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Aluminium alloys in airframes: HSC Engineering Studies Aeronautical Engineering

Syllabus dot point

“Describe the production, heat treatment and key properties of aluminium alloys 2024 and 7075, identify their use in airframe structures, and compare them with structural steel and titanium”

9 min read

Quick answer

Aluminium 2024-T3 (Al-Cu) is the fuselage skin alloy: damage-tolerant and fatigue-resistant, chosen for cyclic pressurisation loading. Aluminium 7075-T6 (Al-Zn) is the wing-spar alloy: highest specific strength but lower fatigue and corrosion resistance, chosen for static bending loads at minimum mass. Both gain their strength from precipitation hardening (solution treatment, quench, ageing), and both are one-third the density of steel with comparable or better specific strength. Titanium (Ti-6Al-4V) beats both on specific strength and corrosion resistance but costs about 10 times more per kilogram, so it is reserved for high-load, high-temperature components like engine pylons and landing gear rather than the whole airframe.

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Exam-style questions
  4. Practice questions
In plain English

Think of precipitation hardening like making rocky road: you melt everything together into a smooth, uniform mixture (solution treatment), then chill it fast so the mixture cannot separate back out into big lumps (the quench), then let it sit at a gentle warm temperature so tiny hard bits crystallise evenly throughout, not as one giant clump (ageing). Those tiny hard bits are what make the aluminium strong, because they get in the way whenever the metal's internal layers try to slide past each other, the same way a handful of fine gravel scattered through soft clay makes it much harder to squash than the clay alone. Choose the wrong "recipe" (too little ageing, or leaving it too long so the bits clump together and coarsen) and you get a much weaker, softer material, which is exactly why aircraft makers specify tempers like T3 and T6 as precisely as a chef specifies oven time and temperature.

What this dot point is asking

NESA wants you to describe how aluminium alloys are produced and heat-treated, identify the specific properties of 2024 and 7075 (the two airframe-grade alloys), explain how the properties dictate selection for fuselage skins versus wing spars, and compare aluminium with structural steel and titanium. This connects to composite materials in aircraft, the alternative modern structural material.

The answer

Production

Aluminium is smelted from alumina (Al2O3Al_2O_3, refined from bauxite by the Bayer process) using the Hall-Heroult electrolytic process at about 950 degrees C in molten cryolite. Australia is the world's largest producer of bauxite (Weipa in Queensland, Boddington in Western Australia) and the second largest producer of alumina. Smelting is energy-intensive: about 14 kWh per kg of aluminium.

The pure metal is then alloyed with copper, zinc, magnesium, silicon and manganese to produce the wrought aluminium alloy families used in aerospace.

Alloy families and tempers

The standard four-digit designation identifies the principal alloying element:

  • 1xxx Pure aluminium (electrical conductors, foil)
  • 2xxx Al-Cu (aerospace, 2024)
  • 5xxx Al-Mg (marine, structural)
  • 6xxx Al-Mg-Si (extrusions, 6061 for general engineering)
  • 7xxx Al-Zn (aerospace, 7075)

The temper designation follows: T3 (solution treated, cold worked, naturally aged), T6 (solution treated and artificially aged), T7 (solution treated and over-aged for stress corrosion resistance).

The precipitation hardening sequence
  1. Solution treatment at about 500 degrees C dissolves alloying elements into the aluminium lattice as a single solid solution.
  2. Rapid quench freezes this as a supersaturated solid solution at room temperature.
  3. Ageing (natural at room temperature for T3; artificial at 120 to 175 degrees C for T6) allows fine intermetallic precipitates to form within the grains, raising yield strength by a factor of 3 to 5 above the annealed state.

Precipitation hardening temperature-time schedule An illustrative temperature versus time schedule for artificially aged 7075-T6. The temperature starts near room temperature, rises and holds at about 500 degrees Celsius during solution treatment, drops sharply during the rapid quench back to near room temperature, then rises again and holds at about 150 degrees Celsius during artificial ageing for several hours before returning to room temperature. Each stage transition is marked with a labelled point sitting on the curve. 500C 375C 250C 125C 20C Solution treatment, ~500C Rapid quench Ageing, ~150C, several hours Temperature versus time for artificial ageing to T6 (illustrative ExamExplained schedule)

Property comparison

Property 2024-T3 7075-T6 Grade 350 steel Ti-6Al-4V
Density (kg/m^3) 2780 2810 7850 4430
Yield strength (MPa) 345 503 350 880
Ultimate (MPa) 485 572 480 950
Specific strength (MPa per kg/m^3) 0.124 0.179 0.045 0.198
Young's modulus (GPa) 73 72 200 114
Fatigue strength at 10710^7 cycles (MPa) 138 159 240 510

Aluminium is one-third the density of steel with comparable yield strength, giving 3 to 4 times the specific strength. Titanium has higher specific strength still but costs about 10 times more per kilogram.

Selecting between 2024 and 7075 in an exam answer

When a question asks you to choose or justify an alloy for a specific component, do not just recite the property table. Match the LOADING TYPE to the property: cyclic/pressurisation loading (fuselage skin) points to fatigue resistance and damage tolerance, so favour 2024-T3; steady bending/static strength at minimum mass (wing spars) points to specific strength, so favour 7075-T6. Always name both the alloy AND the reason tied to the actual loading described in the question, not a generic "it's stronger."

Where each alloy goes

  • Fuselage skin and frames. 2024-T3 sheet (often clad with pure aluminium for corrosion resistance), riveted in place. Boeing 737, 747, 767; Airbus A320, A330. Damage-tolerant under fatigue.
  • Wing spars and ribs. 7075-T6 extrusions and machined parts. Higher strength means smaller cross-section for the same load.
  • Skin around pressurised areas, doors and windows. Doubler plates and reinforcements in 2024 or 7075 depending on local stress.
  • Engine pylons and landing gear. Often high-strength steel or titanium because of higher load and temperature.

Australian context

The Government Aircraft Factories (Port Melbourne and Fishermans Bend, 1936 to 1986) produced aluminium-airframe aircraft including the Avon Sabre (CAC Sabre), the Nomad and the Wirraway trainer. The current Hawker de Havilland operations at Bankstown supply aluminium parts to Boeing under offset agreements. Australian-mined bauxite from Weipa feeds smelters at Tomago (NSW) and Boyne Island (Qld), with much of the aluminium exported as ingot.

Worked example

Example 1. A wing spar made of 7075-T6 has a cross-sectional area of 1500 mm21500 \text{ mm}^2. At a 4 g manoeuvre, the spar carries 600 kN.

σ=600×1031500×10−6=400 MPa\sigma = \frac{600 \times 10^3}{1500 \times 10^{-6}} = 400 \text{ MPa}

This is within the yield strength of 503 MPa but exceeds the fatigue strength of 159 MPa at 10710^7 cycles.

Marker's note: a stress below yield does NOT mean the component is safe indefinitely; always check against the fatigue strength too, and note that the structure relies on a finite design life with scheduled inspections rather than infinite life.

Example 2. Compare the mass of a tension member required to carry 200 kN, made first from 2024-T3 (yield 345 MPa) and then from Grade 350 steel (yield 350 MPa), assuming both are sized to the same safety margin against yield.

Required area for 2024-T3: A=F/σ=200×103/(345×106)=5.80×10−4 m2A = F/\sigma = 200 \times 10^3 / (345 \times 10^6) = 5.80 \times 10^{-4} \text{ m}^2. Mass per metre =5.80×10−4×2780=1.61= 5.80 \times 10^{-4} \times 2780 = 1.61 kg/m.

Required area for steel: A=200×103/(350×106)=5.71×10−4 m2A = 200 \times 10^3 / (350 \times 10^6) = 5.71 \times 10^{-4} \text{ m}^2. Mass per metre =5.71×10−4×7850=4.48= 5.71 \times 10^{-4} \times 7850 = 4.48 kg/m.

Marker's note: even though the two yield strengths are almost identical (345 versus 350 MPa), the steel member is nearly three times heavier because of its much higher density; always finish a comparison like this by stating the mass ratio explicitly, since that is the engineering conclusion the question is really asking for.

Common traps
Confusing the alloy designations
2024 is Al-Cu, 7075 is Al-Zn. The four-digit code communicates the alloy family.
Treating temper as cosmetic
T3 versus T6 affects yield strength by 30 percent or more. Heat treatment is not a finish; it is part of the material specification.
Forgetting fatigue
Aluminium has a less defined endurance limit than steel. Fatigue determines aircraft inspection intervals and service life, and a stress below yield strength does not mean a component is safe indefinitely.
Missing the corrosion issue
2024 in particular has poor corrosion resistance because of the copper. Cladding with pure aluminium or anodising is standard. 7075-T7 (over-aged) trades some strength for better stress-corrosion resistance.
Comparing yield strength alone, ignoring density
Grade 350 steel and 2024-T3 have almost identical yield strength (350 MPa versus 345 MPa), but steel is nearly three times heavier per unit volume, so specific strength (not raw strength) is the correct comparison for aircraft structures.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2019 HSC-style5 marks
Justify the use of aluminium alloy 2024-T3 for the fuselage skin of an aircraft and aluminium alloy 7075-T6 for the wing spar. In your answer, refer to the specific properties of each alloy.
Show worked answer →

Both alloys are precipitation-hardened wrought aluminium, but they have different chemistries (2024 is Al-Cu, 7075 is Al-Zn) and different strengths. The selection is matched to the loading and the failure modes of each structure.

2024-T3 for fuselage skin. 2024 is an aluminium-copper alloy. The -T3 temper means solution-treated, cold-worked and naturally aged.

  • Yield strength about 345 MPa, ultimate about 485 MPa.
  • Excellent fatigue resistance (the key property for a fuselage skin, which experiences pressurisation cycles every flight).
  • Good ductility and damage tolerance: cracks propagate slowly, allowing inspection between flights to catch them.
  • Reasonable corrosion resistance (improved by clad sheet with pure aluminium surface layer).

The fuselage skin is loaded by repeated pressurisation, vibration and gust loads in tension and compression at thousands of cycles per service life. Fatigue and damage tolerance dominate the choice.

7075-T6 for wing spar. 7075 is an aluminium-zinc alloy. The -T6 temper means solution-treated and artificially aged.

  • Yield strength about 503 MPa, ultimate about 572 MPa.
  • Highest specific strength of any common aluminium alloy.
  • Lower fatigue and corrosion resistance than 2024.
  • More notch-sensitive (cracks once started propagate faster).

The wing spar is the main structural member that takes the bending load between the wing tips and the fuselage. Static strength and stiffness at minimum mass dominate the choice. The lower fatigue tolerance is acceptable because the spar carries more steady (1g) and gust loading rather than pressurisation cycles.

Markers reward (1) the alloy chemistries and tempers correctly named, (2) the property comparison (strength versus fatigue and corrosion), and (3) the link between the property requirement and the structure's loading.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks
A wing spar section made of 7075-T6 aluminium has a cross-sectional area of 1200 mm21200 \text{ mm}^2 and carries an axial load of 480 kN in a manoeuvre. Calculate the stress in the spar and state whether it exceeds the alloy's yield strength (503 MPa).
Show worked solution →

Step 1: calculate stress.

σ=FA=480×1031200×10−6=400×106 Pa=400 MPa\sigma = \frac{F}{A} = \frac{480 \times 10^3}{1200 \times 10^{-6}} = 400 \times 10^6 \text{ Pa} = 400 \text{ MPa}

Step 2: compare. 400 MPa is below the yield strength of 503 MPa, so the spar does not yield under this load, though it is well above the fatigue strength (159 MPa), meaning fatigue rather than static yield governs the design life.

Marking criteria: 1 mark for correctly applying σ=F/A\sigma = F/A with consistent units, 1 mark for the correct stress value (400 MPa), 1 mark for correctly comparing against yield strength and stating a conclusion.

foundation3 marks
A structural aluminium component is described as: solution treated at 500 degrees C, quenched, then artificially aged at 150 degrees C for several hours, achieving a yield strength of about 500 MPa in the Al-Zn family. Identify the alloy designation and its temper.
Show worked solution →
Alloy family
Al-Zn corresponds to the 7xxx series, so the alloy is 7075.
Temper
Artificial ageing after solution treatment and quenching is the T6 temper.
Conclusion
The component is 7075-T6, consistent with its high yield strength (about 500 MPa) and use in high-strength structural members such as wing spars.

Marking criteria: 1 mark for correctly identifying 7075 from the Al-Zn chemistry, 1 mark for correctly identifying T6 from the artificial-ageing description, 1 mark for linking the answer to the yield strength given.

core4 marks
The table below gives selected properties of four aerospace structural materials. | Material | Density (kg/m^3) | Yield strength (MPa) | | 2024-T3 | 2780 | 345 | | 7075-T6 | 2810 | 503 | | Grade 350 steel | 7850 | 350 | | Ti-6Al-4V | 4430 | 880 | Using the data, calculate the specific strength (yield strength divided by density) of Grade 350 steel and of 7075-T6, and explain what the comparison shows about why aluminium alloys have historically dominated airframe structures over steel.
Show worked solution →

Specific strength of Grade 350 steel.

3507850=0.0446 MPa per kg/m3\frac{350}{7850} = 0.0446 \text{ MPa per kg/m}^3

Specific strength of 7075-T6.

5032810=0.179 MPa per kg/m3\frac{503}{2810} = 0.179 \text{ MPa per kg/m}^3

Explanation. 7075-T6 has roughly four times the specific strength of Grade 350 steel, meaning a structure built from 7075-T6 can carry the same load as a steel structure at roughly a quarter of the mass, or a much stronger structure at the same mass. Since aircraft performance (fuel burn, payload, range) is extremely sensitive to structural mass, this specific-strength advantage, not raw strength, is why aluminium alloys became the dominant airframe material despite steel having comparable or even similar absolute yield strength.

Marking criteria: 1 mark for the correct steel specific strength, 1 mark for the correct 7075-T6 specific strength, 1 mark for correctly stating the ratio/comparison, 1 mark for explaining the link between specific strength and aircraft mass sensitivity (not just restating the numbers).

core5 marks
An in-service inspection finds pitting corrosion beginning on an unclad 2024-T3 fuselage skin panel near a bilge area where moisture has pooled. Explain why 2024 is particularly susceptible to this failure mode, and describe the standard engineering measure used to prevent it.
Show worked solution →

Why 2024 is susceptible. 2024 is an aluminium-copper alloy, and copper-rich intermetallic precipitates at grain boundaries (formed during precipitation hardening) create local galvanic cells with the surrounding aluminium matrix when moisture and electrolytes (such as condensation in a bilge area) are present. The copper-rich regions act as a cathode relative to the aluminium, driving localised galvanic and pitting corrosion, which is worse in 2024 than in alloys with lower copper content.

Standard preventive measure. Aerospace-grade 2024 sheet is supplied as clad sheet: a thin layer (typically a few percent of total thickness) of high-purity aluminium is roller-bonded onto both surfaces of the 2024 core. The pure aluminium cladding has no copper-rich precipitates, forms a stable protective oxide layer, and is anodic to the core, so it corrodes preferentially and protects the structural 2024 beneath it (sacrificial protection), while also acting as a physical barrier to moisture reaching the core alloy.

Marking criteria: 1 mark for identifying copper-rich precipitates as the cause of galvanic/pitting susceptibility, 1 mark for linking this to the presence of moisture/electrolyte, 1 mark for naming clad sheet as the preventive measure, 1 mark for explaining the sacrificial/barrier protection mechanism, 1 mark for a coherent link from cause to preventive measure.

core4 marks
A wing designer is choosing between 2024-T3 and 7075-T6 for a new access panel located on the lower wing skin, a region that experiences both pressurisation-style cyclic loading from aerodynamic flutter and moderate static bending stress. Recommend an alloy and justify your choice using at least two properties.
Show worked solution →

Recommendation: 2024-T3.

Justification.

  1. Fatigue resistance. The panel experiences cyclic aerodynamic loading (flutter-type loading is repetitive, similar in character to pressurisation cycling), and 2024-T3 has substantially better fatigue strength (about 138 MPa at 10710^7 cycles) than 7075-T6 (about 159 MPa nominal fatigue strength but far more notch-sensitive), making 2024 more damage-tolerant under repeated loading over the aircraft's service life.
  2. Damage tolerance/inspectability. 2024-T3 propagates cracks more slowly and predictably than 7075-T6, so any developing fatigue crack in an access panel (a location prone to stress concentration around fasteners and the panel cut-out) is more likely to be caught during scheduled inspection before it becomes critical.

Because the static bending stress here is described as only moderate, 7075-T6's higher static strength is not the limiting requirement, so the fatigue and damage-tolerance advantage of 2024-T3 should dominate the selection.

Marking criteria: 1 mark for a clear recommendation, 1 mark each for two correctly justified properties (fatigue resistance, damage tolerance/notch sensitivity), 1 mark for linking the properties to the specific loading described (cyclic flutter loading) rather than a generic answer.

exam6 marks
Assess whether titanium alloy Ti-6Al-4V should replace aluminium alloys 2024 and 7075 as the primary structural material for a future single-aisle passenger airframe.
Show worked solution →

This is a 6-mark ASSESS: markers reward a supported judgement, not just a list of properties of each material.

Plan. Thesis: despite titanium's superior specific strength and corrosion resistance, its high cost and processing difficulty make wholesale replacement uneconomic for a high-volume single-aisle airliner, so titanium should stay reserved for high-load or high-temperature local structures. Ti-6Al-4V's yield strength (about 880 MPa) and specific strength (about 0.198 versus 0.179 for 7075-T6) plus better corrosion/heat resistance are attractive, but the Kroll process and machining (poor thermal conductivity causing tool wear) make titanium roughly 10 times more expensive per kilogram, a heavy penalty across a programme building thousands of airframes. Aluminium remains cheap, easy to rivet and machine, and well characterised after a century of service, and adequate for single-aisle loads.

Model paragraph (excerpt). Although Ti-6Al-4V offers a genuine specific-strength advantage over 7075-T6 and markedly better corrosion resistance than 2024, converting a single-aisle airframe to titanium would multiply raw material cost roughly tenfold and increase machining time, because titanium's poor thermal conductivity accelerates cutting-tool wear. For a high-volume programme, this cost penalty outweighs the mass saving, particularly since single-aisle loads are well within what 2024 and 7075 can already carry economically. The more defensible response, and the one industry has adopted, is selective titanium use where its properties are decisive, such as engine pylons and landing gear, while aluminium remains the dominant skin, frame and spar material.

Marker's note: top-band answers (1) use a quantitative comparison (specific strength, cost ratio, yield strength), (2) weigh the trade-off rather than only listing properties, (3) acknowledge titanium's real advantages before rejecting wholesale replacement, and (4) close with an explicit, defensible judgement.

exam5 marks
The table below shows illustrative hardness data for a 7075 aluminium sample artificially aged at 150 degrees C, measured at increasing ageing time. | Ageing time (hours) | Hardness (HB) | | 0 | 95 | | 2 | 130 | | 6 | 160 | | 10 | 175 | | 18 | 172 | | 30 | 155 | Using the data, (a) identify the approximate ageing time that produces peak hardness (the T6 condition), and (b) explain, in terms of precipitation hardening, why hardness falls again after this peak (over-ageing).
Show worked solution →

(a) Peak hardness. Hardness rises from 95 HB at 0 hours to a peak of about 175 HB at 10 hours, then falls at 18 and 30 hours. The approximate ageing time for peak hardness (the T6 condition) is 10 hours at 150 degrees C.

(b) Why hardness falls after the peak (over-ageing). During ageing, fine coherent precipitates nucleate within the aluminium grains and impede dislocation motion, which raises hardness and strength as ageing time increases. Beyond the optimum ageing time, these fine precipitates continue to grow and coarsen (a process related to Ostwald ripening), becoming larger but more widely spaced. Widely spaced, coarse precipitates are less effective at blocking dislocations than the same volume of fine, closely spaced precipitates, so the alloy softens again even though ageing has continued. This over-aged condition is deliberately used for the T7 temper, trading some strength for improved resistance to stress corrosion cracking.

Marking criteria: 1 mark for correctly reading the peak hardness time (10 hours) from the table, 1 mark for correctly identifying this as the T6 condition, 1 mark for describing precipitate coarsening as the mechanism of over-ageing, 1 mark for explaining why coarser precipitates are less effective at blocking dislocations, 1 mark for linking over-ageing to the T7 temper's trade-off.

Practise this

Sources & how we know this

Based on the official NESA syllabus. Not yet individually reviewed by a person; always check NESA and your school's materials for current rules.

ExamExplained