Question

A shaft made of aluminum is 40.0 mm in diameter at room temperature (21°C). Its coefficient of thermal expansion = 24.8 x 10-6 mm/mm per °C. If it must be reduced in size by 0.20 mm in order to be expansion fitted into a hole, determine the temperature to which the shaft must be cooled.

Answer 1

Answer:

Temperature to which the shaft must be cooled, $\theta_2 = -180.61 ^0C$

Explanation:

Diameter of the shaft at room temperature, d₁ = 40 mm

Room temperature, θ₁ = 21°C

Coefficient of thermal expansion, $\alpha = 24.8 * 10^{-6} / ^0 C$

The shaft is reduced in size by 0.20 mm:

Δd = - 0.20 mm

The temperature to which the shaft must be cooled, θ₂ = ?

The coefficient of thermal expansion is given by the equation:

$\alpha = \frac{\triangle d}{d_1 * \triangle \theta}\\\\24.8 * 10^{-6} = \frac{-0.20}{40 * \triangle \theta}\\\\\triangle \theta = \frac{-0.20 }{24.8 * 10^{-6} * 40} \\\\\triangle \theta = - 201.61 ^0 C\\\triangle \theta = \theta_2 - \theta_1\\\\- 201.61 = \theta_2 - 21\\\\\theta_2 = -201.61 + 21\\\\\theta_2 = -180.61 ^0C$