Materials Chemistry (Aerospace)

Question 1 (10)

(a) Three stages (6):

1. Solution treatment — heat to single-phase region (~480 °C for 7075) so all solute (Zn, Mg, Cu) dissolves into the Al solid solution. (2)
2. Quench — rapid cooling traps a supersaturated solid solution (SSSS); solute has no time to precipitate. (2)
3. Ageing (natural at RT or artificial ~120–160 °C) — controlled precipitation of fine coherent particles (GP zones → $\eta'$/MgZn₂). (2)

(b) Peak hardness / over-ageing (4):

• Strength comes from precipitates obstructing dislocation motion. (1)
• Fine, coherent, closely-spaced precipitates give maximum resistance (dislocations must cut/loop past many obstacles) → peak hardness at intermediate time. (2)
• Over-ageing: particles coarsen (Ostwald ripening), become fewer and further apart, lose coherency → dislocations bypass easily (Orowan looping), so strength drops. (1)

Question 2 (10)

(a) (4): W melting point $\approx 3695\ \text{K}$ ($\approx 3422\ °\text{C}$) (2). It fails because W oxidises readily above ~500–600 °C, forming volatile/friable WO₃ that sublimes away in the oxidising exhaust — the surface recedes despite the metal being far below its melting point; hence W needs coatings or inert/reducing environments. (2)

(b) (6): (1 mark each for correct rank order + property; award up to 6)

• W — highest m.p. (~3422 °C); densest, used for throat inserts.
• Re (~3186 °C) — excellent high-T strength & ductility; alloyed with W/Mo, used in throat linings.
• Ta (~3017 °C) — ductile, forms protective/adherent oxide, excellent corrosion resistance, Ta-C coatings.
• Mo (~2623 °C) — lowest of the four; good thermal conductivity but oxidises badly; used with MoSi₂ coatings. Order highest→lowest: W > Re > Ta > Mo.

Question 3 (12)

(a) Iso-strain derivation (5): Assumptions: fibres and matrix perfectly bonded, both strain equally under longitudinal load $\varepsilon_f = \varepsilon_m = \varepsilon_1$; loads carried in parallel. (2) Total load $P = P_f + P_m$; with stress $\sigma = P/A$: $\sigma_1 A = \sigma_f A_f + \sigma_m A_m$ Divide by $A$, area fractions = volume fractions: $\sigma_1 = \sigma_f V_f + \sigma_m V_m$ Apply $\sigma = E\varepsilon$ with common $\varepsilon_1$: $E_1 = E_f V_f + E_m V_m,\quad V_m = 1-V_f$ (3)

(b) Compute (3): $E_1 = 230(0.60) + 3.5(0.40) = 138 + 1.4 = 139.4\ \text{GPa}$

(c) Iso-stress $E_2$ (4): Equal stress, strains add in series: $\frac{1}{E_2} = \frac{V_f}{E_f} + \frac{V_m}{E_m}$ (1) $\frac{1}{E_2} = \frac{0.60}{230} + \frac{0.40}{3.5} = 0.002609 + 0.114286 = 0.116895$ $E_2 = 8.55\ \text{GPa}$ (2) Anisotropy: $E_1/E_2 = 139.4/8.55 \approx 16.3$ — highly anisotropic; transverse is matrix-dominated. (1)

Question 4 (10)

(a) Ablation mechanism (6): (2 marks concept, 1 each for three processes, up to 6) The resin (phenolic) pyrolyses endothermically, absorbing heat, and forms a char layer. Heat-rejection processes:

1. Endothermic pyrolysis / decomposition absorbs energy.
2. Transpiration/blowing — pyrolysis gases percolate outward, thickening the boundary layer and blocking convective heat flux to the wall.
3. Surface radiation / re-radiation from the hot char (high emissivity) radiates energy back to space.
4. Mass loss (surface recession) carries heat away with ablated material. The virgin material beneath stays cool (low conductivity char is a good insulator).

(b) Ablative vs silica tile (4):

Question 5 (10)

(a) SCC (4): SCC = brittle crack growth from combined action of a tensile (static) stress and a corrosive environment on a susceptible material, at stresses well below the yield/normal fracture stress. (2) Three coexisting conditions: (1) susceptible alloy, (2) sustained tensile stress, (3) specific corrosive environment. (2)

(b) Hydrogen embrittlement (6): Atomic hydrogen (from corrosion, plating, welding, or service) diffuses into the lattice and accumulates at high-stress regions / grain boundaries / dislocations, reducing cohesive strength and promoting brittle intergranular fracture at stresses below yield. (4) Mitigation (any two, 1 each): baking after plating to drive out H; using lower-strength / more tolerant alloys or reduced hardness; cadmium/coating with H-barrier; controlling plating processes; avoiding cathodic overprotection. (2)

Question 6 (8)

Parallel (thickness-weighted) average, 50% at $0°$ (contribute $E_1$), 50% at $90°$ (contribute $E_2$): $E_{eff} = 0.5\,E_1 + 0.5\,E_2 = 0.5(140) + 0.5(10) = 70 + 5 = 75\ \text{GPa}$ (4) Comment (4): The real laminate value differs because (i) simple rule-of-mixtures ignores Poisson coupling and shear between plies handled by full laminate (ABD/CLT) theory; (ii) $90°$ plies still carry some load and the plies are constrained to equal strain, so exact classical-laminate result differs slightly; (iii) the cross-ply is far less stiff along $0°$ than a pure UD ply (140 GPa) — this is the trade-off for quasi-balanced properties in both directions. (2 for numeric, +2 reasoning)

[
  {"claim":"E1 rule of mixtures = 139.4 GPa","code":"Ef=230; Em=3.5; Vf=Rational(6,10); Vm=1-Vf; E1=Ef*Vf+Em*Vm; result = abs(float(E1)-139.4)<1e-6"},
  {"claim":"E2 iso-stress = 8.55 GPa (2 dp)","code":"Ef=230; Em=Rational(7,2); Vf=Rational(6,10); Vm=1-Vf; E2=1/(Vf/Ef+Vm/Em); result = abs(float(E2)-8.55)<0.01"},
  {"claim":"Anisotropy ratio E1/E2 approx 16.3","code":"Ef=230; Em=Rational(7,2); Vf=Rational(6,10); Vm=1-Vf; E1=Ef*Vf+Em*Vm; E2=1/(Vf/Ef+Vm/Em); r=E1/E2; result = abs(float(r)-16.3)<0.2"},
  {"claim":"Cross-ply parallel average = 75 GPa","code":"E1=140; E2=10; Eeff=0.5*E1+0.5*E2; result = abs(Eeff-75)<1e-9"}
]