27-302, Microstructure & Properties II, A.D. Rollett

Fall 2002

Homework 4, due Monday, Nov. 25th

Topic: properties dependent on precipitates

1) Strengthening, Coarsening: [50 points][This is essentially question 5.39 in Courtney's book on Mechanical Properties, for thosewho are interested in knowing where it came from] The copper alloy C82500 has acomposition that is approximately 98 wt.% Cu with 2 wt.% Be. This alloy can beprecipitation hardened, see the phase diagram below, and has been traditionally used fortools with special characteristics such as non-sparking hammers. The precipitate isdenoted as g2 and has the composition CuBe. CuBe has a CsCl structure which is a simplecubic lattice with Cu atoms at the corners and a Be atom in the body center position.

Notation:D Diffusion coefficientQ Activation energy for diffusiontc Time required to reach the equilibrium (level rule) volume fraction of precipitate.r particle radiusK constant in the coarsening equation (also degrees Kelvin!)L' mean free path between particlesf volume fraction of particlesecoh misfit parameter in the equation for coherency hardeningG shear modulusb Burgers vector (magnitude)tcoh shear stress (critical resolved shear stress) due to coherency hardening

<M> Average Taylor factor (=3.07)

(a) Describe the steps that you would take to precipitation harden C82500. Stateexplicitly the temperature range that you would use for each treatment.(b) Calculate the time, tc, required to precipitate out the equilibrium volume fraction ofprecipitate at 643 K. Plot the volume fraction as a function of time at 643K; obviously avolume fraction greater than one is not meaningful. Use log-log axes on your plot.Assume that the precipitate volume fraction increases with aging time until theequilibrium volume fraction has been reached assuming that the precipitates grow asspheres. Use the equation for diffusion-controlled 1D growth for the radius of eachprecipitate; this introduces only a minor error in the calculation from ignoring theimpingement of the diffusion fields of neighboring precipitates. Assume site saturatednucleation of the CuBe precipitate with a particle density of 10+23 m-3. Assume thatdiffusion of Be in Cu is governed by D=D0exp-{Q/RT}, where D0 = 3.10-5 m2.s-1, and Q =165 kJ.mol-1.(c) Plot the precipitate radius, r, as a function of aging time for t <tc. Use log-log axeson your plot.(d) Assume that particle coarsening starts immediately at the time at which theequilibrium volume fraction has precipitated; again, in reality coarsening starts at thebeginning of precipitation, so this is a simplifying assumption. Assume that the particlesize varies as

r3 = r3(tc) + KDt

where D is the diffusion coefficient, and K = 2.10-9 m. Plot the particle radius as afunction of annealing time on the same axes as in part (c) using the growth model for t<tc and the coarsening model for t > tc.(e) Plot the bowing and particle shearing stresses as functions of aging time. Assumethat L' = r√(π/f) for dislocation bowing. Assume that the particle shearing stress isgoverned by coherency hardening with ecoh = 0.01. The shear modulus, G = 40 GPa andb = 0.26 nm. The equation that governs coherency hardening is as follows

tcoh = 7 |ecoh|1.5 G √(rf/b)

(f) Deduce aging times that correspond to underaging, peak aging and over aging at 643K. Does the peak aging time correspond to the transition between bowing and shearing?(g) The diagram below shows experimental results for an aging experiment on C82500 at643 K. Compare your calculated results to these experimental results in terms of (i)magnitude of the strength increase at the peak, (ii) time to the peak hardness, and (iii) theshape of the curve. Hint: the formulas for strength give you a shear stress whereas thedata shown below is from a tensile test, so convert your shear stresses to von Misesstresses by multiplying by a Taylor factor, <M>=3.07.

(a) The best answers will address the need to solutionize the material first, then quench,then possibly use a two-step process to first nucleate at a fine scale, followed by a growthstep at higher temperature. A sketch of a TTT diagram would be useful.

(b) BeCu is cubic with a lattice parameter of a=2.698 Å (Pearson’s Handbook).Therefore 1 unit cell of BeCu occupies 6.023.1023 * (2.698.10-10)3 m3 = 6.023.1023 *1.964.10-29 m3 = 1.1828.10-5 m3. Thus the volume per atom of BeCu is 9.816 Å3.Similarly, the volume per atom of Cu is a3/4 = 3.643/4 = 12.057 Å3.

Thus the volume fraction of BeCu is:

f BeCu( ) =0.24 ¥ 9.816

0.24 ¥ 9.816( ) + 0.76 ¥12.057( )= 0.2045

Given the density, we have a volume fraction of{A much cruder estimate is obtained by ignoring the atomic volumes and using the leverrule to obtain the equilibrium volume fraction as 2*0.12 (the latter being the atomicfraction of Be) = 0.24. }Use the equation r=∆X/(Xbeta-Xeq) √(Dt) to obtain the radius as a function of time at370°C. Then the volume fraction is given by:

f = I 4π/3 r3 = 1023 4π/3 {∆X/(Xbeta-Xeq) √(Dt)}3

Diffusion coefficient: D= 3.10-5 exp-{165,000/(8.31*634)} m2.s-1 = 1.182.10-18 m2.s-1.Substituing for the numerical value of the diffusion coefficient:

f = I 4π/3 r3 = 1023 4π/3 {0.12/(0.50- 0.045) √(1.182.10-18 t)}3

The time required to reach the equilibrium volume fraction (ignoring impingement and itseffect on slowing down growth) is tcr = 770 seconds.

Plotting this in Kgraph:

Particle Volume Fraction versus Time

0.0001

0.001

0.01

0.1

1

10 100 1000 104

Q2.Kdata

Frac

tion

Time (s)

An alternative plot with linear axes:

0

0.05

0 .1

0.15

0 .2

0.25

0 200 400 600 800 1000

Fra

cti

on

Time (s)

(c), with (d):The particle radius during coarsening will vary as follows:

r3 = r3(tc) + Kdt = (7.95.10-9)3 + 1.182.10-27 t

Particle Radius versus Time

0.1

1

10

100

10 100 1000 104

Q2.Kdata

GrowthCoarsening

Radi

us (n

m)

Time (s)Alternative plot for (c) and (d) with linear-linear axes:

0

5

10

15

20

25

30

0 2000 4000 6000 8000 1 10 41.2 10 4

Radi

us (

nm)

Time (s)

(e) Coherency Hardening:tcoh = 7 |ecoh|1.5 G √(rf/b)

= 7 0.011.5 40.109 √(r * f / 0.26.10-9)= 280 √(r * f / 0.26.10-9) MPa

tbow = G b / l = Gb/{r√(π/f)} = 40.109* 0.26.10-9 /{r√(π/f)} Pa

(f) The transition occurs at 280 seconds which is clearly a slightly shorter time than theattainment of equilibrium volume fraction (tcr = 770 s). The peak strength does NOTcorrespond to the transition from shearing to bowing but occurs later than this (at ~760 s)and coincides with the point at which coarsening starts.

1

10

100

1000

104

10 100 1000 104

Q2.Kdata

ShearBowing

Shea

r Stre

ss (M

Pa)

Time (s)

(g) Another view but with axes more similar to those used in the question, and with theshear stress converted to von Mises stress by multiplying by 3.07. The peak in stresscomes earlier than observed experimentally. There are various reasons for this differencein outcome: our diffusion data may be inaccurate; the hardening mechanisms may bedifferent than our assumed mechanisms, for example. We are, however, within an orderof magnitude of the time to the peak (approximately 45 minutes or 2700 s) which is quitegood for the simple approximations involved in this calculation.

0

200

400

600

800

1000

1200

0 0.5 1 1.5 2 2.5 3

Q2.Kdata

von

Mis

es s

tress

(MPa

)

Time (hr)

2) Consider the effect of precipitation on creep strength. You will need to do somereading in order to deal with these questions. This will be a challenge in terms of howyou develop your own knowledge in this relatively open-ended exercise. The real-worldrelevance of creep properties is obvious in gas turbine engines and internal combustionengines, but it is also relevant to polymeric materials as well (albeit with a differentscientific basis).

2a. [10 points] What difference does the presence of a precipitate make on creepstrength? Your answer should include at least some discussion of the relationshipbetween strain rate and stress (often described as a power-law equation; strain-rate isproportional to stress^exponent, where exponent takes values between 1 and 20):

†

˙ e = Cs n .

The most obvious difference that adding particles makes on creep strength is that they (ingeneral) increase creep strength. In terms of the deformation kinetics, they increase theexponent very markedly from values around 5 for single phase metals to values of 20 (oreven higher). Some authors postulate a “back-stress” or “threshold stress” that must besubtracted from the applied stress. This allows a smaller value of exponent to be used infitting experimental data. It has to be said, however, that direct physical evidence forsuch a threshold stress is lacking and so there is still some controversy in the literatureabout the basis for this approach. The form of the equation is as follows:

†

˙ e = C s applied -s threshold( )n

The proportionality factor, C, between stress and strain rate is itself highly temperaturedependent (e.g. through the temperature dependence of diffusion). Based on anArrhenius expression, C=exp-{Q/RT}, adding precipitates to a material tends to raise theactivation energy, Q, associated with creep.

2b. [10 points] Find out what you can about creep strength in aluminum alloys and innickel alloys and attempt to answer the question “why can we use nickel alloys at ahigher fraction of their melting point than aluminum alloys?” Your answer shouldinclude some discussion of the value of ordered precipitates.

These are relatively open-ended questions and you may well need to ask the instructorquestions to check that you are on track.

Answer to 2b:

The main difference between the two alloy systems is that in Nickel alloys, it is possibleto alloy with Al (and other elements) to obtain a large volume fraction (>50%) of anordered intermetallic, Ni3Al, that exhibits an increasing resistance to dislocation motionwith increasing temperature. No such possiblity exists for aluminum alloys however andso their creep resistance at high homologous temperatures is always less than for nickelalloys designed for high temperature service.

3. CuNiAl SME alloys (approximate composition in weight %, 14%Al, 8%Ni, balanceCu) have the following characteristics.Single crystal transformation strain (based on pseudoelastic strain): 10%Transformation enthalpy: 8000 J.kg-1

Transformation temperature (Ms): 0°C (arbitrarily chosen from the range quoted inOtsuka & Wayman).(a) [10 points] Plot as stress versus temperature the slope of the boundary between theregion in which either shape memory or pseudoelasticity can be observed and the (highertemperature) region with no effect.(b) [15 points] For a parent phase with a Young’s modulus of 90 GPa, a grain size (parentphase) of 50 µm, and a twin boundary energy of 50 mJ.m-2, estimate the twin spacing inthe martensite based on the concept of self-accommodation. In each grain that is h high,w wide, and d deep there are n lamellae. In order to calculate the elastic energy associatedwith the transformation, assume that each lamella of martensite shears uniformly in thesame direction. The total elastic energy is given by the product of the volume, shown asthe red triangle for each lamella, and the elastic energy per unit volume (Ye2/2 = Ytan2b/2). For each lamella, there is a contribution to the interface energy from thepresence of a twin boundary between adjacent lamellae. You will find that the elasticenergy decreases as the density of twins increases.

h

w

b

(a) Apply the relationship ds/dT = ∆S/e = ∆H/(Tee):

The density of the alloy is 0.14*2.7 + 0.08*8.908 + 0.78*8.92 = 8.048 gm/cc; we mustmultiply the energy per kg by the density in order to obtain the enthalpy in terms ofenergy per unit volume.The slope, ds/dT =8000*8048 / (273 * 0.1) = 293 kPa.K-1

Expressed as the rise in temperature with stress, the slope, dT/ds = 3.41 K.MPa-1.

(b) Assume a depth, d. Call the twin thickness x.Elastic energy = n * (Y tan2b / 2) * ( h2/n2 tanb /2) d = Y d h2 tan3b / 4n = 90e9 * (5e-5)3

* 0.0091176 / 4n = 2.564e-5 / n == 2.564e-5 * x / 5e-5 = 0.51286 * x.

Interfacial energy = n w d g = n * (5e-5)2 0.05 = 1.25e-10 n= 1.25e-10 * 5e-5 / x = 6.25e-15 / x

Differentiate both expressions w.r.t n (the number of lamellae) and set to zero: (Y tan2b / 2) * ( h2/n2 tanb /2) d = w d gOr, n2 = (Y h2 tan3b / 4 w g); n = 453Or,x = h/n = h / √ (Y h2 tan3b / 4 w g) = √ {4 w g / Y tan3b} = 1.1e-7 m = 0.11 µm.

Assume that the grains are equiaxed and have a size of 50 µm, so that h=w=d= 50µm.Then the expression above yields a twin spacing of ~ 0.11 µm, which is a reasonablenumber. Higher modulus will lower the spacing and larger twin boundary energies willincrease it. Note the dependence on grain size.

1.00E-09

1.00E-08

1.00E-07

1.00E-06

1.00E-051.00E-08 1.00E-07 1.00E-06 1.00E-05

elastic

interfacial

total energy

4. Coherency of Precipitate Interfaces4a.[10 points] Given a (cubic) precipitate with lattice parameter a = 3.9 Å, and a matrixwith a = 3.8 Å, shear modulus G = 45GPa (in both phases), estimate the interface energyof a semi-coherent boundary between the two phases that is based on a {100} plane.Keep in mind that there is a lattice misfit in two directions which means that two sets ofdislocations are needed that lie perpendicular to one another.

Answer:We estimate the dislocation energy in the standard manner: T = Gb2/2The misfit parameter is ∂ = (a1 - a2)/ a2 = (3.9-3.8)/3.8 = 0.10 / 3.9 = 0.026315The Burgers vector is the average of the plane spacings, b = (a1+a2)/2 = 3.85 ÅThe spacing (in each direction) is b/∂ = 3.85 / 0.026315 = 146.3 ÅTherefore the energy per unit area of interface is twice the line length on each side of agrid cell, 146.3 Å, multiplied by the line tension, Gb2/2, divided by the area, (146.3 Å)2; = 2 * 146.3 * 45 * 109 * 3.852 / 2 / 146.3 / 146.3 = 45 * 109 * (3.85.10-10)2 / 146.3.10-10

= 0.456 J.m-2.

4b. [10 points] For an observed loss of coherency at a radius r = 5nm, what difference ininterfacial energy would you estimate for incoherent versus coherent interfaces?

Answer:Use the equation that relates the difference in interfacial energy to the elastic energy inthe coherent state: rcrit = 3∆g/4G∂2 <=> ∆ g = rcrit* 4G∂2 /3.The misfit = ∆a/a = = 0.1 / 3.85 = 0.02597.Thus ∆g = 5.10-9* 4 * 45.109 * 0.025972 /3 = 0.202 J.m-2.This is a reasonable value.

4c. [10 points] Based on the results in 4a and 4b above, what estimate for the interfacialenergy of the coherent particle can you make? Clearly your answer for 4a must be largerin magnitude than for 4b in order to be able to obtain a sensible answer.

Answer: we can simply take the difference between the values computed in (a) and (b):coherent interfacial energy = 0.254 J.m-2.

Table of References for Question on Creep

TITLE AUTHOR PUBLISHER

PageNumber fordiscussionof CreepProperties

Mechanical Behavior ofMaterials

T.H. Courtney McGraw-Hill

Ch. 7, p293

Deformation and FractureMechanics of EngineeringMaterials, 3rd ed.

Hertzberg Wiley Ch. 5,p. 145.

The Science of StrongMaterials

J. E. Gordon Penguin(alsoPrinceton)

p 228

Microstructure & Propertiesof Materials – I

J.C.M. Li WorldScientific

Chs. 1 and 2

Mechanical Behavior ofMaterials, 2nd ed, 1999

Norman E.Dowling

PrenticeHall, NJ

Ch. 15, p706

An Introduction to theMechanical Properties ofCeramics, 1998

D.J. Green Cambridge Univ.Press, NY

Ch. 7, p 193

The Mechanical Properties ofMatter, 1964

A.H. Cottrell Wiley,NY

p. 336.

Mechanisms of CreepFracture

H.E. Evans Elsevier entire book!

Mechanical Metallurgy, 3rd

editionG. Dieter McGraw-

HillCh. 13, p.432.