Aluminium–lithium floor beams and the promised weight saving: 22-Mec-B8 Engineering Materials, December 2014, Q7

Question 7: Aluminium–lithium floor beams and the promised weight saving (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. Two candidate floor-beam alloys of comparable mechanical properties, so that the beams may be assumed geometrically identical and the substitution is a pure density exchange.

Given data
Quantity	Value
Incumbent alloy	5 wt% Cu, 1.8 wt% Mg, 1 wt% Mn, balance Al (92.2 wt%)
Proposed alloy	6 wt% Li, 2 wt% Mg, 0.5 wt% Cu, balance Al (91.5 wt%)
Weight of the existing floor beams	10 000 kg
Weight reduction requested	1 500 kg
Densities (g/cm³)	Al 2.70, Cu 8.92, Mg 1.74, Mn 7.47, Li 0.53

Find. The weight saving $W_{s}$ delivered by the substitution, and hence whether Susan's claim that the 1500 kg target can be exceeded is correct.

Approach. Compute each alloy's density as the weight-fraction-weighted average of the element densities, as the question prescribes; note that comparable mechanical properties mean the beams keep the same geometry and therefore the same volume; then scale the mass in proportion to density and compare the saving with the target.

Complete each composition. The aluminium content is whatever is left after the named additions: $$w_{Al,1} = 1 - 0.050 - 0.018 - 0.010 = 0.922$$ $$w_{Al,2} = 1 - 0.060 - 0.020 - 0.005 = 0.915$$ Omitting this step is the commonest way to lose the question, because the balance is more than nine-tenths of the alloy and carries almost all the density.
Apply the prescribed weighted average to the incumbent alloy. Using $\rho = \sum w_{i}\rho_{i}$ as the question instructs, $$\rho_{1} = 0.050(8.92) + 0.018(1.74) + 0.010(7.47) + 0.922(2.70)$$ Evaluating term by term gives $0.4460 + 0.03132 + 0.0747 + 2.4894$, so $$\boxed{\rho_{1} = 3.0414\ \text{g/cm}^{3}}$$ Copper alone adds nearly half a gram per cubic centimetre, which is why the 2xxx alloys are the densest of the common aluminium families.
Do the same for the aluminium–lithium alloy. $$\rho_{2} = 0.060(0.53) + 0.020(1.74) + 0.005(8.92) + 0.915(2.70)$$ which is $0.0318 + 0.0348 + 0.0446 + 2.4705$, giving $$\boxed{\rho_{2} = 2.5817\ \text{g/cm}^{3}}$$ Lithium is the lightest metal in the periodic table at 0.53 g/cm³, so replacing copper with lithium works twice over: it removes a very heavy element and adds a very light one.
Convert the existing weight into a volume. Because the two alloys have comparable mechanical properties, the beams are not resized, so the volume is the invariant: $$V = \frac{W_{1}}{\rho_{1}} = \frac{10\,000}{3.0414} = 3\,287.9\ \text{kg per (g/cm}^{3})$$ The units are unconventional but consistent, since the same conversion factor cancels when the mass is rebuilt in the next step.
Rebuild the mass in the new alloy and take the difference. $$W_{2} = V\rho_{2} = W_{1}\frac{\rho_{2}}{\rho_{1}} = 10\,000\times\frac{2.5817}{3.0414} = 8\,488.5\ \text{kg}$$ so the saving is $$\boxed{W_{s} = 10\,000 - 8\,488.5 = 1\,511.5\ \text{kg}}$$ which is 15.1 per cent of the original beam weight.
Compare with the target and answer the question asked. The requested reduction is 1500 kg and the substitution delivers 1511.5 kg, so $$W_{s} - 1500 = 11.5\ \text{kg} \; (0.8\%\ \text{of the target})$$ Susan is right, but only just. The target is met and exceeded, by about three-quarters of one per cent.

The saving against the customer's request. The two totals are almost indistinguishable on any absolute scale, which is exactly the point: the margin is the third bar, and at 11.5 kg it is smaller than the mass of a single seat.

That margin deserves a comment rather than a celebration. Eleven and a half kilogrammes on a ten-tonne assembly is well inside the scatter of a real weight statement: mill tolerance on the plate thickness, the actual chemistry within the specification band, fastener and sealant weight, and any local reinforcement needed because Al–Li alloys are notoriously anisotropic in the short-transverse direction would all swamp it. The professionally correct answer is therefore that the substitution meets the target on the stated basis, and that the design should not be committed on that basis alone: either accept the result as a demonstration of feasibility and go looking for a second source of saving, or resize the beams to take advantage of the higher specific modulus of the Al–Li alloy, which typically raises Young's modulus by six per cent per weight per cent of lithium and would give a genuine, defensible margin.

Check: the question directs that weighted averages of density be used, so $\rho = \sum w_{i}\rho_{i}$ is the prescribed method and the boxed answers follow it. The volumetrically rigorous rule for a mixture specified by weight fraction is $1/\rho = \sum w_{i}/\rho_{i}$, which gives $\rho_{1} = 2.7873$ and $\rho_{2} = 2.1545\ \text{g/cm}^{3}$ and a saving of 2270 kg. The two conventions differ substantially in magnitude here, because lithium's density is so far from aluminium's, but both clear the 1500 kg target, so the verdict on Susan's claim is robust to the choice. A second-order effect ignored by both is that lithium in solution genuinely lowers the density of the aluminium lattice by more than a simple mixture rule predicts, which is why real 8090 and 2099 alloys measure near 2.54 g/cm³.

Results
Quantity	Symbol	Value
Density of the Al–5Cu–1.8Mg–1Mn alloy	ρ₁	3.0414 g/cm³
Density of the Al–6Li–2Mg–0.5Cu alloy	ρ₂	2.5817 g/cm³
Weight of the beams in the new alloy	W₂	8 488.5 kg
Weight saving	W_s	1 511.5 kg
Saving as a fraction of the beam weight	—	15.1%
Margin over the 1500 kg request	—	+11.5 kg (0.8%)
Verdict on Susan's claim	—	Correct, but with a very thin margin

Topic. Density of an alloy from its composition, and substitution for weight saving at constant geometry.

Key relations. Weighted-average (linear) mixture rule $\rho = \sum w_{i}\rho_{i}$ for weight fractions, as prescribed here; the volumetrically exact rule $1/\rho = \sum w_{i}/\rho_{i}$; from volume fractions the exact rule is instead $\rho = \sum V_{i}\rho_{i}$. At constant geometry, $W_{2} = W_{1}\rho_{2}/\rho_{1}$ and the fractional saving is simply $1 - \rho_{2}/\rho_{1}$, independent of the absolute weight.

Why this works. Two questions have to be separated. The first is what the alloy weighs per unit volume, which is a composition calculation; the second is what the structure weighs, which depends on whether the substitution is at constant geometry or at constant load capacity. The phrase “comparable mechanical properties” in the question is what licenses the constant-geometry assumption, and it is the whole reason the answer reduces to a ratio of densities. Note also that the fractional saving does not depend on the 10 000 kg at all — it is fixed at 15.1 per cent by the two compositions — so the target is met or missed purely on the arithmetic of the density ratio against 15 per cent. Aluminium–lithium alloys exist because each weight per cent of lithium removes roughly three per cent of the density while adding about six per cent to the modulus, an almost unique combination among the light alloys; the practical limit is that above about 2.5 wt% Li ductility and short-transverse toughness fall away sharply, so the 6 wt% of this question is a textbook figure rather than a producible commercial alloy.

Common pitfall. Forgetting the aluminium balance and normalising only the named additions, which returns a density near 5 g/cm³ and an absurd answer. The second pitfall is mixing the two mixture rules within one comparison — using the weighted average for one alloy and the harmonic form for the other — which can reverse the verdict; whichever convention is chosen must be applied to both alloys. The third is to report the saving without checking it against the target, when the question asks a yes-or-no question and the margin turns out to be under one per cent.

Source. Askeland & Wright, The Science and Engineering of Materials, 7th ed., Ch. 14 §14.1–14.3 (aluminium alloys, density and specific strength); Polmear, Light Alloys, 5th ed., Ch. 3 §3.5 (aluminium–lithium alloys); Ashby, Ch. 4 (material property charts).

Question 7 of 8: Aluminium–lithium floor beams and the promised weight saving

Question 7: Aluminium–lithium floor beams and the promised weight saving (20 marks)