Question 6 of 7: Pb–Sn Lever Rule; Coherent Precipitates; Age-Hardening Alloy Selection
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — December 2017. 3 hours,
closed-book examination (approved Casio or Sharp calculator only). Notes on the paper state that
any five questions constitute a complete paper and only the first five questions appearing in the
answer book are marked, with all questions of equal value. All seven questions are solved below
for completeness.
Reference texts: Callister & Rethwisch, Materials Science and
Engineering: An Introduction, 9th ed. (crystal structure and defects, diffusion, mechanical
behaviour and tensile testing, polymers, phase diagrams and the lever rule, precipitation
hardening, corrosion, casting and solidification).
About the figure. The Y–Z binary phase diagram that
part (c) refers to is printed on page 4 of the examination paper, directly above the
question. Its temperature axis carries no numbers — only the three reference lines $T_1$,
$T_2$, $T_3$ — and its composition axis is labelled simply “% Z” running from pure
Y to pure Z. Fig. Q6 below is therefore traced from that printed figure rather than invented: the
four alloy lines, the solvus, the eutectic isotherm and the three temperature lines were read from the printed figure and scaled to a 0–100 % Z axis, giving alloys 1–4 at approximately 5, 15, 25 and 45 % Z, a maximum solid solubility of about 33 % Z at the eutectic isotherm, and a solvus that has fallen to about 11 % Z by $T_3$. Note the order the paper actually prints: $T_1$ lies above the eutectic isotherm while $T_2$ and $T_3$ both lie below it, so the three lines are reference temperatures for the diagram as a whole, not the solution-treatment, quench and aging temperatures of the heat treatment in part (c).
Given. (a) At $200^\circ$C: solubility of Sn in solid $\alpha$ (Pb-rich) $=18$%
Sn; solubility of Pb in the liquid $=43$% Pb (i.e. the liquid is 57% Sn); alloy is 60% liquid + 40%
solid $\alpha$ by mass. (c) The printed Y–Z system of page 4: a terminal $\alpha$ solid
solution whose solubility limit (solvus) falls from a maximum of about 33 % Z at the eutectic
isotherm to about 11 % Z at the lowest marked temperature $T_3$, with the four alloys marked at
about 5, 15, 25 and 45 % Z; $\beta$ forms as a coherent precipitate in $\alpha$.
Find. (a) The overall composition of the 60%L/40%$\alpha$ alloy. (b) The
distinction between coherent and incoherent precipitates. (c) Which of alloys 1–4 can be
age-hardened, and the complete age-hardening procedure.
Approach
(a) is an inverse lever-rule problem: instead of computing phase fractions from a known
overall composition, the phase fractions are given and the overall composition is the
unknown, found from the same lever-rule relation solved the other way. (c) hinges on two
requirements that only some alloy compositions on a partial-solubility diagram satisfy
simultaneously: the alloy must (i) be single-phase $\alpha$ at a solution-treatment temperature just
below the eutectic isotherm — i.e. lie to the left of the maximum solid solubility (so it can
be homogenized) — and (ii) exceed the room-temperature solubility limit (so that
quenching traps a supersaturated solid solution that can precipitate on aging).
(a) Overall composition by the lever rule (inverse form). With $C_\alpha=18$%Sn
and $C_L=100-43=57$%Sn as the tie-line endpoints, and mass fractions $f_L=0.60$, $f_\alpha=0.40$:
$$C_0=f_LC_L+f_\alpha C_\alpha=0.60(57)+0.40(18)=34.2+7.2$$
$$\boxed{C_0=41.4\%\ \text{Sn}\ (58.6\%\ \text{Pb})}$$
Check via the standard lever-rule fractions using this $C_0$:
$f_\alpha=(C_L-C_0)/(C_L-C_\alpha)=(57-41.4)/(57-18)=15.6/39=0.40$ ✓, matching the
given 40% solid.
(b) Coherent vs. incoherent precipitates. A coherent
precipitate maintains a continuous, unbroken crystal lattice across the precipitate/matrix
interface — the atomic planes of the precipitate line up with (and are elastically strained
to match) the planes of the surrounding matrix, so there is no true interface, only a smoothly
distorted lattice region. This coherency strain field is what obstructs dislocation motion so
effectively, which is why coherent precipitates provide strong precipitation hardening. An
incoherent precipitate has its own distinct crystal structure and lattice
orientation, separated from the matrix by a true, discrete phase boundary (like a grain boundary
between two different crystals) with no continuity of atomic planes across it; incoherent
precipitates (typically larger, formed by over-aging a coherent precipitate past its peak) create a
much weaker, less continuous strain field and therefore give much less strengthening, even though
the total volume fraction of second phase may be similar.
(c) Which alloys can be age-hardened. Age hardening requires an alloy
composition that is (i) single-phase $\alpha$ at a solution-treatment temperature just
below the eutectic isotherm, i.e. left of the maximum solid solubility (so the whole specimen can be
homogenized into one solid solution), and (ii) outside the $\alpha$ solubility
limit at the lowest temperature $T_3$ (so quenching traps a supersaturated solid solution with a
driving force to precipitate $\beta$ on aging). Measured off the printed diagram, the two limits are
a maximum solubility of $\approx33$ % Z and a $T_3$ solubility of $\approx11$ % Z, so the
age-hardenable window is roughly $11\%alloy 1 ($\approx5$ % Z) is far too dilute — it lies inside the
$\alpha$ solvus at every temperature down to $T_3$, so it is never supersaturated and has nothing to
precipitate; it cannot be age-hardened. Alloy 4 ($\approx45$ % Z)
lies beyond the maximum solubility — it is two-phase $\alpha+\beta$ at every temperature below
the eutectic isotherm, so it can never be solutionized into a single $\alpha$ phase without heating
into the $L+\alpha$ region (partial melting); it too cannot be conventionally age-hardened.
Alloys 2 and 3 ($\approx15$ and $\approx25$ % Z) both fall inside the window:
each is single-phase $\alpha$ just below the eutectic isotherm but exceeds the much lower $T_3$
solubility limit, so each can be age-hardened (alloy 3, being closer to the
maximum-solubility composition, develops a larger volume fraction of $\beta$ precipitate on aging
than alloy 2 does; its single-phase window is correspondingly narrower, so it needs the tighter
solution-treatment temperature control).
[Figure not reproduced: Fig. Q6 — The Y–Z phase diagram printed on page 4, traced from the printed figure and rescaled to a 0–100 % Z composition axis; the temperature axis is unnumbered on the source, so only its three marked lines $T_1$, $T_2$ and $T_3$ are shown, in the order the paper prints t. See the official exam paper.]
(c) The complete age-hardening procedure (for alloy 2 or 3).
Solution heat treatment: heat to a temperature just below the eutectic isotherm (high
enough to be inside the single-phase $\alpha$ field for the chosen composition, low enough to avoid
incipient melting) and hold until the alloy is fully homogenized into a single-phase $\alpha$ solid
solution (all $\beta$ dissolved).
Quench: cool rapidly (e.g. water quench) to room temperature. The cooling is too
fast for $\beta$ to nucleate and grow by normal diffusion-controlled precipitation, so the result is
a supersaturated solid solution (SSSS) — a metastable single $\alpha$ phase
containing far more solute than the room-temperature solvus allows.
Age: reheat to a moderate aging temperature (well below the solution-treatment
temperature, chosen to balance
nucleation rate against diffusion rate) and hold for a controlled time. This allows $\beta$ to
precipitate out of the supersaturated $\alpha$ as very fine, closely-spaced, coherent
particles, whose strain fields are what impede dislocation motion and raise the strength/hardness
(natural aging occurs slowly at room temperature; artificial aging uses a controlled elevated
temperature for a predictable, optimized peak-strength schedule). Aging must be stopped at or near
the peak-hardness time — over-aging coarsens the precipitate and eventually converts it to the
much less effective incoherent form from part (b), softening the alloy again.
