A thin steel ring is to be fitted tightly onto a brass shaft…

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A thin steel ring is to be fitted tightly onto a brass shaft. At $20\ ^\circ\text{C}$ the shaft is $5.000\ \text{cm}$ across and the inside of the ring is $4.995\ \text{cm}$ across. The ring's cross-section is a rectangle, $1.0\ \text{mm}$ thick (in the radial direction) and $1.0\ \text{cm}$ wide (along the shaft), as shown in the figure. For steel, $\alpha = 1.2 \times 10^{-5}\ \text{K}^{-1}$ and Young's modulus is $2.0 \times 10^{11}\ \text{Pa}$ ; for brass, $\alpha = 1.9 \times 10^{-5}\ \text{K}^{-1}$ . Treat the shaft as rigid when the ring squeezes it, and the ring as thin, with the shaft's radius as its radius. Take the expansion coefficients to stay the same at all the temperatures involved.

A thin steel ring next to the end of a brass shaft. The shaft is 5.000 cm across and the inside of the ring 4.995 cm across at 20 degrees Celsius. The ring's cross-section is 1.0 mm thick radially and 1.0 cm wide.

The ring alone is heated so that it just slides onto the shaft at $20\ ^\circ\text{C}$ . To what temperature must it be heated?
The ring then cools back to $20\ ^\circ\text{C}$ on the shaft. Find the stress in the ring and the tension in it (the pull across any cross-section of the ring).
What pressure does the ring then exert on the shaft?
The ring and the shaft are now cooled together. At what temperature does the ring become loose? Would heating them together loosen it instead?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Use $\Delta D = D\alpha\Delta T$ for both diameters. The stretched ring obeys Hooke's law; the pressure follows from cutting the ring in half.

Step 1. Heating the ring (a). The ring's inside must grow by $0.050\ \text{mm}$ :

\displaystyle \Delta T = \frac{\Delta D}{d\,\alpha_s} = \frac{5.0 \times 10^{-5}}{5.9940 \times 10^{-7}}\ \text{K} = 83.4\ \text{K}, \qquad \text{so } T = 103.4\ ^\circ\text{C}

Step 2. The stretched ring (b). On the shaft at $20\ ^\circ\text{C}$ , its circumference is longer than its natural one by the factor $\dfrac{D}{d}$ , so its strain is $\dfrac{5.0 \times 10^{-5}}{4.9950 \times 10^{-2}} = 1.0010 \times 10^{-3}$ . Then

\displaystyle \sigma = Y \times \text{strain} = 2.002 \times 10^{8}\ \text{Pa}, \qquad T = \sigma \times (1.0\ \text{mm} \times 1.0\ \text{cm}) = 2.002 \times 10^{8} \times 1.0 \times 10^{-5}\ \text{N} = 2002\ \text{N}

Step 3. The pressure (c). Cut the ring across a diameter. The shaft pushes outwards on one half with $p \times (2R \times w)$ , the projected area, and the two cut ends pull back with $2T$ :

\displaystyle p = \frac{T}{Rw} = \frac{2002\ \text{N}}{0.0250\ \text{m} \times 0.010\ \text{m}} = 8.008 \times 10^{6}\ \text{Pa}

Step 4. Cooling together (d). The ring is loose when its free inner diameter equals the shaft's. Brass shrinks faster than steel, so cooling helps:

\displaystyle 4.9950 \times 10^{-2}\,(1 + \alpha_s\Delta T) = 5.0000 \times 10^{-2}\,(1 + \alpha_b\Delta T) \quad\Rightarrow\quad \Delta T = -\frac{5.0 \times 10^{-5}}{9.5000 \times 10^{-7} - 5.9940 \times 10^{-7}}\ \text{K} = -142.6\ \text{K}

that is, at $-122.6\ ^\circ\text{C}$ . Heating them together would make the shaft grow faster than the ring, so the fit would get tighter, not looser.

Answer: (a) $103.4\ ^\circ\text{C}$ . (b) $2.00 \times 10^{8}\ \text{Pa}$ ; $2002\ \text{N}$ . (c) $8.01 \times 10^{6}\ \text{Pa}$ . (d) At $-122.6\ ^\circ\text{C}$ ; heating would only tighten it.

Check. The stress, about $200\ \text{MPa}$ , is below the stress at which ordinary steel starts to give way, so treating the ring as elastic is fair. The loosening temperature is far below room temperature but above absolute zero, so it can be reached with liquid nitrogen.

Watch out. A hole in a plate or a ring grows on heating, exactly as the material around it does. It does not shrink.

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