1Grain Boundaries

Point Defects

2

Point Defects

• A Point Defect is a crystalline defect associated with one or, at most, several atomic sites.

• These are defects at a single atom position.– Vacancies

– Self-interstitials

• Vacancies are the most important form.

3

Vacancies• Vacancies :a normally occupied lattice site from which

an atom or ion is missing

• Thermodynamics says that there is an equilibrium number of vacancies:

– Why is this number of vacancies thermodynamically stable?

RT

QNN V

V exp

Vacancydistortion of planes

4

kT

QNN V

V expOR(depending on units of Q)

5

Thermodynamics of Vacancies

• Introduction of a small number of vacancies lower the “free energy”.– From chemistry: G = H – TS– Entropy (S) is the key.

• A crystal only has one perfectconfiguration.– “configurational” entropy is zero

• Vacancies (on the other hand) give rise to many configurations.– add n vacancies to a lattice with N atoms– so as n↑ ⟹ S ↑ ⟹ G ↓

• Vacancies are created until G is at a minimum (note that if too many vacancies are added, G increases because the material will then contain too many broken bonds).

G

n*nN

n

(not covered in textbook)

6

Equilibrium Vacancy Concentration

• Consider Cu: Qv = 96.1 kJ/mol

– at 1000 K, Xv 10-5

– at 1358K ( TM ), Xv 2 x 10-4

– typically Xv 10-4 at TM

7

• Increasing temperature causes a surface island of atoms to grow.

• Why? The equilibrium vacancy concentration increases by atom motion from the crystal to the surface, where they join the island.

Island grows/shrinks to maintain equil. vancancy conc. in the bulk.

Reprinted with permission from Nature (K.F. McCarty, J.A. Nobel, and N.C. Bartelt, "Vacancies in

Solids and the Stability of Surface Morphology",

Nature, Vol. 412, pp. 622-625 (2001). Image is

5.75 m by 5.75 m.) Copyright (2001) Macmillan

Publishers, Ltd.

Observing the Equilibrium Vacancy Concentration

Low energy electron microscope view of a (110) surface of NiAl.

(Click to Play)

Self-Interstitials

• Self-interstitial: an atom from the crystal that is crowed into an interstitial site.

• The term “interstitial” refers to the empty space between atoms.

• Form a lot of distortion

8

9

Solute Atoms (Solid Solutions)• A solute atom is defined as an atom which different from

the host atoms (e.g. Cu atom in Ni). Sometimes we refer to solutes as impurities.

Substitutional or Interstitial

• Solid solubility

kT

QX sol

sol exp

Factors Affecting Solubility in Substitutional Solid Solutions

• Atomic size factor

• Crystal Structure

• Electronegativity

• Valences

10The tale of the engineering student and the philosopher. . .

11

Solute Mobility

• Which kind of solute (substitutional or interstitial) will be easier to move around (diffuse) in a crystalline solid. Why?

Interstitial solutes move around more easily because:

They are much smaller atoms.Most of the surrounding interstitial spaces are empty.

12

A Note on Specifying Composition

• Adding solute atoms creates an alloy.– We need a way to specify its composition.

– Typically it's done as a weight or atomic percent.

• e.g. steel may contain Fe with 0.4 wt% C

• weight percent (wt%)m1= weight or mass of element 1

• atom percent (at%)

nm1 = m1’/A1

m1‘ = mass in gram

A1= atomic weight

10021

1

1

mm

mC

1001 i

m

m

(binary)

1001'

1

im

m

n

nC

13

• Sn islands move along the surface and "alloy" the Cu with Sn atoms to make "bronze".

• The islands continually move into "unalloyed“ regions and leave tiny bronze particles in their wake.

• Eventually, the islands disappear.

Reprinted with permission from: A.K. Schmid, N.C. Bartelt, and R.Q. Hwang, "Alloying at Surfaces by the Migration of Reactive Two-Dimensional Islands", Science, Vol. 290, No. 5496, pp. 1561-64 (2000). Field of view is 1.5 m and the temperature is 290C.

Low energy electron microscope view of a (111) surface of Cu.

Alloying a Surface

(Click to Play)

Line Defects

14

Dislocations- Linear Defects• Dislocation: linear or one-

dimensional defect around which some of the atoms are misaligned

• Edge Dislocation:

– A linear crystal line defect

– associated with the lattice distortion produced in the vicinity of the end of an extra half-plane of atoms

– Burgers vector perpendicular to the dislocation line

15

Dislocations- Linear Defects

• Screw dislocations:– Associated with the

lattice distortion created when normally parallel planes are joined together to form a helical ramp

– Burgers vector is parallel to the dislocation line

16

17

Dark lines are dislocations

Representation of edge, screw and mixed dislocations

18

Thermodynamics of Dislocations

• Unlike point defects, dislocations are alwaysthermodynamically unstable

• Most dislocations will disappear at high enough temperatures

19

Dislocation Motion

• Dislocations move by breaking and remaking atomic bonds.

• The net result is a shear displacement in the crystal.

Courtesy P. M. Anderson

(Click to Play)

Planar Defects

20

Planar Defects

• Boundaries are two dimensions defects that separate regions of the materials that have different crystal structure or crystallographic orientations

– External surfaces

– Grain boundaries

– Twin boundaries (special type of grain boundary)

– Stacking faults

– Phase boundaries

21

22

Planar Defects

• Interfaces divide crystal regions

• External surfaces – crystal edge

– unbonded electronsexcess energy

“surface” energy

γs (J/m2)

External Surfaces

• Atoms on the free surface of the specimen are in a higher state of energy than atoms in the interior positions

• Surfaces have more dangling bonds

• To reduce this energy, materials try to minimize the total surface energy

23

24

Grain Boundaries

• Most solids are polycrystalline:

– Many grains (a single crystal region within a polycrystalline material) separated by grain boundaries

– There is atomic mismatch in a transition region where two grains meet.

– Boundaries (where grains come together) are called grain boundaries.

grain boundaries

Twin Boundaries

• Specific mirror lattice symmetry

• The mirror plane is termed a twin

• Produced by applied mechanical shear forces or during annealing heat treatments (chapter 7)

• Occurs on a definite crystallographic plane and in a specific direction

25

26

Demo: Crystal model with faults

Point Defects Linear Defects - Dislocations Bulk defects

Model of a perfect single crystal Model of a poly-crystal with many defects

Area defects

Microscopic Examination

27

28

Microstructure

• Structural materials can be inhomogeneous.

– Grain size variation

– Texture

• Metal castings – Solidification involves nucleation, growth and impingement micro

and macrostructure formation

29

Microstructure of annealed Al-3%Mg Alloy(material of the pop-can)

Microstructure Examples

30Al-3%Mg after large deformation and 3 hrs of annealing at 180oC

31Twins in Fe-Ni superalloy (FCC)

32

Solidification

• Nucleation

– small cluster (~1nm) of atoms in crystalline order

• Growth

– atoms are added to the nucleus which now grows to consume the liquid.

• Impingement

– solid grains grow until they meet one another and form a grain-boundary

33

Driving Force

• Thermodynamics

T < TM, Gsolid < Gliquid

• Driving force is...

𝜟Gsol = Gsolid - Gliquid

TMT

G

solid

liquid

Spontaneous transformation

Negative ΔG, ΔG<0

34

Nucleation Barrier• When T is lower than Tm,

changing a volume of liquid into a solid lowers the energy of the material. If the decrease in energy per unit volume is ΔGsol

then for a sphere of radius r, the decrease in energy is:

• At the same time, the solid cluster which forms will have a surface and surface energy, s

Again, for a sphere of radius r, the increase in energy is:

solGr 3

3

4

sr 24

liquid

35

Nucleation Barrier• Solid cluster has a surface and surface energy s

• Total energy

Gtot=VGsol+4r2 s

• Because the surface terms is

larger than the volume term at

small values of r, there is a

nucleation barrier.

• Find r* from

ssolrGr 23 4

3

4

0

r

Gtot

-

36

Effect of temperature on ΔG* and r*

Nucleation• There are 2 types of nucleation:

– Homogeneous nucleation• A solid particle begins to form when atoms in

the liquid cluster together growth of the cluster will continue if the cluster reaches the critical radius (r*)

• Embryo – a cluster of radius < r* will shrink and redissolve

• Nucleus – a cluster of radius ≥ r*

– Heterogeneous nucleation• Activation energy is lowered when nuclei form

on pre-existing surfaces

• A phase transformation will occur spontaneously when ∆G (free energy) has a negative value

• ∆G* activation free energy- energy required for the formation of a stable nucleus

Free-energy vs. Radius plot for homo/heterogeneous nucleation

38

Forces Controlling Nucleation and Growth

• During an undercooling process (during which the temperature drops below TE)…

– … the available reaction energy increases

– … the rate of diffusion decreases

• These are competing effects

-ΔG

T TE

D

Video of Nucleation, Growth and Impingement

40

Solidification in Metals

• Crystal growth into a temperature gradient leads to the formation of dendrites

41

Solidification Structure

• Three zones - chill, columnar, equiaxed.

heat flow

42

Solidification & Continuous Casting

43

Micro to Macro Defects in Solids

• Crystals are never perfect• Many scales of defect - atomic to macro• Macro defects include:

– Pores– Cracks– Foreign inclusions– Other phases

Grain Size Determination

• Grain size – average grain diameter, as determined from a random cross section.

• N Number of grains per square inch (magnified 100X)

• n Grain size number

44

12 nN

Chapter 4

Practice Problems

Practice Problems

1. Consider a metal with a vacancy formation energy of 0.8 eV/atom. If the material is originally at 1031 K, calculate the temperature rise (in K) needed to increase the vacancy fraction by a factor of 3.5:

a) 123 K

b) 143 K

c) 167 K

d) 443 K

e) 567 K

Practice Problems

Answer:

c)or 1677.16610317.1197

7.1197

./1062.8

/8.0exp10232.15.3

:get to3.5by Multiply

10232.11031./1062.8

/8.0exp

5

4

4

51031

KKT

KT

TKatomeV

atomeV

KKatomeV

atomeVX K

Practice Problems

2. Experimental observations on the solidification of a hypothetical metal show that the liquid-solid surface energy is 0.8 J/m2 and the driving force for solidification, ΔGsol, is -8x109 J/m3. If the critical radius for homogenous nucleation is 0.2 nm, calculate the activation energy for homogenous nucleation [Hint: 1 eV =1.602x10-19 J]

a) 0.74 eVb) 0.84 eVc) 0.34 eVd) 0.14 eVe) 0.24 eV

Practice Problems

Answer:

b)or 84.0101.34G

:solve tonm 0.2r and ablesknown vari Substitute

43

4

19-

23

eVJ

rGrG sol