Question 8 of 8: Three aluminium alloys for transport-aircraft floor beams

Question 8: Three aluminium alloys for transport-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 either of two candidate alloys to the same geometry:

Find. (a) the density of each of the three alloys on the weighted-average rule the question prescribes, (b) the volume occupied by the floor beams, (c) the weight saving each candidate delivers at unchanged geometry and which one meets the 8000 N request, and (d) which of the three ranks highest on a strength-to-density selection index.

Approach. Compute each alloy’s density as the weight-fraction-weighted average the question prescribes; get the beam volume from the incumbent weight and density; note that re-making the same beams in a different alloy preserves the volume rather than the mass, so weight scales with the density ratio; then rank all three on the specific-strength index, which is a different question from the weight-saving one and does not have the same answer.

1. Part (a) — write out the weight fractions. The alloying additions are quoted in weight per cent and aluminium makes up the balance: $$\text{A:}\quad w_{Cu} = 0.055,\quad w_{Mg} = 0.015,\quad w_{Al} = 0.93$$ $$\text{B:}\quad w_{Li} = 0.04,\quad w_{Cu} = 0.01,\quad w_{Al} = 0.95$$ $$\text{C:}\quad w_{Li} = 0.03,\quad w_{Mg} = 0.03,\quad w_{Al} = 0.94$$ Each set sums to unity, which is the check to make before going any further.
2. Compute the three densities. Taking the weighted average of density that the question prescribes, $\rho = \sum w_i\rho_i$, $$\rho_A = 0.93(2700) + 0.055(8920) + 0.015(1740) = 2511 + 490.6 + 26.1$$ $$\rho_B = 0.95(2700) + 0.04(530) + 0.01(8920) = 2565 + 21.2 + 89.2$$ $$\rho_C = 0.94(2700) + 0.03(530) + 0.03(1740) = 2538 + 15.9 + 52.2$$ so that $$\boxed{\ \rho_A = 3027.7,\quad \rho_B = 2675.4,\quad \rho_C = 2606.1\ \text{kg/m}^{3}\ }$$ Alloy A is the heaviest because copper, at more than three times the density of aluminium, dominates its additions. Both candidates are lighter, and lithium is why: at 530 kg/m³ it is the lightest metallic element, so every weight per cent of it displaces aluminium with something five times lighter.
3. Part (b) — find the volume of the floor beams. The beams weigh 70 000 N in alloy A, so their mass is $$m_A = \frac{W_A}{g} = \frac{70\,000}{9.81} = 7135.6\ \text{kg}$$ and dividing by the density of alloy A, $$V = \frac{m_A}{\rho_A} = \frac{7135.6}{3027.7}$$ $$\boxed{\ V = 2.3568\ \text{m}^{3}\ }$$ Equivalently $V = W_A/(\rho_A g)$ in one step. That volume is the quantity carried forward: the beams are re-made to the same drawing, so their geometry does not change.
4. Part (c) — weigh the beams in each candidate alloy. Filling the same volume with a lighter alloy, $$W_i = \rho_i V g = W_A\frac{\rho_i}{\rho_A}$$ so $$W_B = 70\,000\times\frac{2675.4}{3027.7} = 61\,855\ \text{N}, \qquad W_C = 70\,000\times\frac{2606.1}{3027.7} = 60\,253\ \text{N}$$ Note that g cancels in the ratio, so neither answer depends on the value taken for it. 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. Evaluate the two savings against the request. Subtracting, $$\Delta W_B = 70\,000 - 61855 = \boxed{8145\ \text{N}}, \qquad \Delta W_C = 70\,000 - 60253 = \boxed{9747\ \text{N}}$$ that is 11.64 % and 13.92 % of the beam weight respectively. Against the 8000 N the customer asked for, both candidates qualify: alloy B clears the target by 145 N, or 1.8 % of the request, and alloy C clears it by 1747 N, or 21.8 %.

Selection for part (c). Both candidates meet the customer requirement as stated, so the choice is made on what else changes. Alloy B is the one to select. It matches the incumbent’s strength to within 2 MPa — 368 against 370, a difference of 0.5 % and well inside the scatter of any real allowable — so the beams can be re-made to the existing drawing with no re-analysis, no re-substantiation of the joints and no change to the loads the surrounding structure sees, and it still removes 8145 N. Alloy C removes 1602 N more, but it buys that with an 8.1 % drop in strength, from 370 to 340 MPa, and that has to be paid for somewhere. The size of the payment settles the matter. If any part of the beam is strength-critical at its present section, restoring the strength means scaling the sections by 370/340, which raises the volume to 2.5647 m³ and the weight to 65 569 N and leaves a saving of only 4431 N — a clear failure of the target. The same correction applied to alloy B, needing only 370/368, still returns 7809 N, within 2.4 % of the request. Alloy B is therefore the robust choice and alloy C the fragile one, and the honest answer records the one reservation attaching to B: its margin over the request is 145 N, which is smaller than the scatter of a typical weight statement, so the saving should be confirmed by a weight audit of the finished beams rather than taken on the density arithmetic alone.

6. Part (d) — rank the three on strength to density. The selection index for a tension member of prescribed length and load, minimising mass, is the specific strength $\sigma/\rho$: $$\frac{\sigma}{\rho}\Big|_A = \frac{370}{3027.7} = 0.12220, \qquad \frac{\sigma}{\rho}\Big|_B = \frac{368}{2675.4} = 0.13755$$ $$\frac{\sigma}{\rho}\Big|_C = \frac{340}{2606.1} = 0.13046$$ in MPa per kg/m³, that is $$\boxed{\ 122.2,\ 137.5\ \text{and}\ 130.5\ \text{kN}\cdot\text{m/kg for A, B and C}\ }$$ so Alloy B is the best material on this criterion, 12.6 % above the incumbent against alloy C’s 6.8 %.

It is worth being explicit about why parts (c) and (d) agree here, because they need not. Part (c) is a fixed-geometry substitution, decided by density alone, and on that criterion alloy C is the lighter material and removes the most weight. Part (d) is a re-design question, decided by the index, and there strength and density trade against each other. The two answers coincide only because alloy C’s extra 2.6 % of density reduction over alloy B is bought with an 8.1 % loss of strength, whereas alloy B gives up 0.5 % of the strength for 11.6 % of the density — so C wins the raw weight contest but loses the efficiency one, and since B already satisfies the customer the raw contest no longer decides anything. Had the request been 9000 N instead of 8000 N, only alloy C would have qualified and the two parts would have returned different alloys. Recognising which regime the question sits in is the substance of the marks.