For a new train line, steel rails are being laid in 30 m lengths on a day when the temperature is 20 °C. What is the appropriate gap allowance required for each rail if they could reach 56 °C in the middle of summer? The expansion coefficient for steel is 1.2 × 10-5 K-1. Give your answer to the nearest mm.

Question

BlackTom AI · Accepted Answer

The linear thermal expansion relation ΔL = α L ΔT is used, where α is the coefficient of linear expansion, L the original length, and ΔT the temperature change.

1) Relevant knowledge points:
- Linear expansion: ΔL = α L ΔT.
- Keep consistent units (meters for L and ΔL, or convert at the end).
- Round final result to nearest mm as requested.

2) Step-by-step reasoning:
- Given: α = 1.2 × 10^(-5) K^(-1), L = 30 m, initial temperature 20 °C, final 56 °C.
- Temperature change: ΔT = 56 − 20 = 36 K.
- Apply formula to find increase in length for one 30 m rail.

3) Calculation with intermediate steps:
- Compute α × L: 1.2 × 10^(-5) × 30 = 3.6 × 10^(-4).
- Multiply by ΔT: ΔL (in meters) = 3.6 × 10^(-4) × 36 = 1.296 × 10^(-2) m.
- Convert to millimetres: 1.296 × 10^(-2) m = 12.96 mm.
- Round to nearest millimetre: 12.96 mm → 13 mm.

4) Why this answer is correct:
- The formula directly gives the increase in length of one 30 m rail for the specified temperature rise. The computed expansion 12.96 mm is the gap that must be left for each rail to avoid compressive stress or buckling at the higher temperature; rounding to the nearest mm yields 13 mm.

5) Why other plausible answers would be incorrect:
- A gap of 0 mm would allow no expansion and risk buckling. Smaller values like 5–10 mm are insufficient because they are less than the computed 12.96 mm. Reporting 12 mm would under-provide (it is truncation rather than rounding) and could still cause stress in service.

6) Conclusion:
The appropriate gap allowance per 30 m rail is 13 mm, which matches the provided answer.

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