physical metallurgy 10 th lecture ms&e 410 d.ast dast@ccmr.cornell.edu 255 4140

Physical Metallurgy10 th Lecture

MS&E 410

D.Ast

dast@ccmr.cornell.edu

255 4140

Microstructure

Length scales

• Angstrom: Atoms (wave functions, d-electrons, ionization etc)

• Tens of Å: Unit cells, molecules, lattices

• to mm :Grains and what comes with it

• mm ..: Phase distribution, texture distribution etc

Modern buzz word: Multiscale modeling.

Each of these tends to be its own specialty

The roller is for texture, equivalent stress-strain and the like

Vacancies:

• Dominant in metals

• At melting points, c ~ 10-4 (measured via Simmons-Balluffi)

• Formation energy ) 0.5… 1.5 eV

• Needed for non-conservative motion of dislocation

• Frenkel pairs generated by neutron damage in reactor materials . (well studied)

• Vacancies in semiconductors - covalent - higher formation E . . much lower concentration (excluding ion implantation)

• Vacancies (covalent, ionic) have charge (how can this be ??)

You had this before - remember ?

HW 10-1

The formation energy of vacancies in Bismuth is 0.35 eV and the formation entropy is 0.3 k

1. Find the melting point of bismuth

2. Calculate the vacancy concentration at the melting point in a) percent and b) vacancies per cm3

3. Repeat above for a temperature that is 1/2 of the melting temperature, on the absolute scale (I.e. melting temperature in degree K)

Note: we will need the 1/2 Tm rule when we get to metal failures that are diffusion driven

How vacancy concentrations are measured

• Almost all experiments (say growth and shrinking of stacking faults), electrical resistivity changes (we know contribution per vacancy) measure the sum of the formation and migration free energy

• To split these values, we need the Simmons Balluffi experiment

Bi

A few things which you might to find out by yourself - is you are born metallurgist ;-)

a) How does the Simmons Balluffi experiment work ??

b) Why is there a picture of Balluffi in our MS&E lounge ?

c) Why has the formation of a vacancy an entropy term ??

Notes

• A vacancy does not jump… atoms do

• Atoms jump when a vacancy is next.

• While vacancies do a random walk, atoms do not

• To correct, we need the (auto) correlation factor f

.

Tracer diffusion

Assume you want to measure the diffusion coefficient of vacancies.

Vacancies are hard to follow ! Tracers can be followed easily.

One dimensional chain.

No matter how many vacancies zip from left to right or right to left, the tracer atom will not move as long as it is equally likely that vacancies arrive from the left or right !

This is annoying, because if we want to calculate diffusion for an impurity, we start with the diffusivities of vacancies. Also for many other problems (dislocation climb, transient diffusion etc) we would like to know how vacancies diffuse *

The tracer and vacancy diffusion is linked by a “fudge factor” f, called the correlation factor, such that

D (tracer) = f D (vacancy)

In the 1-D chain, f = 0. For fcc f= 0.7814 for bcc f= 0.727

The higher the coordination number, the closer f => 1.

You can look at f as the penalty for using figments of imagination., called vacancies, while what really moves are atoms.

Tracer atom

After a jump, the vacancy has equal probability to jump to any nearby site - cause its surrounding looks symmetrical

But it does not to the tracer atom. It is more likely to jump to position 6 than to 1,2,3,4,5 because to get to these it will have to wait for the arrival of another vacancy !

Original position

After one jump

Illustration

Probability to jump if vacancy next Probability of vacancy being next

Interstititial diffuser don’t need a vacancy. That is why they are so much faster in Si which has cvacancy, too low to be measured, at Tm

Notes

• Dislocations are shifts that restore the crystal lattice

• The distance of these shifts is given by b (burgers vector)

• Since E ~ G b2 (G shear modulus), the shortest b rules

• Shifts “can go in steps” => partial dislocations

• Shifts in ordered alloys go in steps using “regular” dislocations

• The burgers vector is an invariant relative to the line direction

• If the line direction changes, dislocation can go from edge to mixed to screw and vice versa.

• Dislocations are vortex lines in the strain tensor field and as all vortex lines either need to start and end at the surface or must make closed loops.

The burgers vector is always the same (invariant) but the line direction in metals is highly variable. Hence above the dislocation goes from edge to screw and back to edge.

The screw part has no compressive or tensile field - pure shear. It, therefore, interacts much weaker with “impurities” (Cotrell

atmosphere etc for the specialists)

The Volterra construction does not use lattice planes and works equally well for glasses. It is pure continuum theory. No b circuits!

In the Volterra description of the screw, there is only uz (r,).

Thus, you can calculate the strain field very easily

The solution has a singularity at r=0

which can’t be…. We need to look at the core with atomistic theory

For those who know anisotropic elastic theory:

You need to pay attention what shear modulus to use.

Is it C44 or 1/2(C12-C11)?

Depends on the line direction..

Because the Volterra dislocation has nothing to do with a lattice, it is a very handy tool to calculate more complicated continuums mechanic problems

Shear bands in a metallic glass. Cylindrical indenture.

One easy way to model this is to make parallel polishing scratches before indenting and then to measure the offset along the shear band.

Then distribute edge dislocations along the shear band to match the measured displacement and add their stress field

The pic is from the Schuh group at MIT.

Notes

The dislocation does not move “by switching atoms along the dislocation line at once”

The dislocation propagates a kink (atomic step in glide plane) sidewise.

Usually such kinks exist naturally.

If not, the dislocation will nucleate a double kink

Notes

Dislocations move on the plane with the maximum resolved shear stress. In addition, they need to move in their glide plane (generally the densest packed plane) which is (111) in fcc.

But when they encounter an obstacle to glide (more later) they will move on a plane with less resolved shear stress - provided they can glide there

Cross slip

An edge dislocation can not cross slip

Because its Burgers vector would be not in lie in the new plane.

A screw dislocation can cross slip at the drop of a hat

•The screw dislocation is dissociated into two partial dislocations, separated by a stacking fault (yellow)

• The partial dislocations contain both a screw and an edge component

The edge component can not glide onto the slip plane. Hence the screw dislocation needs to constrict into a

single dislocations with a/2<110> to move onto the cross slip plane.

The influence of the Stacking Fault Energy on Deformation properties of Metals, in particular Fatigue of Metals

• A low stacking fault energy results results in widely separated partial dislocations.

• Dislocations can not cross slip, no dislocation cell walls form

• Crucial influence on fatigue, cavitation wear etc

Dislocation Cell wall in Cu.

Mughrabi, Nature, 2006

HW 10-2

Google

“Metal fatigue, stacking fault energy, dislocation cell size”

And answer the following questions

a) To get good fatigue resistance, do you want to have a high or low stacking fault energy ?

b) Why ? Explanation limited to 4 sentences

You really should know those by now. If not consult lecture on stacking in fcc and bcc

Dislocation obstacles

• Other dislocation (Forrest dislocations)

• Precipitates

• Grain Boundaries

Forrest dislocation

• A dislocation “moving through” an other will introduce a “kink” or “jog” equal to the moving dislocation’s Burgers vector

• Jogs can have a different glide plane than the “cut” dislocation, which is therefore “pinned at the jog.

For a dislocation to move (glissile) both its Burgers vector and its line direction must lie in a glide plane. Plus there must be shear stress

Precipitates

Dislocations can get around obstacles by climb

Since vacancies increase exponentially with temperature, bypassing obstacles becomes easier at high temperatures.

In general, dislocations act as both source and sinks for point defects

In that view, the dislocation on the left climbs because entropy calls for more vacancies as with increasing T.

HW 10-3

You have, at 30 degree C, 1 cm3 of a metal with a dislocation density of 2.108/cm2. Half of the dislocations are screws, the other one are edge. The metal is simple cubic and, defect free, has density of 1023 atoms/cm3. The thermal expansion coefficient is 20 ppm/degree C. The melting point is 1040 degree C

a) What will be the volume of your sample at 1030 degree C if the material has a very high formation energy for vacancies of 100eV ? Answer in cm3, with 6 decimals

b) If, at 1030 C, the vacancy concentration is 10-4 what will be the volume of your sample ? Answer in cm3, with 6 decimals

c) If, upon cool down to 30 C, the vacancies all disappear by sinking at dislocations, what will the distance, in Angstrom, be the dislocations have to climb?

d) What will the volume of the sample be ? Answer in cm3, with 6 decimals

e) If all the vacancies disappear at the surface, what will the volume of the sample be at 30 C? Answer in cm3, with 6 decimals

Grain Boundaries

• Effective dislocation obstacles

• Dislocation pile up against grain boundary

• Force on leading dislocation is n times larger than on a single dislocation, where n is number of dislocations in pile up

• At a critical value, dislocations will pass. Thus smaller grains => higher yield stress

• Hall - Petch

Much of practical metallurgy is “grain refinement”

Make grain size small and keeping it small during thermal processing.

The smaller the better down to about 100 Angstrom. Below 100 Angstroms, grain boundary sliding will take over

Basic improvement:

Grain size

Icing on the cake:

All others….

High Strength Low Alloy Steels dominate modern Automobiles

And we can push this further!

Optimal Homework for Born Metallurgists

a) If we could process steels economically (!) with “nanotechnology” to a grain size of 20 nm, what would the yield strength be ?

And for those who like to think “green”

b) ~ 1/2 of the weight of a car is steel. How far could we reduce the weight of a 3000 lb car without sacrificing strength ?

c) if the milage 25 mpg, what would the new milage be?*

* Gas milage varies linearly with weight … See appendix to this lecture.

Grain boundary dislocations

A grain boundary is the border between two crystals of different crystal orientation.

A small displacement of the yellow crystal along the arrow will restore the same periodic structure in the grain boundary as before. The displacement is much smaller than the usual burgers vector. Because the energy of dislocation is b2 , a lattice dislocation entering a grain boundary will dissociate into a bundle of “grain boundary dislocations” The result is a small change in the tilt angle of the boundary

Bollmann has developed a beautiful theory about the geometry of grain boundaries and g.b. dislocations.

We need at least 5 independent slip systems to deform a poly crstalline material without opening gaps between grains

In hcp, the prismatic planes may (fingers crossed) be slip planes

Notes

• Because dislocations must either end at surfaces or be closed loops, interior dislocation nucleation requires forming a dislocation loop. Generally very difficult

• Dislocations can nucleate by the condensation of point defects.

=> Provided the point defects do not reach the surface

=> Loop may not be able to glide , check b vector !

=> A common occurrence in Si, because there are few sinks . . for point defects in a perfect crystal (swirl defects etc)

HW 10-5

Vacancies like to cluster out on densest packed planes. Assume this is the case in the dislocation loop shown on the proceeding page. The loop is circular in shape

a) Can any part of glide ? If yes, why? If not, why not ? (Limit yourself to one sentence)

There are countless other ways, but lets look at the Frank Read source

You need

• Shear stress (the crystal wants to deform)

• A dislocation pint at both ends

• The longer the dislocations, the easier to bow out - hence once trick to make metals strong is not to have long dislocation segments !

•Why the loop “closes and peels off” is a mystery to many students ….

• It does and the source goes and goes and goes… forever

Note

• Both are screw dislocations but they are off opposite sign !

• Opposite sign dislocations attract

• Opposite sign dislocation annihilate

The “cut” segments have the same sign and recombine. The source can go thousands of times.

In metals, dislocations can multiply like rabbits. There are many more other schemes to generate new dislocations from existing ones.

In dislocation free semiconductors, such as Si, it is quite another matter.

THE END