Question

A rail track made of steel having length 10 m is clamped on a railway line at its two ends

on a summer day due to rise in temperature by `20^@C,` it is deformed as shown in fig. Find x (displacement of the centre) if `alpha_(steel) = 1.2 xx 10^(-5)//^@C.`

Answer 1

From the given fing
`x = sqrt(((L + Delta L)/(2))^2 - ((L)/(2))^2) = (1)/(2) sqrt((L + Delta L)^2 - L^2) = (1)/(2)[ DeltaL^2+ 2L Delta L]^(1//2)` Since `Delta L` is a small quantity the term wich `Delta L^2` being very small can be neglected. `:. X =(1)/(2) [ 2 L Delta L]^(1//2)`
Now `Delta L = L alpha Delta theta`
`x = (1)/(2) [2 L xx L alpha Delta theta]^(1//2) = (10)/(2) [2xx(1.2 xx 10^(-5))xx 20]^(1//2) = 10[1.2 xx10^(-4)]^(1//2)`
` = 10 xx 10^(-2) xx 1.1m ~~ 0.11 m`