Advanced Materials & Metallurgical interview questions

125 questions with answers (MME-301 to MME-425). Try each one aloud before reading the answer.

Crystallography & Defects

MME-301 · Calculate the equilibrium vacancy concentration in copper at 1000 °C (Qv = 0.9 eV, N ≈ 8.0 × 10²⁸ atoms/m³).

Nv/N = exp(−0.9/(8.62×10⁻⁵ × 1273)) = exp(−8.20) ≈ 2.7 × 10⁻⁴. So Nv ≈ 8.0×10²⁸ × 2.7×10⁻⁴ ≈ 2.2 × 10²⁵ vacancies/m³.

MME-302 · Why is the yield strength of BCC metals strongly temperature-dependent while that of FCC metals is not?

BCC screw dislocations have non-planar cores and a high Peierls–Nabarro stress, so they need thermal activation (kink-pair nucleation) to move. FCC dislocations dissociate on planar {111} close-packed planes with low lattice friction, so FCC strength is governed by athermal obstacles.

MME-303 · Estimate the flow stress contribution from a dislocation density of 10¹⁴ m⁻² (α = 0.4, G = 80 GPa, b = 0.25 nm).

Taylor relation: Δτ = αGb√ρ = 0.4 × 80×10⁹ × 0.25×10⁻⁹ × 10⁷ = 80 MPa shear. Multiplying by a Taylor factor M ≈ 3 gives about 240 MPa tensile.

MME-304 · What stress operates a Frank–Read source of length 1 µm in steel (G = 80 GPa, b = 0.25 nm)?

τ ≈ Gb/L = 80×10⁹ × 0.25×10⁻⁹ / 10⁻⁶ = 20 MPa. Shorter pinning segments need higher stress, which is why fine precipitate spacing strengthens alloys.

MME-305 · Show that an FCC perfect dislocation dissociates into Shockley partials.

On a (111) plane, (a/2)[1̄01] → (a/6)[2̄11] + (a/6)[1̄1̄2], and the vector sum checks. By Frank's b² criterion, the energy falls from a²/2 to a²/6 + a²/6 = a²/3, so dissociation is favourable, with a stacking fault between the partials.

MME-306 · A single FCC crystal is pulled along [001]. Find the Schmid factor for (111)[1̄01] and the yield stress if CRSS = 1 MPa.

cosφ = 1/√3 and cosλ = 1/√2, so m = 1/√6 ≈ 0.408. Yield stress σ = τCRSS/m = 1/0.408 ≈ 2.45 MPa.

MME-307 · How does crystallographic texture affect sheet formability?

Texture makes properties anisotropic. For deep drawing, a high Lankford r-value (width-to-thickness strain ratio), obtained from a strong {111}∥ND γ-fibre texture in IF steels, resists thinning. Variation of r with direction (Δr) causes earing in drawn cups.

MME-308 · What is grain boundary engineering?

Thermomechanical processing that raises the fraction of low-Σ coincidence site lattice boundaries, especially Σ3 annealing twins, in low-SFE FCC alloys. These special boundaries resist intergranular corrosion, sensitization, creep cavitation and cracking.

MME-309 · How does the c/a ratio influence deformation in HCP metals?

The ideal c/a is 1.633. Mg (1.624) and Zn (1.856) slip mainly on the basal plane, while Ti (1.587) and Zr favour prismatic slip. Few independent slip systems mean twinning and ⟨c+a⟩ pyramidal slip are needed for c-axis strain, causing anisotropy and poor room-temperature formability.

MME-310 · Carbon diffuses faster in α-Fe than in γ-Fe. Why is carburizing done in the austenite range?

Ferrite dissolves at most 0.022% C, so it cannot build a useful carbon gradient. Austenite dissolves up to 2.14% C, and at 900–950 °C diffusion is still fast enough. The high surface carbon then forms high-carbon martensite on quenching.

MME-311 · Why does Hall–Petch strengthening break down at very fine grain sizes?

Below roughly 10–30 nm, grains are too small to hold dislocation pile-ups. Deformation shifts to grain-boundary sliding, rotation and boundary-mediated dislocation emission, so strength plateaus or falls (inverse Hall–Petch), and thermal stability becomes a problem.

Phase Diagrams & Phase Transformations

MME-312 · Find the microconstituent fractions in a slowly cooled 1.2% C steel.

Proeutectoid cementite = (1.2 − 0.76)/(6.7 − 0.76) ≈ 0.074, so about 7.4% grain-boundary cementite and 92.6% pearlite.

MME-313 · What are the ferrite and cementite fractions in pearlite?

Lever rule at 727 °C: Fe₃C = (0.76 − 0.022)/(6.7 − 0.022) ≈ 0.11, and ferrite ≈ 0.89. The ratio of ferrite to cementite lamella thickness is roughly 8:1.

MME-314 · What is the JMAK (Avrami) equation and how is it used?

f = 1 − exp(−k·tⁿ) gives the transformed fraction under isothermal conditions. k depends on nucleation and growth rates, and n (about 1–4) reflects the nucleation mode and growth dimensionality. Plotting ln(−ln(1 − f)) against ln t gives n and k, used to build TTT curves.

MME-315 · Why do TTT curves have a "nose"?

Near the eutectoid temperature, undercooling and driving force are small, so nucleation is slow. At low temperatures, diffusion is slow. The fastest transformation occurs at intermediate temperature (about 540 °C for eutectoid plain carbon steel), where both factors balance.

MME-316 · How do alloying elements shift the eutectoid of steel?

Austenite stabilizers (Ni, Mn) lower the eutectoid temperature, while ferrite stabilizers (Cr, Mo, Si, W, Ti) raise it. Almost all alloying elements lower the eutectoid carbon content, so alloy steels can be hypereutectoid below 0.76% C.

MME-317 · What does the Scheil equation predict?

Cs = k·C₀·(1 − fs)^(k−1). It assumes no diffusion in the solid, complete mixing in the liquid and local equilibrium at the interface. It predicts microsegregation and the non-equilibrium eutectic that forms even in alloys below the solubility limit.

MME-318 · How do lath and plate martensite differ?

Lath martensite forms below about 0.6% C as packets of parallel laths with high dislocation density, giving good toughness. Plate (lenticular) martensite forms above about 1% C as twinned plates with midribs, harder but brittle and prone to microcracks. Mixed structures form in between.

MME-319 · Carburize a 0.20% C steel at 950 °C with 1.0% C surface for 10 h (D = 1.1 × 10⁻¹¹ m²/s). Find the depth where C = 0.60%.

(Cx − C₀)/(Cs − C₀) = 1 − erf(z), so 0.5 = 1 − erf(z), erf(z) = 0.5 and z ≈ 0.477. √(Dt) = √(1.1×10⁻¹¹ × 36000) ≈ 6.3×10⁻⁴ m, so x = 2z√(Dt) ≈ 0.60 mm.

MME-320 · What is Ostwald ripening and why does it matter?

Larger precipitates grow at the expense of smaller ones to reduce total interfacial energy. LSW theory gives r³ − r₀³ = kt. Coarsening raises interparticle spacing and lowers strength, which is why slowly diffusing, low-misfit precipitates such as γ′ and oxides are chosen for high-temperature alloys.

MME-321 · What are order–disorder transformations?

Below a critical temperature, atoms in some solid solutions take up specific sublattice sites, as in Cu₃Au, CuZn (β′) and Ni₃Al. Ordering raises strength and electrical resistivity, creates superlattice XRD peaks and antiphase boundaries, and can reduce ductility.

MME-322 · How does eutectic lamellar spacing depend on growth rate?

Jackson–Hunt theory gives λ²·V ≈ constant, so faster solidification produces finer lamellae and higher strength. The spacing balances diffusion distance against α/β interfacial energy.

Heat Treatment

MME-323 · A quenched 4340 shaft shows soft spots with ferrite at mid-radius. What went wrong?

Likely causes are an austenitizing temperature below Ac₃ or too short a soak, a delayed or insufficiently agitated quench (vapour blanket), or a section too large for the steel's hardenability. Check furnace temperature, quench agitation and the Jominy data against the section size.

MME-324 · Design a heat treatment route for a heavily loaded automotive gear.

Use a low-carbon alloy steel (e.g. 8620-type). Gas carburize to about 0.8–1.0% surface C and a case depth matched to tooth size (often 0.8–1.2 mm), oil- or press-quench, and temper at 150–200 °C. The result is a 58–62 HRC case with compressive residual stress and a tough 30–45 HRC core, followed by shot peening.

MME-325 · How does a designer use DI to choose a steel?

Calculate the required ideal critical diameter from the actual section size and the quench severity H, using Grossmann charts. Then pick the cheapest steel whose DI, calculated from C, grain size and alloy multiplying factors, meets it with margin. Extra hardenability raises cost and quench cracking risk.

MME-326 · Tempering at 600 °C for 1 h is specified, but the furnace is limited to 550 °C. What time gives the same effect?

Hollomon–Jaffe: P = T(20 + log t). At 873 K and 1 h, P = 17,460. At 823 K, 20 + log t = 21.215, so t ≈ 16 h.

MME-327 · How can quench distortion be minimized in a long, thin part?

Use symmetric design and uniform sections, normalize or stress-relieve before hardening, load the part vertically, and use controlled agitation. Martempering, polymer or high-pressure gas quenching, or press (die) quenching help. Choosing a higher-hardenability steel allows a milder quench.

MME-328 · A D2 tool steel die is below the target hardness after hardening and one temper. What should be done?

High austenitizing temperature probably left excessive retained austenite. Use multiple tempers (double or triple) at the secondary-hardening temperature, or a cryogenic treatment between quench and temper, to transform it. Confirm the austenitizing temperature is not too high.

MME-329 · What is secondary hardening in high-speed steels?

Tempering at about 540–570 °C precipitates fine alloy carbides (Mo₂C, W₂C, VC) and transforms retained austenite on cooling. Hardness rises above the as-quenched value, giving the red hardness needed for cutting tools.

MME-330 · How is dual-phase steel produced?

Intercritical annealing of a low-C steel between Ac₁ and Ac₃ forms ferrite plus carbon-enriched austenite, followed by rapid cooling to turn the austenite into martensite (about 10–30%). The result has continuous yielding, a low yield-to-tensile ratio and high work hardening.

MME-331 · What is quenching and partitioning (Q&P)?

Quench to between Ms and Mf to form a controlled amount of martensite, then hold at the same or a higher temperature so carbon partitions from martensite into retained austenite and stabilizes it. The TRIP effect of the stable austenite gives high strength with good ductility in third-generation AHSS.

MME-332 · How is carbon potential controlled in a gas carburizing furnace?

Carbon potential depends on the CO/CO₂ and H₂/H₂O equilibria of the endothermic gas. It is measured by an in-situ oxygen probe (most common), dew point or infrared CO₂ analysis, and controlled by adjusting enriching gas (natural gas or propane) or air. Boost-diffuse cycles avoid carbides.

MME-333 · How does induction frequency control hardened depth?

Reference depth δ = 503√(ρ/(μr·f)) m. Above the Curie point (μr = 1, ρ ≈ 1.2 μΩ·m), 10 kHz gives about 5.5 mm and 100 kHz about 1.7 mm. Use high frequency for shallow cases and small parts, and low frequency for deep hardening of large shafts.

Mechanical Behavior & Testing

MME-334 · What are the main strengthening mechanisms in metals?

Work hardening (dislocation density), grain refinement (Hall–Petch, σy = σ₀ + k·d^−½), solid-solution strengthening (lattice misfit), and precipitation or dispersion hardening (particles blocking dislocations). Martensitic transformation and texture also contribute. All work by making dislocation motion harder.

MME-335 · What is the ductile-to-brittle transition and why does it matter?

BCC and HCP metals change from ductile to brittle cleavage fracture as temperature falls, because yield stress rises sharply. It contributed to failures of Liberty ships and the Titanic. Designers specify Charpy energy at the minimum service temperature and lower the transition with fine grains, Ni and low P and S.

MME-336 · A wide plate (KIC = 50 MPa√m) has a central crack 2a = 20 mm under 200 MPa. Is it safe, and what is the critical crack length?

K = σ√(πa) = 200√(π × 0.01) ≈ 35.4 MPa√m, which is below 50, so it is safe (safety factor about 1.4 on K). Critical a = (1/π)(KIC/σ)² = (1/π)(0.25)² ≈ 19.9 mm, so 2a ≈ 40 mm.

MME-337 · How is the Paris law used in damage-tolerant design?

da/dN = C(ΔK)^m, with m ≈ 3 for steels. Integrating from the detectable crack size to the critical size gives the remaining life, which sets inspection intervals. With m = 3, doubling the stress range cuts life by a factor of 8.

MME-338 · What is the leak-before-break criterion?

A pressure vessel or pipe is designed so that a through-wall crack, with length about twice the wall thickness, is still below the critical size. It leaks detectably before breaking catastrophically. This needs high toughness relative to stress: KIC > σ√(πt).

MME-339 · Using the Goodman relation, find the allowable stress amplitude for σm = 200 MPa, Su = 600 MPa, Se = 300 MPa.

σa/Se + σm/Su = 1, so σa = 300(1 − 200/600) = 200 MPa before a safety factor. Mean tensile stress reduces the allowable alternating stress.

MME-340 · What specimen thickness gives a valid plane-strain KIC for a steel with KIC = 50 MPa√m and σy = 500 MPa?

B ≥ 2.5(KIC/σy)² = 2.5 × (0.1)² = 0.025 m, so at least 25 mm. Crack length and ligament must meet the same size requirement.

MME-341 · How do creep mechanisms affect turbine blade design?

Power-law dislocation creep (ε̇ ∝ σⁿ, n ≈ 3–8) dominates at high stress. Nabarro–Herring creep (∝ 1/d²) and Coble creep (∝ 1/d³) dominate at low stress and fine grain size. Blades therefore use directionally solidified or single-crystal superalloys, eliminating transverse grain boundaries, with γ′ to block dislocations.

MME-342 · For a steel with σ₀ = 70 MPa and k = 0.74 MPa·m½, what is the yield strength at d = 25 µm and at 4 µm?

d^−½ = 200 m^−½ at 25 µm, so σy = 70 + 148 = 218 MPa. At 4 µm, d^−½ = 500 m^−½, so σy = 70 + 370 = 440 MPa. Grain refinement also improves toughness.

MME-343 · Why does shot peening improve fatigue life?

It plastically deforms the surface, creating a compressive residual stress layer (typically 0.1–0.5 mm deep) and work hardening. The compression lowers the effective mean stress where fatigue cracks start, and benefits are largest for notched or high-strength parts.

MME-344 · A failed crankshaft shows beach marks and ratchet marks on a smooth zone and a rough fibrous zone. How do you interpret it?

It is a fatigue failure. Ratchet marks point to multiple origins, often from a stress concentration such as a fillet. The size of the final fracture zone relative to the fatigue zone indicates the stress level: a large final zone means high nominal stress. Next, check the fillet radius, surface finish, hardness and loading.

Iron & Steelmaking

MME-345 · How much hematite ore (80% Fe₂O₃) is needed per tonne of pig iron with 94% Fe?

Fe₂O₃ contains 69.94% Fe, so Fe₂O₃ needed = 940/0.6994 ≈ 1344 kg. Ore needed = 1344/0.80 ≈ 1680 kg per tonne, assuming no Fe loss to slag.

MME-346 · Estimate the oxygen needed to decarburize 1 t of hot metal from 4.3% to 0.05% C (all C to CO).

C removed = 42.5 kg = 3.54 kmol. C + ½O₂ → CO needs 1.77 kmol O₂ ≈ 39.7 Nm³. Oxidation of Si, Mn, Fe and some CO₂ formation raise real consumption to about 50–60 Nm³/t.

MME-347 · What are the technical challenges of hydrogen-based DRI for low-CO₂ steel?

Fe₂O₃ + 3H₂ → 2Fe + 3H₂O is endothermic, unlike CO reduction, so heat must be supplied. The DRI has no carbon, which affects EAF foaming slag and carburization. Gangue must be low, ideally DR-grade pellets. Large supplies of low-cost green hydrogen are also needed.

MME-348 · Outline a clean-steel strategy for bearing-quality steel.

Use low-oxygen Al deoxidation with slag control (high basicity, low FeO), vacuum degassing, gentle Ar stirring for flotation, Ca treatment or careful avoidance of it, shrouded transfers, tundish flow control devices, and electromagnetic stirring. Aim for total oxygen at the ppm level and few large hard oxides.

MME-349 · Continuously cast microalloyed slabs show transverse corner cracks. What are the cause and the fix?

Straightening occurs in the hot-ductility trough (about 700–900 °C), where AlN or Nb(C,N) precipitates and thin ferrite films form at austenite grain boundaries. Adjust secondary cooling so corners straighten above or below the trough, reduce oscillation-mark depth, and control N, Al and Nb levels.

MME-350 · How does TMCP produce high-strength, tough HSLA plate?

Microalloying (Nb, Ti, V) plus controlled rolling below the recrystallization stop temperature produces pancaked austenite with many nucleation sites. Accelerated cooling then gives very fine ferrite or bainite. Strength comes from grain refinement and precipitation, not carbon, so weldability stays good.

MME-351 · What controls the sulfur partition ratio between slag and metal?

LS = (%S)/[%S] increases with slag sulfide capacity (higher basicity), lower oxygen potential (deoxidized steel, low FeO) and higher temperature. Ladle desulfurization therefore uses Al-killed steel with CaO-Al₂O₃ slag and strong stirring.

MME-352 · What are the trade-offs between scrap-based and DRI-based EAF charges?

Scrap is cheaper and uses less energy but carries tramp elements (Cu, Sn, Ni) that cannot be refined out and cause hot shortness. DRI dilutes residuals but adds gangue that increases slag and power, about 15–20% more electricity per tonne depending on metallization.

MME-353 · How is interstitial-free (IF) steel made and why?

RH vacuum decarburization reaches C below about 30 ppm, and Ti and/or Nb tie up the remaining C and N. With no free interstitials, the steel is non-ageing and develops a strong {111} texture, giving very high r-values for deep-drawn car body panels.

MME-354 · How do DP, TRIP and TWIP steels differ?

DP (ferrite + martensite) offers high work hardening at moderate cost. TRIP (ferrite + bainite + retained austenite) gains uniform elongation as austenite transforms under strain. TWIP (high-Mn austenite, low SFE) twins under deformation for very high strength and ductility but is costly and hard to weld.

MME-355 · Why do CRI and CSR matter for blast furnace coke?

CRI (coke reactivity index) measures reactivity with CO₂, and CSR (coke strength after reaction) measures the strength left afterwards. Weak, reactive coke degrades in the lower shaft, reducing permeability and raising coke rate. High-PCI furnaces need low-CRI, high-CSR coke.

Non-Ferrous Metallurgy

MME-356 · Explain precipitation hardening.

Solution-treat to dissolve the solute, quench to retain a supersaturated solid solution, then age naturally or artificially to form fine coherent or semi-coherent precipitates that block dislocations. Al 2xxx and 7xxx alloys are classic examples. Overaging coarsens precipitates and lowers strength.

MME-357 · Why would an aerospace designer specify 7075-T73 instead of T6?

T73 is overaged in two steps, which coarsens grain-boundary η precipitates and widens precipitate-free zones. This greatly improves resistance to stress corrosion cracking and exfoliation, at a cost of roughly 10–15% in strength. T76 and T74 are intermediate compromises.

MME-358 · What is the precipitation sequence in Al–Cu alloys?

Supersaturated solid solution → GP zones (Cu-rich discs) → θ″ (coherent) → θ′ (semi-coherent) → θ (Al₂Cu, incoherent). Peak hardness is associated with θ″ and θ′, and overaging gives coarse θ.

MME-359 · Why are single-crystal superalloys used for turbine blades?

Removing grain boundaries eliminates grain-boundary creep and cracking. Grain-boundary strengtheners (C, B, Zr, Hf) can then be removed, which raises the incipient melting temperature and allows higher solution treatments and more γ′. Re and Ru additions further slow diffusion.

MME-360 · How does Ti-6Al-4V microstructure affect performance?

Equiaxed or bimodal structures (processed in the α + β field) give the best ductility, low-cycle fatigue and crack initiation resistance. Fully lamellar structures (β-processed) give higher fracture toughness, slower fatigue crack growth and better creep. The choice depends on whether initiation or propagation dominates.

MME-361 · Select a material for seawater-cooled condenser tubes.

90/10 or 70/30 Cu–Ni alloys resist seawater corrosion and biofouling and handle moderate velocities (about 2.5–3.5 m/s). Titanium tolerates high velocities, polluted water and cleaning and has very long life, but costs more and needs galvanic care with Cu-alloy tubesheets. Admiralty brass is limited to cleaner, slower water.

MME-362 · What are the advantages and challenges of Al–Li alloys?

Each 1 wt% Li lowers density by about 3% and raises modulus by about 6%, through δ′ (Al₃Li) and T1 (Al₂CuLi) precipitates. Older alloys suffered anisotropy, low short-transverse toughness and Li loss in processing. Third-generation alloys use lower Li with Cu, Mg and Ag.

MME-363 · Why are TCP phases avoided in superalloys?

σ, μ and Laves phases are brittle plates that form during long exposure when refractory elements (Cr, Mo, W, Re) are excessive. They take strengtheners out of the matrix and start cracks. Compositions are screened with electron-vacancy (PHACOMP) or d-electron (Md) methods.

MME-364 · Why is oxygen-free copper used instead of ETP copper in hydrogen atmospheres?

ETP copper contains Cu₂O. At high temperature, hydrogen diffuses in and reacts (Cu₂O + H₂ → 2Cu + H₂O), forming high-pressure steam that causes grain-boundary fissures (hydrogen sickness). Oxygen-free (OF/OFE) or phosphorus-deoxidized copper avoids this.

MME-365 · Compare the specific strengths of Ti-6Al-4V, 7075-T6 and quenched-and-tempered 4340.

Approximate values: Ti-6Al-4V 950 MPa/4.43 g/cm³ ≈ 214 kN·m/kg; 7075-T6 570/2.81 ≈ 203; 4340 at 1450/7.85 ≈ 185. They are similar, so selection depends on temperature, stiffness, corrosion, fatigue and cost.

Extractive Metallurgy & Thermodynamics

MME-366 · Estimate the decomposition temperature of CaCO₃ at 1 atm CO₂ (ΔH° ≈ 178 kJ/mol, ΔS° ≈ 160 J/mol·K).

ΔG° = ΔH° − TΔS° = 0 at T = 178,000/160 ≈ 1110 K, about 840 °C. Above this temperature, calcination is spontaneous at 1 atm CO₂. Lower CO₂ partial pressure lowers it.

MME-367 · Estimate the specific energy for Hall–Héroult aluminium at 4.2 V cell voltage and 95% current efficiency.

Charge per kg = (1000/26.98) × 3 × 96485 ≈ 1.073×10⁷ C ≈ 2980 Ah. Energy = 2980 × 4.2/0.95 ≈ 13.2 kWh/kg, which matches modern smelters at about 13–14 kWh/kg.

MME-368 · What are the advantages of flash smelting for copper concentrates?

Dry concentrate burns in oxygen-enriched air in suspension, and sulfide oxidation supplies most of the heat (nearly autogenous). This cuts fuel use, gives high-grade matte and produces a strong continuous SO₂ stream for acid plants. The trade-offs are higher Cu losses in slag and the need for slag cleaning.

MME-369 · When is pressure (autoclave) leaching chosen?

For refractory sulfide gold ores, nickel laterites (HPAL at about 250 °C in H₂SO₄) and zinc concentrates. High temperature and oxygen pressure speed kinetics, oxidize sulfides to elemental S or sulfate without SO₂ emissions, and reject iron as hematite. Capital cost and scaling are high.

MME-370 · How does gold cyanidation work and how is gold recovered?

Elsner reaction: 4Au + 8CN⁻ + O₂ + 2H₂O → 4Au(CN)₂⁻ + 4OH⁻, at pH 10–11 to avoid HCN. Gold is adsorbed on activated carbon (CIP/CIL), then eluted and electrowon or recovered by Merrill–Crowe zinc cementation. Cyanide detoxification is required.

MME-371 · How is iron removed from zinc leach solutions?

Iron dissolved during hot acid leaching of zinc ferrites is precipitated as jarosite, goethite or hematite by controlling pH, temperature and oxidation. Jarosite is simplest but makes large residues. Hematite gives a saleable, smaller residue but needs autoclaves.

MME-372 · How would you optimize a sulfide flotation circuit?

Grind to liberation size, select collectors (xanthates, dithiophosphates), frothers and depressants (lime for pyrite, cyanide or SO₂ for Zn), and control pH. Use rougher–cleaner–scavenger stages with regrinding. Manage the grade–recovery trade-off with mass balances and economics.

MME-373 · 1000 t of ore at 1% Cu gives a 28% Cu concentrate at 90% recovery. Find the concentrate mass and tailings grade.

Cu recovered = 1000 × 0.01 × 0.9 = 9 t, so concentrate = 9/0.28 ≈ 32.1 t. Tailings hold 1 t Cu in about 967.9 t, about 0.10% Cu. The ratio of concentration is about 31.

MME-374 · Why does vacuum favour some refining reactions?

By Le Chatelier's principle, lowering pressure favours reactions that produce gas, such as [C] + [O] → CO(g) and [H] → ½H₂(g), and evaporation of volatile impurities. That allows deep decarburization, degassing and refining of reactive metals (vacuum arc remelting, electron beam melting) without oxidation.

MME-375 · How does zone refining achieve ultra-high purity?

A narrow molten zone is passed along a bar many times. Impurities with distribution coefficient k = Cs/CL < 1 concentrate in the liquid and are swept to one end. Smaller k gives more effective refining, reaching very high purity for Si, Ge and some metals.

Corrosion & Protection

MME-376 · What is sensitization in stainless steel and how is it prevented?

Heating austenitic stainless steel to about 450–850 °C, such as in a weld HAZ, precipitates Cr₂₃C₆ at grain boundaries. This depletes adjacent Cr below the passivity level and causes intergranular corrosion. Prevent it with low-carbon L grades, stabilized grades (321 with Ti, 347 with Nb), or a solution anneal and quench.

MME-377 · What corrosion rate of steel corresponds to a corrosion current density of 1 µA/cm²?

CR (mm/y) = 3.27×10⁻³ × i × EW/ρ = 3.27×10⁻³ × 1 × 27.92/7.87 ≈ 0.0116 mm/y. So 1 µA/cm² corresponds to about 11.6 µm/y for iron.

MME-378 · Size zinc anodes for 20-year protection of 10,000 m² of coated pipe (95% coating efficiency, 10 mA/m² for bare steel).

I = 10,000 × 0.05 × 0.01 = 5 A. Charge = 5 × 20 × 8760 = 876,000 Ah. Zn capacity ≈ 780 Ah/kg, so 1123 kg, or about 1.3 t at 85% utilization. Verify protection at −850 mV or more negative versus Cu/CuSO₄.

MME-379 · Select a stainless steel for warm seawater piping.

316L (PREN ≈ 24) pits and crevice-corrodes in warm seawater. Choose super duplex (e.g. 2507, PREN > 40) or 6Mo superaustenitic grades, or titanium for critical service. Control crevices and chlorination, and avoid galvanic coupling to carbon steel.

MME-380 · How do Evans diagrams and mixed potential theory explain corrosion?

The corrosion potential is where the total anodic current equals the total cathodic current on polarization (E vs log i) curves. Their intersection gives Ecorr and icorr. Adding oxidant or a more efficient cathode raises icorr, and inhibitors or cathodic protection lower it.

MME-381 · A 304 stainless hot-water tank cracks near welds after 2 years. Diagnose and recommend.

Branched transgranular cracks point to chloride SCC, driven by chlorides concentrating at hot surfaces (above about 60 °C) and weld residual stress. Switch to duplex 2205 or a higher-Ni alloy, reduce chlorides, avoid evaporation crevices and stress-relieve or shot-peen welds.

MME-382 · Why is hardness limited for steels in sour (H₂S) service?

H₂S poisons the hydrogen recombination reaction, so much more atomic hydrogen enters the steel, causing sulfide stress cracking in hard microstructures. NACE MR0175/ISO 15156 limits carbon and low-alloy steels to 22 HRC (about 250 HV), including weld HAZs.

MME-383 · How is icorr found from linear polarization resistance?

Stern–Geary: icorr = B/Rp, with B = βaβc/(2.303(βa + βc)). With βa = βc = 0.12 V/decade, B ≈ 0.026 V. Rp = 1000 Ω·cm² then gives icorr ≈ 26 µA/cm².

MME-384 · What do the kinetic laws of high-temperature oxidation indicate?

Parabolic growth (x² = kp·t) means growth is controlled by diffusion through a protective scale, as with Cr₂O₃ and Al₂O₃. Linear growth means a porous or cracked scale or interface control. Logarithmic growth applies to thin films at low temperature. Alloys rely on selective oxidation of Cr, Al or Si.

MME-385 · Why are duplex coating systems (galvanizing plus paint) so effective?

The paint protects the zinc from consumption, while zinc prevents rust creep from damaged paint. Combined life typically exceeds the sum of the individual lives by about 1.5–2.5 times. Surface preparation (sweep blasting or a tie coat) is needed for paint adhesion on zinc.

Powder Metallurgy

MME-386 · An iron compact pressed to 6.8 g/cm³ (full density 7.87) sinters to 7.2 g/cm³. Find the initial porosity and linear shrinkage.

Relative green density = 6.8/7.87 = 86.4%, so porosity is 13.6%. Assuming isotropic shrinkage, linear shrinkage = 1 − (6.8/7.2)^(1/3) ≈ 1.9%.

MME-387 · Why are powder metallurgy and HIP used for nickel superalloy turbine discs?

Highly alloyed discs segregate badly when cast and wrought. Gas-atomized powder plus HIP or extrusion and isothermal forging gives a fine, homogeneous structure. The key risks are ceramic inclusions and prior particle boundaries (oxide or carbide films), controlled by fine sieving, clean atomization and hot working.

MME-388 · How can a powder metallurgy gear approach wrought-steel fatigue performance?

Raise density above about 7.4 g/cm³ by warm or double press–double sinter. Use surface densification by rolling to near full density at the tooth root and flank. Add case carburizing or sinter hardening and shot peening. Porosity at the surface is the main fatigue-limiting factor.

MME-389 · What is spark plasma sintering (FAST) and when is it useful?

Pulsed DC current passes through a graphite die and the powder under uniaxial pressure, giving very fast heating and densification in minutes. It retains nanostructures and metastable phases and densifies hard-to-sinter ceramics, but is limited to simple shapes and modest sizes.

MME-390 · How do cobalt content and WC grain size affect cemented carbide?

More Co (e.g. 6% to 15%) raises toughness and lowers hardness and wear resistance. Finer WC grains (submicron) raise hardness and edge strength. Roughing and impact tools use more Co and coarse grains, while finishing tools use less Co and fine grains, often with coatings.

MME-391 · How do laser powder bed fusion parameters control porosity?

Volumetric energy density E = P/(v·h·t). Too low gives lack-of-fusion pores (irregular, between layers), and too high gives keyhole pores (spherical, deep). For example, 200 W, 1000 mm/s, 0.1 mm hatch and 30 µm layer give about 67 J/mm³. Gas pores from the powder need HIP.

MME-392 · Why does MIM need careful dimensional design?

After debinding, sintering to about 95–99% density shrinks the part about 15–20% linearly. Any non-uniform powder loading, gravity sag or friction causes distortion, so tooling is oversized precisely and supports are designed in.

MME-393 · What are key controls when sintering stainless steel powder?

Use high-purity H₂ or vacuum with a low dew point to reduce Cr₂O₃, and avoid N₂-rich atmospheres that form Cr₂N. Delube properly to avoid carbon pickup. Cool fast enough through about 850–450 °C to avoid sensitization. These steps preserve corrosion resistance.

MME-394 · How do ODS alloys differ from γ′-strengthened superalloys?

Fine, stable oxide dispersoids (e.g. Y₂O₃ or Y–Ti–O clusters), introduced by mechanical alloying, pin dislocations by the Orowan mechanism and do not dissolve or coarsen near the melting point. That gives creep strength above about 1000 °C and radiation tolerance, but the alloys are costly and anisotropic.

MME-395 · How can green density above 7.3 g/cm³ be achieved in ferrous powder metallurgy?

Warm compaction (powder and die at about 100–150 °C with special lubricants), die-wall lubrication with less internal lubricant, high-compressibility powders, high-velocity compaction, or double press–double sinter.

Welding & Joining Metallurgy

MME-396 · Calculate the heat input for SMAW at 25 V, 200 A and 300 mm/min (η = 0.8).

HI = 0.8 × 25 × 200 × 60/(1000 × 300) = 0.8 kJ/mm. Compare with the qualified range in the WPS. Higher heat input lowers HAZ hardness but coarsens grains and reduces toughness.

MME-397 · Calculate the IIW CE for a steel with 0.18 C, 1.40 Mn, 0.10 Cr, 0.05 Mo, 0.02 V, 0.20 Ni, 0.10 Cu and decide on preheat.

CE = 0.18 + 1.40/6 + (0.10 + 0.05 + 0.02)/5 + (0.20 + 0.10)/15 = 0.18 + 0.233 + 0.034 + 0.020 ≈ 0.47. Above about 0.45, low-hydrogen consumables and preheat are normally needed, with the level set by thickness, restraint and hydrogen.

MME-398 · How should carbon steel be welded to austenitic stainless steel?

Use an overalloyed filler such as 309L, so that dilution with carbon steel still gives austenite with some ferrite on the Schaeffler or WRC diagram and avoids martensite. For high-temperature service, use Ni-based fillers (ENiCrFe) to limit carbon migration and thermal-expansion stresses.

MME-399 · Why is t8/5 used to characterize weld cooling?

t8/5, the cooling time from 800 to 500 °C, spans the range where austenite transforms in steels. It links heat input, preheat, thickness and joint geometry to HAZ hardness and toughness. Short times give martensite, and long times give coarse ferrite or upper bainite with poor toughness.

MME-400 · What is reheat (stress-relief) cracking?

Intergranular cracking in the coarse-grained HAZ of Cr–Mo–V, some Ni-based and certain stainless steels during PWHT or high-temperature service. Intragranular carbide precipitation strengthens grain interiors, so relaxation strain concentrates at weak boundaries. Mitigate it by grinding toes, controlling composition and minimizing coarse grains.

MME-401 · What are a WPS, PQR and WPQ?

A WPS (Welding Procedure Specification) gives the parameters to use. A PQR (Procedure Qualification Record) documents the test weld and the mechanical tests that qualify it. A WPQ qualifies the individual welder. Changing essential variables requires requalification.

MME-402 · What makes aluminium alloys difficult to weld?

A tenacious Al₂O₃ film (melting point about 2050 °C) needs AC or DCEP cleaning action. High conductivity needs high heat input. Hydrogen causes porosity, and the wide freezing range causes hot cracking (in 6xxx without 4xxx or 5xxx filler). Heat-treatable alloys soften in the HAZ, so design allowables are reduced.

MME-403 · What are the key controls for welding duplex stainless steel?

Keep ferrite at about 35–65% in weld and HAZ. Too low a heat input gives excessive ferrite and Cr₂N, and too high a heat input or interpass temperature (above about 150 °C) causes σ phase. Use Ni-enriched fillers and N-containing shielding or backing gas, and avoid PWHT unless it is a full solution anneal.

MME-404 · How can weld distortion be controlled in a large fabrication?

Minimize weld volume with proper joint design (double-V, narrow gap), balance welds about the neutral axis, use back-step or skip sequences, preset or pre-camber parts, use strongbacks and fixtures, and use low-heat-input processes. Post-weld flame straightening or stress relief may still be needed.

MME-405 · A Q&T steel weld shows toe cracks discovered 24 hours after welding. What is the likely cause and fix?

Delayed cracking points to hydrogen-induced cold cracking in a hard HAZ. Use H4/H5 low-hydrogen consumables, baked and stored properly. Apply preheat and maintain interpass temperature, add post-heating for hydrogen bake-out, reduce restraint, and inspect after a sufficient delay, often 48 h.

Materials Characterization

MME-406 · When would you use SEM, TEM or XRD?

SEM images surfaces and fractures, typically to a few nm, with EDS chemistry. TEM images internal structure (dislocations, precipitates) down to atomic scale on thin foils, with diffraction. XRD identifies crystal structures and phases and measures lattice parameter, residual stress and texture on bulk samples.

MME-407 · A cubic FCC metal shows its first peak at 2θ = 43.3° with Cu Kα (0.15406 nm). Find the lattice parameter.

The first FCC peak is (111). d = λ/(2 sinθ) = 0.15406/(2 × sin 21.65°) ≈ 0.2088 nm, so a = d√3 ≈ 0.3617 nm. This matches copper.

MME-408 · How are cubic XRD patterns indexed?

The ratios of sin²θ for successive peaks follow h² + k² + l². BCC gives 2, 4, 6, 8, 10, 12, a ratio of 1:2:3:4:5:6. FCC gives 3, 4, 8, 11, 12, a ratio of 1:1.33:2.67:3.67:4. Simple cubic includes 1, 2, 3, 4, 5, 6 but skips 7.

MME-409 · How is residual stress measured with the XRD sin²ψ method?

The lattice spacing d of a high-angle reflection is measured at several tilt angles ψ. Stress is proportional to the slope of d against sin²ψ, multiplied by E/(1 + ν) divided by d₀. It measures surface stress to a depth of a few µm. Deeper profiles need layer removal or neutron diffraction.

MME-410 · Outline a systematic failure analysis of a fractured component.

Collect service history and preserve the evidence. Then do visual and stereo examination, NDT, fractography (SEM), chemical analysis, metallography, hardness and mechanical tests, and stress analysis. Identify the mode and root cause, then recommend corrective action. Avoid cleaning or mating the fracture surfaces before documenting them.

MME-411 · How is EBSD used to assess plastic strain?

Kernel average misorientation (KAM), grain orientation spread and geometrically necessary dislocation density maps show local lattice rotations from plastic deformation. Strained regions appear with high misorientation, and recrystallized grains appear near-zero. Pattern quality also drops with strain.

MME-412 · What can atom probe tomography reveal?

Three-dimensional atom-by-atom chemical maps at near-atomic resolution, from field evaporation of a needle specimen. It quantifies solute clusters, segregation at grain boundaries and dislocations, and the chemistry of nanoscale precipitates, including light elements such as C, B and H isotopes.

MME-413 · When is XPS the right technique?

For chemistry of the top few nanometres: elemental composition and chemical (oxidation) state from binding-energy shifts. Examples are passive films on stainless steel, contamination, coatings and adhesion failures. Depth profiling uses ion sputtering.

MME-414 · How do phased-array UT and TOFD improve weld inspection?

Phased array steers and focuses beams electronically, giving sector scans with fast coverage and imaging. TOFD times diffracted signals from flaw tips to size through-wall height accurately, independent of orientation. Combined, they can replace radiography for many thick-section welds.

MME-415 · An XRD peak at 2θ = 40° (Cu Kα) has instrument-corrected FWHM of 0.5°. Estimate the crystallite size.

β = 0.5° = 0.00873 rad and θ = 20°. t = 0.9 × 0.15406/(0.00873 × cos 20°) ≈ 16.9 nm, ignoring microstrain broadening (separated by a Williamson–Hall analysis).

Composites, Ceramics & Polymers

MME-416 · Find the longitudinal and transverse moduli of a CFRP with Vf = 0.6, Ef = 230 GPa and Em = 3.5 GPa.

Longitudinal: E∥ = 0.6 × 230 + 0.4 × 3.5 ≈ 139.4 GPa. Transverse: 1/E⊥ = 0.6/230 + 0.4/3.5 = 0.1169, so E⊥ ≈ 8.6 GPa. This shows strong anisotropy.

MME-417 · Find the critical length for a carbon fibre with σf = 3.5 GPa, d = 7 µm and τc = 50 MPa.

lc = σf·d/(2τc) = 3500 × 7×10⁻⁶/(2 × 50) ≈ 245 µm. Chopped fibres should be well over about 3.7 mm (15lc) for near-continuous efficiency.

MME-418 · A ceramic has Weibull m = 10 and σ₀ = 400 MPa. Find its survival probability at 300 MPa and the effect of a 10× larger volume.

Ps = exp[−(300/400)¹⁰] = exp(−0.056) ≈ 0.945. For the same failure probability, strength scales as (V₁/V₂)^(1/m) = 10^(−0.1) ≈ 0.79, so a larger part is about 21% weaker.

MME-419 · How would you lay up a CFRP panel loaded in several directions?

Use a balanced, symmetric quasi-isotropic layup such as [0/±45/90]s, which avoids bending–extension and shear coupling and warping during cure. Add 0° plies along principal loads and ±45° for shear and buckling. Avoid grouping many identical plies, to limit delamination, and check first-ply failure (Tsai–Wu) and damage tolerance.

MME-420 · Estimate the thermal shock parameter of alumina (σf = 350 MPa, ν = 0.22, E = 380 GPa, α = 8 × 10⁻⁶ /K).

R = σf(1 − ν)/(E·α) = 350×10⁶ × 0.78/(380×10⁹ × 8×10⁻⁶) ≈ 90 K. This first-order estimate ignores heat-transfer (Biot) effects. Si₃N₄ scores much higher because of lower α and higher strength.

MME-421 · Select a ceramic for high-speed hybrid bearing balls versus a cutting tool for hardened steel.

For bearing balls, use Si₃N₄: low density, high toughness for a ceramic, good rolling-contact fatigue and electrical insulation. For hard turning of steel, use cBN or alumina–TiC composites. Diamond is unsuitable because carbon dissolves into iron at cutting temperatures.

MME-422 · How is long-term polymer creep predicted from short tests?

Time–temperature superposition: tests at several temperatures are shifted along the log-time axis to form a master curve. The WLF equation, log aT = −C₁(T − Tr)/(C₂ + T − Tr), gives the shift above Tg, and Arrhenius shifts are used below Tg.

MME-423 · What material indices guide selection for minimum mass in stiffness-limited design?

Tie in tension: maximize E/ρ. Beam in bending: E^(1/2)/ρ. Panel in bending: E^(1/3)/ρ. Use σf/ρ, σf^(2/3)/ρ and σf^(1/2)/ρ for strength-limited cases. Plot on Ashby charts with selection lines of the matching slope.

MME-424 · What is the trade-off in fibre–matrix interface strength?

A strong interface transfers load efficiently, giving high strength and transverse properties, but lets cracks run straight through, so the composite is brittle. A weaker interface allows debonding, crack deflection and fibre pull-out, giving toughness. In CMCs a weak interphase (BN, pyrolytic C) is essential.

MME-425 · Outline the design of a SiC/SiC CMC turbine shroud.

Use SiC fibres with a BN interphase for crack deflection, and a SiC matrix by CVI, melt infiltration or PIP. Apply an environmental barrier coating (e.g. rare-earth silicates) to prevent recession from water vapour. Design for pseudo-ductile behaviour, matrix cracking stress limits, through-thickness weakness and attachment thermal stresses.

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