Question

A rail track made of steel having length 10 m is clamped on a railway line at its two ends (Fig.). on summer day due to rise in temperature by 20ºC, it is deformed as shown in figure. Find x (Displacement of the centre) if α_steel= 1.2 × 10^-5/ºC.

Answer 1

Let us consider the diagram below,

Given: α = 1.2 × 10^-5ºC^-1 , L = 10m, ∆T = 20ºC.

By Pythagoras theorem,

x ² = [(L+ ∆L) \(\frac{1}{2}\) ]² − \((\frac{L}{2})\) ²

= \(\frac{1}{4}\) [L² + ∆L² + 2L∆L] − \(\frac{L^2}{4}\)

x = \(\frac{1}{2}\sqrt{∆L^2+2L∆L}\)

∆L² ≪ L,

∴ neglecting ∆L²

x = \(\frac{1}{2}\) × \(\sqrt{2L∆L}\)

But ∆L = Lα∆t

∴ x = \(\frac{1}{2}\sqrt{2L\times L\alpha∆t}\)

= \(\frac{1}{2}L\sqrt{2\alpha∆t}\)

= \(\frac{10}{2}\times\sqrt{2\times1.2\times10^{-5}\times20}\)

= 5 × \(\sqrt{4\times1.2\times10^{-4}}\) 

= 5 × 2 × 1.1 × 10⁻² 

= 0.11 m

x = 11 cm