Help please thermal expansion

In summary, the band has a coefficient of linear expansion of 17.3X10^-6 degrees C, a youngs modulus of 18X10^10 N/m^2, and a thickness of .50mm. If the band is heated to 80 degrees Celsius, it fits snugly over a tooth. If cooled to 37 degrees Celsius, the tension in the band is 270 N.
  • #1
ballahboy
34
0
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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  • #2
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
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.

[tex] \Delta l = 2.34*10^{-5}~m [/tex] 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 [itex]A=thickness*height= 0.5*10^{-3}*4*10^{-3} m^2 [/itex]
 
Last edited:
  • #4
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
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 [itex]L = 2\pi r = 6.28*.005 m= .0314[/itex] 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)[tex]\Delta L /L = \alpha \Delta T[/tex]

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

From (1) I get

[tex]\Delta L = .0314*17.3E(-6)*43 = 2.34E(-5) m.[/tex]

From (2) I get

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

AM
 
  • #6
ok i got it now.. i did the area part wrong
Thanks to both of you for the help :D
 

What is thermal expansion?

Thermal expansion is the tendency of a substance to expand when heated and contract when cooled. This is due to the increase and decrease in the average distance between its atoms or molecules.

What causes thermal expansion?

Thermal expansion is caused by the increase in the kinetic energy of particles when they are heated. This leads to an increase in the space between particles, resulting in an expansion of the substance.

How is thermal expansion measured?

Thermal expansion is typically measured using the coefficient of thermal expansion, which is the fractional change in length or volume per degree change in temperature. It is represented by the symbol α (alpha).

What are some real-world applications of thermal expansion?

Thermal expansion is a common phenomenon that is observed in many everyday objects. Some examples include the expansion of metal bridges in hot weather, the cracking of sidewalks due to extreme temperature changes, and the use of bimetallic strips in thermostats.

Can thermal expansion be controlled?

While thermal expansion cannot be completely eliminated, it can be controlled to some extent. This can be achieved through the use of materials with low coefficients of thermal expansion, designing structures to allow for expansion and contraction, and using expansion joints or flexible materials in construction.

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