27-302, Microstructure & Properties II, A.D. Rollett. Fall Homework 4, due Monday, Nov. 25 th. Topic: properties dependent on precipitates

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1 27-302, Microstructure & Properties II, A.D. Rollett Fall 2002 Homework 4, due Monday, Nov. 25 th Topic: properties dependent on precipitates 1) Strengthening, Coarsening: [50 points] [This is essentially question 5.39 in Courtney's book on Mechanical Properties, for those who are interested in knowing where it came from] The copper alloy C82500 has a composition that is approximately 98 wt.% Cu with 2 wt.% Be. This alloy can be precipitation hardened, see the phase diagram below, and has been traditionally used for tools with special characteristics such as non-sparking hammers. The precipitate is denoted as g 2 and has the composition CuBe. CuBe has a CsCl structure which is a simple cubic lattice with Cu atoms at the corners and a Be atom in the body center position. Notation: D Diffusion coefficient Q Activation energy for diffusion t c Time required to reach the equilibrium (level rule) volume fraction of precipitate. r particle radius K constant in the coarsening equation (also degrees Kelvin!) L' mean free path between particles f volume fraction of particles e coh misfit parameter in the equation for coherency hardening G shear modulus b Burgers vector (magnitude) shear stress (critical resolved shear stress) due to coherency hardening t coh

2 <M> Average Taylor factor (=3.07) (a) Describe the steps that you would take to precipitation harden C State explicitly the temperature range that you would use for each treatment. (b) Calculate the time, t c, required to precipitate out the equilibrium volume fraction of precipitate at 643 K. Plot the volume fraction as a function of time at 643K; obviously a volume 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 the equilibrium volume fraction has been reached assuming that the precipitates grow as spheres. Use the equation for diffusion-controlled 1D growth for the radius of each precipitate; this introduces only a minor error in the calculation from ignoring the impingement of the diffusion fields of neighboring precipitates. Assume site saturated nucleation of the CuBe precipitate with a particle density of m -3. Assume that diffusion of Be in Cu is governed by D=D 0 exp-{q/rt}, where D 0 = m 2.s -1, and Q = 165 kj.mol -1. (c) Plot the precipitate radius, r, as a function of aging time for t <t c. Use log-log axes on your plot. (d) Assume that particle coarsening starts immediately at the time at which the equilibrium volume fraction has precipitated; again, in reality coarsening starts at the beginning of precipitation, so this is a simplifying assumption. Assume that the particle size varies as r 3 = r 3 (t c ) + KDt where D is the diffusion coefficient, and K = m. Plot the particle radius as a function of annealing time on the same axes as in part (c) using the growth model for t <t c and the coarsening model for t > t c. (e) Plot the bowing and particle shearing stresses as functions of aging time. Assume that L' = r (π/f) for dislocation bowing. Assume that the particle shearing stress is governed by coherency hardening with e coh = The shear modulus, G = 40 GPa and b = 0.26 nm. The equation that governs coherency hardening is as follows t coh = 7 e coh 1.5 G (rf/b) (f) Deduce aging times that correspond to underaging, peak aging and over aging at 643 K. 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 at 643 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) the shape of the curve. Hint: the formulas for strength give you a shear stress whereas the data shown below is from a tensile test, so convert your shear stresses to von Mises stresses by multiplying by a Taylor factor, <M>=3.07.

3 (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 growth step 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 * ( ) 3 m 3 = * m 3 = m 3. Thus the volume per atom of BeCu is Å 3. Similarly, the volume per atom of Cu is a 3 /4 = /4 = Å 3. Thus the volume fraction of BeCu is: f ( BeCu) = ( ) + ( ) = Given the density, we have a volume fraction of {A much cruder estimate is obtained by ignoring the atomic volumes and using the lever rule to obtain the equilibrium volume fraction as 2*0.12 (the latter being the atomic fraction of Be) = } Use the equation r= X/(X beta -X eq ) (Dt) to obtain the radius as a function of time at 370 C. Then the volume fraction is given by: f = I 4π/3 r 3 = π/3 { X/(X beta -X eq ) (Dt)} 3 Diffusion coefficient: D= exp-{165,000/(8.31*634)} m 2.s -1 = m 2.s -1. Substituing for the numerical value of the diffusion coefficient: f = I 4π/3 r 3 = π/3 {0.12/( ) ( t)} 3 The time required to reach the equilibrium volume fraction (ignoring impingement and its effect on slowing down growth) is t cr = 770 seconds.

4 Plotting this in Kgraph: Particle Volume Fraction versus Time 1 Q2.Kdata 0.1 Fraction An alternative plot with linear axes: Time (s) Fraction Time (s) (c), with (d): The particle radius during coarsening will vary as follows: r 3 = r 3 (t c ) + Kdt = ( ) t

5 Particle Radius versus Time Growth Coarsening 100 Q2.Kdata Radius (nm) Time (s) Alternative plot for (c) and (d) with linear-linear axes: Radius (nm) Time (s) (e) Coherency Hardening: t coh = 7 e coh 1.5 G (rf/b) = (r * f / ) = 280 (r * f / ) MPa t bow = G b / l = Gb/{r (π/f)} = * /{r (π/f)} Pa

6 (f) The transition occurs at 280 seconds which is clearly a slightly shorter time than the attainment of equilibrium volume fraction (t cr = 770 s). The peak strength does NOT correspond to the transition from shearing to bowing but occurs later than this (at ~760 s) and coincides with the point at which coarsening starts Shear Bowing Q2.Kdata Shear Stress (MPa) Time (s) (g) Another view but with axes more similar to those used in the question, and with the shear stress converted to von Mises stress by multiplying by The peak in stress comes earlier than observed experimentally. There are various reasons for this difference in outcome: our diffusion data may be inaccurate; the hardening mechanisms may be different than our assumed mechanisms, for example. We are, however, within an order of magnitude of the time to the peak (approximately 45 minutes or 2700 s) which is quite good for the simple approximations involved in this calculation.

7 1200 Q2.Kdata 1000 von Mises stress (MPa) Time (hr) 2) Consider the effect of precipitation on creep strength. You will need to do some reading in order to deal with these questions. This will be a challenge in terms of how you develop your own knowledge in this relatively open-ended exercise. The real-world relevance of creep properties is obvious in gas turbine engines and internal combustion engines, but it is also relevant to polymeric materials as well (albeit with a different scientific basis). 2a. [10 points] What difference does the presence of a precipitate make on creep strength? Your answer should include at least some discussion of the relationship between strain rate and stress (often described as a power-law equation; strain-rate is proportional 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 (in general) increase creep strength. In terms of the deformation kinetics, they increase the exponent very markedly from values around 5 for single phase metals to values of 20 (or even higher). Some authors postulate a back-stress or threshold stress that must be subtracted from the applied stress. This allows a smaller value of exponent to be used in fitting experimental data. It has to be said, however, that direct physical evidence for such a threshold stress is lacking and so there is still some controversy in the literature about 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 temperature dependent (e.g. through the temperature dependence of diffusion). Based on an Arrhenius expression, C=exp-{Q/RT}, adding precipitates to a material tends to raise the activation energy, Q, associated with creep.

8 2b. [10 points] Find out what you can about creep strength in aluminum alloys and in nickel alloys and attempt to answer the question why can we use nickel alloys at a higher fraction of their melting point than aluminum alloys? Your answer should include some discussion of the value of ordered precipitates. These are relatively open-ended questions and you may well need to ask the instructor questions 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 possible to alloy with Al (and other elements) to obtain a large volume fraction (>50%) of an ordered intermetallic, Ni 3 Al, that exhibits an increasing resistance to dislocation motion with increasing temperature. No such possiblity exists for aluminum alloys however and so their creep resistance at high homologous temperatures is always less than for nickel alloys designed for high temperature service. 3. CuNiAl SME alloys (approximate composition in weight %, 14%Al, 8%Ni, balance Cu) have the following characteristics. Single crystal transformation strain (based on pseudoelastic strain): 10% Transformation enthalpy: 8000 J.kg -1 Transformation temperature (M s ): 0 C (arbitrarily chosen from the range quoted in Otsuka & Wayman). (a) [10 points] Plot as stress versus temperature the slope of the boundary between the region in which either shape memory or pseudoelasticity can be observed and the (higher temperature) region with no effect. (b) [15 points] For a parent phase with a Young s modulus of 90 GPa, a grain size (parent phase) of 50 µm, and a twin boundary energy of 50 mj.m -2, estimate the twin spacing in the 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 associated with the transformation, assume that each lamella of martensite shears uniformly in the same direction. The total elastic energy is given by the product of the volume, shown as the red triangle for each lamella, and the elastic energy per unit volume (Ye 2 /2 = Y tan 2 b/2). For each lamella, there is a contribution to the interface energy from the presence of a twin boundary between adjacent lamellae. You will find that the elastic energy decreases as the density of twins increases. h b w (a) Apply the relationship ds/dt = S/e = H/(T e e):

9 The density of the alloy is 0.14* * *8.92 = gm/cc; we must multiply the energy per kg by the density in order to obtain the enthalpy in terms of energy 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 tan 2 b / 2) * ( h 2 /n 2 tanb /2) d = Y d h 2 tan 3 b / 4n = 90e9 * (5e-5) 3 * / 4n = 2.564e-5 / n = = 2.564e-5 * x / 5e-5 = * x. Interfacial energy = n w d g = n * (5e-5) = 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 tan 2 b / 2) * ( h 2 /n 2 tanb /2) d = w d g Or, n 2 = (Y h 2 tan 3 b / 4 w g); n = 453 Or, x = h/n = h / (Y h 2 tan 3 b / 4 w g) = {4 w g / Y tan 3 b} = 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 reasonable number. Higher modulus will lower the spacing and larger twin boundary energies will increase it. Note the dependence on grain size. 1.00E E E E E E E-07 elastic interfacial total energy 1.00E E Coherency of Precipitate Interfaces 4a. [10 points] Given a (cubic) precipitate with lattice parameter a = 3.9 Å, and a matrix with a = 3.8 Å, shear modulus G = 45GPa (in both phases), estimate the interface energy of 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 of dislocations are needed that lie perpendicular to one another.

10 Answer: We estimate the dislocation energy in the standard manner: T = Gb 2 /2 The misfit parameter is = (a 1 - a 2 )/ a 2 = ( )/3.8 = 0.10 / 3.9 = The Burgers vector is the average of the plane spacings, b = (a 1 +a 2 )/2 = 3.85 Å The spacing (in each direction) is b/ = 3.85 / = Å Therefore the energy per unit area of interface is twice the line length on each side of a grid cell, Å, multiplied by the line tension, Gb 2 /2, divided by the area, (146.3 Å) 2 ; = 2 * * 45 * 10 9 * / 2 / / = 45 * 10 9 * ( ) 2 / = J.m -2. 4b. [10 points] For an observed loss of coherency at a radius r = 5nm, what difference in interfacial energy would you estimate for incoherent versus coherent interfaces? Answer: Use the equation that relates the difference in interfacial energy to the elastic energy in the coherent state: r crit = 3 g/4g 2 <=> g = r crit * 4G 2 /3. The misfit = a/a = = 0.1 / 3.85 = Thus g = * 4 * * /3 = J.m -2. This is a reasonable value. 4c. [10 points] Based on the results in 4a and 4b above, what estimate for the interfacial energy of the coherent particle can you make? Clearly your answer for 4a must be larger in 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 = J.m -2. Table of References for Question on Creep TITLE AUTHOR PUBLISH ER Mechanical Behavior of Materials Deformation and Fracture Mechanics of Engineering Materials, 3 rd ed. The Science of Strong Materials Microstructure & Properties of Materials I Mechanical Behavior of Materials, 2 nd ed, 1999 An Introduction to the Mechanical Properties of Ceramics, 1998 Page Number for discussion of Creep Properties Ch. 7, p293 T.H. Courtney McGraw- Hill Hertzberg Wiley Ch. 5, p J. E. Gordon Penguin (also Princeton) J.C.M. Li World Scientific Norman E. Prentice Dowling D.J. Green Hall, NJ Cambridg e Univ. Press, NY p 228 Chs. 1 and 2 Ch. 15, p 706 Ch. 7, p 193

11 The Mechanical Properties of Matter, 1964 Mechanisms of Creep Fracture Mechanical Metallurgy, 3 rd edition A.H. Cottrell Wiley, p NY H.E. Evans Elsevier entire book! G. Dieter McGraw- Hill Ch. 13, p. 432.