NivaarExam PrepOfficial exam papers ↗

21-Mat-A6 Materials Selection and Design for Materials Processing · May 2015

Question 1 of 8: Metastable Precipitation Sequences, Spinodal Decomposition and Ordered Domains

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

10-Met-A6 — Phase Transformation and Thermal Treatment of Metals and Alloys — National Exams, May 2015 — 3 hours — 8 questions printed, first 5 as answered are marked (all 8 answered below as a complete study resource).

Reference texts: Callister & Rethwisch, Materials Science and Engineering: An Introduction, 10th ed.; Porter, Easterling & Sherif, Phase Transformations in Metals and Alloys, 3rd ed.; ASM Handbook Vol. 4, Heat Treating; Krauss, Steels: Processing, Structure, and Performance.

Every question below is genuine phase-transformation/heat-treatment content.

Question 1: Metastable Precipitation Sequences, Spinodal Decomposition and Ordered Domains (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

1.1 — (a) A sequence of metastable precipitates from one composition, read off the equilibrium diagram

Take an Al–4 wt.%Cu alloy as the chosen precipitation-hardening system. The equilibrium Al–Cu phase diagram shows a single-phase α (Al solid solution) field at high temperature that narrows sharply as temperature falls, with a solvus line separating α from the two-phase (α+θ) field, $\theta$ being the equilibrium intermetallic Al2Cu. Solution-treating at, say, 540 °C (inside the single-phase α field for 4% Cu) dissolves all the copper; quenching to room temperature crosses the solvus far faster than diffusion can respond, freezing in a supersaturated solid solution (SSSS) of α at a composition the diagram says should already be two-phase. Aging that SSSS does not jump straight to the equilibrium $\theta$; because the diagram only fixes the START and END states, not the PATH, the system instead descends through a sequence of progressively more stable, progressively less coherent intermediate structures, each nucleating more easily than the next because each has a lower interfacial-energy barrier even though each also sits at a higher bulk free energy than its successor:

GP zones (fully coherent Cu-rich clusters, no equilibrium counterpart) → $\theta''$ (coherent, ordered, tetragonal) → $\theta'$ (semi-coherent platelets) → $\theta$ (equilibrium Al2Cu, incoherent)

The controlling idea is that at each stage the system takes whichever path has the lowest NUCLEATION barrier available to it, not the lowest final free energy: GP zones, being fully coherent, cost almost no interfacial energy (all elastic strain, no broken bonds) and so nucleate essentially instantly and homogeneously throughout the matrix; as aging continues, the zones themselves become the heterogeneous nucleation sites for the next, less-coherent phase, and so on down the sequence, with the equilibrium $\theta$ only appearing once the earlier phases have dissolved to feed its (larger, incoherent, widely spaced) growth. Peak strength is reached partway down this sequence — typically at $\theta''$ or early $\theta'$ — where particle spacing is still fine enough to force dislocations to bow between particles (Orowan looping) at a high stress, and it is lost again (overaging) once $\theta$ coarsens and the spacing grows.

1.2 — (b) Precipitation versus spinodal decomposition

Both start from a supersaturated single-phase solid solution, but they differ in every mechanistic respect because of WHERE inside the miscibility gap the alloy is quenched, relative to the spinodal curve (the locus where $\partial^2 G/\partial c^2 = 0$):

In short: precipitation needs a barrier-crossing nucleation event and produces sharp-interfaced discrete particles; spinodal decomposition is barrier-free and produces a continuously modulated, interconnected two-phase structure — the difference traces entirely to the sign of the curvature of the free-energy-versus-composition curve at the alloy's own composition and aging temperature.

1.3 — (c) Why ordered domains form, and a sketch of one ordered structure

Certain binary systems have a NEGATIVE heat of mixing between UNLIKE atom pairs — the A–B bond energy is lower (more favourable) than the average of the A–A and B–B bond energies, $\Omega = z(\varepsilon_{AB}-\tfrac12(\varepsilon_{AA}+\varepsilon_{BB})) < 0$. Below a critical ordering temperature $T_c$ (set by the size of $\Omega$ relative to $k_BT$), the equilibrium state therefore maximizes the number of unlike-neighbour bonds by segregating the two atomic species onto two distinct, crystallographically distinguishable sublattices — a long-range-ordered superlattice — rather than remaining randomly (disordered-solid-solution) mixed. Because ordering can nucleate independently, and with equal probability, at many spatially separated points in the same crystal, and because the ordered structure typically has LOWER symmetry than the disordered parent lattice (so there is more than one crystallographically equivalent way to assign the two sublattices), regions that start ordering from different nucleation points can end up "out of step" with each other — correct in their own local ordering pattern but shifted relative to their neighbour by a lattice vector that is a symmetry operation of the disordered parent but NOT of the ordered structure. Where two such regions meet, an antiphase boundary (APB) forms, and each self-consistently ordered region bounded by APBs is an ordered DOMAIN.

Au (corners, 1/8 site each)Cu (face centres, 1/2 site each)Cu3Au (L1_2) ordered superlattice1 Au + 3 Cu per unit cell
Cu3Au (L12) ordered superlattice, sketched as one example: Au occupies the cube corners (1/8 site each → 1 Au/cell) and Cu the six face centres (1/2 site each → 3 Cu/cell), giving the 1:3 stoichiometry. Below $T_c\approx390\,{}^{\circ}\text{C}$ this ordered arrangement is favoured over a random FCC solid solution of the same composition because Au–Cu near-neighbour bonds are energetically preferred over Au–Au and Cu–Cu bonds.
← Paper overview