grain boundaries 1 - mcmaster...
TRANSCRIPT
1Grain Boundaries
Point Defects
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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.
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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
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kT
QNN V
V expOR(depending on units of Q)
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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)
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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
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• 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
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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. . .
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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.
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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
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• 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
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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
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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
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Dark lines are dislocations
Representation of edge, screw and mixed dislocations
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Thermodynamics of Dislocations
• Unlike point defects, dislocations are alwaysthermodynamically unstable
• Most dislocations will disappear at high enough temperatures
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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
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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
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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
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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
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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
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Microstructure
• Structural materials can be inhomogeneous.
– Grain size variation
– Texture
• Metal castings – Solidification involves nucleation, growth and impingement micro
and macrostructure formation
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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)
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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
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Driving Force
• Thermodynamics
T < TM, Gsolid < Gliquid
• Driving force is...
𝜟Gsol = Gsolid - Gliquid
TMT
G
solid
liquid
Spontaneous transformation
Negative ΔG, ΔG<0
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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
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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
-
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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
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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
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Solidification in Metals
• Crystal growth into a temperature gradient leads to the formation of dendrites
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Solidification Structure
• Three zones - chill, columnar, equiaxed.
heat flow
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Solidification & Continuous Casting
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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
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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
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4
19-
23
eVJ
rGrG sol