A steel bar of cross sectional area 1 cm square and 50 cm long at 30^(0)C fits into the space between two fixed supports. If the bar is now heated to 280^(0)C, what force will it exert against the supports ? (alpha " for steel " = 11 xx 10^(-6)//^(0)C " and Young's modulus for steel " = 2 xx 10^(11) N//m^(2))

A steel bar of cross sectional area 1 cm square and 50 cm long at 30^(

1.	A steel bar of cross sectional area 1 cm square and 50 cm long at 30^(0)C fits into the space between two fixed supports. If the bar is now heated to 280^(0)C, what force will it exert against the supports ? (alpha " for steel " = 11 xx 10^(-6)//^(0)C " and Young's modulus for steel " = 2 xx 10^(11) N//m^(2))
Answer» Solution :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.