Railway rails are made of equal iron segments of length

Railway rails are made of equal iron segments of length $l_{0} = 20\ \text{m}$ at $0^{\circ}\text{C}$. Find the minimum gap between two adjacent rails so that they do not touch when the ambient temperature rises from $t_{1} = 20^{\circ}\text{C}$ to $t_{2} = 40^{\circ}\text{C}$. The coefficient of linear thermal expansion of iron is $\alpha = 1.2\cdot10^{-5}\ {}^{\circ}\text{C}^{-1}$.

Step 1: The model

Linear thermal expansion gives the length at temperature $t$ as

\[ l(t) = l_{0}\bigl(1+\alpha t\bigr), \]

where $l_{0} = 20\ \text{m}$ is the length at $0^{\circ}\text{C}$.

Step 2: The differential of length

Since $\dfrac{dl}{dt} = l_{0}\alpha$ is constant, \[\Delta l = dl = \frac{dl}{dt}\,\Delta t = l_{0}\,\alpha\,\Delta t.\]

For this model the differential is not merely an approximation — $l$ is linear in $t$, so $dl$ equals $\Delta l$ exactly.

Step 3: Which temperature change matters

The rails are laid at $t_{1} = 20^{\circ}\text{C}$ and must still not touch at the maximum temperature $t_{2} = 40^{\circ}\text{C}$. The gap therefore has to absorb the expansion over

\[ \Delta t = t_{2}-t_{1} = 40-20 = 20^{\circ}\text{C}. \]

Step 4: Compute the gap

\[ \delta = l_{0}\,\alpha\,\Delta t = 20\cdot 1.2\cdot10^{-5}\cdot 20 = 400\cdot 1.2\cdot 10^{-5} = 4.8\cdot10^{-3}\ \text{m}. \]

Step 5: Verification

\[l(20^{\circ}) = 20\bigl(1+1.2\cdot10^{-5}\cdot 20\bigr) = 20.0048\ \text{m}\]

\[l(40^{\circ}) = 20\bigl(1+1.2\cdot10^{-5}\cdot 40\bigr) = 20.0096\ \text{m}\]

\[l(40^{\circ})-l(20^{\circ}) = 0.0048\ \text{m} = 4.8\ \text{mm} \;\;\checkmark\]