Question

The two ends of a cylindrical rod of nickel 1 2 0 . 0 0 mm long and 1 2 . 0 0 0 mm in diameter are maintained rigid. If the rod is initially at 6 0 ° c , to what temperature must it be cooled in order to have a 0 . 0 2 4 - mm reduction in diameter? The values of Poisson's ratio and for nickel are 0 . 3 1 and 1 3 . 3 × 1 0 - 6 ( ° c ) - 1 , respectively.

Answer 1

Therefore to accomplish the specified diameter decrease, the rod must be cooled to -189948.6°C.

To solve this equation

Given:

Length of the rod (L) = 120.00 mm

Initial diameter of the rod (D₀) = 12.000 mm

Initial temperature (T₀) = 60.00 °C

Reduction in diameter (ΔD) = 0.024 mm

Poisson's ratio (ν) = 0.31

Coefficient of thermal expansion (α) = 13.3 × 10⁻⁶ (°C⁻¹)

Objective:

Determine the final temperature (T) to achieve the desired reduction in diameter

Calculations:

Using the starting diameter, compute the initial cross-sectional area (A0):

A₀ = π * (D₀/2)² = π * (12.000 mm/2)² = 113.0973 mm²

Using the reduction in diameter, calculate the change in cross-sectional area (A):

A = -D * D0 = -0.024 mm * 12.000 mm = -0.90478 rad2.

Using the change in cross-sectional area and the length of the rod, calculate the change in volume (V):

V = A * L = -108.574 mm3 = -0.90478 rad2 * 120.00 mm

Calculate the temperature change (T) using the volume change, starting cross-sectional area, coefficient of thermal expansion, and Poisson's ratio:

T = V / (A₀ * α * (1 - 2ν)) = -108.574 mm³ / (113.0973 mm² * 13.3 × 10⁻⁶ (°C⁻¹) * (1 - 2 * 0.31)) = 190008.6°C

Calculate the final temperature (T) by subtracting the change in temperature from the initial temperature:

T = T₀ - T = 60.00 °C - 190008.6°C = -189948.6°C