22-Mec-A4 Design and Manufacture of Machine Elements · May 2016

Question 4 of 6: Thermal Stress in a Magnesium Tube and Aluminium Rod Separated by a Gap

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

Notes on this paper

Paper format. National Examinations, May 2016 — 07-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-process theory) and Part B (Q4–Q6, machine-element analysis). Candidates answer two from Part A and two from Part B; four questions constitute a complete paper and all questions carry equal value (25 %). All six are solved here.

Reference texts. Kalpakjian & Schmid, Manufacturing Engineering and Technology (Part A); Groover, Fundamentals of Modern Manufacturing (Part A); Budynas & Nisbett, Shigley's Mechanical Engineering Design, 11th ed. (Part B); Hibbeler, Mechanics of Materials, 10th ed. (Q4, Q5); Norton, Machine Design: An Integrated Approach, 6th ed. (Q6).

Check: two corrections carried through Part B. (1) Q5 prints the shaft modulus as “E = 30 ksi”; a 30 ksi modulus is physically impossible for steel and would make the shaft a rubber band, so it is read as the standard E = 30 × 10⁶ psi (30 Mpsi). (2) Q6 supplies the linkage dimensions but not the cross-section of the levers, so the stress in members 2 and 3 is worked for an explicitly stated assumed section and the required section is also reported. Both readings are flagged where they are used, in the spirit of Note 1 on the cover page.

Question 4: Thermal Stress in a Magnesium Tube and Aluminium Rod Separated by 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	$L_{mg}$	300 mm
Magnesium tube AB — outside / inside diameter (Section a–a)	$d_o / d_i$	25 mm / 20 mm
Aluminium rod CD — length	$L_{al}$	450 mm
Aluminium rod CD — diameter	$d$	25 mm
Initial gap between cap E and end C	$\delta_g$	0.2 mm
Temperature rise	$\Delta T$	$80 - 30 = 50\ ^\circ\text{C}$
Coefficients of thermal expansion	$\alpha_{mg},\ \alpha_{al}$	$26\times10^{-6}$, $24\times10^{-6}\ /^\circ\text{C}$
Elastic moduli	$E_{mg},\ E_{al}$	44.7 GPa, 68.9 GPa

Find. The normal stress developed in the magnesium tube and in the aluminium rod after the temperature rises to 80 °C, given that the two members may close the 0.2 mm gap and then push on one another.

Q4 arrangement. The magnesium tube AB is built into the left wall at A and capped at B by the rigid plate E; the aluminium rod CD is built into the right wall at D. At 30 °C a 0.2 mm gap separates E from C.

Approach. Let both members expand freely; if their combined free growth exceeds the gap they must touch, after which a single common compressive force $F$ acts through the contact (the two members are in series), and the compatibility condition is that the total free growth minus the total elastic shortening equals the gap.

Compute the two cross-sectional areas. The tube is an annulus and the rod is solid: $$A_{mg} = \frac{\pi}{4}\left(d_o^{2} - d_i^{2}\right) = \frac{\pi}{4}\left(25^{2} - 20^{2}\right) = 176.71\ \text{mm}^{2}$$ $$A_{al} = \frac{\pi}{4}d^{2} = \frac{\pi}{4}(25)^{2} = 490.87\ \text{mm}^{2}$$ The rod is 2.78 times the area of the tube, which is why the tube will carry the higher stress for the same force.
Compute the free thermal growth of each member and confirm contact. Each member is free to grow toward the gap before contact, so $$\delta_T = \alpha\,\Delta T\,L$$ $$\delta_{T,mg} = (26\times10^{-6})(50)(300) = 0.390\ \text{mm}, \qquad \delta_{T,al} = (24\times10^{-6})(50)(450) = 0.540\ \text{mm}$$ Their sum is $0.390 + 0.540 = 0.930\ \text{mm}$, which comfortably exceeds the 0.2 mm gap. Contact therefore occurs well before 80 °C is reached, and a contact force develops.
Write the compatibility condition. Once E bears on C, one and the same compressive force $F$ passes through the contact into both members (they are in series between two immovable walls). Each is shortened elastically by $FL/(AE)$, and what remains of the free growth must exactly close the gap: $$\delta_{T,mg} + \delta_{T,al} - \left(\frac{FL_{mg}}{A_{mg}E_{mg}} + \frac{FL_{al}}{A_{al}E_{al}}\right) = \delta_g$$ This is the whole of the mechanics; everything that follows is arithmetic.
Evaluate the two flexibilities. Working in newtons and metres, $$\frac{L_{mg}}{A_{mg}E_{mg}} = \frac{0.300}{(176.71\times10^{-6})(44.7\times10^{9})} = 3.7981\times10^{-8}\ \text{m/N}$$ $$\frac{L_{al}}{A_{al}E_{al}} = \frac{0.450}{(490.87\times10^{-6})(68.9\times10^{9})} = 1.3306\times10^{-8}\ \text{m/N}$$ The magnesium tube is nearly three times the more flexible member, so it absorbs most of the elastic take-up.
Solve for the contact force. Substituting the excess growth $0.930 - 0.200 = 0.730\ \text{mm}$, $$F = \frac{0.730\times10^{-3}}{(3.7981 + 1.3306)\times10^{-8}} = 14\,234\ \text{N}$$ $$\boxed{F = 14.23\ \text{kN (compressive, common to both members)}}$$
Convert the force to normal stresses. Each member carries the same force over its own area: $$\sigma_{mg} = \frac{F}{A_{mg}} = \frac{14\,234}{176.71\times10^{-6}} = 80.6\ \text{MPa}, \qquad \sigma_{al} = \frac{F}{A_{al}} = \frac{14\,234}{490.87\times10^{-6}} = 29.0\ \text{MPa}$$ $$\boxed{\sigma_{mg} = 80.6\ \text{MPa (C)}, \qquad \sigma_{al} = 29.0\ \text{MPa (C)}}$$
Check the answer independently. Recovering the gap from the strained lengths: the tube grows $0.390 - (14\,234)(3.7981\times10^{-8})\times10^{3} = 0.390 - 0.5406 = -0.1506\ \text{mm}$ and the rod grows $0.540 - (14\,234)(1.3306\times10^{-8})\times10^{3} = 0.540 - 0.1894 = 0.3506\ \text{mm}$. Their sum is $0.200\ \text{mm}$, exactly the original gap, so the compatibility condition is satisfied. Note that the tube ends up shorter than it started: the compressive strain more than offsets its thermal growth, which is the physically correct result for the softer, more flexible member.

Both stresses are safely elastic. AM1004-T61 magnesium has a tensile yield strength of roughly 150 MPa and 6061-T6 aluminium roughly 255 MPa, so the tube runs at about 54 % of yield and the rod at about 11 %. The assembly survives the temperature rise, but the tube is clearly the critical member and a further 40 °C would bring it close to yielding.

Quantity	Symbol	Result
Free thermal growth of the tube	$\delta_{T,mg}$	0.390 mm
Free thermal growth of the rod	$\delta_{T,al}$	0.540 mm
Common contact force	$F$	14.23 kN (compression)
Normal stress in the magnesium tube	$\sigma_{mg}$	80.6 MPa (compressive)
Normal stress in the aluminium rod	$\sigma_{al}$	29.0 MPa (compressive)

Topic. Statically indeterminate axial members with thermal loading and an initial clearance.

Key relations. Free thermal elongation $\delta_T = \alpha\,\Delta T\,L$; axial flexibility $\delta_F = FL/(AE)$; superposition compatibility $\sum \delta_T - \sum \delta_F = \delta_g$ once contact is established; normal stress $\sigma = F/A$.

Why this works. The problem is statically indeterminate to the first degree: equilibrium alone cannot find the contact force, because the walls can supply whatever reaction is needed. The missing equation is geometric. The trick that makes it easy is recognising that once the gap closes, the two members carry the same force — they are in series through the rigid cap, exactly like two springs end to end — so a single unknown $F$ suffices and the two flexibilities simply add. The gap enters as a one-off deduction from the free growth, nothing more. If the sum of the free growths had turned out to be less than the gap, no contact would occur and both stresses would be zero: that check is not optional, it is the first thing to compute.

Common pitfall. Assuming that each member carries a different force, or writing separate compatibility equations for each and thereby double-counting the gap. A second frequent error is forgetting to subtract the gap at all, which here would give $F = 18.1\ \text{kN}$ and stresses about 27 % too high. A third is using the tube's outside diameter as if it were solid, which would understate the tube stress by nearly a factor of three.

Source. Hibbeler, Mechanics of Materials, 10th ed., §4.4–4.6 (statically indeterminate axially loaded members and thermal stress); Budynas & Nisbett, Shigley's Mechanical Engineering Design, 11th ed., §4.1–4.2.

← Question 3 Next: Question 5 →