- #1

Karol

- 1,380

- 22

## Homework Statement

A steel ring has a diameter of 75 mm at 20 deg C. how much must it be heated in order to mount on a brass shaft of dia 75.05

Then both are cooled. how must they be cooled so the ring will fall off?

## Homework Equations

Thermal elongation coefficient of steel: 12E-6

Thermal elongation coefficient of brass: 20E-6

Expansion of area under heating/cooling: ##A=A_0(1+2\alpha \Delta t)##

## The Attempt at a Solution

Heating:

$$\pi\cdot 75.05^2=\pi\cdot 75^2 (1+2\cdot 12E-6\cdot\Delta t)\rightarrow \Delta t=55.6^0C$$

It's too low, isn't it?

For cooling i consider as if the ring isn' t mounted on the shaft, it's aside of it:

$$\pi\cdot 75.05^2(1-2\cdot 20E-6\cdot\Delta t)=\pi\cdot 75^2 (1-2\cdot 12E-6\cdot\Delta t)\rightarrow \Delta t=83.1^0C$$

Also too small, isn't it?

Is it correct what i have done, taking the ring aside? i didn't know how to solve it with the ring mounted, i don't know if to calculate efforts or not and how to do it. i know the formula for stress due to heating in a straight bar.