Microstructure and Properties II

1. Microstructure and Properties II MSE 27-302 Fall, 2002 (2nd mini-course) Prof. A. D. Rollett http://neon.mems.cmu.edu/rollett/27302/ 27302.html

2. Course Content • 27-302 is the second of a pair of (mini-)courses that describe the relationship between materials microstructure and properties. • This course deals mainly with multi-phase microstructures. There is a strong emphasis on phase transformations as the basis for understanding the origin of (useful) microstructures. • 27-301 dealt mainly with single phase microstructures and their properties. • Multi-phase materials made through natural processes will be contrasted with (man-made) composite materials and biomaterials. • Students are expected to learn a set of technical skills in addition to improving various attributes of scientist/engineers (communications, ethics, how to design experiments, …)

3. Topics • Where does microstructure come from? Phase transformations, kinetics of transformations, the Kolmogorov-Johnson-Mehl-Avrami equation. • Properties of Composite materials: background material on glass-ceramics for Lab 1. • Phase transformations: driving forces, thermodynamics of nucleation (precipitation reactions). • Transformations: kinetics of growth: a simple TTT diagram. How to calculate and predict TTT and CCT diagrams. • The role of interfaces in heterogeneous nucleation; example of the Al-Cu system; sequences of metastable precipitates. • The age-hardening curve; methods of measuring transformations. The similarities between mechanical hardness and magnetic hardness. • Impact of precipitation on complex properties: example of fatigue as a microstructure-sensitive property. • More complex diffusive transformations: example of Fe-C system for eutectoid reactions. • Continuous transformations: spinodal decomposition. • Coarsening of two-phase structures; effect of two-phase structures on creep properties (Ni-alloys as an example). • Competition between transformation mechanisms; discussion of non-diffusive transformations; massive transformations, martensitic transformations; exploitation of martensitic reactions for shape-memory alloys. • The ultimate in complicated microstructures: introduction to welding and joining.

4. Technical topics • Technical Issues: • S olid state transformations • Differences between transformations from the liquid state and transformations starting from the solid state: the influence of crystalline structure • Driving forces - should the reaction take place? • Nucleation and growth (thermodynamics, kinetics): the rate at which reaction takes place • Influence of defects on transformations • Prediction of temperature-time-transformation (TTT) curves (and/or continuous-cooling-transformation, CCT, curves) • Military transformations • Precipitate coarsening • These topics are some of the underpinnings for understanding various phenomena that are important for microstructure-property relationships.

5. Material Properties, Phenomena • Examples of phenomena for which microstructure-property relationships are significant: • Age Hardening • Shape memory effect, alloys • Alloy optimization • Multiphase materials and creep • Energy absorption in structures • Fatigue resistance • Exploitation of nanostructured, amorphous materials • Optimization of Materials Design • All the technical topics are relevant to understanding and engineering the phenomena. • Certain material systems are important examples.

6. Materials Systems • Clearly there are too many material systems to study in one course. Certain systems are very useful as examples, however. • Al-Cu: precipitation, metastable phases, age hardening, effects of crystal structure, coarsening • Fe-C-X (steel): allotropic transformations, eutectoids, military transformations, tempering, hardenability

7. Student Input for 302 • In 27-301, each student was required to make a short presentation in class. • In 27-302, student input will be sought through discussion sessions. The objective is to learn how to apply the understanding of microstructure-property relationships to a specific system(s). • The culmination of the student input exercise will be a discussion of the pros and cons and changing a given material (for a specific application). • Discussions will be held in the second half of the Weds class. • Next we discuss the sequence of steps required.

8. Materials Design • The sequence of steps leading towards the discussion of materials design: • Each student chooses an application for which a material is critical in at least one component (Oct. 23rd, Weds). • The application is analyzed to determine which component is materials-critical (Oct. 30th, Weds). • The material is analyzed to determine its microstructure and likely processing history (Oct. 30th, Weds). • The microstructure-property relationships are analyzed (Nov. 13th, Weds). • Possible changes to the microstructure are analyzed for their effect on properties (Nov. 20th, Weds). • Discussion between a “pro-change” group and a “status-quo” group on the merits of optimization of the material (Nov. 25th, Monday). • Each student writes up a report on materials optimization.

9. Applications • Stents • Sutures • Bone substitute • Stealth aircraft (Low Observable materials) • Nuclear reactors (fuels) • Solar cells • Light weight armor (ceramic armor)

10. 302 Jeopardy: 1 1. Rank is sum of the rank of the quantities on each side 4. -RT ln{X0/Xe}. Q1. How is the rank of a property tensor determined from the rank of each related quantity? Q4. What is the formula for the driving force for precipitation in a simple 2-phase system? 2. Free energy 5. No nucleation barrier Q2. What thermodynamic quantity should we use to predict whether or not a reaction will occur? Q5. Name a key difference between discontinuous and continuous phase transformation. 3. 2-fold symmetry axes (diads) 6. Approximately 3 times the yield stress. Q3. Which symmetry element is found on <110> directions in fcc materials? Q6. How much greater is the hardness than the yield stress (same units)?

11. 302 Jeopardy: 2 1. Proportional to undercooling 4. -Hf (∆T/Tmelt). Q1. How is driving force related to undercooling? Q4. What is the formula for the driving force for solidification? 2. Difference between the temp. at which the composition intersects the solvus (liquidus…) and the current temp. 5. Two phases in a composite generally expand/contract at different rates with ∆T. Q2. How is the undercooling defined? Q5. What is one cause of residual stress in a composite material? 3. Einstein notation 6. Differentiate the total energy. Q3. What is the name for the convention that states that repeated indices are summed over? Q6. How do we determine the point at which an energy release rate is zero?

12. 302 Jeopardy: 3 1. Balance between rates of adding surface energy and gaining free energy from transformation 4. Because large interfacial energies mean high barriers to nucleation (and heterogeneous sites, if available, operate first). Q1. How does one determine the barrier to nucleation? Q4. Why is homogeneous nucleation only observed in a few cases? 2. In precipitation of pro-eutectoid ferrite, the thermodynamic term involves the log of a ratio of terms in (1-X). 5. 16πg3/∆GV2. Q2. Why is the driving force for a eutectoid decomposition small compared to decomposition of a simple solid solution (e.g. pro-eutectoid decomposition of austenite)? Q5. What is the formula for the critical free energy of nucleation? 6. It is a volumetric energy and is subtracted off the chemical free energy for transformation. 3. 2g/∆GV Q3. What is the formula for the critical radius? Q6. What is the role of elastic energy in nucleation?

13. 302 Jeopardy: 4 4. The rate increases because of increasing driving force but then decreases because of decreasing diffusion rate. 1. 16πg3/{∆GV-∆GS} 2 Q1. What is the free energy barrier in the presence of an elastic energy? Q4. Why does the growth rate first increase with undercooling and then decrease? 2. Al2Cu platelets aligned with {100} planes. 5. D 2C = C/t. Q2. What effect does elastic anisotropy have on precipitation in the Al-Cu system? Q5. What is the diffusion equation (w/o source terms)? 3. Matching of close-packed planes, e.g. {110}bcc// {111}fcc 6. Linearized gradients. Q3. What impact does atomic matching have on the orientation relationship between parent and product phases? Q6. What approximation can we make to solve the diffusion equation for ppt growth in 1D?

14. 302 Jeopardy: 5 4. Solute diffuses from small precipitates to large ones. 1. The change in concentration around one precipitate affects the concentration around adjacent precipitates. Q1. What is the cause of “impingement” of concentration fields? Q4. What causes coarsening of precipitates? 2. Grain boundaries act as “short circuit” diffusion paths for transport of solute to precipitates. 5. <R3(t)> - <R3(t=0)> = k t. Q2. Why do precipitates grow more rapidly on grain boundaries than in the bulk (at low temperatures)? Q5. What is the relationship between radius and time for coarsening? 3. Decreasing radius of a precipitate raises its solubility. 6. x = ∆C0/ (Cb - Ce) √(Dt). Q3. What does the Gibbs-Thomson effect do to precipitates? Q6. What is the relationship between ppt size and time for diffusion controlled growth in 1D?

15. Office hours, CA’s • Office hours will be as in 301: 3:30-5 Monday, 11:30-12:30 Weds/Fri. • The CA for the Lab is Ms. Mitra Taheri.

16. Exam Rules • No books; no lecture notes; no computers • One “cheat sheet” with notes (both sides if you like); hand in the the cheat sheet with the exam paper/book. You must write the notes yourself: copying and pasting is OK, but not literal cut and paste. The idea of the “cheat sheet” is for you to go through the course material and extract the most important ideas, equations, etc. • Calculator OK (but not a device, such as a Palm Pilot, in which you can store lecture notes etc.)

17. 27-302, Labs • Lab 1 = Investigation of precipitation in glass-ceramics. Purpose: to demonstrate the effect of phase transformation on mechanical and optical properties. • Lab 2 = Short experiments on crystallization of amorphous metals, magnetic domain imaging and age hardening curves.

18. Calendar: 302 Please consult the separate file posted on the website.

19. Topic List: 302 • Where does microstructure come from? Phase transformations, kinetics of transformations, the Kolmogorov-Johnson-Mehl-Avrami equation. • Properties of Composite materials: rule of mixtures. Background material on glass-ceramics for Lab 1. • Phase transformations: driving forces, thermodynamics of nucleation (precipitation reactions). • Transformations: kinetics of growth: a simple TTT diagram. • The role of interfaces in heterogeneous nucleation; example of the Al-Cu system; sequences of metastable precipitates. • The age-hardening curve; methods of measuring transformations. • Impact of precipitation on complex properties: example of fatigue as a microstructure-sensitive property. • More complex diffusive transformations: example of eutectoid reactions. • Continuous transformations: spinodal decomposition. • Coarsening of two-phase structures; effect of two-phase structures on creep properties (Ni-alloys as an example). • Competition between transformation mechanisms; discussion of non-diffusive transformations; massive transformations, martensitic transformations; exploitation of martensitic reactions for shape-memory alloys. • Parallels between magnetic and mechanical hardness. • The ultimate in complicated microstructures: introduction to welding and joining: not addressed in 2001. • Cellular structures: foams, wood, bread(!), bone, composites. • Guest Lecture (Prof. E. Towe): quantum dot structures in semiconductors.

20. Sample problem: 1 • KJMA equation (transformation kinetics): an alloy is recrystallized at 2 different temperatures, 400 and 500°C. The KJMA exponent is found to be 2. By interpolating the f vs time data, the time required for 50% recrystallization is found to be 30s and 5m, respectively. Estimate the activation energy for the process. • Answer: use the form of the equation from the homework:t={-ln(.5)/k}1/n=> k = -ln(.5)/ tnAssume k=k0exp-Q/RT; => -Q/RT=ln(k/k0)So, -Q=RT1ln(k1/k0) =RT2ln(k2/k0)-Q/R*(1/T1-1/T2) = ln(k1/k2)Q=R ln(t12/t22) /(1/T1-1/T2)Q= 8.31 ln(302/3002) (1/673 - 1/773)Q = 199,086 J/mole

21. Sample Problem: 2 • Composites: a certain type of (cheap) plywood is made up of two thin outer sheets of a high density wood with a lower density filler material inside. If the modulus of the cladding is 10 Gpa, and each sheet is 2mm thick, and the modulus of the filler layer is 100 Mpa with a thickness of 10mm, what is the stiffness of the plywood, measured through the thickness? • Apply elementary isostress theory. Modulus = VA(1/EA) + VB(1/EB) = 1/14 (4/10 + 10/0.1) GPa = 0.14 GPa. • The composite is dominated by the softer filler layer because of the loading method. It would be much stiffer if it were loaded on its edge.

22. Sample Problems: 3 • Nucleation: in a problem on solidification, the latent heat is 50,000 J/mole and the melting point (liquidus) is 850°C. The molar volume is 106m-3. No appreciable nucleation is observed in a carefully controlled experiment in which only homogeneous nucleation can occur. What is the volumetric driving force for an undercooling of 50°C? • Answer - use the expression for driving force where the latent heat (enthalpy) is known. ∆GV=(∆H∆T/Te)/Vm = 50,000.50/(850+273)/10-6= 2.23.109 J.m-3 • Based on this information, what is the apparent interfacial energy? • Answer: assume that ∆G* ~ 60kT at the point where nucleation occurs; 60*1.38.10-23*1123=16πg3/3 /(2.23.109)2so, g = 3√(0.276) = 0.65 J.m-2.

23. Sample Problem: 4 • Precipitate growth rates: for a precipitate that is pure element B, and a solvus line described by log10(XB) = 2.853 - 2.875.10+3/T, where XB is the composition in atomic %,what is the growth rate at T=600°C for a matrix composition X0B=1.5% 1 minute after nucleation has taken place? Assume 1D growth (e.g. of a slab of precipitate nucleated on a grain boundary). The pre-factor and activation energy for diffusion of B in A are 7.4.10-5 m2.s-1 and Q=217.2 kJ.mole-1, respectively. • Answer - first calculate the equilibrium concentration of matrix (alpha) in equilibrium with the precipitate (beta): XB = 0.36Then the growth rate is given by v=∆X/2(Xb-Xe) * √(D/t)= (1.5-0.36)/2/(100-0.36)*√(7.4.10-5*exp-{217,200/8.31/873}/60)= 2.0 10-12 m.s-1, or 7nm per hour! • Pretty slow!

24. Sample Problem: 5 • Coherency Loss: show how the following expression can be derived for the critical size of a precipitate at which coherency is lost.rcrit = 3∆g/4µd2. • Answer: recall that ∆Gcoherent= 4µd2 * 4πr3/3 + 4πr2gcoherent ∆Gnon-coherent = 4πr2gnon-coherentAt the transition size, the two free energies will be the same, and above this size, the precipitate with incoherent interface will have the lower energy. Therefore we can write that 4µd24πrcrit3/3 + 4πrcrit2gcoherent= 4πrcrit2gnon-coherentWrite∆g=(gnon-coherent - gcoherent)Thus rcrit = 3∆g/4µd2.

25. Sample Problem: 6 • Coherency Loss, contd.: for the problem outlined in number 5, given a (cubic) precipitate with lattice parameter 3.9 Å, and a matrix with a=3.8 Å, shear modulus µ=45GPa, and an observed loss of coherency at r=5nm, what difference in interfacial energy would you estimate for incoherent versus coherent interfaces? • Answer: turn the equation around and estimate the difference:rcrit = 3∆g/4µd2 <=> ∆g =rcrit* 4µd2 /3.The misfit = ∆a/a = 0.1/3.8 = 0.0263.Thus ∆g =5.10-9*4* 45.109 * 0.02632 /3 = 0.21 J.m-2. • This is a reasonable value.

26. Sample Problem: 7 • Spinodal Decomposition: how can we represent the phenomenology of spinodal decomposition? One key is to postulate a function for the dependence of free energy on composition. The simplest form that will yield a G(X) curve with a central “hump” is this: G(X) = 25,000 * {4(X-0.5)4 - (X-0.5)2} J. mole-1 • Based on this constitutive description, we can now ask, for example, what the limits of the chemical spinodal are? • Answer: differentiate the formula to find the curvature and set it equal to zero to locate the inflection points: d2G/dX2 = 25,000 * {4*4*3(X-0.5)2 - 2} = 0 48(X-0.5)2 = 2 X = 0.5 ± √(2/48) = 0.704 or 0.296 • We can also easily obtain the miscibility gap because of the symmetry of the function about X=0.5: dG/dX=0 =>dG/dX = 25,000 * {4*4(X-0.5)3 - 2(X-0.5)} = 0 =>(X-0.5)2 = 1/8 => X = 0.146 or 0.854

27. Sample Problem 7: graph • A plot of G(X) = {4(X-0.5)4 - (X-0.5)2} Chemical Spinodal Miscibility Gap

28. Sample Problem: 8 • Heterogeneous Nucleation versus Homogeneous: Consider problem 5.5 from P&E and estimate the ratio between the homogeneous and heterogeneous nucleation rates. The critical free energy for homogeneous nucleation is 10-19 J and the temperature is 500°C. Assume that the effective grain boundary thickness is ~0.4nm and the grain size ~25µm; gAA = 500, gAB = 600 mJ.m-2. • Answer: First calculate the contact angle: gAA = 2gAB cosqq = cos-1 (gAA/ 2gAB) = 53.1° • Then calculate the shape factor, S(q):S(q) = 0.5 (2 + cosq)(1 - cosq)2 = 0.208 • The ratio in nucleation rates is given by P&E Eq. 5.25: Nhet/Nhomo = C1/C0 exp-{(∆G*homo-∆G*hetero)/kT}

29. Sample Problem: 8, contd. • Nhet/Nhomo = C1/C0 exp{(∆G*homo-∆G*hetero)/kT} = ∂/D exp{(∆G*homo- S(q) ∆G*homo)/kT} = ∂/D exp{((1- S(q)* ∆G*homo)/kT} = 0.4/25,000 exp{(1-0.208)*10-19/(1.38. 10-23 *773) = 0.027 • Note the sign of the exponential which gives a large number. The ratio of the (effective) grain boundary thickness to grain size decreases the ratio quite significantly. In practical terms, heterogeneous nucleation is most significant at (or adjacent) to the nucleation sites (boundaries, dislocations etc.). gab a Grainboundary in alpha q gaa b

30. Sample Problem, no. 9 • From Dieter, p219 (adapted): • Question: Al-4%Cu (by wt.) has a yield stress of 600MPa. Estimate the particle size and spacing. • Solution: recognize that this stress relates to age hardening beyond the peak hardness. Therefore use the Orowan bowing stress to estimate the stress.s = <M> tcrss = <M> Gb/l • G=27.6GPa; b=0.25nm; <M>=3.1:spacing = 3.1*27,600*0.25.10-9/ 600= 35.7 nm • Now we must estimate the volume fraction of particles for which we use the phase diagram, assuming that we are dealing with the equilibrium phase, q, which is 54 w/o Cu, and the a in equilibrium with it, 0.5 w/o Cu. • Wt. % Al = (54-4)/(54-0.5) = 93.5; wt. % q = 4-0.5/(54-0.5)=6.5 • Volume of a = 93.5gm/2.7 gm/cm3 =34.6 cm3 • Volume of q = 6.5/ 4.443 gm/cm3 = 1.5 cm3 • Volume fraction of a= 0.96; volume fraction of q = 0.04. • Use l=4r(1-f)/3f (slide 22): r =3*0.04*35.7/4/(1-0.04) = 1.12 nm.

31. Cheating Policy • Students are referred to the University Policy About Cheating and Plagiarism (Organization Announcement No. 297, 6116/80). It shall be the policy in this course to discourage cheating to the extent possible, rather than to try to trap and to punish. On the other hand, in fairness to all concerned, cheating and plagiarism will be treated severely. • Cheating includes but is not necessarily limited to: • 1.Plagiarism, explained below. • 2.Submission of work that is not the student's own for reports or quizzes. • 3.Submission or use of falsified data. • Plagiarism includes (but is not limited to) failure to indicate the source with quotation marks or footnotes, where appropriate, if any of the following are reproduced in the work submitted by a student: • 1.A graph or table of data. • 2. Specific language. • 3.Exact wording taken from the work, published or unpublished, of another person."

32. Test, Exams, Grading Policy • Homeworks: 1 per week 100 points • Quizzes: 1 per week 20 points • Exams: two: see weighting below • Grading Policy A > 90% B > 80% C > 70% D > 55% • The instructor will request an Oral exam in borderline cases. • Weighting (%):Homeworks 15Quizzes 5Lab 30Exams 50 • Notes: the distribution between the two exams is to be determined. The quizzes are mainly there to encourage students to stay on top of the material. The 30% weighting for the Lab (or Project) reflects the number of units assigned to the Lab part of the class.