Weight saving from an aluminium–lithium floor beam: 22-Mec-B8 Engineering Materials, May 2013, Q8

Question 8: Weight saving from an aluminium–lithium floor beam (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. Floor beams weighing 9800 kg in an Al–5Cu–1.5Mg alloy, to be replaced by an Al–4Li–1Cu alloy in the same beams; the customer asks for 2000 kg off the aircraft and the engineer claims about 50 % of that from this substitution. Constituent densities:

Given data
Element	Al	Cu	Mg	Li
Density (g cm⁻³)	2.70	8.92	1.74	0.53
Existing alloy (wt %)	93.5	5.0	1.5	—
Proposed alloy (wt %)	95.0	1.0	—	4.0

Find. The weight saved by re-making the same floor beams in the Al–Li alloy, and hence whether the engineer’s claim of about half the 2000 kg objective is achievable.

Approach. The beams are the same beams — same geometry, so the same volume of metal. Compute the weighted-average density of each alloy from its weight fractions as the question directs, get the beam volume from the known 9800 kg and the existing density, and re-weigh that volume in the new alloy.

Complete each alloy composition. The balance in each case is aluminium:$$w_{\mathrm{Al}}^{(1)} = 100 - 5.0 - 1.5 = 93.5\ \text{wt\%}, \qquad w_{\mathrm{Al}}^{(2)} = 100 - 4.0 - 1.0 = 95.0\ \text{wt\%}$$
Take the weighted-average density of the existing alloy. Following the instruction in the question,$$\rho_1 = \sum_i w_i \rho_i = 0.935(2.70) + 0.050(8.92) + 0.015(1.74)$$$$\rho_1 = 2.5245 + 0.4460 + 0.0261 = 2.997\ \text{g}\,\text{cm}^{-3}$$The 5 % of copper is what lifts this above the 2.70 of pure aluminium — copper is more than three times as dense as the base metal.
Take the weighted-average density of the proposed alloy. Likewise,$$\rho_2 = 0.950(2.70) + 0.040(0.53) + 0.010(8.92)$$$$\rho_2 = 2.5650 + 0.0212 + 0.0892 = 2.675\ \text{g}\,\text{cm}^{-3}$$Two changes act in the same direction here: lithium is by far the lightest metallic element and pulls the average down, and the copper content has been cut from 5 % to 1 %, removing the heaviest constituent.
Find the volume of metal in the beams. The existing beams weigh 9800 kg, so$$V = \frac{W_1}{\rho_1} = \frac{9800\ \text{kg}}{2.997\ \text{g}\,\text{cm}^{-3}} = 3.271\times 10^{6}\ \text{cm}^{3} = 3.27\ \text{m}^{3}$$Because the substitution is a like-for-like change of material in an existing design, this volume is unchanged.
Re-weigh the same volume in the new alloy.$$W_2 = V\rho_2 = W_1\frac{\rho_2}{\rho_1} = 9800\left(\frac{2.675}{2.997}\right) = 9800(0.8928) = 8749.6\ \text{kg}$$and therefore the saving is$$\boxed{\ \Delta W = 9800 - 8749.6 = 1050\ \text{kg}\ }$$
Test the claim. The engineer promised about 50 % of the 2000 kg objective, which is 1000 kg. The substitution delivers 1050 kg, so$$\frac{\Delta W}{2000} = \frac{1050}{2000} = 52.5\ \% > 50\ \%$$$$\boxed{\ \textbf{Yes} \text{ --- the suggestion is achievable, with a small margin}\ }$$

The saving actually delivered by the substitution, against the engineer’s claim and the customer’s full request. Plotting the two absolute beam weights (9800 against 8750 kg) would make the whole decision invisible; the decision is about the difference.

It is worth being clear about what this result does and does not say. The substitution meets the engineer’s specific claim, but it delivers only about half of what the customer asked for; the remaining 950 kg must come from elsewhere in the aircraft. The margin over the claim is also thin — 50 kg on 1000, about 5 % — so it would not survive much erosion. In practice a designer would not simply substitute material and keep every dimension: the Al–Li alloy is also about 6 % stiffer per unit lithium added, so a re-sized beam could bank a further saving on a stiffness-limited design. Against that, Al–Li costs three to five times as much per kilogram, is anisotropic in the short-transverse direction, and demands tighter control of machining and chemical processing.

Check: the question instructs “assume weighted averages of density”, and the arithmetic mean weighted by weight fraction has been used accordingly. Strictly, density mixes by volume rather than by mass, so the rigorous form is 1/ρ = Σw_i/ρ_i, which gives ρ₁ = 2.774 and ρ₂ = 2.334 g cm⁻³ and a saving of 1554 kg. Both routes exceed the 1000 kg claim, so the verdict — yes, it is possible — is unchanged; the figure quoted above follows the method the question prescribes. A further real-world caveat: the calculation assumes an unchanged beam geometry and takes no credit for, and no penalty from, the differing strength of the two alloys.

Weight-saving assessment
Quantity	Symbol	Value
Density, Al–5Cu–1.5Mg	ρ₁	2.997 g cm⁻³
Density, Al–4Li–1Cu	ρ₂	2.675 g cm⁻³
Volume of the floor beams	V	3.27 m³
Beam weight in the new alloy	W₂	8750 kg
Weight saved	ΔW	1050 kg
Fraction of the 2000 kg objective met	—	52.5 % (claim was 50 %)
Verdict	—	Yes — achievable

Topic. Alloy density from composition, and why lithium is the only practical alloying element that lightens aluminium.

Key relations. Weighted average by weight fraction, ρ = Σw_iρ_i (the form the question prescribes); rigorous volumetric mixing, 1/ρ = Σw_i/ρ_i. For a substitution at fixed geometry, W₂/W₁ = ρ₂/ρ₁.

Why this works. Lithium is the lightest metal in the periodic table, with a density of 0.53 g cm⁻³ — about one fifth that of aluminium — and it is one of very few elements with appreciable solid solubility in aluminium. Each 1 wt % of lithium lowers the alloy density by roughly 3 % and, unusually, raises the elastic modulus by roughly 6 %, because the Al₃Li precipitate that forms on ageing is itself stiff. Both effects push the specific stiffness E/ρ the right way at once, which is why Al–Li has become the aluminium answer to carbon-fibre competition in airframes. The trade-off is cost, anisotropy, and a reactive melt that must be handled under cover.

Common pitfall. Assuming the beams keep the same weight rather than the same volume, which makes the saving come out as zero. Almost as common is forgetting that the balance of each alloy is aluminium and normalising the given solute percentages to 100 % on their own. Finally, note that the copper reduction from 5 % to 1 % contributes a substantial part of the saving here, not the lithium alone.

Source. Askeland & Wright, Ch. 13 (non-ferrous alloys: aluminium alloys, age hardening); Callister & Rethwisch, Ch. 11.

Question 8 of 8: Weight saving from an aluminium–lithium floor beam

Question 8: Weight saving from an aluminium–lithium floor beam (20 marks)