how much would the machinist have to lower the temperature (in °C) of the rod to make it fit the hole?

Question

A machinist wishes to insert a steel rod with a diameter of 9 mm into a hole with a diameter of 8.998 mm. By how much would the machinist have to lower the temperature (in °C) of the rod to make it fit the hole?

Answer 1

The cross sectional area of the hole is found using:

A = πd²/4 = π(8.998)²/4 = 63.5890 mm²

The cross sectional area of the rod is:

A = πd²/4 = π(9)²/4 = 63.6173 mm²

This means you have to shrink the rod by 0.0283 mm².

The standard equation for length changes using the coefficient of thermal expansion is ∆ℓ = αℓ₀∆T however area changes twice as much as length so: ∆A = 2αA₀∆T

You don't say what type of steel it is but for 1018 carbon steel the alpha is about 12x10^-6 m/m/K making the equation:

0.0283 = 2(12x10^-6)(63.6173)∆T

∆T = 0.0283/2/12x10^-6/63.6173 = 18.52 K or 18.52 °C so he would need to cool the rod more than that.