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Thermal expansion of bimetallic strip

  1. Dec 24, 2017 #1
    1. The problem statement, all variables and given/known data
    Thailand IJSO 14th Camp 2 No 4.png

    A bimetallic strip is made of metal which has coefficient of thermal expansion is equal to α1 and the other's is equal to α2 at the temperature of T0. The temperature is increased to T0 + ΔT (ΔT > 0). The strip curves as shown in the figure. If both strip have the same width of d. Find r1 measured from the center of the curve to the center of the strip as shown in the figure

    2. Relevant equations
    1) radian is (Length of the curve)/(Radius)
    2) Using the following estimation
    (1 + x)-1 ≈ 1-x and (1 + ax)(1+bx) ≈ 1+(a+b)x for x << 1

    3. The attempt at a solution
    α1 strip expansion is L0(1 + α1ΔT) for the length
    and d(1 + α1ΔT) for the width
    α2 strip expansion is L0(1 + α2ΔT) for the length
    and d(1 + α2ΔT) for the width

    Multiplying radian with radius, the result should be the curve length

    r1θ = L0(1 + α1ΔT)
    r2θ = L0(1 + α2ΔT)

    r2 = r1 + d(2 + α1ΔT + α2ΔT)/2

    Is that correct?
     
    Last edited: Dec 24, 2017
  2. jcsd
  3. Dec 24, 2017 #2
    What is the ratio of r2 to r1? What is the difference between r2 and r1?
     
  4. Dec 24, 2017 #3
    The difference is d(2 + α1ΔT + α1ΔT)/2 because of the width expansion.

    Is that correct?
     
  5. Dec 24, 2017 #4
    Neglecting the change in thickness, it is just d.
     
  6. Dec 24, 2017 #5
    Then the difference should be d. And the ratio should be (r1 + d)/r1 or (1 + α2ΔT)/(1+α1ΔT)

    Is that correct?
     
  7. Dec 24, 2017 #6
    Yes. To terms of 1st order in ##\alpha \Delta T##, this gives $$\frac{d}{r_1}=(\alpha_2-\alpha_1)\Delta T$$
     
  8. Dec 24, 2017 #7
    Thank you.
    Which mean r1 is equal to d/((α2 - α1)ΔT)

    What about the estimation? Or do I have to estimate them before the calculation?
     
  9. Dec 24, 2017 #8
    I don't know what you mean by "the estimation." Are you talking about retaining only terms of first order in ##\alpha \Delta T##?
     
  10. Dec 24, 2017 #9
     
  11. Dec 24, 2017 #10
    Ok. That’s what we did with the equation in post #5.
     
  12. Dec 24, 2017 #11
    Then it should be 1/(1 - α2ΔT)(1 + α1ΔT) and then 1/(1+(α1 - α2)ΔT)

    Is that correct?
     
  13. Dec 24, 2017 #12
    Yes. So, from that, what is d/r1?
     
  14. Dec 24, 2017 #13
    ##\frac{d}{r_1} = \frac{-(\alpha_1 - \alpha_2) \Delta T}{1 + (\alpha_1 - \alpha_2) \Delta T}##

    Then ##r_1 = (\frac{1}{(\alpha_2 - \alpha_1) \Delta T} - 1) d##

    Is that correct?
     
  15. Dec 25, 2017 #14
    No. Check your math.
     
  16. Dec 25, 2017 #15
    From these equations
    ##r_1 \theta = L_0 (1 + \alpha_1 \Delta T)##
    ##r_2 \theta = L_0 (1 + \alpha_2 \Delta T)##

    Then substitute ##r_2## with ##r_1 + d##

    ##\frac{r_1 + d}{r_1} = \frac{1}{1+(\alpha_1 - \alpha_2) \Delta T}##

    Where am I going wrong?
     
  17. Dec 25, 2017 #16
    $$\frac{r_1 + d}{r_1} = 1+(\alpha_2 - \alpha_1) \Delta T$$
    $$1+\frac{d}{r_1}= 1+(\alpha_2 - \alpha_1) \Delta T$$
     
  18. Dec 25, 2017 #17
    Oh! I should estimate the equation into

    ##1 + (\alpha_2 - \alpha_1) \Delta T##

    ##r_1 = \frac{d}{(\alpha_2 - \alpha_1) \Delta T}##

    Is that correct?
     
  19. Dec 25, 2017 #18
    Yes. It all involves knowing how to make the correct mathematical approximations.
     
  20. Dec 25, 2017 #19
    Thank you very much and Merry Christmas.
     
  21. Dec 25, 2017 #20
    Back at ya.
     
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