

Burgers

 

 

 

Screw: b is parallel or antiparallel to u: this deter-

 

gested by the two simple

 

vector

of

dislocation

1

 

 

mines whether the spiral goes up or down (clockwise

examples above, but they

 

 

becomes −b1) on the left-

 

 

or anticlockwise).

 

 

 

 

 

can always

be

resolved

 

 

 

 

 

 

 

hand side, and then Eq.

 

 

 

 

 

 

 

 

into a combination of edge

 

 

 

 

 

 

 

12.1 becomes

 

 

 

 

 

and screw components.

 

 

 

 

 

 

 

 

Σbi = −b1 + b2 + b3 = 0

 

 

 

 

 

 

 

Dislocation lines can end at the surface of a crystal

 

 

 

 

 

 

 

 

(12.2)

and at grain boundaries, but never inside a crystal lattice.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Thus, dislocations must either form closed loops or branch

 

Elasticity

 

 

 

 

 

 

 

 

 

 

 

into other dislocations. The meeting of three or more dis-

 

We usually assume that elasticity (how stress is related to

locations

at a

point, or

node,

is a necessary condition

 

 

strain) is linear. The assumption is that strains are small

indicating that the Burgers vector is conserved. Consider

 

 

so that stresses are linearly related to strains (Hooke’s

the dislocation b1 in Figure 12.3, which branches into two

 

 

law). We then think of the crystal as a continuum (a struc-

dislocations with Burgers

 

 

 

 

 

 

 

 

 

 

tureless

“sponge”)

 

with

vectors b2 and b3. A Burgers

 

 

 

 

 

 

 

 

 

R FOR A SCREW DISLOCATION

two

 

independent

elastic

circuit has

been

drawn

 

 

around

each

 

dislocation

The displacement of any point on this surface from its

constants (for example, μ,

according to the line sense,

“perfect-crystal position” is given by

 

the shear modulus, and ν

u, indicated, and it follows

 

 

 

 

 

 

 

Poisson’s ratio At the core

 

 

 

bφ

 

 

 

of

the

dislocation,

 

the

from the diagram that b is

 

R =

 

 

 

 

 

 

 

 

strains are

too

large

for

conserved.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

this assumption to be valid

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b1 = b2 + b3

 

(12.1)

 

is a fraction of 2π). The strain in

the complete

so we exclude this region

 

 

 

 

 

 

 

 

from

our

calculation

and

 

 

 

 

 

 

 

 

unwrapped surface is then

 

 

 

 

 

The large circuit on the

 

 

 

 

 

replace it by a fudge factor

 

 

 

 

 

 

 

right-hand side

of

Figure

 

γ =

 

b

 

involving

 

ro, the

 

core

12.3 encloses two disloca-

 

 

 

radius (not ideal because it

 

r

 

 

tions, but since it passes

 

 

 

assumes

nothing

varies

through

the

 

same

good

The shear stress due to the screw dislocation is then

across the core and we do

material as the b1 circuit

not know what ro is). One

 

 

 

 

 

 

 

on the left-hand side the

 

τ = μγ

 

consequence

of

linear

Burgers vector must be the

 

 

elasticity

 

is

that

when

1 2 .1 A Q U I C K R E V I E W O F D I S L O C A T I O N S ....................................................................................................................

203

Burgers vectors are sepa-

STRESSES FOR AN EDGE DISLOCATION

 

will exert a force on any

rated into screw and edge

 

other neighboring disloca-

The stresses are referred to xyz corrdinates for a disloca-

 

components, the

stresses

 

tion. The force is given by

tion parallel to z. This dislocation does not have cylin-

 

due

to

these

separate

 

 

 

 

 

 

 

 

drial symmetry.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

F = σb

 

 

 

 

components can be treated

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(12.3)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

individually

and

then

σ11 = −

 

 

μb

 

 

 

x2 (3x12

+ x22 )

Box 12.1

 

 

 

 

 

 

 

 

added to give the total

2π(1 −

ν)

 

 

(x12 + x22 )2

 

 

If we were being rigorous

stress.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

we would express the force

We do not usually think

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

as bσ since b is a vector

of ceramics as being elastic

σ22

=

 

 

μb

 

 

 

x2

(x12 x22 )

Box 12.2

 

and σ is a tensor.

 

 

 

 

but

do

not

confuse

local

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2π(1

ν) (x12 + x22 )2

 

F = (b · σ) × u

 

 

 

 

stresses with macroscopic

 

 

 

 

(12.4)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

behavior. Macroscopically

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The term b · σ repre-

the ceramic may be brittle,

 

σ33 = ν(σ11 + σ22)

 

 

 

 

 

Box 12.3

 

but if the atoms move

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sents a force per unit

slightly off their perfect-

σ12

=

 

 

μb

 

 

 

x1

(x12 x22 )

Box 12.4

 

length, F, which acts in

crystal sites because of an

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a direction that is nor-

2π(1

ν) (x12 + x22 )2

 

applied force, that is an

 

 

 

 

mal to b.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

elastic

deformation.

We

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

can also have anisotropic

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

An important result of

 

THE STRAIN ENERGY

 

 

elasticity (which becomes

 

 

 

applying Eq. 12.4

is that

 

For a screw dislocation

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

important

for

noncubic

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

the force acting on disloca-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ceramics)

and

nonlinear

 

dE =

1

μ

 

b 2

lrdr

 

 

tion 1 due to the stress field

elasticity, but usually ap-

 

 

 

 

 

 

 

 

of dislocation 2 depends on

 

 

2

r

 

 

proximate

to

the

simple

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b1 · b2: if b1 · b2 is positive,

case.

 

 

 

 

 

 

 

 

dE = μb2 l

dr

 

 

 

 

 

 

 

 

 

 

then the dislocations repel

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

one another; if it is nega-

Displacement Fields

 

 

 

 

 

 

 

 

 

 

r

 

 

 

 

 

 

 

tive, then the interaction is

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

attractive (but be

careful

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

We usually derive the dis-

 

 

 

 

R

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

about the line direction).

placement field of a screw

 

E = dE = μb2

ln

R

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dislocation

because

it is

 

 

 

 

r0

 

 

 

 

 

 

 

r0

 

 

 

 

 

 

 

Dislocation Strain

much easier than deriving

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

For an edge dislocation this equation becomes

 

 

Energy

 

 

 

 

the edge dislocation.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The screw dislocation

 

 

R

 

 

 

 

 

 

 

μb2

 

 

 

 

 

R

 

 

 

 

 

Since a dislocation creates

is formed from a cylinder

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a strain field, which in turn

 

E = dE =

 

 

 

 

 

 

ln

 

 

 

of crystal by making a cut

 

 

r0

 

 

 

 

 

4π(1 − ν)

 

 

r0

 

 

creates a stress field, each

along the cylinder and dis-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

placing one side relative to

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

the other by a vector b where b is the same value every-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where on the cut and is parallel to the axis of the cylinder;

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

we first have to cut out a cylinder at the core so as to make

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

this process physically possible. We take the screw dislo-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cation in Figure 12.4 and “unwrap” the surface at a dis-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

tance r away from the dislocation core; remember that b

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

is parallel to the dislocation line so it does not matter

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where we start the unwrapping process. We then calculate

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

θ

 

 

 

 

 

 

 

 

r

the strain, γ, in the unwrapped surface and then calculate

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

γ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

the shear stress due to the screw dislocation. As you move

 

 

 

 

 

 

 

 

 

r

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

z

 

 

 

 

 

 

 

 

 

 

 

 

 

 

away from the center of the cylinder, the strain decreases

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

as r−1 and, as noted earlier, when r becomes very small

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

( r0), the assumptions of elasticity break down, so this

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

relationship does not hold in the core region.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Remember μ depends on the interatomic bonding and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b depends on the crystal lattice.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

For the edge dislocation, we will just quote the result

 

 

 

 

FIGURE 12.4 Deducing strain by “unwrapping” a screw

 

 

 

 

for the stress field (Eqs. Box 12.1–12.4). This stress field

 

 

 

 

dislocation.

 

 

 

 

 

 

 

 

 

204 ................................................................................................................................

A R E D I S L O C A T I O N S U N I M P O R T A N T ?

dislocation

will

have

 

an

 

 

 

FREE ENERGY OF A DISLOCATION

We normally just take the

associated

strain

energy.

 

 

 

constant α to be between

 

In Chapter 11 we showed how

 

G could be determined

This strain energy is termed

 

 

0.5 and

1.0.

This result

 

for the formation of point defects in a crystal using

the self-energy of the dis-

 

provides

a

criterion

for

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

location. We can then deter-

 

 

 

 

 

 

 

 

G = E T S

what

dislocations

can

be

mine the strain energy per

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

formed in a given crystal.

unit volume by integrating

 

S was obtained using statistical mechanics methods.

Those with

 

the smallest

over the volume

affected

 

allowed

Burgers

vector

 

What are the contributions to the entropy of a disloca-

by the strain field (i.e., by

 

have

the

lowest

strain

 

tion? There are three.

 

 

 

 

 

determining the area under

 

 

 

 

 

 

energy and, consequently,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

the

triangle in

the stress

 

Positional

entropy—similar to the configurational

are the most likely to form.

versus strain plot shown in

 

(This will be very impor-

 

entropy we described for point defects.

Figure 12.5). For the screw

 

tant for a ceramic with a

 

Entropy due to vibration of the dislocation line about

dislocation,

l2πrdr is

 

an

 

large unit cell; e.g., YAG.)

 

 

its equilibrium position.

 

 

 

 

 

element of volume at radius

 

 

 

 

 

 

The crystal determines

 

Changes in the vibrational entropy of the surround-

r around a length l of the

 

the actual value of b. If b

 

ing lattice because of the large strains near the dis-

dislocation.

 

 

 

 

 

 

is a perfect

lattice

vector,

 

 

 

 

 

 

location core.

 

 

 

 

 

 

At the center of a dislo-

 

 

 

 

 

 

then the structure of the

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cation the crystal is highly

 

The

total entropy contribution is only 0.5 eV/atom.

crystal will not be changed

strained with

atoms

dis-

 

Less than E, but not negligible.

by the passage of a disloca-

placed

from their normal

 

tion. If b is not a lattice

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sites and we cannot use

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

vector, then a planar defect

linear

elasticity

so

we

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(a stacking

fault) will be

 

 

 

 

 

 

 

 

DISSOCIATION

again

exclude

this region

 

 

 

 

 

 

 

 

present on one side of the

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

from

the

calculation

 

by

 

 

 

 

 

 

 

 

b b1 + b2 + b3 + · · · · b10

dislocation

and a

second

making the inner limit r0.

 

 

 

 

 

 

 

 

so-called partial (or imper-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

There must be a contribu-

 

 

 

 

 

 

 

 

b2 b12 + b22 + b23 + · · · · + b102

fect)

dislocation will

be

tion to the self-energy of

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

required

to

remove

the

the

dislocation

from

this

 

For dissociation into two Shockley partial dislocations

stacking fault.

 

 

core, but we need atomis-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The important point is

tic

modeling to

estimate

 

 

 

 

 

a

 

 

a

 

 

a

 

 

 

that

the

elastic

energy

 

 

b =

[110] →

[121] +

 

[21 1] = bP1 + bP2

the

value;

it

is

usually

 

 

 

depends on three factors:

 

 

 

 

6

assumed to be about 10%

 

 

 

2

 

6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

An

elastic

constant,

μ,

of the total strain energy of

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

the dislocation. As before, we just quote the result for the

 

 

because the energy of the dislocation depends on the

edge dislocation. The only difference compared to the

 

 

stress and strain that it causes in the lattice, and ν if

screw dislocation is the 1 − ν factor. We can express

 

 

there is an edge component.

 

 

 

 

 

 

the energy of a mixed dislocation as

 

 

 

 

 

 

 

 

The radii, R and r (and r0), because the total energy of

 

 

 

 

 

μb2

sin2θ

 

 

 

 

R

 

 

 

 

 

the dislocation depends on the volume of crystal it

 

 

 

 

 

2

 

 

 

 

 

 

effects.

 

 

 

 

 

 

 

 

 

 

E =

 

 

+ cos

θ ln

 

 

 

 

(12.5)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 − ν

 

 

 

 

 

 

 

The Burgers vector, b, because this determines the

magnitude of the strain.

The important result is that the strain energy of a dislocation is proportional to the square of the Burgers vector b and the shear modulus of the material.

E = αμb2

(12.6)

σ

12 σε

ε

FIGURE 12.5 Elastic energy of a screw dislocation from a plot of stress versus strain.

We will assume linear isotropic elasticity throughout this discussion.

Compact Core versus Spread Core

If a dislocation dissociates into partial dislocations, the self-energy will be reduced providing the angle between the Burgers vectors is acute (they are parallel not antiparallel). As a simple demonstration, consider the possibility that 10 partial dislocations each with Burgers vector b/10 are formed. The strain energy is reduced by a factor of 10! Once formed, these dislocations will repel one another since they have parallel Burgers vectors. However, since the Burgers vectors of these partial dislocations are not

1 2 .1 A Q U I C K R E V I E W O F D I S L O C A T I O N S ....................................................................................................................

205

lattice vectors, a planar defect is created between each pair of partial dislocations. If the energy required to form these planar defects is greater than the reduction in the strain energy, then the dissociation will not occur. The bestknown example of dislocation dissociation is found in face-centered cubic (fcc) metals. The dissociation produces two so-called Shockley partial dislocations, which do not have parallel Burgers vectors.

In an fcc crystal, the special feature of this dislocation dissociation is that the planar defect formed between the two partial dislocations can be regarded as a mistake in the stacking of the close-packed layers as shown in Figure 12.6. The term stacking fault is then applied to all such planar defects and the energy per unit area is the stack- ing-fault energy (SFE or γ, but do not confuse this γ with strain). Of course, for the general concept, the planes do not need to be close-packed planes; you can stack any planes incorrectly!

When calculating the SFE, we do not equate the SFE to the energy required to create the fault; instead we con-

(A)

(B)

FIGURE 12.6 The intrinsic stacking fault in an fcc crystal.

sider the force balance. The two partial dislocations will try to repel each other (bP1 · bP2 > 0), but they cannot separate too much because of the SF. Although we have deduced this dissociation for fcc metals using an energy argument, it is also important for the behavior of the dislocations. In fcc metals, these partial dislocations allow the atoms to move over “valleys” instead of over “hills” as the dislocation glides, and they restrict glide to the plane of the stacking fault. In Cu and Si, dislocations are widely ( 4 nm) dissociated, but the same dislocation in Al has a rather narrow core.

If the dislocation changes its line direction for a short segment, this segment is called a kink if it lies in the glide plane or a jog if it causes the dislocation to step out of its glide plane.

In general, a dislocation will itself contain many such defects along its length.

12.2 SUMMARY OF DISLOCATION PROPERTIES

We have not reviewed all the properties of dislocations, but have concentrated on those you should know when considering dislocations in ceramics.

Dislocations cannot end inside the crystal lattice.

We almost always assume that strains are so small that linear elasticity is a good approximation; so, Hooke’s law holds.

The displacement field for any dislocation falls off as r −1 as we move away from the dislocation.

The strain energy (or self energy) of a dislocation actu-

ally depends on the character of the dislocation, but setting E = αGb2 is a good estimate, where α is 0.5.

Parallel dislocations repel one another if the angle between their Burgers vector is less than 90°. (Be careful with u.)

A dislocation can always lower its strain energy by spreading its core or dissociating. Whether this will lower the total energy depends on the energy required to form the distorted region between the “partial dislocations.” If dissociation occurs (forming identifiable partial dislocations) then this region is called a stacking fault.

Dislocations glide by the movement of kinks and climb by the movement of jogs. Since climb requires changing the number of point defects (reacting or absorbing them), we call it nonconservative motion.

12.3 OBSERVATION OF DISLOCATIONS

We can divide the techniques used to “see” dislocations into two categories:

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A R E D I S L O C A T I O N S U N I M P O R T A N T ?

1.Direct

2.Indirect

Direct techniques allow us to see the arrangement of atoms around the dislocation. The most useful direct technique is high-resolution transmission electron microscopy (HRTEM). The resolution of commercial instruments allows direct observation of columns of atoms and edge dislocations can be identified as terminating planes of atoms as shown in Figure 12.7. These images look very much like the schematics we draw to illustrate edge dislocations, and the Burgers vectors can be determined directly from the image using the Burgers circuit construction (unless there is a screw component present).

Indirect techniques rely on the fact that dislocations create a strain field and are regions of high energy. The most widely used of the indirect methods is again TEM. We will show you several examples of TEM images of dislocations in ceramics in later parts of this chapter. Figure 12.8 shows how planes near a dislocation may be bent into an orientation where they are strongly diffracting even when the rest of the crystal is not. The intensity of the direct beam will be reduced (and that of the diffracted beam increased) and the dislocation will appear as a dark line in the bright-field image. Images of dislocations are often quite difficult to interpret and the position of the dislocation in the TEM image will generally not correspond exactly to its actual position in the crystal. However, this approach is extremely useful in answering many of

FIGURE 12.7 HRTEM of the core of a dislocation in Al2O3.

FIGURE 12.8 Diffraction contrast from dislocations in Al2O3.

the questions that arise about the nature of dislocations in

amaterial.

Is the dislocation interacting with other dislocations, or with other lattice defects?

Is the dislocation jogged, kinked, or straight?

What is the density of dislocations in the sample?

Has the dislocation adopted some special configuration, such as a helix?

By tilting our specimen in the TEM to a condition in which the dislocations are invisible (or at least appear faint in the image), if we know the diffracting conditions (specifically the diffraction vector g), then we can obtain b using the invisibility criterion (generally needing three independent g vectors).

g · b = 0

(12.7)

Another indirect (but lower resolution) approach is the etch-pit method. In this technique the surface of the sample is polished and immersed in a suitable chemical etchant. The rate of removal of material around the dislocation (where the crystal is strained) is usually more rapid than in the surrounding crystal and pits are formed where the dislocation line intersects the surface. The clearest demonstration that etch pits are associated with dislocations is shown for Si in Figure 12.9a where the dislocations can be seen in cross section due to absorption of IR illumination by the Cu decorating the dislocation; today, this task might be more easily accomplished using a focused ionbeam (FIB) instrument. Figure 12.9b shows a plan-view image of etch pits in a single crystal of Nd-doped YAG. The etch pits were revealed by immersing the polished crystal in concentrated phosphoric acid at 250°C for 8 minutes.

The etch pit method is usually best when dislocation densities are low (<104 mm−2). If the dislocation densities are too high the etch pits overlap each other and it is very difficult to resolve their shape and count densities.

1 2 . 3 O B S E R VA T I O N O F D I S L O C A T I O N S ...........................................................................................................................

207

(A)

(B)

FIGURE 12.9 Etch pits at dislocations emerging at a surface in Si.

(a)The cross-sectional view in Si; (b) the plan view in YAG.

12.4DISLOCATIONS IN CERAMICS

In discussing dislocations in ceramic materials, the principle is always the same. We deduce the possible Burgers vectors first, then the glide planes. We will begin by asking if there is anything special about dislocations in ceramics; we can preempt the answer by saying yes, as usual, the bonding and charge can add their own effects. The second best-known special feature of dislocations in ceramics is actually that the unit cell of such materials is usually larger than for the simple metals, so we will see similarities to other materials such as the intermetallics that also have large unit cells.

The structure of the dislocation core in ceramics depends on three factors

1.Charge of the ions

2.Size of ions

3.Presence of directional bonds

Perhaps the more important question is: why learn about dislocations in ceramics when ceramics do not deform plastically as easily as metals? This question then leads us to question why this statement is true. Is it always true? Can we change anything? Modern materials often involve interfaces. Interfaces are directly related to dislocations. The growth of thin films often involves dislocations. Dislocations also play an important role in radiation damage of ceramics. So there are many reasons for understanding dislocations in ceramics.

12.5 STRUCTURE OF THE CORE

Not much is really known about the core of dislocations in ceramics. For the examples we will show, you should remember that we have usually chosen one Burgers vector (usually the most important one), one line direction, and thus one glide plane. Furthermore, we usually draw the edge dislocation because it is easiest to draw, not because it is the most important. In this section, we will assume that the dislocation core is compact. The examples are chosen to illustrate particular features.

NaCl: it is relatively simple and illustrates the effect of charge.

Si: it illustrates the effect of directional (covalent) bonding.

Al2O3 and olivine: they are noncubic materials.

Although NaCl and MgO structures are both based on the cubic-F Bravais lattice, like Cu, there is no evidence for any dislocation dissociation. There are detailed atomistic calculations that confirm that the compact core is preferred. A schematic diagram of such a dislocation viewed along the [001] direction is shown in Figure 12.10. All the ions seen here lie in the same (001) plane. If we remove this plane of atoms the structure looks the same, but all the ions are reversed in sign. So charge is balanced as long as there are no jogs or kinks on the dislocation.

The glide plane for dislocations in NaCl and MgO is {110} rather than the {111} you might expect for fcc. You may read that the glide plane is {110} because this plane is electrically neutral and motion on this plane avoids charged layers gliding over one another. However, ErAs, an exotic semimetal (no ionic charge), with the NaCl structure, has the same glide system. PbS, a semiconductor with this structure, shows glide occurring on {001} planes and in TiC dislocations glide occurs on {111} planes. The real reason for the {110} glide plane is still being studied. The suspicion is that the core actually does spread on different planes, but that still does not tell us why the spreading depends on the material.

The simplest covalently bonded ceramics are Si and Ge. The covalent bond formed by two atoms sharing electrons is a localized and directional bond. Cubic ZnS has the same structure, but the Si–Si basis is replaced by Zn–

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A R E D I S L O C A T I O N S U N I M P O R T A N T ?

 

 

+ A

 

 

 

 

[110]

(001)

 

+

+

+

+

+

 

+

+

+

+

+

B

+

+

 

+

+

 

+

+

+

+

C

 

+

+

+

+

[110]

D

BÕ +

 

 

 

 

+

+

+

+

+

+

+

+

+

+

+

+

+ DÕ

 

+

 

+

 

+

 

+

+

 

+

+

+

+

qA = + 18 e qB = – 18 e qC = – 18 e qD = – 18 e

q= – 18 e q= + 18 e q= + 18 e q= + 18 e

FIGURE 12.10 The core of a dislocation in NaCl.

S, although the bond still has a large covalent component. This feature is important in determining the characteristics of dislocations in covalent materials. Dislocations in these materials tend to be immobile at low temperatures. Extensive slip occurs only at elevated temperatures.

Of the many covalent crystals, the diamond-cubic and c-ZnS structures are among the simplest and most widely studied. Since the crystal lattice is fcc, perfect dislocations

1

2

always wins). Like

have the fcc Burgers vector –<110> (b

 

2

 

 

dislocations in fcc metals, dislocations in Si, Ge, and c-

ZnS glide on {111} planes. Figure 12.11a shows a ¯ (110)

projection of c-ZnS; you can recognize the {111} planes (also shown by the projected tetrahedron). We make a dislocation by cutting out half a plane, as shown by the box labeled 1564. Then we merge the atom at P and Q; since these are on different levels (one is a large circle and the other is small), the dislocation will have a component of b into the page: it is a 60° perfect dislocation. However, we could have removed the slice of material 1234 and joined atoms R and S to make exactly the same b. Therefore, there are two possible {111} slip planes, type I and type II, and two possible dislocation cores. [These slip planes are usually called the glide (I) and shuffle (II) planes, which can be confusing since the dislocation can glide on both!] The 60° dislocation is shown with its extra plane ending at a type II plane. Imagine removing material 1234: there is then one unpaired bond per atom along the dislocation core as shown in Figure 12.11b. This defect is called a dangling bond. This type of dislocation does exist as illustrated in Figure 12.11c. Here, the dislocation alternates between the compact core and a dissociated structure. The explanation for this change is that the dislocation

steps from a type I plane to a type II plane: it dissociates on one but not on the other because the SFE is different on the two planes. The step from the shuffle plane to the glide plane is a special jog. The dangling bonds can then reconstruct; now the bonds are distorted but not dangling! However, this dislocation cannot move without rebreaking the reconstructed bonds so the dislocation is sessile.

If we left the half-plane 1234 where it is but removed the other half, we would create a dislocation with the opposite b, but instead of the last atom being R(S) it would be T(U): Zn instead of S! The core of the dislocation is fundamentally different: it is still a shuffle dislocation. These two dislocations are fundamentally different because the zinc blende structure does not have a center of symmetry. Similar considerations will hold for materials such as AIN and GaN, which also lack a center of symmetry but have the wurtzite structure crystal structure. The stacking fault in the diamond-cubic structure is described as AbBbCcBbCcAaBb: the pair of planes Aa behaves just as if it were one fcc plane in this case. Two possible SFs are shown in Figure 12.11d.

The thoroughly studied metals are either cubic or they have the closely related hcp structure. Many ceramic materials are neither cubic nor hcp. Olivine and sapphire both have oxygen sublattices that can be thought of as distorted hexagonal close packed (hcp), but the distribution of cations makes them very different.

Olivine. In the olivine group of minerals (important orthorhombic silicates), the energies of the [100], [010], and [001] dislocations are all different. In fact, the energy of the [010] dislocation will be much greater since it has a much larger Burgers vector.

1 2 . 5 S T R U C T U R E O F T H E C O R E ......................................................................................................................................

209

14

 

+

+

 

+

+

+

+

+

(111)

+

 

 

+

(111)

T

+

U

 

+

 

 

+

23

 

+ 5

6 +

[112]

+ R

S +

P

Q

 

[111]

+

+

 

+

+

B

+

a

+

A

+

c

+

C

+

b

+

B

+

a

+

A

+

c

(A)

(B)

(C)

A

C

B

A

I

B

A

(D)

(E)

U

 

S

R

Q

 

B

A

C

I

A

I

B

A

FIGURE 12.11 Models for forming dislocations and stacking faults in sphalerite. (a) Two cuts in which to make an edge dislocation; (b) dislocation made by removing 1234; (c) two structures for a dislocation with Burgers vector b in Si; (d and e) intrinsic and extnnsic in Si.

For forsterite (Mg2SiO4): a = 0.475 nm, b = 1.020 nm, and c = 0.598 nm.

For fayalite (Fe2SiO4): a = 0.482 nm, b = 1.048 nm, and c = 0.609 nm.

For monticellite (CaMgSiO4): a = 0.4815 nm, b = 1.108 nm, and c = 0.637 nm.

The added complication is that olivine is not an isotropic material so using μ and ν (which automatically implies elastic isotropy) would be a simplification.

Sapphire. Al2O3 is also very anisotropic although we can use pseudoisotropic values of μ and ν when dislocations are confined to lie on the basal plane. The most common perfect dislocations do have the shortest

Burgers vector

1

¯

–<1120> as seen in Figure 12.12, but

 

3

 

other perfect dislocations have been reported including [0001]. Even when they have the shortest b, they may

¯

glide (if they do at all) on other planes such as {1100}. Examples of dislocations in Al2O3 are shown in Figure 12.13.

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