Question

A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10¯⁵ K¯¹, steel = 1.2 × 10¯⁵K¯¹ ).

Answer 1

Given: The initial temperature is 40 °C, the final temperature is 250 °C, the length of two rods at initial temperature is 50 cm, the diameter of two rods at initial temperature is 3.0 mm, the coefficient of linear expansion of brass is 2.0× 10 −5   K −1 and the coefficient of linear expansion of steel is 1.2× 10 −5   K −1 .

The change in length of the brass rod is given as,

Δ l 1 l 1 = α 1 ×ΔT Δ l 1 = l 1 × α 1 ×ΔT

Where, the change in length of the brass rod is Δ l 1 , length of the brass rod at initial temperature is l 1 , coefficient of linear expansion of brass is α 1 and the temperature difference is ΔT.

By substituting the given values in the above expression, we get

Δ l 1 =50×2.0× 10 −5 ×( 250−40 ) =0.21 cm

The change in length of the steel rod is given as,

Δ l 2 l 2 = α 2 ×ΔT Δ l 2 = l 2 × α 2 ×ΔT

Where, the change in length of the steel rod is Δ l 2 , length of the steel rod at initial temperature is l 2 , coefficient of linear expansion of steel is α 2 and the temperature difference is ΔT.

By substituting the given values in the above expression, we get

Δ l 2 =50×1.2× 10 −5 ×( 250−40 ) =0.126 cm

Total change in length is given as,

Δl=Δ l 1 +Δ l 2

By substituting the values in the above expression, we get

Δl=0.13+0.21 =0.336 cm

Thus, the total change in length of combined rod is 0.336 cm.

Since, both the ends of combined rod are free to expand therefore, there is no thermal stress developed at the junction.