Question 3 of 7: Substitutional Solid Solutions; ASTM Grain Size Number; Recognising Fatigue Failure

Question 3: Substitutional Solid Solutions; ASTM Grain Size Number; Recognising Fatigue Failure (3 of 5)

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) $18$ grains counted in a $2\ \text{in}\times2\ \text{in}$ micrograph area at $400\times$ linear magnification; page-1 relation $N=2^{n-1}$.

Find. (a) Definition and favouring factors for substitutional solid solutions. (b) ASTM grain size number $G$. (c) Diagnostic features of a fatigue fracture surface.

Approach

(a)–(c) are largely descriptive; (b) requires converting the observed grain count/area/ magnification into the ASTM-standard count (grains per square inch at $100\times$), then solving $N=2^{n-1}$ for $n$.

(a) Substitutional solid solution and its favouring factors. A substitutional solid solution forms when solute atoms directly replace solvent atoms on the parent crystal lattice sites, as opposed to squeezing into the interstitial spaces between them. The Hume–Rothery rules give the factors that favour extensive substitutional solubility: (1) atomic size — the atomic radii of solute and solvent should differ by less than about $15\%$, or lattice strain limits solubility; (2) crystal structure — both elements should share the same crystal structure for complete (all-proportions) solubility; (3) electronegativity — the two elements should have similar electronegativity, or they tend to form an intermetallic compound instead of a solid solution; (4) valence — other factors equal, a metal of lower valence more readily dissolves one of higher valence than the reverse.

1. (b) Grains per square inch of micrograph at $400\times$. The counted area is already in inches: $A=2\times2=4$ in$^2$, so $$N_{400} = \frac{18\ \text{grains}}{4\ \text{in}^2} = 4.5\ \text{grains/in}^2\ \text{(at }400\times\text{)}$$
2. (b) Rescale to the ASTM-standard $100\times$ magnification. The count per unit area at magnification $M$ converts to the equivalent count at $100\times$ by the square of the magnification ratio (at $100\times$ the same real area is imaged onto $16\times$ less micrograph area): $$N_{100} = N_{400}\left(\frac{M}{100}\right)^2 = 4.5\left(\frac{400}{100}\right)^2 = 4.5\times16$$ $$N_{100} = 72\ \text{grains/in}^2\ \text{(at }100\times\text{)}$$
3. (b) Solve the ASTM relation $N=2^{n-1}$ for the grain size number $n$. $$n - 1 = \log_2 N_{100} = \frac{\ln 72}{\ln 2} = 6.17$$ $$\boxed{n \approx 7.2\ \ (\text{ASTM grain size number} \approx 7)}$$ (ASTM No. 7 is a medium grain size, typical of ordinary wrought and annealed metals.)

(c) Recognising a fatigue failure. A fatigue fracture surface has a characteristic appearance that distinguishes it from a single-overload ductile or brittle fracture. Macroscopically, the surface typically shows smooth, semi-elliptical beach marks (concentric arcs also known as conchoidal marks) that trace the successive positions of the slowly advancing crack front, radiating outward from a single initiation site — almost always a surface stress concentrator (a notch, machining mark, corrosion pit, or inclusion). Microscopically (SEM), the slow-growth region shows fine, closely-spaced striations, each corresponding to one load cycle. The crack-growth region is typically smooth/burnished (from the crack faces rubbing together under cyclic loading) in contrast to the final, rapid-overload region, which is rough, granular, and fibrous — the sudden ductile or brittle tearing that occurs once the remaining uncracked ligament can no longer support the peak load. A key diagnostic overall: fatigue fractures show little to no gross plastic deformation elsewhere on the part (the part looks “brittle” even in a normally ductile material), because the applied stresses are well below the material’s yield strength — failure occurs entirely by progressive, localized crack growth over many cycles, not by general yielding.