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22-Mec-B8 Engineering Materials · Undated paper

Question 7 of 8: Three aluminium alloys for transport-aircraft floor beams

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams — 16-Mec-B8 Engineering Materials, undated sitting (the printed footer reads 16-Mec-B8/May 2019). Three hours; any non-communicating calculator permitted. Eight problems, all of equal value; any FIVE constitute a complete paper, so each problem is worth 20 marks. Candidates are urged to submit a clear statement of any assumptions made. All eight problems are solved below.

Reference texts (22-Mec-B8 Engineering Materials).

  • Askeland & Wright, The Science and Engineering of Materials, 7th ed. — the primary syllabus text.
  • Callister & Rethwisch, Materials Science and Engineering: An Introduction, 10th ed.
  • Dieter, Mechanical Metallurgy, 3rd ed. — Considère's construction and plastic instability.
  • Shackelford, Introduction to Materials Science for Engineers, 8th ed.
  • Fontana, Corrosion Engineering, 3rd ed. — Faraday's law, corrosion rates and cathodic protection.
  • Ashby, Materials Selection in Mechanical Design, 5th ed. — selection criteria and material indices.
  • Polmear, Light Alloys, 5th ed. — aluminium tempers and Al–Li alloys.

Note on this sitting. Five of the eight problems restate standing 22-Mec-B8 archetypes with fresh or unchanged data — the magnesium sacrificial anode, the FRP consolidation essay (problem 5, now asking for three routes and for their applicability to primary versus secondary structure rather than four routes and their trade-offs alone), the ABS-versus-phenolic selection (problem 6), the aluminium–lithium floor-beam substitution (problem 7, restated in kilograms rather than newtons) and Considère necking (problem 8). Problems 1, 2 and 3 are new to the subject: they are Faraday's-law corrosion rate, Fick's first law applied to carburizing, and an atom-counting exercise on a silicon wafer. Every calculation has been re-worked from this paper's own numbers.

Question 7: Three aluminium alloys for transport-aircraft floor beams (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. One set of floor beams, to be re-made to the same drawing in either of two candidate alloys:

Given data
QuantitySymbolValue
Alloy A (incumbent)—Al − 5 wt% Cu − 2 wt% Mg, 370 MPa
Alloy B (candidate)—Al − 4 wt% Li − 1 wt% Cu, 368 MPa
Alloy C (candidate)—Al − 3 wt% Li − 3 wt% Mg, 340 MPa
Mass of the existing floor beamsmA7000 kg
Mass reduction requestedΔmreq800 kg
Density of aluminiumρAl2700 kg/m3
Density of copperρCu8920 kg/m3
Density of magnesiumρMg1740 kg/m3
Density of lithiumρLi530 kg/m3

Find. (a) the density of each of the three alloys on the weighted-average rule the question prescribes, (b) the volume occupied by the floor beams, (c) the mass saving each candidate delivers at unchanged geometry and which one meets the 800 kg request, and (d) which of the three ranks highest on a strength-to-density selection index.

Check: the 800 kg reduction is read here as 800 kg to be taken out of the 7000 kg of floor beams, since the beams are the only mass the question states and the only structure being re-made. The question says the aircraft's total mass is to fall by 800 kg, and the engineer's proposal is that re-making the beams alone will achieve it; that is exactly the claim parts (b) and (c) test.

Approach. Compute each alloy's density as the weight-fraction-weighted average the question prescribes; get the beam volume from the incumbent mass and density; note that re-making the same beams in a different alloy preserves the volume rather than the mass, so mass scales with the density ratio; then rank all three on the specific-strength index, which is a different question from the mass-saving one and does not have the same answer.

  1. Part (a) — write out the weight fractions. The alloying additions are quoted in weight per cent and aluminium makes up the balance: $$\text{A:}\quad w_{Cu} = 0.05,\quad w_{Mg} = 0.02,\quad w_{Al} = 0.93$$ $$\text{B:}\quad w_{Li} = 0.04,\quad w_{Cu} = 0.01,\quad w_{Al} = 0.95$$ $$\text{C:}\quad w_{Li} = 0.03,\quad w_{Mg} = 0.03,\quad w_{Al} = 0.94$$ Each set sums to unity, which is the check to make before going any further.
  2. Compute the three densities. Taking the weighted average of density that the question prescribes, $\rho = \sum w_i\rho_i$, $$\rho_A = 0.93(2700) + 0.05(8920) + 0.02(1740) = 2511 + 446 + 34.8$$ $$\rho_B = 0.95(2700) + 0.04(530) + 0.01(8920) = 2565 + 21.2 + 89.2$$ $$\rho_C = 0.94(2700) + 0.03(530) + 0.03(1740) = 2538 + 15.9 + 52.2$$ so that $$\boxed{\ \rho_A = 2991.8,\quad \rho_B = 2675.4,\quad \rho_C = 2606.1\ \text{kg/m}^{3}\ }$$ Alloy A is the heaviest because copper, at more than three times the density of aluminium, dominates its additions. Both candidates are lighter, and lithium is why: at 530 kg/m3 it is the lightest metallic element, so every weight per cent of it displaces aluminium with something five times lighter.
  3. Part (b) — find the volume of the floor beams. The beams have a mass of 7000 kg in alloy A, so $$V = \frac{m_A}{\rho_A} = \frac{7000}{2991.8}$$ $$\boxed{\ V = 2.340\ \text{m}^{3}\ }$$ That volume is the quantity carried forward: the beams are re-made to the same drawing, so their geometry does not change.
  4. Part (c) — find the mass of the beams in each candidate alloy. Filling the same volume with a lighter alloy, $$m_i = \rho_i V = m_A\frac{\rho_i}{\rho_A}$$ so $$m_B = 7000\times\frac{2675.4}{2991.8} = 6259.7\ \text{kg}, \qquad m_C = 7000\times\frac{2606.1}{2991.8} = 6097.6\ \text{kg}$$ This is the pivot of the whole question: a candidate who scales masses directly by the weight fractions, rather than through the common volume, gets a meaningless answer.
  5. Evaluate the two savings against the request. Subtracting, $$\Delta m_B = 7000 - 6259.7 = \boxed{740.3\ \text{kg}}, \qquad \Delta m_C = 7000 - 6097.6 = \boxed{902.4\ \text{kg}}$$ that is 10.58 per cent and 12.89 per cent of the beam mass respectively. Against the 800 kg the customer asked for, alloy B reaches 92.5 per cent of the objective and falls 59.7 kg short, while alloy C exceeds it by 102.4 kg.
Mass removed from the floor beams (kg) — savings, not beam masses02004006008001 0001 200Alloy B (4Li–1Cu)740 kgAlloy C (3Li–3Mg)902 kgrequested800 kgB short by 59.7 kgC clears by 102 kgOnly Alloy C meets the 800 kg request; Alloy B reaches 92.5 % of it.
The saving each candidate delivers, set against the saving the customer asked for. Plotting the three savings rather than the three beam masses is what makes the 59.7 kg shortfall and the 102 kg surplus visible at all — on a scale of 7000 kg they would be a pixel wide.

Selection for part (c). Only Alloy C meets the customer requirement as stated. Alloy B is the more attractive material in almost every other respect — it matches the incumbent's strength to within 2 MPa, so the beams could be re-made to the existing drawing with no structural re-analysis at all — but it removes only 740.3 kg and misses the target. Alloy C clears the target with 102 kg in hand, at the cost of an 8.1 per cent drop in strength, from 370 to 340 MPa. The honest engineering answer therefore carries a condition: select Alloy C provided the beams are not strength-critical at their present sections. If they are, restoring the strength by scaling the sections in proportion to 370/340 raises the volume to 2.546 m3 and the mass to 6635.6 kg, leaving a saving of only 364 kg — which fails the target by a wider margin than alloy B did. Alloy B, needing a resize of only 370/368, would still deliver 706 kg. That reversal is the substance of the question: whether C is genuinely the better choice depends entirely on whether the floor beams are sized by strength or by stiffness, deflection and minimum-gauge rules, and a real answer would say so to the customer rather than quote a bare number.

  1. Part (d) — rank the three on strength to density. The selection index for a tension member of prescribed length and load, minimising mass, is the specific strength $\sigma/\rho$: $$\frac{\sigma}{\rho}\Big|_A = \frac{370}{2991.8} = 0.12367, \qquad \frac{\sigma}{\rho}\Big|_B = \frac{368}{2675.4} = 0.13755$$ $$\frac{\sigma}{\rho}\Big|_C = \frac{340}{2606.1} = 0.13046$$ in MPa per kg/m3, that is $$\boxed{\ 123.7,\ 137.5\ \text{and}\ 130.5\ \text{kN}\cdot\text{m/kg for A, B and C}\ }$$ so Alloy B is the best material on this criterion, 11.2 per cent above the incumbent against alloy C's 5.5 per cent.
Strength-to-density index σ/ρ (kN·m/kg)115120125130135140Alloy A Al–5Cu–2Mg 370 MPa123.7Alloy B Al–4Li–1Cu 368 MPa137.5Alloy C Al–3Li–3Mg 340 MPa130.5Alloy B wins on specific strength, though only Alloy C meets the mass request.Axis truncated at 115 so the 13.8 kN·m/kg spread between the three alloys is legible.
Strength-to-density ranking of the three alloys. Alloy B wins because it buys nearly all of alloy C’s density reduction while giving up almost none of alloy A’s strength — the opposite verdict to part (c), and deliberately so.

The two halves of this question deliberately disagree, and saying so is part of the answer. Part (c) is a fixed-geometry substitution, where only density matters and the lightest alloy wins; part (d) is a re-design question, where the section is free to change and strength and density trade against each other. Alloy C is the lighter material but has given up 8.1 per cent of the strength to get there, whereas alloy B gives up 0.5 per cent of the strength for 10.6 per cent of the density. On a specific-strength basis that makes B the better material, even though it is C that satisfies this particular customer request at the drawing as it stands.

Results for the three-alloy comparison
QuantityAlloy AAlloy BAlloy C
Composition (balance Al)5Cu–2Mg4Li–1Cu3Li–3Mg
Strength σ (MPa)370368340
(a) Density ρ (kg/m3)2991.82675.42606.1
Mass of the beams (kg)70006259.76097.6
(c) Mass saving (kg)—740.3902.4
Saving as % of the 800 kg request—92.5 %112.8 %
(d) Specific strength σ/ρ (kN·m/kg)123.7137.5130.5
Verdict—Best on σ/ρ; misses the request by 59.7 kgSelected for (c): clears the request by 102 kg
(b) Volume of the floor beams2.340 m3 (unchanged by the substitution)

Check: the exam directs that the alloy density be taken as a simple weighted average of its constituents, ρ = Σwiρi, and that prescription is followed above. The rigorous volumetric mixture rule, 1/ρ = Σwi/ρi, gives ρA = 2765.9, ρB = 2334.0 and ρC = 2369.7 kg/m3 and hence savings of 1093 kg and 1003 kg. The choice of rule changes the answer to part (c): under the prescribed weighted average alloy B falls 59.7 kg short of the target, whereas under the volumetric rule it clears the target comfortably and both candidates would qualify. The exam prescribes the weighted average, so that is the answer given above, but the sensitivity should be stated rather than hidden — the shortfall is under one per cent of the beam mass and is well inside the scatter of a real weight statement. The part (d) ranking, by contrast, is robust: alloy B leads on specific strength under either convention. Two engineering caveats belong in any real report as well: the calculation assumes the beams are re-made to identical geometry, and it assumes the whole 800 kg is to be taken out of the floor beams alone.