Question 2 of 8: Substitutional Solid Solutions; ASTM Grain Size; Fatigue Failure

Question 2: Substitutional Solid Solutions; ASTM Grain Size; Fatigue Failure (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.

Given. (b) $G=18$ grains counted in a $2\times2$ in micrograph area at linear magnification $M=400\times$.

Find. (a) Definition and Hume-Rothery favourability factors for substitutional solid solutions. (b) ASTM grain size number $n$. (c) Fatigue-failure fracture-surface identification.

Approach

Part (b) is a direct application of the ASTM grain-size standard, which is defined at 100× magnification: a grain count made at a different magnification must first be rescaled to an equivalent count at 100× (area scales as the square of linear magnification) before applying $N=2^{n-1}$.

1. (a) Substitutional solid solutions. A substitutional solid solution forms when solute atoms directly replace solvent atoms on the parent crystal lattice's regular lattice sites (as opposed to squeezing into the interstitial spaces between them). The Hume-Rothery rules govern how much solute can dissolve this way: (1) atomic size factor — the atomic radii of solute and solvent should differ by less than ≈15%, or the lattice strain becomes too large to accommodate; (2) crystal structure factor — extensive (complete) solubility requires solute and solvent to share the same crystal structure; (3) electronegativity factor — the two elements should have similar electronegativity, or they will instead tend to form an intermetallic compound; (4) valence factor — for a given solvent, a solute of higher valence is more soluble than one of lower valence than the solvent. Cu–Ni is the textbook example of a system satisfying all four rules and showing complete (all-proportions) solubility.
2. (b) Convert the count to grains per square inch at 100×. At the viewing magnification $M=400\times$, the observed count is $$n_{400}=\frac{G}{A}=\frac{18}{2\times2}=4.5\ \text{grains/in}^2.$$ Since the same physical sample area appears magnified $(M/100)^2$× larger in area when viewed at $M$ rather than at the ASTM-standard 100×, the number of grains per unit area apparently falls by that same factor at higher magnification — so rescaling to 100× requires multiplying back up: $$N_{100}=n_{400}\times\left(\frac{M}{100}\right)^2=4.5\times\left(\frac{400}{100}\right)^2 =4.5\times16=72\ \text{grains/in}^2\ \text{(at 100$\times$)}.$$
3. Apply the ASTM formula. With $N=2^{n-1}$ (page-1 formula), $$n-1=\log_2N_{100}=\log_2(72)=6.17\ \Rightarrow\ \boxed{n\approx7.2}.$$ An ASTM grain size number of ≈7 corresponds to a moderately fine-grained structure.
4. (c) Recognising a fatigue failure. Fatigue fractures have a visually distinctive macroscopic appearance quite unlike a ductile overload or brittle fracture. The fracture surface typically shows: (i) a smooth, often shiny, concentric "beach-mark" region with rings radiating from the crack initiation site (usually a surface stress concentrator — a notch, tool mark, corrosion pit, or inclusion) — these beach marks record successive positions of the slowly-advancing crack front, often corresponding to load-interruption events (e.g. shutdown/restart); (ii) at a finer scale within the beach-mark region, closely spaced striations (visible under SEM), each representing one load cycle's worth of crack advance; and (iii) a final, rougher, more granular or fibrous fast-fracture zone where the remaining, progressively shrinking uncracked ligament could no longer support the load and failed suddenly by ductile or brittle overload — often showing shear lips. The near-total absence of gross plastic deformation (necking) elsewhere on the part, combined with these three features together, is the standard diagnostic for a fatigue origin.