Aluminium–lithium substitution for aircraft floor beams: 22-Mec-B8 Engineering Materials, December 2016, Q6

Question 6: 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 − 4.5 wt% Cu − 1.5 wt% Mg
Proposed alloy	—	Al − 3 wt% Li − 1 wt% Cu
Weight of the existing floor beams	W₁	95 000 N
Weight reduction requested	ΔW_req	10 % of W₁ = 9500 N
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 lithium	ρ_Li	0.53 g/cm³

Find. The weight saving delivered by substituting the Al–Li alloy for the incumbent alloy at unchanged beam geometry, and hence whether the engineer’s suggestion can deliver the 10 per cent reduction the customer has asked for.

Check: the only weight the question states is the 95 000 N of floor beams, so the 10 per cent reduction is read here as 10 per cent of that figure, i.e. 9500 N to be taken out of the beams. If instead the 10 per cent referred to the whole aircraft’s gross weight, no substitution confined to the floor beams could approach it, and the answer would be a trivial “no”.

The saving the substitution delivers, set against the saving the customer asked for. Plotting the two savings rather than the two beam weights is what makes the 902 N shortfall visible at all — on a scale of 95 000 N it would be one pixel.

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 weight 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: $$\text{incumbent:}\quad w_{Cu} = 0.045,\quad w_{Mg} = 0.015,\quad w_{Al} = 0.940$$ $$\text{proposed:}\quad w_{Li} = 0.030,\quad w_{Cu} = 0.010,\quad w_{Al} = 0.960$$ 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, $$\rho_1 = \sum w_i\rho_i = 0.940(2.70) + 0.045(8.92) + 0.015(1.74)$$ $$\rho_1 = 2.5380 + 0.4014 + 0.0261 = 2.9655\ \text{g/cm}^3$$ The alloying additions raise the density about 9.8 % above pure aluminium, essentially all of it from the copper, which is more than three times as dense as the base metal. The magnesium, lighter than aluminium, pulls very slightly the other way.
Compute the density of the Al–Li alloy. By the same rule, $$\rho_2 = 0.960(2.70) + 0.030(0.53) + 0.010(8.92)$$ $$\rho_2 = 2.5920 + 0.0159 + 0.0892$$ $$\boxed{\ \rho_2 = 2.6971\ \text{g/cm}^3\ }$$ Lithium is the lightest metallic element, so a 3 wt % addition buys a 9.1 % density reduction relative to the incumbent alloy even after allowing for the copper that goes with it.
Recognise that the substitution preserves volume, not weight. The beams are re-made to the same drawing, so their geometry — and therefore their volume — is unchanged. Taking $g = 9.81\ \text{m/s}^2$, the incumbent beams have a mass of $95\,000/9.81 = 9684$ kg, and $$V = \frac{9684\ \text{kg}}{2965.5\ \text{kg/m}^3} = 3.266\ \text{m}^3$$ This is the pivot of the whole question. A candidate who scales weights directly by the weight fractions, rather than through the common volume, gets a meaningless answer.
Find the weight of the substituted beams. Filling that same volume with the lighter alloy, $$W_2 = \rho_2 V g = W_1\frac{\rho_2}{\rho_1} = 95\,000\times\frac{2.6971}{2.9655} = 95\,000(0.90949)$$ $$\boxed{\ W_2 = 86\,401.8\ \text{N}\ }$$ Note that the gravitational constant cancels, so the result does not depend on the value taken for $g$; the intermediate mass and volume are quoted only to make the physical picture concrete.
Evaluate the saving. Subtracting, $$\Delta W = W_1 - W_2 = 95\,000 - 86\,401.8$$ $$\boxed{\ \Delta W = 8598.2\ \text{N}\ = 9.05\ \%\ \text{of the beam weight}\ }$$ That saving is what the question asks to be determined first, and the verdict follows directly from it.
Compare the saving with what the customer asked for. The requested reduction is 10 per cent of 95 000 N, that is 9500 N. Expressing the delivered saving as a fraction of it, $$\frac{\Delta W}{\Delta W_{req}} = \frac{8598.2}{9500} = 0.9051$$ $$\boxed{\ 90.5\ \%\ \text{of the objective, short by }901.8\ \text{N}\ }$$ So the honest answer to “is this possible?” is no, not by this substitution alone. Changing the floor-beam alloy removes 8598 N, which is nine tenths of the target and by far the largest single contribution available, but it leaves 902 N — just under one per cent of the beam weight — still to be found. A candidate who answers a bare “yes” has not compared 9.05 % with the 10 % asked for; a candidate who answers a bare “no” and stops has not told the customer that the proposal gets 90 per cent of the way there.

Results for the Al–Li substitution
Quantity	Symbol	Value
Density of Al–4.5Cu–1.5Mg	ρ₁	2.9655 g/cm³
Density of Al–3Li–1Cu	ρ₂	2.6971 g/cm³
Volume of the floor beams (unchanged)	V	3.266 m³
Weight of the substituted beams	W₂	86 401.8 N
Weight saving delivered	ΔW	8598.2 N
Weight saving as a percentage of beam weight	ΔW/W₁	9.05 %
Weight saving requested	ΔW_req	9500 N (10 %)
Fraction of the objective achieved	ΔW/ΔW_req	90.5 % (short by 901.8 N)
Verdict	—	Not by itself — 9.05 % against the 10 % requested

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 = 2.7639 and ρ₂ = 2.4197 g/cm³ and hence a saving of 11 831 N, or 12.45 %. The choice of rule changes the verdict here: the prescribed weighted average falls 902 N short of the 10 per cent target while the volumetric rule clears it comfortably. The exam prescribes the weighted average, so that is the answer given above, but the sensitivity should be stated rather than hidden — it is exactly the kind of convention that a real weight statement has to fix in writing before anyone quotes a number to a customer. Two engineering caveats belong in any real report as well: the calculation assumes the beams are re-made to identical geometry, whereas Al–Li also has a higher specific modulus and could be re-sized for further saving; and it assumes the whole reduction is to be taken out of the floor beams alone.

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 the 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 weight, 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 here, and the exam explicitly prescribes the former. A third is to forget that the weights are given in newtons and to mix them with densities in g/cm³ without going through mass — although, as step 5 shows, the ratio form of the answer sidesteps the conversion entirely.

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 6 of 8: Aluminium–lithium substitution for aircraft floor beams

Question 6: Aluminium–lithium substitution for aircraft floor beams (20 marks)