# Homework Help: Thermal elongation-ring on a shaft

1. Sep 22, 2014

### Karol

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Thermal elongation coefficient of steel: 12E-6
Thermal elongation coefficient of brass: 20E-6
Expansion of area under heating/cooling: $A=A_0(1+2\alpha \Delta t)$

3. 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.

2. Sep 22, 2014

### Staff: Mentor

How much does the steel ring have to be heated such that it's diameter will be 75.05 mm?

Chet

3. Sep 22, 2014

### Karol

I wrote, it has to be heated with more 55.6C

4. Sep 22, 2014

### Staff: Mentor

Oh, sorry. The full line didn't show up on my iPhone, so I missed most of your equation.
In the second part, your answer also looks correct, if that 83 C is the amount the temperature is lowered below 20C. I did this part a little differently, by writing:

$$75.05(1-20\times 10^{-6} \Delta t)=75.0(1-12\times 10^{-6} \Delta t)$$

5. Sep 22, 2014

### Karol

I don't think it's correct because you made linear elongation and i read in a book that i have to calculate area expansion, like i did

6. Sep 22, 2014

### Staff: Mentor

Either way, you get the same answer (solve my equation and see). I used linear expansion because all line segments and arcs of the material must increase in length by the same fraction. I just wanted to show you that the problem could be done using linear expansion also.

Chet

7. Sep 22, 2014

### Karol

So there is no difference if the ring is mounted on the shaft or aside of it, the same temperature is needed, right? how can you explain that these conditions are the same?

8. Sep 22, 2014

### Staff: Mentor

Why would you think they wouldn't be, especially if, in the end, for the mounted case, the shaft and the ring are not even touching.

Chet