A steel ring with a 2.5000-in. inside diameter at 20.0o C is to be warmed and slipped over a brass shaft with a 2.5020-in. outside diameter at 20.0o C. (a) To what temperature should the ring be warmed? (b) If the ring and the shaft together are cooled by some means such as liquid air, at what temperature will the ring just slip off the shaft?

Solution 89P Step 1: Equation for linear thermal expansion, L/L = T L Where, L - Change in length L - Original length L Coefficient of linear thermal expansion T - Change in temperature Step 2: a) nitial diameter of the steel ring, D = 2.5000 inches Final diameter needed after expansion, D’ = 2.5020 inches Therefore, change in diameter, D = 2.5020 inches - 2.5000 inches = 0.0020 inches So, we can rewrite the equation, D/D = T L(steel) -6 -1 Coefficient of linear expansion of steel, L(steel) 12 × 10 Initial temperature of the ring, T = 20 1 We have to find out the temperature corresponding to 2.5020 inches expansion. Put that temperature as T . 2 Therefore, we can write, T = (T - T ) = (T 220)1 2 That is, 0.0020 inches / 2.5000 inches = 12 × 10 (T - 20) -6 -1 2 -6 -1 -5 -6 -1 0.0008 = (12 × 10 × T )- (26 ×10 2= (12 × 10 × T ) - 0.00024 2 0.0008 + 0.00024 = (12 × 10 × T ) -6 -1 2 0.00104 = (12 × 10 × T )-6 -1 2 T2 0.00104 / 12 × 10 -6 -1 T2 86.67