Question

a steel ring with a 2.500 in diameter at 20 degrees Celsius is to he warned and slipped over a brass shafts with a 2.5030in outside diameter at 20 degrees Celsius
To what temperature should the ring be warmed?
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

Part A.

In this case steel ring is need to be slipped over a brass shaft, So Using relation between linear expansion and temperature:

dL = L0*_steel*dT

L0 = Initial diameter of steel ring = 2.500 in

Change in diameter for this steel ring to be slipped over brass shaft = dL = 2.5030 - 2.5000 = 0.0030 in

_steel = thermal expansion coefficient of steel = 1.2*10^-5 /C

So,

dT = dL/(L0*_steel)

dT = 0.0030/(2.500*1.2*10^-5) = 100 C

dT = Tf - Ti = 100 C

Tf = 100 + Ti = 100 + 20 = 120 C

So ring should be warmed till 120 C

Part B.

Now when ring and shaft both are cooled together, then since initial diameter of both are equal, So for same dT, value of final length of both ring and shaft will be equal

Ls = Lb

L_s0*(1 + _steel*dT) = L_b0*(1 + _brass*dT)

L_s0 + L_s0*_steel*dT = L_b0 + L_b0*_brass*dT

dT = (L_b0 - L_s0)/(L_s0*_steel - L_b0*_brass)

_brass = thermal expansion coefficient of brass = 2.0*10^-5 /C

Using known values:

dT = (2.5030 - 2.5000)/(2.5000*1.2*10^-5 - 2.5030*2.0*10^-5)

dT = -149.55 C = Tf - Ti

Tf = -149.55 + 20 = -129.55 C

So both ring and shaft should be cooled by -129.55 C, for ring to slip off the shaft