Level 4 — ApplicationMaterials Chemistry (Aerospace)

Materials Chemistry (Aerospace)

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Level 4 — Application (novel/unseen problems, no hints) Time: 60 minutes | Total: 60 marks

Answer ALL questions. Data provided where needed. Show all working.

Question 1 — Precipitation-hardened aluminium airframe (12 marks)

A wing spar is machined from 7075 aluminium, whose strength derives from precipitation hardening of the Al–Zn–Mg–Cu system.

(a) The processing sequence is: solution treatment → water quench → artificial ageing. Explain the microstructural purpose of each of the three steps, and why the order cannot be reversed. (6)

(b) A batch is accidentally aged at too high a temperature for too long, and the yield strength drops below specification. Name this phenomenon and explain, in terms of precipitate size and spacing, why strength falls. (3)

(c) 7075 in the peak-aged (T6) temper is susceptible to stress-corrosion cracking, whereas the T73 (over-aged) temper is not. Explain the trade-off engineers accept when specifying T73 over T6. (3)

Question 2 — Rocket nozzle refractory selection (12 marks)

A designer must choose a throat-liner material for a bipropellant engine where the throat reaches ~2800 °C in an oxidising exhaust. Candidate data:

Metal	Melting point (°C)	Density (g cm⁻³)	Oxidation behaviour
W	3422	19.3	oxide volatile above ~800 °C
Mo	2623	10.2	oxide (MoO₃) sublimes ~795 °C
Ta	3017	16.7	forms adherent Ta₂O₅ but embrittles
Re	3186	21.0	oxide volatile, but ductile

(a) On melting point alone, rank the four metals for the 2800 °C throat and state which are immediately eliminated. (3)

(b) Melting point is not the deciding factor here. Identify the property that actually governs the choice in this oxidising environment, and explain why a bare refractory metal will fail regardless of melting point. (4)

(c) Propose a practical engineering solution that lets a refractory-metal liner survive, and name a specific coating chemistry/technique from surface-treatment practice. (3)

(d) Re is often alloyed with W in nozzle throats. Give two reasons, one mechanical and one relating to fabrication. (2)

Question 3 — Thermal protection heat-flux problem (14 marks)

During re-entry, the underside of an orbiter reaches a surface heat flux while a silica tile protects the aluminium structure beneath (max allowable 175 °C).

Data:

Silica tile thermal conductivity $k = 0.050\ \text{W m}^{-1}\text{K}^{-1}$
Tile outer surface temperature (steady) $T_{\text{hot}} = 1260\ ^\circ\text{C}$
Structure interface must stay $\le 175\ ^\circ\text{C}$
Assume 1-D steady conduction, $q = k\,\Delta T/L$

(a) Compute the minimum tile thickness $L$ required if the steady-state conductive heat flux through the tile is limited to $q = 3500\ \text{W m}^{-2}$ . (4)

(b) Using that same thickness, and now taking $q = k\,\Delta T/L$ with the given $T_{\text{hot}}$ and $T_{\text{cold}} = 175\ ^\circ\text{C}$ , compute the actual steady-state flux $q$ and comment on whether the thickness from (a) is adequate. (4)

(c) The silica tile is a reusable, non-ablative system, whereas PICA on a capsule is single-use. Explain the fundamental difference in the physical mechanism of heat rejection between the two, and why the shuttle could reuse tiles but a PICA heatshield cannot. (4)

(d) UHTCs such as $\text{ZrB}_2$ and $\text{HfB}_2$ are used on sharp leading edges rather than silica tiles. Give the property reason. (2)

Question 4 — CFRP laminate lay-up (12 marks)

A flat control-surface skin is a symmetric CFRP laminate with the stacking sequence $[0/45/-45/90]_s$ .

(a) Write out the full ply sequence implied by the subscript $s$ , and state how many plies the laminate contains. (3)

(b) Explain, in load-transfer terms, why the $0°$ plies are placed on the outside of a laminate loaded primarily in bending, rather than at the mid-plane. (3)

(c) A quasi-isotropic response is desired. State whether $[0/45/-45/90]_s$ qualifies and justify with the ply-angle criterion. (3)

(d) A technician mistakenly lays up $[0/45/-45/90]$ (not symmetric). State one mechanical consequence during cure/loading and why symmetry avoids it. (3)

Question 5 — Corrosion failure diagnosis (10 marks)

A high-strength steel landing-gear bolt (electroplated with cadmium) fractures in service. Fractography shows brittle intergranular cracking with no macroscopic yielding.

(a) Two hydrogen-related mechanisms are candidates: hydrogen embrittlement introduced during plating, versus stress-corrosion cracking in service. Give one distinguishing feature that would help you tell them apart. (3)

(b) Explain the chemical origin of hydrogen entering the steel during electroplating, and state the standard post-plating heat treatment used to mitigate it (including approximate purpose/temperature effect). (4)

END OF PAPER

Answer keyMark scheme & solutions

Question 1 (12)

(a) [6]

Solution treatment (~465–480 °C): dissolves the Zn/Mg/Cu solute into a single-phase supersaturated solid solution — homogenises the alloy so alloying elements are available for later precipitation. (2)
Quench: rapid cooling traps solute in a supersaturated solid solution at room T, preventing equilibrium coarse precipitation at grain boundaries — freezes in the excess solute + vacancies needed for ageing. (2)
Artificial ageing (~120–160 °C): controlled decomposition forms fine, coherent GP zones / η′ precipitates that impede dislocation motion → strength. (1)
Order fixed: must dissolve solute before you can trap it; must trap it (quench) before you can precipitate it in a fine dispersion. Reversing gives coarse equilibrium phases and no strengthening. (1)

(b) [3] Over-ageing. (1) Precipitates coarsen (Ostwald ripening): fewer, larger, more widely spaced particles. (1) Dislocations bypass them by Orowan looping at increasing spacing → strengthening contribution ∝ 1/spacing falls, so yield strength drops. (1)

(c) [3] T73 over-ages deliberately; SCC resistance improves because coarser precipitate structure/grain-boundary precipitate distribution reduces anodic path susceptibility and H-assisted cracking. (2) Trade-off: T73 sacrifices ~10–15% peak strength for markedly better SCC resistance. (1)

Question 2 (12)

(a) [3] By melting point (high→low): W (3422) > Re (3186) > Ta (3017) > Mo (2623). (1) At 2800 °C, Mo (2623 °C) is eliminated — below throat temperature. (1) W, Re, Ta all melt above 2800 °C so survive on Tm alone. (1)

(b) [4] The governing property is high-temperature oxidation resistance (in the oxidising exhaust). (2) Even W/Re, which have adequate Tm, form volatile oxides that sublime/evaporate above ~800 °C → continuous recession of the throat surface; the metal is consumed by oxidation long before melting, so a bare refractory metal fails. (2)

(c) [3] Apply a protective oxidation-resistant coating, e.g. an iridium or silicide (MoSi₂/disilicide) or HfC/ZrC coating, applied by CVD / plasma spraying / pack cementation. (2) The coating isolates the metal from the oxidising gas. (1)

(d) [2] Mechanical: Re raises ductility / lowers ductile-brittle transition of W (relieves W's brittleness). (1) Fabrication: W–Re resists recrystallisation embrittlement and is more workable/weldable. (1)

Question 3 (14)

(a) [4] $\Delta T = 1260 - 175 = 1085\ \text{K}$ . (1) $q = k\Delta T/L \Rightarrow L = k\Delta T/q$ . (1) $L = (0.050 \times 1085)/3500 = 54.25/3500 = 0.0155\ \text{m} \approx 15.5\ \text{mm}$ . (2)

(b) [4] With $L = 0.0155\ \text{m}$ , $q = k\Delta T/L = 0.050\times1085/0.0155 = 3500\ \text{W m}^{-2}$ (by construction equals the limit). (2) Since computed flux equals the design limit exactly, the thickness is the minimum adequate value; any thinner tile exceeds allowable flux/interface temperature, so 15.5 mm is just adequate (add margin in practice). (2)

(c) [4] Silica tile: passive insulation + radiative re-emission — high emissivity coating radiates heat back to atmosphere while the low-conductivity ceramic limits conduction to structure; no material is consumed → reusable. (2) PICA: ablation — the material pyrolyses/chars and the sacrificial mass is carried away (blowing gas absorbs energy and blocks convective flux); material is chemically consumed, so it cannot be reused. (2)

(d) [2] UHTCs have very high melting points AND high thermal conductivity + oxidation resistance, letting sharp (small-radius) leading edges, which see the highest heat flux, conduct/spread heat without ablating; silica tiles would recede/fail at such fluxes and cannot form sharp edges. (2)

Question 4 (12)

(a) [3] Symmetric about mid-plane: $[0/45/-45/90/90/-45/45/0]$ . (2) 8 plies. (1)

(b) [3] In bending, stress is highest at the outer surfaces (max distance from neutral axis). (1) Placing stiff/strong $0°$ fibres there maximises bending stiffness (I contribution $\propto$ distance²) and carries the bending tensile/compressive load efficiently. (2)

(c) [3] Yes, quasi-isotropic. (1) Criterion: plies at equal angular increments covering $0,\pm45,90$ (i.e. $180°/n$ spacing with equal numbers of each) — here $0,45,-45,90$ present in equal proportion → in-plane stiffness independent of direction. (2)

(d) [3] Non-symmetric laminate has bending–extension coupling (non-zero B matrix): (1) it warps/twists on cure (thermal residual stresses) and distorts under in-plane load. (1) Symmetry makes B = 0, so extension and bending decouple and the part stays flat. (1)

Question 5 (10)

(a) [3] Distinguishing feature (any one, well explained, 3):

HE from plating is present before service and shows delayed brittle failure under sustained load with no corrosion products; SCC requires a corrosive environment during service and shows corrosion/oxide on crack faces.
SCC is time/environment-dependent and progresses with an active electrolyte; internal HE occurs even in dry/clean conditions.

(b) [4] During cadmium electroplating, cathodic reactions (H⁺ reduction / water reduction) at the steel surface liberate atomic hydrogen; some atomic H diffuses into the steel lattice instead of forming H₂ gas. (2) Mitigation: post-plate baking (hydrogen relief), typically ~190–220 °C for several hours, which drives dissolved hydrogen out by diffusion before it embrittles. (2)

(c) [3] High-strength steels have high hardness/tensile strength and low fracture toughness; (1) their martensitic microstructure and high internal stresses make them highly susceptible to H-assisted intergranular cracking at low absorbed-H levels, (1) whereas ductile mild steels tolerate hydrogen and yield rather than crack. (1)

[
  {"claim":"Q3a minimum tile thickness ≈ 0.0155 m",
   "code":"k=0.050; dT=1260-175; q=3500; L=k*dT/q; result = abs(L-0.0155)<1e-3"},
  {"claim":"Q3b flux with that thickness returns 3500 W/m^2",
   "code":"k=0.050; dT=1260-175; L=k*dT/3500; qcalc=k*dT/L; result = abs(qcalc-3500)<1e-6"},
  {"claim":"Q4a symmetric [0/45/-45/90]s has 8 plies",
   "code":"n=4*2; result = n==8"},
  {"claim":"Q2a Mo melting point below 2800C throat",
   "code":"result = 2623 < 2800"}
]