# Help please thermal expansion

1. Jan 9, 2005

### ballahboy

The band below is stainless steel (coefficient of linear expansion=17.3X10^-6 degrees C, youngs modulus=18X10^10 N/m^2). It is essentially circular with an intial mean radius of 5.0mm, a height of 4.0mm and a thickness of 0.50mm. If the band just fits snugly over the tooth when heated to a temperature of 80degrees C, what is the tension in the band when it cools to a temperature of 37degrees C?

hmm.. first i used (change in length)=(coef. of linear expansion)(original length)(change in temp). I used 0.0314m for the original length but not sure if its right. Then u plugged everything into the youngs modulus formular.. plugged in 0.0628m^2 for area and 2.34X10^-5 for change in length. Again not sure in the value are correct. Then i solved for F in the forumla and got something totally off. The answer is suppose to be 270N. Is there i step that i skipped or did i do everything totally wrong.. someone help please!!
THanks!!

2. Jan 9, 2005

### Gokul43201

Staff Emeritus
The dimensions of the band that are given - what temperature are they specified at ?

The question, as phrased is poorly written, because it leaves an ambiguity about the temperature at which the given dimensions are valid.

Nevertheless, I'll give it a shot and see what I come up with...

3. Jan 9, 2005

### Gokul43201

Staff Emeritus
I get about ~268 N, so I think that answer is correct.

Show your working, and I (or someone lese) can show you what's wrong.

$$\Delta l = 2.34*10^{-5}~m$$ is correct.

What next ?

Edit : I see an error in the number you used for the area. What you want is the cross-section area $A=thickness*height= 0.5*10^{-3}*4*10^{-3} m^2$

Last edited: Jan 9, 2005
4. Jan 9, 2005

### ballahboy

hmm.. after you got the change in length, what did you do after? Did you plug it into the Young's modulus formula? If so, what did you use as the area? I was confused about this part of the problem.. Thanks for the response :D

5. Jan 9, 2005

### Andrew Mason

It would be helpful to have a diagram. I assume, as you did, that the length of the band is $L = 2\pi r = 6.28*.005 m= .0314$ and A = 2E-6m^2. I assume that in cooling it is prevented by the tooth from shortening that length.

It appears to me that you understand the physics and have just made a math error.

First find the amount that it would shrink in going to 37C and then find the tension required to stretch it back to the original length.

(1)$$\Delta L /L = \alpha \Delta T$$

To find the tension, :
$$E = \frac{T/A}{\Delta L/L}$$
(2)$$T = E\Delta LA/\L$$

From (1) I get

$$\Delta L = .0314*17.3E(-6)*43 = 2.34E(-5) m.$$

From (2) I get

$$T = 18E10*2.34E(-5)*2E(-6)/.0314 = 268.2 N = 270 N$$ (2 sign. fig)

AM

6. Jan 9, 2005

### ballahboy

ok i got it now.. i did the area part wrong
Thanks to both of you for the help :D

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