Aluminium–lithium substitution for aircraft floor beams: 22-Mec-B8 Engineering Materials, May 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 − 5.5 wt% Cu − 1.6 wt% Mg − 1.2 wt% Mn
Proposed alloy	—	Al − 5 wt% Li − 1 wt% Cu
Mass of the existing floor beams	W₁	10 500 kg
Weight 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 weight saving delivered by substituting the Al–Li alloy for the incumbent alloy at unchanged beam geometry, expressed as a percentage, and hence whether the engineer is right that it delivers “almost all” of the customer’s 1500 kg requirement.

The saving the substitution delivers set against the saving the customer asked for. At full scale the two are almost the same length, which is the answer in one glance; the magnified panel underneath is what makes the residual 35 kg shortfall — 2.3 % 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: $$\text{incumbent:}\quad w_{Cu} = 0.055,\quad w_{Mg} = 0.016,\quad w_{Mn} = 0.012$$ $$w_{Al} = 1 - 0.055 - 0.016 - 0.012 = 0.917$$ $$\text{proposed:}\quad w_{Li} = 0.050,\quad w_{Cu} = 0.010,\quad w_{Al} = 0.940$$ 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, $$\rho_1 = \sum w_i\rho_i = 0.917(2.70) + 0.055(8.92) + 0.016(1.74) + 0.012(7.47)$$ $$\rho_1 = 2.4759 + 0.4906 + 0.0278 + 0.0896 = 3.0840\ \text{g/cm}^3$$ The alloying additions raise the density about 14 % above pure aluminium. The copper accounts for most of that on its own — it is more than three times as dense as the base metal — with the manganese contributing about a fifth as much again and the magnesium, lighter than aluminium, pulling very slightly the other way.
Compute the density of the Al–Li alloy. By the same rule, $$\rho_2 = 0.940(2.70) + 0.050(0.53) + 0.010(8.92)$$ $$\rho_2 = 2.5380 + 0.0265 + 0.0892$$ $$\boxed{\ \rho_2 = 2.6537\ \text{g/cm}^3\ }$$ Lithium is the lightest metallic element, so a 5 wt % addition buys a 14.0 % density reduction relative to the incumbent alloy even after allowing for the 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: $$V = \frac{W_1}{\rho_1} = \frac{10\,500\ \text{kg}}{3084.0\ \text{kg/m}^3} = 3.405\ \text{m}^3$$ 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, $$W_2 = \rho_2 V = W_1\frac{\rho_2}{\rho_1} = 10\,500\times\frac{2.6537}{3.0840} = 10\,500(0.86048)$$ $$\boxed{\ W_2 = 9035.0\ \text{kg}\ }$$
Evaluate the saving and answer the question asked. Subtracting, $$\Delta W = W_1 - W_2 = 10\,500 - 9035.0$$ $$\boxed{\ \Delta W = 1465.0\ \text{kg}\ = 13.95\ \%\ \text{of the beam mass}\ }$$ That percentage is what the question asks for, and the second percentage follows from it directly.
Compare the saving with what the customer asked for. Expressing the delivered saving as a fraction of the requested 1500 kg, $$\frac{\Delta W}{\Delta W_{req}} = \frac{1465.0}{1500} = 0.9766$$ $$\boxed{\ 97.7\ \%\ \text{of the objective, short by }35.0\ \text{kg}\ }$$ So the engineer’s claim is upheld as stated: almost all of the weight-saving objective can indeed be met by the substitution, but not the whole of it. The honest report says the floor-beam change delivers 1465 kg and that the remaining 35 kg — a fifth of one per cent of the beam mass — has to be found somewhere else in the airframe, or negotiated away. A candidate who answers a bare “yes” has not read the word almost in the question; a candidate who answers “no” because 1465 < 1500 has missed it just as badly in the other direction.

Results for the Al–Li substitution
Quantity	Symbol	Value
Density of Al–5.5Cu–1.6Mg–1.2Mn	ρ₁	3.0840 g/cm³
Density of Al–5Li–1Cu	ρ₂	2.6537 g/cm³
Volume of the floor beams (unchanged)	V	3.405 m³
Mass of the substituted beams	W₂	9035.0 kg
Weight saving delivered	ΔW	1465.0 kg
Weight saving as a percentage of beam mass	ΔW/W₁	13.95 %
Weight saving requested	ΔW_req	1500 kg
Fraction of the objective achieved	ΔW/ΔW_req	97.7 % (short by 35.0 kg)
Verdict	—	“Almost all” is correct — 97.7 % of the target, not 100 %

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 = 2.8043 and ρ₂ = 2.2542 g/cm³ and hence a saving of 2060 kg, or 19.6 %. Here the choice of rule changes the verdict: the prescribed weighted average falls 35 kg short of the 1500 kg target while the volumetric rule clears it by 560 kg. 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. Two further 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 it assumes the entire 1500 kg of airframe reduction is to come from 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 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 here, and the exam explicitly prescribes the former.

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)