A steel bar of cross sectional area 1 cm square and 50 cm long at `30^(@)C` fits into the space between two fixed supports. If the bar is now heated to `280^(@)C`, what force will it exert against the supports ? (`alpha` for steel = `11xx10^(-6)//""^(@)C` and Young's modulus for steel = `2xx10^(11)N//m^(2)`.

Explanation

Theory : We have coefficient of linear expansion
`alpha=(l_(2)-l_(1))/(l_(1)(t_(2)-t_(1)))rArr(l_(2)-l_(1))/l_(1)=alpha(t_(2)-t_(1))`
But `(l_(2)-l_(1))/l_(1)` is the change in length per unit original length of the bar which is equal to the strain `[(Deltal)/l]`
`therefore` strain = `alpha(t_(2)-t_(1))" "...(1)`
By the definition of Young.s modulus, `Y=("Stress")/("Strain")`
Stress = Y `xx` Strain = `Yalpha(t_(2)-t_(1))." "....(2)`
This is called thermal Stress
Stress = `("Force")/("Area of cross section")`
`therefore` Force exerted on the supports = Stress `xx` Area of cross section
`=Yalpha(t_(2)-t_(1))A=YalphaA(t_(2)-t_(1))." "...(3)`
In this problem, `Y=2xx10^(11)N//m^(2),alpha=11xx10^(-6)//""^(@)C,`
`A=1xx1cm^(2)=1xx10^(-4)m^(2),t_(1)=30^(@)C,t_(2)=280^(@)C`.
The force exerted on the supports
= `2xx10^(11)xx11xx10^(-6)xx10^(-4)xx250=55000N`.