22-Mec-A4 Design and Manufacture of Machine Elements · December 2018

Question 6 of 6: Thermal Stress in a Magnesium Tube and Aluminium Rod Across a Gap (25 marks)

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

Notes on this paper

Paper format. National Examinations, December 2018 — 16-Mec-A4 Design and Manufacture of Machine Elements. Three hours, open book, any non-communicating calculator. Six questions in two parts: Part A (Q1–Q3, manufacturing processes) and Part B (Q4–Q6, machine-element analysis). The rubric asks for two questions from each part — four questions constitute a complete paper, each worth 25 %. All six are solved here, since the set is intended as a study resource.

Reference texts.

Kalpakjian, S. & Schmid, S., Manufacturing Engineering and Technology, 8th ed. — Ch. 11–12 (casting processes and design), Ch. 16 (sheet-metal forming), Ch. 30 (fusion welding).
Groover, M. P., Fundamentals of Modern Manufacturing, 7th ed. — Ch. 10–11 (casting), Sec. 20.5 (stretch forming).
Budynas, R. & Nisbett, K., Shigley's Mechanical Engineering Design, 11th ed. — Ch. 9 (welding, weld-as-a-line, Tables 9-2 and 9-3).
Hibbeler, R. C., Engineering Mechanics: Statics, 14th ed. — Ch. 6 (frames and machines), Ch. 8 (dry friction).
Hibbeler, R. C., Mechanics of Materials, 10th ed. — Sec. 4.6 (thermal stress in statically indeterminate axial members).

Check: every boxed result in this paper, which also carries the independent closure checks noted in each question (moment closure on Q4, volume conservation on Q2, gap compatibility on Q6).

Question 6: Thermal Stress in a Magnesium Tube and Aluminium Rod Across a Gap (25 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.

Quantity	Symbol	Value
Magnesium tube AB: length, outside and inside diameter	$L_{mg}$, $d_o$, $d_i$	300 mm, 25 mm, 20 mm
Aluminium rod CD: length, diameter	$L_{al}$, $d$	450 mm, 25 mm
Initial gap between cap E and rod end C	$\delta_g$	0.25 mm at 35 °C
Temperature rise	$\Delta T$	80 − 35 = 45 °C
Coefficients of thermal expansion	$\alpha_{mg}$, $\alpha_{al}$	$26\times10^{-6}$, $24\times10^{-6}$ /°C
Elastic moduli (AM1004-T61, 6061-T6)	$E_{mg}$, $E_{al}$	44.7 GPa, 68.9 GPa

Find. The normal stress in the magnesium tube and in the aluminium rod after the temperature rise.

Both members are built into rigid walls and grow toward one another. Once they meet, the rigid cap forces them to share a single contact force, so their flexibilities add in series.

Approach. Test first whether the free thermal growths close the gap; if they do, the two members act in series through the rigid cap under one common force, found by requiring that the combined thermal elongation minus the combined elastic shortening equals the gap.

Compute the cross-sectional areas. $$A_{mg} = \frac{\pi}{4}\left(25^{2}-20^{2}\right) = 176.7\ \text{mm}^{2} \qquad A_{al} = \frac{\pi}{4}(25)^{2} = 490.9\ \text{mm}^{2}$$
Test whether contact actually occurs. Released from the constraint, each member grows by $\delta = \alpha\,\Delta T\,L$: $$\delta_{mg} = (26\times10^{-6})(45)(300) = 0.351\ \text{mm} \qquad \delta_{al} = (24\times10^{-6})(45)(450) = 0.486\ \text{mm}$$ Their sum is 0.837 mm, which comfortably exceeds the 0.25 mm gap, so the cap does contact the rod and forces develop. Had the sum been smaller, both stresses would be zero and the problem would end here.
Recognise the series arrangement. The cap is rigid and the walls at A and D are rigid, so the only load path is tube → cap → rod. A single compressive force $F$ therefore acts in both members — not one force per member — and the gap is a single one-off deduction from the combined growth, not a term applied to each member.
Write the compatibility equation. The total free growth, less the elastic shortening of each member under $F$, must equal the gap that has been closed: $$\delta_{mg}+\delta_{al} - F\!\left(\frac{L_{mg}}{A_{mg}E_{mg}} + \frac{L_{al}}{A_{al}E_{al}}\right) = \delta_g$$
Evaluate the series flexibility. With $E_{mg} = 44.7$ GPa and $E_{al} = 68.9$ GPa, $$\frac{L_{mg}}{A_{mg}E_{mg}} = \frac{300}{(176.7)(44{,}700)} = 3.798\times10^{-5}\ \text{mm/N}$$ $$\frac{L_{al}}{A_{al}E_{al}} = \frac{450}{(490.9)(68{,}900)} = 1.331\times10^{-5}\ \text{mm/N}$$ so the sum is $f = 5.128\times10^{-5}$ mm/N. The thin magnesium tube supplies nearly three-quarters of the total flexibility, and will therefore take most of the strain.
Solve for the common force. $$F = \frac{(\delta_{mg}+\delta_{al}) - \delta_g}{f} = \frac{0.837-0.25}{5.128\times10^{-5}}$$ $$\boxed{F = 11.45\ \text{kN (compression in both members)}}$$
Convert to normal stresses. $$\sigma_{mg} = \frac{F}{A_{mg}} = \frac{11{,}446}{176.7} \qquad \sigma_{al} = \frac{F}{A_{al}} = \frac{11{,}446}{490.9}$$ $$\boxed{\sigma_{mg} = 64.8\ \text{MPa (C)} \qquad \sigma_{al} = 23.3\ \text{MPa (C)}}$$
Close the compatibility check. The elastic shortenings are $F(3.798\times10^{-5}) = 0.4347$ mm in the tube and $F(1.331\times10^{-5}) = 0.1523$ mm in the rod. Subtracting both from the free growths: $$0.837 - 0.4347 - 0.1523 = 0.250\ \text{mm} \quad\checkmark$$ which is exactly the gap, confirming the solution. Note that the tube's net movement, $0.351-0.435 = -0.084$ mm, is negative: the compressive strain outruns its own thermal growth, so end B finishes slightly to the left of where it started. That is a correct result, not a sign error — the rod does all of the closing and more.

Both stresses are safely below yield — AM1004-T61 magnesium yields near 152 MPa and 6061-T6 aluminium near 255 MPa — so the assembly stays elastic and the linear superposition used above is valid.

Result	Value
Free thermal growth, tube / rod	0.351 mm / 0.486 mm (sum 0.837 mm > 0.25 mm gap)
Cross-sectional area, tube / rod	176.7 mm² / 490.9 mm²
Common contact force	11.45 kN (compressive)
Normal stress in the magnesium tube	64.8 MPa, compressive
Normal stress in the aluminium rod	23.3 MPa, compressive

Topic: Statically indeterminate axial members with a thermal load and an initial clearance.

Key relations: $\delta_T = \alpha\,\Delta T\,L$; $\delta_F = FL/AE$; contact test $\sum\delta_T > \delta_g$; series compatibility $\sum\delta_T - F\sum\!\left(L/AE\right) = \delta_g$; $\sigma = F/A$.

Why this works: With a rigid cap between them, the tube and rod form a single load path, so force is common and deformations add — the mechanical analogue of two springs in series. The gap simply relaxes the compatibility condition by a fixed amount: instead of the total deformation being zero, it must equal $\delta_g$. Recasting the problem as “springs in series with a preset slack” makes the algebra one equation in one unknown regardless of how many members are chained.

Common pitfall: Writing a separate compatibility equation for each member and subtracting the gap in both, which double-counts the clearance and returns a force roughly 20 % low here. The second trap is skipping the contact test entirely — on a colder problem or a wider gap the correct answer is zero stress, and a solver who charges straight into the compatibility equation will produce a spurious tensile force.

Source: Hibbeler, Mechanics of Materials, 10th ed., Sec. 4.4 and 4.6 (statically indeterminate axially loaded members; thermal stress).

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