Question

At a temperature of 24.00°C the hole in a steel plate has a diameter of 1.055 cm. If a steel rod with a diameter of 1.062 cm has to just slip through this hole, to what temperature should the plate be heated? The coefficient of area expansion of steel is 24 ✕ 10−6/°C.

Answer 1

If the follow figure we can see the expansion of the steel plate the original diameter (D_0) of the hole in the plate is 1.055 cm, so the initial Area is given by: A_0 = pi D^2_0/4 = pi (1.055 cm)^2/4 A_0 = 0.874168 cm^2 the required diameter (D_1), for the steel rod can just slip through the hole of the plate is 1.062 cm, so the final Area is given by: A_1 = pi D^2_1/4 = pi (1.062 cm)^2/4 A_1 = 0.885807 cm^2 using the equation of thermal expansion, for a surface we have: delta A/A_0 = 2a. delta T (A_1 - A_0)/A_0 = 2a . (T_1 - T_0) 2a = thermal coefficient of area expansion of steel = 24 times 10^-6 0_C^-1. T_0 = temperature before the heating = 24 degree c. T_1 = temperature after the heating. So, rt_1 is equal to: T_1 = (A_1 - A_0)/2a . A_0 + T_0 T_1 = (0.885807 - 0.874168)cm^2/24 times 10^-6