Aluminium–lithium substitution for aircraft floor beams: 22-Mec-B8 Engineering Materials, May 2018, Q4

Question 4: Aluminium–lithium substitution for 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 in a second alloy to the same geometry:

Given data
Quantity	Symbol	Value
Incumbent alloy	—	Al − 5.5 wt% Cu − 1.8 wt% Mg − 1.2 wt% Mn
Proposed alloy	—	Al − 5 wt% Li − 0.5 wt% Cu
Mass of the existing floor beams	W₁	10 500 kg
Mass reduction requested	ΔW_req	1500 kg
Density of aluminium	ρ_Al	2.70 g/cm³
Density of copper	ρ_Cu	8.92 g/cm³
Density of magnesium	ρ_Mg	1.74 g/cm³
Density of manganese	ρ_Mn	7.47 g/cm³
Density of lithium	ρ_Li	0.53 g/cm³

Find. The mass saving delivered by substituting the Al–Li alloy for the incumbent alloy at unchanged beam geometry, and hence whether the proposal that the 1500 kg objective can be met entirely by this change is true or false.

The saving the substitution delivers set against the saving the customer asked for. At full scale the two are nearly the same length, which is the answer at a glance; the magnified number line underneath is what makes the 65 kg surplus — 4.4 % of the target — legible at all.

Approach. Compute each alloy’s density as the weight-fraction-weighted average the question prescribes, note that re-making the same beams in a different alloy preserves the volume rather than the mass, and scale the mass by the density ratio.

Write out the weight fractions of both alloys. The alloying additions are quoted in weight per cent and aluminium makes up the balance: $$\text{incumbent:}\quad w_{Cu} = 0.055,\quad w_{Mg} = 0.018,\quad w_{Mn} = 0.012$$ $$w_{Al} = 1 - 0.055 - 0.018 - 0.012 = 0.915$$ $$\text{proposed:}\quad w_{Li} = 0.050,\quad w_{Cu} = 0.005,\quad w_{Al} = 0.945$$ Each set sums to unity, which is the check to make before going any further.
Compute the density of the incumbent alloy. Taking the weighted average of density that the question prescribes, $$\rho_1 = \sum w_i\rho_i = 0.915(2.70) + 0.055(8.92) + 0.018(1.74) + 0.012(7.47)$$ $$\rho_1 = 2.4705 + 0.4906 + 0.0313 + 0.0896 = 3.0821\ \text{g/cm}^3$$ The alloying additions raise the density about 14 % above pure aluminium. Copper accounts for most of that on its own — it is more than three times as dense as the base metal — with manganese contributing about a fifth as much again, while magnesium, being lighter than aluminium, pulls very slightly the other way.
Compute the density of the Al–Li alloy. By the same rule, $$\rho_2 = 0.945(2.70) + 0.050(0.53) + 0.005(8.92)$$ $$\rho_2 = 2.5515 + 0.0265 + 0.0446$$ $$\boxed{\ \rho_2 = 2.6226\ \text{g/cm}^3\ }$$ Lithium is the lightest metallic element, so a 5 wt % addition buys a 14.9 % density reduction relative to the incumbent alloy even after allowing for the half per cent of copper that goes with it.
Recognise that the substitution preserves volume, not mass. The beams are re-made to the same drawing, so their geometry — and therefore their volume — is unchanged: $$V = \frac{W_1}{\rho_1} = \frac{10\,500\ \text{kg}}{3082.1\ \text{kg/m}^3} = 3.4068\ \text{m}^3$$ 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.
Find the mass of the substituted beams. Filling that same volume with the lighter alloy, $$W_2 = \rho_2 V = W_1\frac{\rho_2}{\rho_1} = 10\,500\times\frac{2.6226}{3.0821} = 10\,500(0.85092)$$ $$\boxed{\ W_2 = 8934.7\ \text{kg}\ }$$
Evaluate the mass saving, which is what the question asks for. Subtracting, $$\Delta W = W_1 - W_2 = 10\,500 - 8934.7$$ $$\boxed{\ \Delta W = 1565.3\ \text{kg} = 14.91\ \%\ \text{of the beam mass}\ }$$
Test the saving against the customer’s requirement and answer true or false. Expressing the delivered saving as a fraction of the requested 1500 kg, $$\frac{\Delta W}{\Delta W_{req}} = \frac{1565.3}{1500} = 1.0435$$ $$\boxed{\ 104.4\ \%\ \text{of the objective, a surplus of }65.3\ \text{kg}\ }$$ The proposal is therefore true: replacing the floor-beam alloy alone removes 1565 kg, which covers the whole of the 1500 kg the customer asked for and leaves 65 kg of margin. That margin is only 0.6 % of the beam mass, so the honest engineering report says the objective is met with very little to spare rather than met comfortably — a weight statement that later grows by even one per cent on the beams would erase it.

Results for the Al–Li substitution
Quantity	Symbol	Value
Density of Al–5.5Cu–1.8Mg–1.2Mn	ρ₁	3.0821 g/cm³
Density of Al–5Li–0.5Cu	ρ₂	2.6226 g/cm³
Volume of the floor beams (unchanged)	V	3.4068 m³
Mass of the substituted beams	W₂	8934.7 kg
Mass saving delivered	ΔW	1565.3 kg
Saving as a percentage of beam mass	ΔW/W₁	14.91 %
Mass reduction requested	ΔW_req	1500 kg
Fraction of the objective achieved	ΔW/ΔW_req	104.4 % (surplus 65.3 kg)
Verdict	—	TRUE — the objective is met entirely, with 65 kg to spare

Check: the exam directs that weighted averages of density be used, i.e. ρ = Σw_iρ_i, and that prescription is followed above. The rigorous volumetric mixture rule, 1/ρ = Σw_i/ρ_i, gives ρ₁ = 2.8011 and ρ₂ = 2.2477 g/cm³, hence a saving of 2074.4 kg or 19.76 %. On this paper the verdict is robust to the choice of rule — both conventions clear the 1500 kg target — but that is a fact about these particular compositions and has to be recomputed for every such question rather than assumed. Two engineering caveats belong in any real report: the calculation assumes the beams are re-made to identical geometry, whereas Al–Li also has a 5–10 % higher specific modulus and could be re-sized for further saving; and if the beams turn out to be strength-critical rather than stiffness-critical, the substitution has to be re-checked against the Al–Li alloy’s own allowables before the 65 kg margin can be believed.

Topic. Density of a substitutional alloy from its composition, and the constant-volume basis for a material substitution.

Key relations. Weighted-average (prescribed) rule ρ = Σw_iρ_i; rigorous volumetric rule 1/ρ = Σw_i/ρ_i; constant-geometry substitution W₂/W₁ = ρ₂/ρ₁. For stiffness-driven aerospace structure the governing merit index is the specific modulus E/ρ, and for a panel in bending E^1/3/ρ.

Why this works. Alloying elements dissolve substitutionally in the aluminium lattice, so to a first approximation each contributes its own mass and its own volume in proportion to how much of it is present. That is why a density estimate can be made from composition alone, without any knowledge of the phase diagram. Lithium is the pivotal addition: at 0.53 g/cm³ it is the lightest metal, and every 1 wt % dissolved in aluminium lowers density by roughly 3 % while raising elastic modulus by about 6 %, so it improves the specific modulus twice over. This is exactly why Al–Li alloys such as 2090, 2195 and 2099 displaced conventional 2000-series alloys in launch-vehicle tankage and airframe structure.

Common pitfall. Working the substitution at constant mass, which produces the absurd conclusion that no weight is saved at all — the beams keep their dimensions, not their weight. A second trap is silently switching between the weighted-average and the reciprocal mixture rules; they give noticeably different densities, and the exam explicitly prescribes the former. A third is answering a bare “true” without quoting the 1565 kg the question actually asks to be calculated.

Source. Askeland & Wright, Ch. 13 (non-ferrous alloys: aluminium and its alloys); Callister & Rethwisch, Ch. 11; Ashby, Materials Selection in Mechanical Design, Ch. 5 for the specific-modulus index.

Question 4 of 8: Aluminium–lithium substitution for aircraft floor beams

Question 4: Aluminium–lithium substitution for aircraft floor beams (20 marks)