Question

A steel ring with a 2.5000-in. inside diameter at$\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$is to be warmed and slipped over a brass shaft with a 2.5020-in. outside diameter at $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$. (a) To what temperature should the ring be warmed? (b) If the ring and the shaft together are cooled by some means such as liquid air, at what temperature will the ring just slip off the shaft?

Short answer

(a) The temperature to which the ring should be warmed is 87ºC

(b) The temperature at which the ring will just slip off the shaft is -80ºC

Step-by-step solution

(a) Determination of the temperature to which the ring should be warmed.

The linear expansion relation is,

$∆L=L_0\alpha ∆T$

...(i)

$∆T=\frac{∆L}{L_0\alpha }$

...(ii)

$$\begin{gathered} ∆L=2.5020‐2.5000 \\ =0.0020in \end{gathered}$$

Substitute all the values in equation (ii),

$$\begin{gathered} ∆T=\frac{0.0020in}{\left(2.5000in\right)\left(1.2\times {10}^{-5}{\left(°C\right)}^{-1}\right)} \\ =66.7°C \end{gathered}$$

Therefore, the temperature to which the ring should be warmed is,

$$\begin{gathered} T=T_0+∆T \\ =20.0°C+66.7°C \\ =87°C \end{gathered}$$

(b) Determination of temperature at which the ring will just slip off the shaft.

The linear expansion equation for the diameter of the brass shaft is,

$L_b=L_{0b}\left(1+{\alpha }_b∆T\right)$

The linear expansion equation for the diameter of the hole in the steel ring,

$L_s=L_{0s}\left(1+{\alpha }_s∆T\right)$

For same rise in temperature $∆T$,

$$\begin{gathered} L_{0b}\left(1+{\alpha }_b∆T\right)=L_{0s}\left(1+{\alpha }_s∆T\right) \\ {L}_{0b}+L_{0b}{\alpha }_b∆T=L_{0s}+L_{0s}{\alpha }_s∆T \end{gathered}$$

Rearranging for ,

$$\begin{gathered} ∆T=\frac{L_{0b}-L_{0s}}{L_{0s}{\alpha }_s-L_{0b}{\alpha }_b} \\ =\frac{2.5020in.-2.5000in.}{\left(2.5000in.\right)\left(1.2\times {10}^{-5}{\left(°C\right)}^{-1}\right)-\left(2.5020in.\right)\left(2.0\times {10}^{-5}{\left(°C\right)}^{-1}\right)} \\ =-100°C \end{gathered}$$

Thus, the temperature at which the ring will just slip off the shaft is,

$$\begin{gathered} T=T_0+∆T \\ =20.0°C-100°C \\ =-80°C \end{gathered}$$