21-Mat-A3 Structure and Characterization of Materials · Dec-12-Mtl-A3 2018
Question 4 of 8
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2018 — 12-Mtl-A3, Structure and Characterization of Materials. Three hours, open book, any non-communicating calculator permitted. Eight questions constitute a complete exam paper; all eight are solved here.
Reference texts. The answers below are keyed to the standard undergraduate materials-science references recommended for this syllabus code:
W. D. Callister Jr. and D. G. Rethwisch, Materials Science and Engineering: An Introduction, 10th ed. — crystal structure and defects (Ch. 4), mechanical properties and fracture (Ch. 6, 8), phase diagrams (Ch. 9), diffraction (Ch. 3), polymer structure and mechanical behaviour (Ch. 14–15).
B. D. Cullity and S. R. Stock, Elements of X-Ray Diffraction, 3rd ed. — structure-factor selection rules for cubic lattices.
Part (a). Plane-strain fracture toughness $K_{Ic}$ generally increases with rising temperature. Higher temperature increases the material's capacity for plastic deformation ahead of the crack tip (lower yield strength, more active slip systems, larger crack-tip plastic zone), which absorbs more energy before unstable fracture. This is most dramatic for BCC metals (e.g. ferritic steels), which show a pronounced ductile-to-brittle transition: $K_{Ic}$ is low and roughly constant below the transition temperature, then rises sharply as temperature increases through it.
Part (b) Given.
Quantity
Aluminum alloy component
Titanium alloy component
$K_{Ic}$
40 MPa$\sqrt{\text{m}}$
50 MPa$\sqrt{\text{m}}$
Failure stress $\sigma$
200 MPa
150 MPa
Max. surface crack length $a$
4.0 mm
10.0 mm
Find. Whether the titanium-alloy version of this same component (same geometry, hence same crack-geometry factor $Y$) also fails under its stated stress/crack combination.
Approach. Fracture occurs when $K_I = Y\sigma\sqrt{\pi a} = K_{Ic}$. The aluminum failure data calibrates $Y$ for this component's geometry; apply the same $Y$ to the titanium scenario and compare the resulting $K_I$ to titanium's $K_{Ic}$.
Calibrate the geometry factor from the Al failure. Since the Al component is observed to fail exactly at $\sigma=200$ MPa, $a=4.0$ mm, this is the critical condition $K_I=K_{Ic,Al}$: $$Y = \frac{K_{Ic,Al}}{\sigma_{Al}\sqrt{\pi a_{Al}}}$$
Apply the same Y to the Ti scenario and take the ratio (Y cancels). Because "this same component" is used for both alloys, $Y$ is identical in both expressions, so it can be eliminated directly: $$K_{I,Ti} = Y\,\sigma_{Ti}\sqrt{\pi a_{Ti}} = K_{Ic,Al}\cdot\frac{\sigma_{Ti}}{\sigma_{Al}}\cdot\sqrt{\frac{a_{Ti}}{a_{Al}}}$$ Substituting: $$K_{I,Ti} = 40\times\frac{150}{200}\times\sqrt{\frac{10.0}{4.0}} = 40\times0.75\times1.581$$ $$\boxed{K_{I,Ti} \approx 47.4\ \text{MPa}\sqrt{\text{m}}}$$
Compare to the titanium alloy's own toughness. $K_{I,Ti}\approx47.4\ \text{MPa}\sqrt{\text{m}} < K_{Ic,Ti}=50\ \text{MPa}\sqrt{\text{m}}$, so the applied stress-intensity factor stays below the material's critical value.
Check
Assumes the crack-length convention (whether "maximum length" is used directly as $a$, or as $2a$/$2c$ for a surface flaw) is the same in both scenarios, which the problem statement implies by calling it "this same component." Under that assumption the geometry factor $Y$ cancels exactly and the pass/fail conclusion is convention-independent.
