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22-Mec-B8 Engineering Materials · May 2017

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

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2017 — 16-Mec-B8 Engineering Materials. Three hours, open book; any non-communicating calculator permitted. Eight problems, all of equal value; any FIVE constitute a complete paper, so each problem is worth 20 marks. Candidates are urged to submit a clear statement of any assumptions made. All eight problems are solved below, because the set as a whole is the study resource.

Reference texts (22-Mec-B8 Engineering Materials).

  • Askeland & Wright, The Science and Engineering of Materials, 7th ed. — the primary syllabus text.
  • Callister & Rethwisch, Materials Science and Engineering: An Introduction, 10th ed.
  • Dieter, Mechanical Metallurgy, 3rd ed. — Considère's construction and plastic instability.
  • Shackelford, Introduction to Materials Science for Engineers, 8th ed. — ceramics, glasses and glass-ceramics.
  • Fontana, Corrosion Engineering, 3rd ed. — galvanic series, sacrificial protection and Faraday's law.
  • Ashby, Materials Selection in Mechanical Design, 5th ed. — selection criteria and material indices.
  • Polmear, Light Alloys, 5th ed. — aluminium tempers, Al–Li alloys and maraging steels.

Note on this sitting. Seven of the eight problems restate archetypes seen in 2013–2016 with fresh data, and every calculation below has been re-worked from this paper's numbers. Question 8 is the genuinely new one: the familiar aluminium–lithium substitution has been expanded into a three-alloy selection problem that also asks for the beam volume and for a strength-to-density ranking, and the two halves of it do not pick the same alloy.

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:

Given data
QuantitySymbolValue
Alloy A (incumbent)—Al − 5 wt% Cu − 2 wt% Mg, 370 MPa
Alloy B (candidate)—Al − 4 wt% Li − 1 wt% Cu, 368 MPa
Alloy C (candidate)—Al − 3 wt% Li − 3 wt% Mg, 340 MPa
Weight of the existing floor beamsWA70 000 N
Weight reduction requestedΔWreq8000 N
Density of aluminiumρAl2700 kg/m3
Density of copperρCu8920 kg/m3
Density of magnesiumρMg1740 kg/m3
Density of lithiumρLi530 kg/m3

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.

Check: the 8000 N reduction is read here as 8000 N to be taken out of the 70 000 N of floor beams, since the beams are the only weight the question states and the only structure being re-made. Gravity is taken as g = 9.81 m/s2 for part (b); as step 5 shows, g cancels out of the weight-saving answers, so nothing in (c) or (d) depends on it.

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: $$\text{A:}\quad w_{Cu} = 0.05,\quad w_{Mg} = 0.02,\quad w_{Al} = 0.93$$ $$\text{B:}\quad w_{Li} = 0.04,\quad w_{Cu} = 0.01,\quad w_{Al} = 0.95$$ $$\text{C:}\quad w_{Li} = 0.03,\quad w_{Mg} = 0.03,\quad w_{Al} = 0.94$$ 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, $\rho = \sum w_i\rho_i$, $$\rho_A = 0.93(2700) + 0.05(8920) + 0.02(1740) = 2511 + 446 + 34.8$$ $$\rho_B = 0.95(2700) + 0.04(530) + 0.01(8920) = 2565 + 21.2 + 89.2$$ $$\rho_C = 0.94(2700) + 0.03(530) + 0.03(1740) = 2538 + 15.9 + 52.2$$ so that $$\boxed{\ \rho_A = 2991.8,\quad \rho_B = 2675.4,\quad \rho_C = 2606.1\ \text{kg/m}^{3}\ }$$ 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, $$V = \frac{m_A}{\rho_A} = \frac{7135.6}{2991.8}$$ $$\boxed{\ V = 2.3850\ \text{m}^{3}\ }$$ 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 $$W_B = 70\,000\times\frac{2675.4}{2991.8} = 62\,597\ \text{N}, \qquad W_C = 70\,000\times\frac{2606.1}{2991.8} = 60\,976\ \text{N}$$ 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, $$\Delta W_B = 70\,000 - 62597 = \boxed{7403\ \text{N}}, \qquad \Delta W_C = 70\,000 - 60976 = \boxed{9024\ \text{N}}$$ that is 10.58 % and 12.89 % of the beam weight respectively. Against the 8000 N the customer asked for, alloy B reaches 92.5 % of the objective and falls 597 N short, while alloy C exceeds it by 1024 N.
Weight removed from the floor beams (N) — savings, not beam weights02 0004 0006 0008 00010 00012 000Alloy B (4Li–1Cu)7 403 NAlloy C (3Li–3Mg)9 024 Nrequested8 000 NB short by 597 NC clears by 1024 NOnly Alloy C meets the 8000 N request; Alloy B reaches 92.5 % of it.
The saving each candidate delivers, set against the saving the customer asked for. Plotting the three savings rather than the three beam weights is what makes the 597 N shortfall and the 1024 N surplus visible at all — on a scale of 70 000 N they would be a pixel wide.

Selection for part (c). Only Alloy C meets the customer requirement as stated. Alloy B is the more attractive material in almost every other respect — it matches the incumbent’s strength to within 2 MPa, so the beams could be re-made to the existing drawing with no re-analysis — but it removes only 7403 N and misses the target. Alloy C clears the target with 1024 N in hand, at the cost of an 8.1 % drop in strength, from 370 to 340 MPa. The honest engineering answer therefore carries a condition: select Alloy C provided the beams are not strength-critical at their present sections. If they are, restoring the strength by scaling the sections in proportion to 370/340 raises the volume to 2.5955 m3 and the weight to 66 356 N, leaving a saving of only 3 644 N — which fails the target by a wider margin than alloy B did. Alloy B, needing a resize of only 370/368, would still deliver 7 063 N. That reversal is the substance of the question: whether C is genuinely the better choice depends entirely on whether the floor beams are sized by strength or by stiffness, deflection and minimum-gauge rules, and a real answer would say so to the customer rather than quote a bare number.

  1. 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 $\sigma/\rho$: $$\frac{\sigma}{\rho}\Big|_A = \frac{370}{2991.8} = 0.12367, \qquad \frac{\sigma}{\rho}\Big|_B = \frac{368}{2675.4} = 0.13755$$ $$\frac{\sigma}{\rho}\Big|_C = \frac{340}{2606.1} = 0.13046$$ in MPa per kg/m3, that is $$\boxed{\ 123.7,\ 137.5\ \text{and}\ 130.5\ \text{kN}\cdot\text{m/kg for A, B and C}\ }$$ so Alloy B is the best material on this criterion, 11.2 % above the incumbent against alloy C’s 5.5 %.
Strength-to-density index σ/ρ (kN·m/kg)115120125130135140Alloy A Al–5Cu–2Mg 370 MPa123.7Alloy B Al–4Li–1Cu 368 MPa137.5Alloy C Al–3Li–3Mg 340 MPa130.5Alloy B wins on specific strength, though only Alloy C meets the weight request.Axis truncated at 115 so the 13.9 kN·m/kg spread between the three alloys is legible.
Strength-to-density ranking of the three alloys. Alloy B wins because it buys nearly all of alloy C’s density reduction while giving up almost none of alloy A’s strength.

The two halves of this question deliberately disagree, and saying so is part of the answer. Part (c) is a fixed-geometry substitution, where only density matters and the lightest alloy wins; part (d) is a re-design question, where the section is free to change and strength and density trade against each other. Alloy C is the lighter material but has given up 8.1 % of the strength to get there, whereas alloy B gives up 0.5 % of the strength for 10.6 % of the density. On a specific-strength basis that makes B the better material, even though it is C that satisfies this particular customer request at the drawing as it stands.

Results for the three-alloy comparison
QuantityAlloy AAlloy BAlloy C
Composition (balance Al)5Cu–2Mg4Li–1Cu3Li–3Mg
Strength σ (MPa)370368340
(a) Density ρ (kg/m3)2991.82675.42606.1
Weight of the beams (N)70 00062 59760 976
(c) Weight saving (N)—7 4039 024
Saving as % of the 8000 N request—92.5 %112.8 %
(d) Specific strength σ/ρ (kN·m/kg)123.7137.5130.5
Verdict—Best on σ/ρ; misses the request by 597 NSelected for (c): clears the request by 1024 N
(b) Volume of the floor beams2.3850 m3 (unchanged by the substitution)

Check: the exam directs that the alloy density be taken as a simple weighted average of its constituents, ρ = Σwiρi, and that prescription is followed above. The rigorous volumetric mixture rule, 1/ρ = Σwi/ρi, gives ρA = 2765.9, ρB = 2334.0 and ρC = 2369.7 kg/m3 and hence savings of 10 930 N and 10 027 N. The choice of rule changes the answer to part (c): under the prescribed weighted average alloy B falls 597 N short of the target, whereas under the volumetric rule it clears the target comfortably and both candidates would qualify. The exam prescribes the weighted average, so that is the answer given above, but the sensitivity should be stated rather than hidden. The part (d) ranking, by contrast, is robust: alloy B leads on specific strength under either convention. Two engineering caveats belong in any real report as well: the calculation assumes the beams are re-made to identical geometry, and it assumes the whole 8000 N is to be taken out of the floor beams alone.

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