## Task 1/62

A steel tire is to be fitted onto a wooden wheel with a diameter $d=75 \mathrm{~mm}$. The tire is to be fitted at a temperature $t \approx 500^{\circ} \mathrm{C}$. After cooling, the diameter $D$ of the tire should be $0.5 \mathrm{~mm}$ smaller than the diameter $d$ of the wheel, so that the tire presses the wheel firmly together.

How large must the diameter $D^{\prime}$ of the tire be at the time of manufacture, so that it has the required size after cooling?

The expansion coefficient of wrought iron is assumed to be $\alpha \approx 0.000012$ per degree.
The diameter $D^{\prime}$ of the tire must be $D^{\prime}=74.5 \, \text{mm}$, as the tire is manufactured at normal temperature. However, the temperature $t \approx 500^{\circ} \text{C}$ is not sufficient to enlarge it enough to fit it onto the wheel. If $U=\pi D \approx 233.8 \, \text{mm}$, then

$$
U+U_{z}=U\left(1+t_{z}\right)=233.8 \cdot (1+0.000012 \cdot 500) \approx 235.2 \, \text{mm}
$$

thus $U_{z} \approx 1.4 \, \text{mm}$, where $U_{z}$ denotes the increase in circumference $U$ and $t_{z}$ denotes the temperature increase (in this case, $t_{z}=t$ was set). If we also denote the increase in diameter $D$ with $D_{z}$, then from $D+D_{z}=\frac{U+U_{z}}{\pi}$, it follows that $D_{z} \approx 0.47 \, \text{mm}$.

However, $D_{z}>0.5 \, \text{mm}$ would be required for $D+D_{z}>75 \, \text{mm}$.