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)))`
`(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 straim `[ (Delta l)/(l ) ] `
`therefore` Strain = `alpha (t_(2) - t_(1)) " "` ............. (1)
By the definition of Young.s modulus, `Y = ("Stress")/("Strain")`
Stress = Y `xx ` Strain = Y `alpha(t_(2) - t_(1)) .` ........ (2)
This is called thermal Stress
Stress = `("Force")/("Area of cross section")`
`therefore ` Force exerted on the supports = Stress x Area of cross section
= Y `alpha (t_(2) - t_(1)) A = Y alpha A (t_(2) - t_(1))` .......... (3)
In this problem,
`Y = 2 xx 10^(11) N//m^(2), alpha = 11 xx 10^(-6) //^(0)` C,
A = ` 1 xx 1cm^(2) = 1 xx 10^(- 4) m^(2)`,
`t_(1) = 30^(0)C , t_(2) = 280^(0) C `.
The force exerted on the supports
`= 2 xx 10^(11) xx 11 xx 10^(-6) xx 10^(-4) xx 250 = 55000` N.