How Does Thermal Expansion Affect the Tension in a Stainless Steel Band?

AI Thread Summary
The discussion revolves around calculating the tension in a stainless steel band as it cools from 80°C to 37°C. Participants clarify the calculations involving the coefficient of linear expansion and Young's modulus, with specific attention to the correct area of the band. There is some confusion regarding the original length and area used in the calculations, leading to discrepancies in the tension results. Ultimately, the correct tension is confirmed to be approximately 270 N, with participants assisting each other to resolve calculation errors. The importance of accurate dimensions and understanding the physics behind thermal expansion is emphasized throughout the conversation.
ballahboy
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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 formula.. 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!
 
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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...
 
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:
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
 
ballahboy said:
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 formula.. 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!
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
 
ok i got it now.. i did the area part wrong
Thanks to both of you for the help :D
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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