Question

A steel ring with a 2.5000-in. inside diameter at 20.0 ?C is to be warmed and slipped over a brass shaft with a 2.5030 in. outside diameter at 20.0 ?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?

Answer 1

Coefficient of linear expansion of steel = 13x10^-6 deg C
Coefficient of linear expansion of brass = 19x10^-6 deg C

A) Change in length = (length)*()*(change in tempreture)
    To slip steel ring over Brass shaft, increase in diameter of ring = 0.0030 In.
    0.0030 = 2.5000*13x10^-6*(T)
     (T) = 92.3 deg C
Ring should be warmed to 20+92.3 = 112.3 deg C

B) Tempreture at which, ring diameter , becomes just equal to diameter of brass shaft, it slips off the shaft.
Hence decrease in diameter of shaft is 0.0030 In. more than that of ring
2.5030*(19x10^-6)(T) = 0.0030 + 2.5000*(13x10^-6)(T)
(T) = 199.24 deg C
Since this is decrease in tempreture,
Final tempreture = 20 - 199.2 = -179.2 deg C