25-Nav-A6 Advanced Strength of Materials (25-Mec-A6) · May 2016
Question 4 of 6: Thermal Stress in a Restrained Tube-and-Rod Assembly
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Kalpakjian & Schmid, Manufacturing Engineering and Technology, 8th ed.; Groover, Fundamentals of Modern Manufacturing, 7th ed.; Hibbeler, Mechanics of Materials, 10th ed.; Hibbeler, Engineering Mechanics: Statics, 14th ed.; Shigley's Mechanical Engineering Design, 11th ed.
Check: the exam is headed "98-Mar-A6, Design and Manufacture of Machine Elements" and Part A (Q1–Q3) is entirely machining/casting/forming content with zero naval-architecture material. Solved here as the paper actually printed; reference texts above are chosen for the real content. Part B (Q4–Q6) is genuine strength-of-materials/machine-design content.
Check: Q5 states the shaft modulus as "E = 30 ksi," which is off by three orders of magnitude for steel (actual E ≈ 29,000–30,000 ksi = 30×106 psi); the shaft's computed self-weight, w = γA = 0.283 × (π/4)(32) = 2.00 lb/in, comes out to a clean round number confirming the 3-in-diameter reading, and the printed "30 ksi" is treated as a dropped exponent (E = 30×106 psi used throughout).
Question 4: Thermal Stress in a Restrained Tube-and-Rod Assembly (equal value, Part B)
Given. Tube AB (Mg alloy) fixed at wall A, rod CD (Al alloy) fixed at wall D, with a 0.2 mm gap between the tube's rigid end cap and rod end C at 30 °C; temperature rises to 80 °C.
Given data
Quantity
Tube AB (Mg, AM1004-T61)
Rod CD (Al, 6061-T6)
Length
300 mm
450 mm
Cross-section
25 mm OD × 20 mm ID (annular)
25 mm dia. (solid)
Modulus $E$
44.7 GPa
68.9 GPa
Expansion coeff. $\alpha$
$26\times10^{-6}/^{\circ}\text{C}$
$24\times10^{-6}/^{\circ}\text{C}$
Initial gap = 0.2 mm at 30 °C; final temperature = 80 °C ($\Delta T=50^{\circ}\text{C}$)
Find. The normal stress developed in the tube and in the rod once the temperature rises to 80 °C.
Tube AB (Mg) and rod CD (Al), both fixed at their outer ends, separated by a 0.2 mm gap at the rigid cap; heating closes the gap and puts both members into compression.
Approach. First check, using unrestrained free thermal expansion, whether the 0.2 mm gap actually closes; if it does, the assembly becomes a single indeterminate series system between two rigid walls, and one compatibility equation (free expansion of both members, minus the gap, equals their combined elastic compression under a single common force $P$) gives $P$, from which both stresses follow directly.
Check whether the gap closes. Free (unrestrained) thermal growth of each member over $\Delta T=50^{\circ}\text{C}$:
$$\delta_{tube}=\alpha_{mg}L_{AB}\Delta T=26\times10^{-6}(300)(50)=0.390\text{ mm}\qquad \delta_{rod}=\alpha_{al}L_{CD}\Delta T=24\times10^{-6}(450)(50)=0.540\text{ mm}$$
Total free growth $=0.390+0.540=0.930\text{ mm}$, which exceeds the 0.2 mm gap, so the gap closes and both members are compressed by a common force $P$ once contact is made.
Compatibility. The excess growth beyond the gap must be absorbed as elastic compression of both members in series:
$$(\delta_{tube}+\delta_{rod})-\text{gap}=P\left(\frac{L_{AB}}{A_{tube}E_{mg}}+\frac{L_{CD}}{A_{rod}E_{al}}\right)$$
Excess $=0.930-0.200=0.730\text{ mm}$. Flexibility $=\dfrac{300}{176.71\times44{,}700}+\dfrac{450}{490.87\times68{,}900}=5.128\times10^{-5}\text{ mm/N}$, so
$$\boxed{P=\frac{0.730}{5.128\times10^{-5}}=14{,}234\text{ N}\approx14.23\text{ kN (compressive, common to both members)}}$$
Normal stresses. With the same compressive $P$ carried by both members in series:
$$\sigma_{tube}=\frac{P}{A_{tube}}=\frac{14{,}234}{176.71}=80.6\text{ MPa (C)}\qquad \sigma_{rod}=\frac{P}{A_{rod}}=\frac{14{,}234}{490.87}=29.0\text{ MPa (C)}$$
