Question

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

Answer 1

Consider the diagram

Applying Pythagoras theorem in the right-angled triangle in figure

${\displaystyle {\left(\frac{L+\Delta L}{2}\right)}^2={\left(\frac{L}{2}\right)}^2+x^2}$

⇒ ${\displaystyle x=\sqrt{{\left(\frac{L+\Delta L}{2}\right)}^2={\left(\frac{L}{2}\right)}^2}}$

 = ${\displaystyle \frac{1}{2}\sqrt{{(L+\Delta L)}^2-L^2}}$

= ${\displaystyle \frac{1}{2}\sqrt{(L^2+\Delta L^2+2L\Delta L)-L^2}}$

= ${\displaystyle \frac{1}{2}\sqrt{(\Delta L^2+2L\Delta L)}}$

 As increase in length ΔL is very small, therefore, neglecting (ΔL)², we get

${\displaystyle x=\frac{1}{2}\times \sqrt{2L\Delta L}}$  ......(i)

But ΔL = LαΔt  ......(ii)

Substituting the value of ΔL in equation (i) from equation (ii)

${\displaystyle x=\frac{1}{2}\sqrt{2L\times L\alpha \Delta t}}$

= ${\displaystyle \frac{1}{2}L\sqrt{2\alpha \Delta t}}$

= ${\displaystyle \frac{10}{2}\times \sqrt{2\times 1.2\times {10}^{-5}\times 20}}$

= ${\displaystyle 5\times \sqrt{4\times 1.2\times {10}^{-4}}}$

= ${\displaystyle 5\times 2\times 1.1\times {10}^{-2}}$

= 0.11

= 11 cm