Trying to find how much you would have to lower the temp to make the rod fit the hole

Question

A machinist wishes to insert a copper rod with a diameter of 8 mm into a hole with a diameter of 7.998 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

I am assuming the following formula for superficial expansion:

When temperature increases by ΔT then area A increases by

ΔA = 2αAΔT

where α is the coefficient of linear expansion.

For copper

α = 16.7×10^-6°C^-1

In your example,

A = 7.998²π/4

ΔA = 8²π/4 - A

ΔT = ΔA/2αA = ((8/7.998)² - 1)/(2×16.7×10^-6) = 15°C