Question 5 of 8: Aluminium–lithium substitution for aircraft floor beams

Question 5: 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:

Find. The weight saving delivered by substituting the Al–Li alloy for the incumbent alloy at unchanged beam geometry, and hence whether the customer’s 750 kg requirement can be met.

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.

1. Write out the weight fractions of both alloys. The alloying additions are quoted in weight per cent and aluminium makes up the balance: $$w_{Cu} = 0.050,\quad w_{Mg} = 0.015,\quad w_{Al} = 1 - 0.050 - 0.015 = 0.935 \quad\text{(incumbent)}$$ $$w_{Li} = 0.040,\quad w_{Cu} = 0.010,\quad w_{Al} = 1 - 0.040 - 0.010 = 0.950 \quad\text{(proposed)}$$ Each set sums to unity, which is the check to make before going any further.
2. Compute the density of the incumbent alloy. Taking the weighted average of density that the question prescribes, $$\rho_1 = \sum w_i\rho_i = 0.935(2.70) + 0.050(8.92) + 0.015(1.74)$$ $$\rho_1 = 2.5245 + 0.4460 + 0.0261 = 2.9966\ \text{g/cm}^3$$ The 5 % copper alone raises the density about 8 % above pure aluminium, because copper is more than three times as dense.
3. Compute the density of the Al–Li alloy. By the same rule, $$\rho_2 = 0.950(2.70) + 0.040(0.53) + 0.010(8.92) = 2.5650 + 0.0212 + 0.0892$$ $$\boxed{\ \rho_2 = 2.6754\ \text{g/cm}^3\ }$$ Lithium is the lightest metallic element, so a 4 wt % addition buys a 10.7 % density reduction even after allowing for the copper that goes with it.
4. 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: $$V = \frac{W_1}{\rho_1} = \frac{7000\ \text{kg}}{2996.6\ \text{kg/m}^3} = 2.336\ \text{m}^3$$ 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.
5. Find the mass of the substituted beams. Filling that same volume with the lighter alloy, $$W_2 = \rho_2 V = W_1\frac{\rho_2}{\rho_1} = 7000\times\frac{2.6754}{2.9966} = 7000(0.89281)$$ $$\boxed{\ W_2 = 6249.6\ \text{kg}\ }$$
6. Evaluate the saving and answer the question asked. Subtracting, $$\Delta W = W_1 - W_2 = 7000 - 6249.6$$ $$\boxed{\ \Delta W = 750.4\ \text{kg}\ (10.7\ \%)\ }$$ The requested reduction was 750 kg, so the answer to “is this possible?” is yes — but only just. The substitution clears the requirement by 0.4 kg, a margin of 0.05 %. Any engineer reporting this should say so explicitly: a proposal that meets its target to within one part in two thousand has no reserve against alloy tolerance, machining allowance or a late design change, and the honest recommendation is that the substitution be combined with a second weight-saving measure rather than relied on alone.