Question 3 of 8: Aluminium–lithium floor beams and the promised weight saving

Question 3: Aluminium–lithium floor beams and the promised weight saving (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 second aluminium alloy:

Find. The weight saving delivered by substituting the Al–Li alloy at unchanged beam geometry, and hence whether the engineer’s claim that it supplies more than half of the customer’s 15 % objective can be sustained.

Approach. Compute each alloy’s density as the weight-fraction-weighted average that the question prescribes, recognise that re-making the same beams in a different alloy preserves the volume and not the mass, scale the mass by the density ratio, and compare the saving against both the engineer’s 50 % threshold and the customer’s full objective.

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 five per cent of copper alone raises the density about eight per cent 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: $$V = \frac{W_1}{\rho_1} = \frac{8000\ \text{kg}}{2996.6\ \text{kg/m}^3} = 2.670\ \text{m}^3$$ This is the pivot of the whole question. A candidate who scales masses directly by the weight fractions, rather than through the common volume, obtains a meaningless answer.
5. Find the mass of the substituted beams. Filling that same volume with the lighter alloy, $$W_2 = \rho_2 V = W_1\frac{\rho_2}{\rho_1} = 8000\times\frac{2.6754}{2.9966} = 8000(0.89278)$$ $$\boxed{\ W_2 = 7142\ \text{kg}\ }$$
6. Evaluate the saving. Subtracting, $$\Delta W = W_1 - W_2 = 8000 - 7142 $$ $$\boxed{\ \Delta W = 858\ \text{kg}\ (10.7\ \%\ \text{of the beam mass})\ }$$
7. Answer the question that was actually asked. The customer’s objective is a 15 % reduction, $$\Delta W_{req} = 0.15(8000) = 1200\ \text{kg},\qquad \text{engineer's threshold} = 0.5(1200) = 600\ \text{kg}$$ and the substitution delivers $$\frac{\Delta W}{\Delta W_{req}} = \frac{858}{1200} = 0.715$$ $$\boxed{\ \text{Yes: the swap supplies } 71.5\ \%\ \text{of the objective, comfortably more than half}\ }$$ So the engineer is right, and by a clear margin rather than marginally. What the answer must also say is that the substitution does not meet the customer’s requirement on its own: it leaves 342 kg, about 29 % of the objective, to be found elsewhere in the airframe. Reporting the 71.5 % without that caveat would misrepresent the proposal.