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Help please thermal expansion

  1. Jan 9, 2005 #1
    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. jcsd
  3. Jan 9, 2005 #2

    Gokul43201

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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...
     
  4. Jan 9, 2005 #3

    Gokul43201

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    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: Jan 9, 2005
  5. Jan 9, 2005 #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
     
  6. Jan 9, 2005 #5

    Andrew Mason

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    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
     
  7. Jan 9, 2005 #6
    ok i got it now.. i did the area part wrong
    Thanks to both of you for the help :D
     
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