A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is unstrained at `20^@C`. Find the longitudinal strain developed in the rod if the temperature rises to `50^@C`. Coefficient of linear expansion of steel `=1.2 xx 10^(-5)C^(-1)`.

Explanation

Given, ` theta _1 = 20^@C theta_2= 50^@C `
` alpha_steel =(1.2 xx 10^-5/ (^@C))`
` Longitudinal strain =?`
Strain = (Delta L) /(L) - (L alpha Delta theta)/ (L) = (alpha Delta theta )`
` =(1.2 xx (10^-5) xx (50-20))`
` = 1.2 xx 10^-5 xx 30 `
` = (36 xx 10^-5) = (3.6 x 10^-4 )`
The strain is opposite to the direction of expansion.