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Materials IA > Course D - Microstructure > Flashcards

Course D - Microstructure Flashcards

1
Q

how should you prepare a sample for reflected light microscopy

A

1) mount, grind, polish the sample to achieve a flat, level surface

2) etch the surface to highlight important features - usually done using chemicals

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2
Q

what does etching do

A
  • etching is treating a sample with a chemical to highlight features, the etching process occurs quicker at grain boundaries and other impurities so they will reflect light differently
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3
Q

what sort of samples are required for transmitted light microscopy

A
  • Transmitted light microscopy relies on detecting the light that has passed through a sample
  • very thin samples required
  • good for showing grain sizes
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4
Q

what causes contrast in transmitted light microscopy

A
  • differences in absorption
  • differences in birefringence
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5
Q

what properties must a material sample have for SEM

A
  • should be prepared through polishing and etching
  • cannot be conductive
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6
Q

what is atomic force microscopy (AFM)

A
  • a very sharp tip is mounted on a cantilever and moved across the surface of a material
  • this measures the height of the sample
  • atomic level resolution can be achieved
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7
Q

what do pressure-temperature phase diagrams show

A
  • for a given composition they show which phase is most THERMODYNAMICALLY stable at some temperature and pressure
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8
Q

why is the most thermodynamically stable phase not necessarily the phase present at a given temp/pres.

A
  • kinetics can prevent a phase from forming
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9
Q

what do temperature-composition phase diagrams show

A
  • they show which phase is most stable at some temperature and composition for a constant, given pressure
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10
Q

what do we mean when we talk about a system

A

“The subject of a thermodynamic analysis”

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11
Q

what are the 4 types of equilibrium (or not equilibrium) that a phase can be in at any given point

A

1) stable equilibrium - in a potential energy minimum, the overall minimum for the system

2) unstable equilibrium - at a potential energy maximum, unstable to any perturbations

3) not in equil - not at a max/min, there’s a driving force

4) metastable equil. - stable to small perturbations but not lowest energy state, local potential energy min. but not overall min.

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12
Q

what are the first and second laws of thermodynamics

A

1st law of thermodynamics: Total energy of universe is conserved

2nd law of thermodynamics: Entropy of the universe cannot decrease

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13
Q

what is the internal energy, U of a system, give the differential form of the equation for U

A

internal energy of a system, U = potential energy + kinetic energy

dU = δq + δw

q = heat
w = work done

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14
Q

define heat, give a differential equation for it

A

“The energy that ‘flows’ across a system boundary in response to a temperature gradient”

δq = CdT

C = heat capacity

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15
Q

define work done, give a differential equation for it

A

“the energy that flows across a system boundary in response to a force moving through a distance”

δw = -p dV

p = pres.
V = vol.

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16
Q

using the differential definitions for heat and work done, redefine the differential equation for internal energy

A

we know
dU = δq + δw
so
dU = CdT - pdV

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17
Q

give the overall equation for enthalpy, H

A

H = U +PV

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18
Q

give a differential form of the equation for enthalpy

A

dH = dU + PdV + VdP

and dU = CdT - pdV
so

dH = CdT + VdP

i.e. enthalpy is the heat transferred at a constant pressure

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19
Q

define entropy, how does it link to the 2nd law of thermodynamics

A

entropy, S, is a measure of disorder

from 2nd law of TD we get
dS(univ) > 0

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20
Q

what is entropy at equilibrium

A

dS = δqrev / T

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21
Q

define Gibbs free energy and give both differential and non-differential eq.’s

A

Gibbs free energy, G, is the energy available to do useful work

  • G allows us to find the equil. state of a system, using only properties of the system, not the surroundings

G = H -TS
dG = dH - TdS - SdT

we know dH = δq + VdP
so
dG = δq + VdP - TdS - SdT

at const. temp. and pres.
dG = δq -TdS

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22
Q

what can we say about G at equil for a single phase

A

dG = 0
Gibbs free energy, G, tends to a minimum at equil.

for spontaneous processes dG<0

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23
Q

for a single phase of constant composition, what can we say about the variation of G with temp

A
  • it will ALWAYS decrease
    G = H-TS

its a straight line
y-int = H
grad at any point (draw tangent) = -S

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24
Q

what can we say about which phase is most stable at a temperature for a system of two phases of const. identical comp (e.g. liquid, solid of same phase)
link to G-T graph

A
  • we have two curves on a G-T graph
  • whichever is the lowest G phase at any point is the more stable phase
  • the point at which the two lines cross is the equil. temp

phase 1: G1 = H1 - TS1
phase 2: G2 = H2 - TS2

difference in G:
ΔG = ΔH - T ΔS

we are interested in the sign of ΔG

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25
Q

how does G change with composition for a mechanical mixture of two phases A and B

A
  • it simply changes as a weighted average of the two phases
  • i.e. at 100% A, mark Ga on the y-axis, at 100% B, mark Gb on the y-axis, draw a straight line between them
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26
Q

what are the three things we must consider when determining the free energy curve for a single phase of A and B (or the change in free energy of mixing)

A

1) an enthalpy change associated with the fact that A-B interactions are different to A-A or B-B interactions, ΔHmix

2) an enthalpy change due to the random mixing of A and B atoms, ΔSmix

3) combining these to form a change in free energy when mixing

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27
Q

derive the expression for ΔHmix when forming a solution of A and B from a solid solution for 1 mole of atoms

A

ΔHmix = Hs - Hmm

let:
XA = fraction of A atom
XB = 1-XA = fraction of B atoms
EAA = interaction energy between A-A as nearest neighbours
EBB = “”
EAB = “”
coordination number of A,B = Z

Hmm = (NAZ / 2) (XA EAA + XB EBB)
Hs = (NAZ / 2) ( XA^2 EAA + XB^2 EBB + 2 XA XB EAB)

combining and rearranging using XA + XB = 1 gives
ΔHmix = Hs - Hmm = (NA*Z /2) XA XB (2EAB - EAA - EBB) = XA XB ψ

ψ = (NA*Z /2) (2EAB - EAA - EBB)

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28
Q

what 3 cases does the expression for ΔHmix imply (regarding the sign of ψ)

A

1) ψ = 0 , EAA = EBB = EAB, totally randomly mixed solid solution

2) ψ < 0, EAB < EAA, EBB, totally ordered alternating A-B structure, ΔHmix < 0

3) ψ > 0, EAA,EBB < EAB, totally separated, unmixed, ΔHmix > 0

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29
Q

derive the expression for ΔSmix when forming a solution of A and B from a solid solution for one mole

A

we start with
S = kln(ω)
where ω is the number of distinguishable ways of arranging atoms on the available sites

Ωmm = 1
(A atoms on A sites, B atoms on B)

Ωs = NA! / ((XANA)! ((1-XA)NA)!)

ΔSmix = kln(Ωs) - kln(Ωmm) = kln(Ωs)

once the Stirling approximation has been used this gives

ΔSmix = -R(XA ln(XA) + XB ln(XB))

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30
Q

derive the expression for ΔGmix when forming a solution of A and B from a mechanical mixture

A

ΔGmix = ΔHmix - TΔSmix

= XA XB ψ + RT(XA ln(XA) + XB ln(XB))

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31
Q

consider the same 3 cases regarding the sign of ψ, explain what this means for when forming a solution is favourable

A

1,2) ψ = 0 or ψ < 0, ΔHmix <= 0, ΔSmix, >0, ΔGmix<0 at ALL TEMPS – always favourable to form solid solution

3) ψ > 0, ΔHmix, ΔSmix >0, ΔGmix sometimes<0 depending on temp, shape of curve depends on temp

BEST TO SEE NOTES FOR THIS

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32
Q

in case 3) (ψ > 0, ΔHmix, ΔSmix >0, ΔGmix sometimes<0 depending on temp, shape of curve depends on temp), how do we know when it’s better in 2 phases or a solid solution

A
  • at high T, the curve will have 1 minimum, it is always favourable to form a solid solution in this case
  • at lower T, the curve will have two minimums, between the two minimums it is NOT favourable to form a solid solution, elsewhere it is

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33
Q

what process can we apply to a two phase mixture with lines plotted on a G against comp diagram, to work out whether it is more favourable to have one of the two phases or a mixture

A
  • draw a common tangent between the two curves of interest
  • mark the compositions of the two phases where the common tangent meets their curve
  • if the composition is between these two then it will split into two different phases of the compositions marked, the proportions are just a weighted average
  • if it is outside of this region, then it will remain as the relevant phase

NOTE: the compositions that the common tangent touches are NOT necessarily the minimum of the curve

BEST TO SEE NOTES FOR THIS

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34
Q

Define a phase

A

A phase can be defined as a portion of a system whose structure, properties and
composition are homogeneous and which is physically distinct from other parts of the
system.

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35
Q

how do G curves relate to a phase diagram

A
  • G-composition plots can be obtained for a range of temperatures
  • for each plot the points where there is the boundary between two phases and one phase can be determined then plotted on a temperature- composition plot as points
  • this can be repeated and phase boundaries determined
  • or the same can be done in reverse to determine the G-Composition plot
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36
Q

using a Temp-comp. phase diagram, if you’re in a two phase region, how can the proportions of each phase be determine

A
  • use the lever rule
  • basically take weighted averages of each phase to give the correct overall composition
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37
Q

what can we say about the free energy curves of the phases in a ‘complete solubility in liquid and solid phases’ phase diagram

give some features of the phase diagram

A
  • this is the case where 3) ψ <= 0 for both liquid and solid phases
  • hence the curves for both phases only have 1 minimum
  • the phase diagram has a solid phase section at bottom, a liquid phase section at top and a tilted disc shape section of (L+S) spanning the whole composition range midway
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38
Q

what is the key feature of a Eutectic system

what does the phase diagram look like

A
  • the defining feature of a Eutectic system is that at some (Eutectic) composition, solidification occurs at a single temp. and we have L —> α + β
  • the phase diagram has thin(ish) p-shaped sections at either extreme of composition, a large liquid section at top, a large α + β section at bottom and then two sections which meet at the Eutectic point of L+α and L+β
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39
Q

describe the process of Eutectic solidification

A
  • as the alpha phase forms in the liquid, B-atoms are rejected out of the α
  • a B-rich section of liquid then surrounds the α phase
  • this causes β to form, generally adjacent to the α
  • α + β forms
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40
Q

describe how the Eutectic microstructure forms

A
  • α, β form simultaneously, forming a microstructure of alternating lamallae (plates) of α, β

As α, β form:
- B-atoms are pushed out the front of the α phase, and vice versa as they advance
- the atoms must redistribute in the liquid to allow solid phases to grow correctly
- the extent to which they can redistribute determines the length-scale (thickness) of the plates

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41
Q

give an example of a Eutectic system and a common use of eutectic alloys

A

Pb-Sn is a eutectic system

Eutectic systems (but not necessarily Pb-Sn) are often used as solders because if not at eutectic comp, they will go mushy before solidifying fully so can still be manipulated on solidifying

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42
Q

how do eutectic systems (once solidified) tend to appear under a microscope

A

stripy due to the lamallae

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43
Q

describe what microstructure forms and how it does when cooling a eutectic system not at the eutectic composition

A
  • initially liquid single phase
  • moves into two phase region e.g α+L
  • on cooling through α+L, some α ‘nuclei’ form, drawing a tie line generally shows more liquid than α is present
  • as the α forms, it pushes more β into solution, the composition of the liquid changes/ becomes more B-rich
  • the α that forms in the α+L section is called primary alpha
  • as the solution approaches the eutectic temp, the liquid approaches the eutectic comp.
  • on passing through the eutectic comp, the remaining liquid undergoes a eutectic transformation and the standard lamallae structure forms between the primary alpha sections
44
Q

how can we determine a phase diagram from a cooling curve

A

Obtain cooling curves for samples at different compositions:
- for a pure sample (or eutectic), phase transformation occurs at a specific temperature
- for a mixture, phase transformation occurs over a range of temperatures
- any change in gradient represents passing over a phase boundary

45
Q

define equilibrium cooling, what compositions will be present

A

equilibrium cooling is when the composition of the solid remains uniform and changes as solidification progresses:

  • the composition of both the solid and liquid at a temperature is directly given by the tie line that can be drawn
46
Q

what must occur to have equilibrium cooling

A
  • a very slow cooling rate (to allow for atomic diffusion)
47
Q

what are the 4 things that can occur when non-equilibrium cooling occurs (2 are only for Eutetic)

A

1) Coring
2) dendrite formation
3) finer Eutectic pattern
4) more Eutectic phase than the lever rule predicts

48
Q

explain how coring occurs/ the effects of it

A
  • if equil. is not maintained, the first solid that forms will retain a higher conc. of B atoms (assuming the β phase is forming)
  • this means the fractions of liquid/solid are no longer given by simple tie lines on the phase diagram
  • as more B is trapped inside the solid, the liquid is more A-rich than is expected at equil.
  • true compositions depend on diffusion rates within the solid
  • if slow diffusion, we have significant conc. grad in solid
49
Q

explain the formation of dendrites in non-equilibrium cooling

A
  • as a solid grows on cooling through the α+l region, it rejects β atoms
  • this gives a B-rich region surrounding the α, there’s a conc. grad.
  • If there are any small, random pertubances on the advancing solid, they find themselves in a less B-rich region
  • this encourages growth and forms a +ve feedback system
50
Q

in what direction do dendrites tend to grow in

A
  • some preferred crystallographic directions
51
Q

state the two main diffusion mechanisms

A
  • interstitial
  • substitutional
52
Q

explain how interstitial diffusion occurs, what can we say about its rate

A
  • smaller atoms can sit in interstitial sites and ‘jump’ from one interstice to another to diffuse
  • generally fast as the atom can move in any direction at any time
53
Q

explain how substitutional diffusion occurs, what can we say about its rate

A
  • atoms move from one site to an adjacent vacant one
  • requires vacancies so generally slower
54
Q

what does Fick’s first law show

A
  • in a concentration gradient, there’s a net flux of solute through the solid
  • Fick’s first law describes this as
    J = -D (∂^2C / ∂x^2)

D(interstitial) > D(substitutional)

55
Q

what does Fick’s second law show

A
  • it shows that when the system is not in a steady state, conc. grad. and flux vary with time, we use Fick’s second law for this

∂C/∂t = D (∂^2C/ ∂x^2)

56
Q

what is the solution to find distance diffused from diffusivity and time

A

x = sqrt(Dt)

note this is only approximate

57
Q

how does the diffusion constant vary with temperature

A
  • it follows an Arrhenius relationship

D = Do exp(-Q/KT)
OR
D = Do exp(-Q/RT)
depending on units

Q = activation energy
T = temp
R = ideal gas const.
K = Boltzmann constant

58
Q

what is the driving force for the nucleation of a new phase, give the equation that describes it

A

“The driving force for the nucleation of a new phase is the change in free energy ΔG”

at equil, Te
ΔG = ΔH-TeΔS = 0
ΔH = TeΔS
so at a given temp

ΔG = (Te-T)ΔS

i.e. driving force is proportional to deviation from equil. temp.

59
Q

what are the two main factors that we should consider when thinking about nucleation (either type)

A

1) the change in energy per unit vol.on transformation to a new phase, ΔGv

2) γ, the surface energy per unit area between phases

60
Q

give the expression for the energy of homogenous nucleation (sphere)

A

Wn = 4/3 π r^3 ΔGv + 4π r^2 γ

61
Q

what is the critical radius for homogenous nucleation, how can we find the expression for it, give the equation

A

we know the graph of Wn against r has a maximum, this is at r*
- for r > r, the nucleus naturally grows
- for r < r the nucleus naturally shrinks

doing dWn / dr = 0 yields
r* = -2γ /ΔGv

and
Wn* = (16π/ 3)(γ^3 /ΔGv^2)

62
Q

what is the issue with the model for homogenous nucleation, how can we edit it

A
  • it assumes no strain term when forming a new phase, this is unlikely, we can add a strain term U

Wn = (4πr^3 / 3)(ΔGv + U) + 4πr^2 γ

  • both r* and Wn* increase

NOTE: in the case of ferromagnetism or ferroelectric materials, this term could represent stray field energy

63
Q

what are the two factors we must consider when thinking about the rate of homogenous nucleation

A

1) population of critical nuclei (r<r), proportional to exp(-Wn / KT)
2) rate at which particles can add to make nuclei post-critical, proportional to exp(-Q/KT)

64
Q

give the overall equation for nucleation frequency (the homogenous case), what can we say about the nucleation frequency close to or far away from Te

A

nucleation frequency = I
I = Cn exp( - (Wn*+Q) / KT)

  • no nucleation at Te
  • increases (due to increased driving force) moving away from Te
  • if too far below Te atomic motion is too low so nucleation rate drops
65
Q

what is heterogenous nucleation

A
  • this is where nucleating is catalysed by the presence of defects in the original phase, spherical ‘caps’ form on defects such as the wall
66
Q

what are the relationships between critical radius and critical work done in homogenous and heterogenous nucleation

A

r(hetero) = r(homo)

Wn(hetero) = Wn(homo) ((2+cosθ)(1-cosθ)^2 /4)

θ = angle between surface and cap at base

67
Q

what can we say about the ease with which heterogenous and homogenous nucleation can occur, which occurs faster

A
  • heterogenous nucleation has a lower critical work done so will generally occur quicker
  • often occurs at grain boundaries, triple points, container surfaces and at defects
68
Q

what do we need to consider when deriving the growth rates of a new phase

A
  • the energy barrier to forming the new vs. old phase / the asymmetry between them

Forward jump rate (old –> new) proportional to exp(-Q/KT)

Backwards jump rate (new –> old) proportional to exp( -Q+VAΔGv) / KT)

hence growth rate, ν, is proportional to the difference between the two

ν = Cg exp(-Q/KT) [1-exp(VAΔGv/KT)]

VA = vol. per atom

69
Q

what can we say about the growth rate at different deviations from Te

A
  • no growth at Te
  • max growth near Te
  • low atomic mobility so little growth a distance from Te
70
Q

what are the 4 types of solid-solid interface

A

1) coherent
2) strained coherent
3) semi-coherent
4) incoherent

71
Q

give the features of a coherent solid-solid interface

A
  • perfect alignment of lattices
  • ideal, lowest energy
72
Q

give the features of a strained coherent solid-solid interface

A
  • it is likely that any coherent interface will have some elastic strain
  • higher energy, strain energy increases with size of particle
  • this causes a transition to semi-coherent
73
Q

give the features of a semi-coherent solid-solid interface

A
  • introduction of dislocations
  • reduces overall elastic strain
  • contributes to energy in other ways
  • occurs where there’s more mismatch or a larger precipitate
74
Q

give the features of an incoherent inteface

A
  • Large mismatch
  • V high energy
75
Q

we tend to model precipitates forming as being spherical, why is this not usually the case

A
  • spherical precipitates minimise surface area but we must consider interfacial energy too
  • atoms are more likely to add to a higher energy incoherent interface so they tend to grow faster
  • this often leads to disk like shapes
76
Q

what are Widmanstätten microstructures

A
  • plate-like precipitates often lie in particular crystallographically related orientations
  • e.g. precipitation of bcc (110) plane on an fcc close packed (111) plane
  • this allows for the matching of close packed lattice rows/planes and is thus lower energy
77
Q

give the main features of the Fe-C system, what phases of steel are (that we are interested in)

A
  • Eutectoid transformation (723°C, 0.8wt% C)
  • Eutectic transformation (1130°C, 4.3%C)

4 main phases:
- α-Fe, Ferrite, bcc, V low C%, up to 910°C
- γ-Fe, Austenite, fcc, up to 1.7wt%C, 723-1492°C
- δ-Ferrite, bcc, higher temp, 1400-1535°C, low wt%C
- Cementite, Fe3C (higher carbon content)

78
Q

describe the Eutectoid transformation of steel, what phases form, what is the resulting microstructure

A

Eutectoid transformation = Fe-0.8wt%C, 723°C

Austenite –> Ferrite + Cementite
γ —> α + Fe3C

  • on the eutectoid transformation, Ferrite rejects C atoms, Fe3C accepts them
  • lamallae structure forms
  • this is called Pearlite
79
Q

explain the hypereuctectoid transformation in steels, give the features of the resulting microstructure

A

Hypereuctectoid transformation from about 1000°C, Fe-1wt%C

1) initially we have large grains of γ
2) on the γ–> γ + Fe3C transformation (about 740°C), Fe3C precipitates at grain boundaries
3) At the eutectoid temp, the remaining γ undergoes γ–> α + Fe3C forming pearlite

80
Q

explain the hypoeuctectoid transformation in steels, give the features of the resulting microstructure

A

Fe - 0.1wt%C

1) Ferrite produced below about 850°C
2) about 10% γ still remains at Eutectoid temp
3) at eutectoid temp, remaining γ undergoes γ–> α + Fe3C forming pearlite

81
Q

what is a time-temperature-transformation (TTT) diagram

A
  • for first-order phase transformation you need nucleation and growth
  • we can describe the overall rate of transformation using t(y), the time taken to transform a fraction y of the material from one phase to another at a given temp.
  • an isothermal transformation diagram shows t(y) as a func. of temperature
  • contours are plotted for given y values, C-shaped curves
  • generally plotted on log-log axis
82
Q

what occurs if the cooling rate of a substance is sufficiently fast to avoid the ‘nose’ of the TTT diagram

A
  • Quenching occurs
  • none of the new phase forms, often a new metastable phase will form
  • technically TTT diagrams are isothermal so should not be used for this purpose but a rough cooling rate can be estimated using

dT/dt = ΔTnose/tnose

83
Q

what is martensite

A
  • martensite is a metastable phase of steel which forms on rapid cooling, sufficiently quick that no pearlite forms from austenite
  • it forms through a diffusionless mechanism
84
Q

what occurs in a martenistic transformation

A
  • it is a displacive phase transformation, it does not require diffusion
  • in fcc austenite a carbon atom sits in an octahedral interstice
  • on slow cooling usually pearlite is formed so the carbon forms Fe3C precipitates but on rapid cooling a displacive phase transition occurs and a bct structure forms
  • the carbon atom now sits in the centre of the bct martensite
  • there is much more strain
85
Q

explain why martensite forms and is brittle

A
  • austenite and ferrite have massively different solubilities for carbon so when austenite is cooled rapidly, it forms a very super-saturated solution
  • it effectively forms a ferrite structure but with too much carbon contained
  • the carbon sits in the centre of the unit cell and changes it from bcc to bct
  • this causes strain fields and locks dislocations
  • making it very hard and brittle
86
Q

what can occur when martensite is tempered/ annealed

A
  • the carbon diffuses and forms precipitates

i.e. ferrite and cementite form
NOTE: the cementite does not have the pearlite structure in this case

  • the resultant structure after tempering is still hard but much less brittle
87
Q

what are the 3 categorisations of cooling rate, give examples of how each is achieved

A

1) slow cooling e.g. leave to cool in air
2) moderate cooling e.g. cool in oil
3) rapid/ fast cooling e.g. cool in water

88
Q

what are the 2 general consequences of faster cooling rates

A
  • stronger driving force for nucleation so many smaller nuclei form
  • diffusion is very slow at lower temperatures so smaller grains form
89
Q

explain how casting works, describe the 3 ‘sections’/ grain forms of a cast structure

A
  • it is a common metal processing route, liquid metal is poured into a mould
  • solid nucleates and grows, often heterogenously from the walls of the container

1) small equiaxed grains
- occur at edge
- fast cooling rates near edges and sites for heterogenous nucleation means nucleation is favoured over growth so lots of small grains form

2) columnar grains
- results from grain growth along preferred directions
- along temperature gradient

3) large equiaxed grains
- slowest cooling rate in centre means growth is favoured over nucleation

90
Q

what occurs in glass formation

A
  • if a fast enough cooling rate can be achieved then crystallisation can be avoided altogether
  • for metals it is difficult to form but small quantities can be made by dropping metal onto a spinning, cooled wheel
91
Q

what can we say about the effect of cooling rate on eutectic and eutectoid microstructures

A
  • faster cooling rate leads to finer lamallae structure, finer spacing λ
  • the rate at which a eutectic/oid phase grows is dependent on how fast diffusion in front of growing lamallae can occur
  • at low temps, this rate is low so λ is smaller
92
Q

how can we estimate the minimum lamallae spacing λ at a given temperature

A
  • consider the solidification on 1m^3 of liquid
  • forms eutectic lamallae
  • driving force ΔGv at T
  • 2/λ m^2 of α-β interface for each m^3 of eutectic, let γ(α-β) be the surface energy for this interface

hence

ΔGtot = ΔGv + 2γ(α-β)/λ

min is at ΔGtot=0
so
λmin = -2γ(α-β)/ΔGv

93
Q

what is hot rolling

A
  • material is pushed through rollers whilst still hot
  • deformation process
94
Q

what is forging

A
  • deformation process
  • the material is squashed in a press between die
95
Q

what is extrusion

A
  • deformation process
  • the material is pumped through a shape
96
Q

what is deep drawing

A
  • deformation process
  • a sample is held between die and ‘punched’
97
Q

in the Al-Cu system (important in aircraft) what is the general method to make commercial alloys (4wt% Cu)

A

1) anneal at 540°C to have all Cu in solution
- homogenises solution, removes coring

2) quench to room temperature to give supersaturated solid solution (SSSS)
- forms SSSS, prevents θ phase from forming

3) anneal at < 200°C to form ppts
- low temps, more nucleation, less growth

98
Q

focusing on the annealing part of forming Al-Cu alloys, what is the first phase to form

A
  • a metastable phase
  • Guinier-Preston Zones (GP zones) form first
  • these are generally single atomic layers of Cu-rich material
  • form by homogeneous nucleation
  • coherent interface
99
Q

why doesn’t the θ phase form immediately when annealing a SSSS of an Al-Cu alloy

A

θ = tetragonal
α = fcc

  • incoherent faces, large barrier to nucleation
100
Q

focusing on the annealing part of forming Al-Cu alloys, what is the second phase(s) to form

A

from GP zones and α1, θ” and α2 forms:
- θ” = tetragonal crystal structure
- layers of Cu atoms separated by 3 atomic layers of Al
- Remains coherent but (001) face is least strained
- (100) and (010) are more strained, grow faster, leads to disc shapes

101
Q

focusing on the annealing part of forming Al-Cu alloys, what is the third phase(s) to form

A
  • if annealing continues then the α2 and θ” from α3 and θ’
  • the (001) is still semicoherent
  • the (010) and (100) now incoherent
  • θ’ nucleates heterogeneously on crystal defects
102
Q

focusing on the annealing part of forming Al-Cu alloys, what is the fourth/final phase to form

A

α3 and θ’ form α4 and θ
- it’s bct so all interfaces are incoherent
- particles nucleate heterogenously

103
Q

what is the general process under annealing to form θ

A

GP zones —> θ” —> θ’ —> θ

104
Q

why does the formation of θ occur in 4 different stages and not in one

A
  • the individual transformations have low Ea compared to a straight change from α to θ
  • so the rate via the intermediates is greater
105
Q

when calculating the compositions and proportions of a hypoeutectic microstructure what temperature should we do it for

A
  • the eutectic temperature
106
Q

what does the cooling curve for a hypoeutectic (or hyper) alloy look like

A
  • initially linear negative
  • change of gradient, still linear negative as entering L+α region
  • then horizontal section where liquid undergoes eutectic transformation
  • then decreasing again

Materials IA (9 decks)

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