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

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 transport-aircraft floor beams, to be re-made to the same drawing in a lighter alloy:

Find. The weight saving delivered by substituting the Al–Li alloy for the incumbent at unchanged beam geometry, and hence whether the engineer’s specific claim — that the substitution realizes more than 90 per cent of the 10 000 N the customer asked for — is true.

Approach. Compute each alloy’s density as the weight-fraction-weighted average the question prescribes, recognise that re-making the same beams in a different alloy preserves the volume rather than the mass, scale the weight by the density ratio, and test the resulting saving against the 90 per cent claim.

1. Write out the weight fractions of both alloys. The alloying additions are quoted in weight per cent and aluminium makes up the balance: $$\text{incumbent:}\quad w_{Cu} = 0.04,\quad w_{Mg} = 0.01,\quad w_{Al} = 0.95$$ $$\text{proposed:}\quad w_{Li} = 0.03,\quad w_{Cu} = 0.01,\quad w_{Al} = 0.96$$ 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, $$\rho_1 = \sum w_i\rho_i = 0.95(2.70) + 0.04(8.92) + 0.01(1.74)$$ $$\rho_1 = 2.5650 + 0.3568 + 0.0174 = 2.9392\ \text{g/cm}^3$$ The additions raise the density about 8.9 % above pure aluminium, almost all of it from the copper, which is more than three times as dense as the base metal; the magnesium, being lighter than aluminium, pulls very slightly the other way.
3. Compute the density of the Al–Li alloy. By the same rule, $$\rho_2 = 0.96(2.70) + 0.03(0.53) + 0.01(8.92)$$ $$\rho_2 = 2.5920 + 0.0159 + 0.0892$$ $$\boxed{\ \rho_2 = 2.6971\ \text{g/cm}^3\ }$$ Lithium is the lightest metallic element, so a 3 wt % addition buys an 8.2 % density reduction relative to the incumbent even after allowing for the one per cent of copper that goes with it.
4. 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 $85\,000/9.81 = 8664.6$ kg, and $$V = \frac{8664.6\ \text{kg}}{2939.2\ \text{kg/m}^3} = 2.948\ \text{m}^3$$ 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.
5. Find the weight of the substituted beams. Filling that same volume with the lighter alloy, $$W_2 = \rho_2 V g = W_1\frac{\rho_2}{\rho_1} = 85\,000\times\frac{2.6971}{2.9392} = 85\,000(0.91763)$$ $$\boxed{\ W_2 = 77\,998.6\ \text{N}\ }$$ Note that $g$ cancels, so the result does not depend on the value taken for it; the intermediate mass and volume are quoted only to make the physical picture concrete.
6. Evaluate the saving. Subtracting, $$\Delta W = W_1 - W_2 = 85\,000 - 77\,998.6$$ $$\boxed{\ \Delta W = 7001.4\ \text{N} = 8.24\ \%\ \text{of the beam weight}\ }$$ That saving is the quantity the question asks to be estimated, and the verdict follows directly from it.
7. Test the engineer’s claim. The requested reduction is 10 000 N. Expressing the delivered saving as a fraction of it, $$\frac{\Delta W}{\Delta W_{req}} = \frac{7001.4}{10\,000} = 0.7001$$ $$\boxed{\ 70.0\ \%\ \text{of the objective, short by }2998.6\ \text{N}\ }$$ So the answer to “is this possible?” is no. The substitution is worth having — 7001 N is a real and substantial saving, seven tenths of what was asked and by some way the largest single reduction available from a change of beam material alone — but it delivers 70 per cent of the objective, not the “more than 90 per cent” the engineer claimed. The claim overstates the benefit by about 30 percentage points, and the remaining 2999 N would have to come from elsewhere in the aircraft.

It is worth seeing why the claim fails by so much, because the arithmetic is unforgiving in a way that is easy to miss. The density ratio is fixed by the two compositions and cannot be argued with: it buys 8.24 per cent of the beam weight. The customer, however, asked for 10 000 N out of beams that weigh 85 000 N — that is 11.76 per cent. A substitution worth 8.24 per cent simply cannot reach an 11.76 per cent target, and no amount of care over the density arithmetic will change that. The engineer’s error is one of scale rather than of method, and the quickest way to expose it in an exam is to convert both figures to percentages of the same base before computing anything at all.