lec 12 and final

8/12/2019 Lec 12 and Final

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NENG 506 Mechanics of

NanomaterialsClass 11

J. Lloyd


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Point Defects and Dislocations

Point defects are those we just looked at Substitutional Impurities Interstitials Vacancies

The point defects have a hydrostatic compressive ortensile stress field associated with them dependingon relative size to the host


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Solution Hardening Two kinds

Substitutional solution hardening Interstitial solution hardening

Not all metal and alloys are suitable for this tooccur

15% rule


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Solution Hardening

Chemical Affinity If they like each other too much you get intermetallic

compounds In other cases you can get complete solid solubility

Valence Higher valence has higher solubility in lower valence Linear relationship with Cu

Cu-Ni


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Vacancies A vacancy is a point defect with a non-zero thermal

equilibrium concentration Can be derived from entropy of mixing

where Hf is the enthalpy of formation of a vacancy and C 0 isthe concentration number of lattice sites

Recall that the enthalpy has the PV term so that thevacancy concentration in the presence of an applied stressis

kT

H C C f

v exp

0

kT C C hvovs

exp


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Vacancies Also consider that since vacancy consists of a

vacant lattice site there is some collapse of thelattice surrounding the vacancy

The vacancy thus is a point source of tensile stress

As such, it will be attracted to compressive side of anedge dislocation This will produce a Cottrell atmosphere of a higher

than normal concentration of vacancies in this region

In addition the vacancies will be attracted to the coreregion So, what happens if a vacancy collapses into the

core


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Dislocation Climb Dislocations can move out of the glide plane,

but with difficulty The process must occur via diffusion of vacancies

to or from the dislocation core There is no significant thermal equilibrium

concentration of self-interstitials


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Dislocation Climb If climb occurs the glide plane changes

Keep in mind that the dislocation has extension into andout of the page

Thus when a single vacancy is absorbed or emitted causing climb,a jog is created

It takes the diffusion of many vacancies to enable a dislocation toclimb out of a crystal

The presence of a climb induced jog in an edge or mixeddislocation does not impede glide

But for a screw dislocation it does


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Dislocation Climb Depends on the diffusion of vacancies to or

from the core Therefore climb is thermally activated

Arrhenius Relation

kT H

D D D exp0


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Homework #5

Consider a dislocation situated near the edgeof a crystal and another one situated in thecenter with no applied stress

Comment on the relative behavior with respect toclimb of those two dislocations

Justify your answers with BOE (back of theenvelope) calculations


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Dislocation climb Let us put an applied stress, xx, on the crystal

We now have a stress at the dislocation core as well as anincreased stress in the dislocation stress field

There will now be a chemical potential given by the stress Recall that the equilibrium concentration of vacancies is thermally

activated with an activation enthalpy Thus the vacancy concentration in equilibrium with the stressed

core is now

so that depending on the sign of the stress the core will either emitor swallow vacancies to maintain equilibrium causing climb

kT C C xx

3exp0


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Dislocation climb The equilibrium vacancy concentration on the

x side of the crystal is increased similarly However at the y face where there is no applied

stress it is not increased We are infinite in z

Therefore there is a chemical potential for thevacancies to be emitted or swallowed by the core

xx xx


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Hydrostatic Pressure Let us place our crystal in under hydrostatic pressure

The vacancy concentration in contact with the core is now

But this increase is also experienced on all the other faces ofthe crystal and there is no net force Just like for yield, hydrostatic pressure does nothing for

climb

kT P

C C

exp0

xx= yy

xx= yy

xx= yy

xx= yy


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Dislocation Climb Force

Consider the applied stress in the x direction andif we remove or add a line of atoms to thedislocation at the core, the total energy change

per unit length is

where h is the distance advanced (lattice constant)

The force is then then simply (per unit length)

The force depends on the sign of the applied stress

bhW xx

b F xx


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The social life of a dislocation Let us have two dislocations of opposite sign on

parallel slip planes opposite one another

Recall

and that the equilibrium vacancy concentration is increased,thus they will annihilate reducing the total strain energy ofthe crystal


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Thermally activated glide

Takes place at jogs and kinks Jogs and kinks are generally not on a glide plane They can be eliminated by climb processes

(thermally activated) freeing a dislocation to glide


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Thermally activated glide Due to kink motion

The kink itself (in a screw dislocation) can beconsidered to diffuse conservatively (without avacancy involved)

The jump frequency is given as

The diffusion coefficient is then

And similar to the climb force

Invoking the Relation Einstein

The kink diffusivity in FCC materials has a very lowactivation energy

Not so for diamond cubic (Si) where thermally activatedglide is relatively slow even close to the melting point.

kT E k expn

kT E

a D k k exp2n

ba F

kT

ba D

kT

F Dv k

k dk


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Thermally activated glide processes


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Dislocation climb Self interstitials can be formed by extreme stress

Let us imagine a dislocation pinned at an impurity or at aprecipitate

Increased shear stresses and perhaps dislocation pileupscan increase the stress at the pinned location to whereinterstitials may be emitted by the climb process

This takes a lot of energy


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Precipitation Hardening Similar to solution hardening but in this case the

dislocation barriers are second phase particles Proper Heat Treating is crucial to the performance At higher temperatures, more solute is soluble, the

metal is heated in a solution anneal The metal is then quenched to a much lower

temperature, trapping the solute in a supersaturatedsolution.

The temperature is then raised slightly to allow thesolute to form precipitates The second phase will act as pins for dislocations

making the metal much stronger


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Precipitation Hardening


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The precipitate can pin dislocations making itMore difficult to yield. The more precipitates thereare, the more resistant the material to dislocationmotion

In addition, the tighter the spacing, the moreenergy required to free the dislocation


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The interparticle spacing, l , can be expressed as

where f is the volume fraction of particles of radius, r The shear stress needed for the dislocation to move

through the sea of particles is (Orowan)

So it is easily seen that a large number of fine particlesis the best thing for strength

f

r f 3

14 l

l Gb


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Precipitation Hardening If we also account for the increase in strain

hardening as well due to the stress fields of allthese dislocation loops we get The Orowan-Ashby equation

br Gb

ln13.0l


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Al/Cu

Duralumin (Al/Cu often with a little Mn and Mg for stabilizing) Aluminum alloy used primarily in

aviation First used extensively in rigid airships

Aluminum by itself is light weight, but relatively weak FCC atomic weight = 27

Alloying the Al with Cu produced an alloy that wasnearly as strong as steel when properly heat treated

It was found to get stronger just by letting it sit around atroom temperature

Discovered in 1903 Became the standard metal to make airplanes and is

still in use


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Precipitation Hardening(Coherency)

Al/Cu system Heat treating will strengthen

the alloy initially, but if theheating is continued the alloybecomes weaker

This is called over-aging There is an optimum

time/temperaturerelationship that must bemaintained to get the bestproperties


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Al/Cu

Guinier Preston Zones (GP Zones)


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Al/Cu The GP I zone is relatively soft (although harder than pure Al) as the solute begins

to cluster with other solute atoms in the supersaturated solution For Al/Cu room temperature is a mild annealing temperature. This is how it was

discovered in 1903, but dislocations were not yet known and the reason for thehardening was not understood

The GP II zone occurs when the precipitates begin to occupy {100} planes in thematrix in a manner coherent with the lattice.

This coherency very effectively pins the dislocations making it very difficult to deformthe crystal

This effectiveness is in part due to the strain in the local lattice near the dislocation. There is a slight reduction in yield point, but strain hardening is greatly enhanced leading

to strength overall rivaling steels


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Al/Cu

The GP III zone occurs when the precipitates grow to the pointwhere coherency is lost and dislocations can more easilymove through the lattice

Al/Cu is kept in a refrigerator until it is used so that it can beformed in the soft state, then strengthened later by annealingat mild temperatures

The use of Al/Cu alloys is not good for very high speed aircraft wherecompression at high Mach increases the temperature to over-agingconditions


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Dispersion Hardening

A similar case can be made for what is known asdispersion hardening Here the second phase is not created by the precipitation

from a supersaturated solution, but is created by the

presence of a different material that has been mixed intothe lattice Internal oxidation Mechanical mixing

Good for high temperature alloys


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NENG 303 Mechanics ofNanomaterials

Class 12

J. Lloyd


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Roscoe Turner


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Homework #5 Consider a dislocation situated near the edge

of a crystal and another one situated in thecenter with no applied stress

Comment on the relative behavior with respect toclimb of those two dislocations

Justify your answers with BOE (back of theenvelope) calculations

K


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Katana The most famous and sophisticated ancient

use of the metallurgical art Precipitation hardening at its finest Also known as a Samurai Sword

In the Russo-Japanese war of 1905, a Japanese Samuraiattacked a Russian machine gun and cut the barrel inhalf with his sword

K t


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Katana First you start with a very high grade ore (Tamahagane jewel

steel) (traditionally from an iron bearing river sand) and smelt

with charcoal for about three days Do not let it melt

This is to ensure that the carbon goes into solution atexactly the proper composition

It varies in the location of the Tamahagane from ~0.5% to 1.5%


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Katana

The steel is separated according to carbon content byits appearance and ease with which it can beseparatedIt is then heated, folded and pounded many times toextract all the impurities in the form of slag and to

disburse the carbon (spheroidization) Typically 16 times

This produces 2 16 65,000 layers


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Katana Then the lower carbon steel is placed in an envelope

of higher carbon steel and forge welded Producing a hard sharp more brittle case surrounding a

tougher more resilient core This produces the desired properties of a fine sharp edge

and resistance to breaking

K


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Katana The blade is then heat treated and properly quenched (cooled),

using a sophisticated process of painting with clay, to produce theproper microstructure and as an added benefit, the difference inthe properties of the two steels makes the blade curve elegantly

It is then polished to show the texture

A proper Katana can cost upwards of $50,000


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The transverse stress is significantly higher in thepipe than the pressure it is containing

This can be used to approximate encapsulated thinfilm conductors

Hoop Stress


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Creep Rupture

C


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Creep If a material is loaded above the yield point, before

catastrophic failure and kept there for a period of time, slowdeformation will take place until eventually failure will occur. Initially creep is rather rapid (Primary Creep), then it proceeds at a

constant rate (Steady State Creep) If held at constant load, the creep rate increases when the cross

sectional area is reduced (Tertiary Creep) At constant stress the creep rate remains constant


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Creep Rate Empirical

Andrade (1932) Cumbersome where 0 is the elastic strain limit and b and k are

adjustable constants Obviously this represents a lack of understanding

whats going on physically

Garofalo (1960) where 0 is the instantaneous strain, t is the limit of

stage 1 creep and dot is the steady state creep rate

t et k b

3

1

0 1

t e s

t r

t 1

0


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Creep Creep rate is temperature and stress dependent

Engineering expressions have 4 material dependentadjustable parameters!

kT H

GT A

elyalternativor kT H

AB

cm

n

s

c s

exp

exp


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Deformation MapHarold Frost (Dartmouth) and Mike Ashby (Harvard)


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Deformation Maps

Stress Temperature Map Strain Rate Stress Mapat a given temperature


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Deformation Maps

The general form is truefor most materials


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Creep Creep takes place by a number of deformation processes

Slip Sub-grain formation Grain boundary sliding Diffusive Creep

Many of the deformation processes only occur at hightemperatures

High temperature is operationally defined as above the meltingtemperature T mp

This temperature is, of course, different for different metals For Al, this would be 140C For Cu, this would be 400C For Fe, this would be 635C For Pb , this would be 27C . Room temperature


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Power Law Creep

Complicated constant temperature

where n varies between 3 and 10 Complex variety of mechanisms operating Recent modeling (2012) shows that

power law creep can be accounted forby dislocation climb

n s

G


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High Temperature Creep Slip

New slip systems can become activated at hightemperature that do not operate at low temperature

Al

At low temperature slip plane is {111} Above 260C, we add {100} and {211} planes The CRSS is reduced with temperature

Zr CRSS = 1kg/mm2 at 77K

CRSS = 0.2kg/mm2

at 575K (300C) CRSS = 0.02 kg/mm 2 at 1,075K (800C) Dislocation sources can be more active due to

increased climb at jogs/kinks etc.

Creep


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Creep Sub-boundary formation

At higher temperatures dislocations can climb andarrange into energetically favorable positions

This provides bending of the crystal Which makes the grain boundaries very uncomfortable

C


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Creep Grain Boundary Sliding

At high temperatures another mechanism candominate called grain boundary sliding

Shear occurs along the grain boundary Mostly at relatively slow creep rates The precise mechanism is controversial

The Thermodynamics of Stressed Solids


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Thought Experiment

Imagine a mobile solute component in a liquid which has nosolubility in the solid (hashed line). The solute, however, hassolubility in the solid. It can be shown that under the samepressure on the different faces the chemical potential is thesame

Perpetual motion machine

The Thermodynamics of Stressed SolidsJ.C.M. Li, R.A. Oriani and L.S. Darken

Zeit. Physik. Chem. Neue Folge, 49, 271 (1966 )

Li Oriani and Darken


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Li, Oriani and Darken For a not necessarily uniformly stressed solid

We will consider a closed isothermal system inequilibrium

Only elastic stresses will be applied Satisfies the conditions reversibility and constant

temperature

The first two terms can be rearranged to produce theexpression for the free energy

Thus the last term can be considered the free energy butaccording to Moutiers Theorem the work to obtain this isindependent of path

Note also that the condition for equilibrium for a chemicalprocess is that the change in free energy is at a minimum

dW dS T dU

Li O i i d D k


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Li, Oriani and Darken Determine the work to add or remove a component by

two methods and equate them since the theoremstates that the work is path independent The first path with be the direct transfer of a quantity of

the solute to the fluid The work done on the body to add or remove the nk of

solute is defined as Wk This is all the work/energy needed to remove or add the quantity of

solute The work done on the fluid is the work of expansion (or

contraction) with the addition of the solute to the fluidat the pressure of the fluid, this is W

k

F Thus the total work for this procedure is

k k F k nW W

Li Oriani and Darken


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Li, Oriani and Darken The second pathway will be to completely relax the stresses in the

solid, but not in the fluid (by magic) and transfer some solute fromthe solid to the fluid, then reapply the stress to its original value

The amount of work done by the relaxation is w. Now the chemical potentials ( = dE/dn k)of the solute on the surface

of the unstressed solid k0 and the pressurized liquid k can beexpressed.

The work of transferring the solute from the solid to the liquid is then(

k-

k

0)nk

Now we reapply the stress and also add the work of expansion for atotal by this procedure

Equating the work from the two procedures and realizing atequilibrium they must vanish overall

k F k

k k k nW n

w m m

0

k

k

k k W n

w 0m m


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Li, Oriani and Darken

Examining this we realize that the one term is thepartial molal strain energy and the other is thenormal stress times the molal volume

where is the chemical potential in the stressedsolid,

0 the chemical potential in the unstressed

solid, w is the strain energy with the addition (orsubtraction) of the solute and is the partialmolal (or atomic) volume (depending on units)

m m w0


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Li, Oriani and Darken What does this tell us

For one, this tells us for a given stress that the chemicalpotential and therefore the solubility will change with thenormal stress component (or pressure)

If the solute increases the volume of the solid (ie thesolute atoms are bigger than solvent atoms) tensile stresswill increase solubility and compressive stress reducesolubility

Vice versa The story of a vacancy is interesting

The creation of a vacancy increases the volume, therefore it ispromoted by tensile stress

But a vacancy is a point source of tensile stress so it is attracted toa compressive stress field.

Diffusive Creep


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Diffusive Creep Diffusion proceeds down a chemical potential

gradient In the absence of all other gradients that is the

concentration gradient

Entropy gradient However, any contributor to the chemical

potential can drive diffusion Stress gradients are most important

m w

St i E g


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Strain Energy The strain energy is given as

The stress is that of a vacancy, but the strain energy is mitigated bythe ratio of the stress to the modulus

Stresses are generally in the range of tens or hundreds of MPa Moduli are in the range of tens or hundreds of Gpa Thus the strain energy term is usually negligible

Thus the chemical potential gradient due to stress is usuallydescribed as

Where the stress is the normal component to the boundary

E w 2

2

m

Stress Gradients


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Stress Gradients Mass diffusion is due to the motion of vacancies

If the vacancies are going to the left, the atoms are goingto the right

Vacancies annihilate when they meet a surface or a sinklike a grain boundary

Consider a polycrystalline metal undergoing a uniaxial tensilestress in the y direction

The equilibrium vacancy concentrationon grain boundaries parallel to the appliedstress in the y direction is

kT

H C C

kT C C

f L

v

exp

exp

0

0


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Diffusive Creep Since the diffusive flux of vacancies (or atoms)

will be

the crystals will grow in thedirection of the appliedtensile stress

This is strictly true at alltemperatures and all stresslevels

But at low temperatures it maybe exceedingly slow

vvvv C D J m


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Final Exam


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Question 1

WaferNormal

SidewallNormal

AlongTrench

Question 1


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Question 1(10 Points)

On the previous page we have EBSD (Electron BackScattered Diffraction) images of a polycrystalline Cumaterial. The images are taken such that the orientation ofthe grains are analyzed with respect to three orthogonaldirections (up, right, and out of the page). The color code is

explained in the pie slice, where the crystal orientationwith respect to the direction is denoted. Ie . If its blue, thecrystal is oriented with (111) plane perpendicular to thatdirection

Why do the red grains not change color when the othersdo?

Question 2


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Question 2(20 points)

We have a single crystal of Al in a tensiletester. We are applying a tensile stress, y perpendicular to the 212 plane and we have

found that slip has occurred. What is the shear stress that produced the

slip?

Question 4


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Given a crack in an Aluminum wing spar that is 2cm long. What is the stress concentration for this crack?

Let us assume atomic dimensions at the crack tip(unrealistic perhaps) but

What size drill would you use to reduce the stressto ten times the applied load with a pilot hole?

(10 points) For three times the applied stress? (5points)

Q15 points)

Question 3


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Question 3(5 points)

Discuss why quasicrystals are brittle Also discuss the modulus of elasticity for a

quasicrystal as compared to a real crystal

lec 12 and final

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