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Find the increase in the length of the rod

  1. May 3, 2014 #1

    utkarshakash

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    Gold Member

    1. The problem statement, all variables and given/known data
    A metal rod of length L at temperature of 0°C is not uniformly heated such that the temperature is given by the distance x along its length measured from one end when:
    [itex]T(x) = T_0 \sin (\pi x/L) [/itex]
    Accordingly, points at x = 0 and x = L are also zero temperature, whereas at x = L/2, where the argument of sine function is π/2, the temperature have the maximum value T0. The coefficient of linear expansion of the rod is α. Find the increase in the length of the rod in function of α and T0.

    3. The attempt at a solution

    Let us consider a differential element dx at a distance x from one end of the rod.

    [itex]Δ(dx) = dx \alpha dT \\
    ΔL = \alpha T_0 \displaystyle \int_0^L \cos \left( \dfrac{\pi x}{L} \right) dx [/itex]

    But the above equation gives me 0! :confused:
    I know something's going wrong here.
     
  2. jcsd
  3. May 3, 2014 #2

    Doc Al

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    Staff: Mentor

    Careful. That should be ΔT, not dT. The temperature difference is a finite function of x, not a differential. (And if it's with respect to 0°, then ΔT = T.)
     
  4. May 3, 2014 #3

    utkarshakash

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    Thanks!
     
  5. May 3, 2014 #4
    What is the answer ?
     
  6. May 3, 2014 #5

    utkarshakash

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    [itex]2L \alpha T_0 / \pi [/itex]
     
  7. May 3, 2014 #6
    Could you please show how you got this answer.
     
  8. May 3, 2014 #7

    utkarshakash

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    Gold Member

    Sure.

    Let's consider a differential element dx at a distance x from one end of the rod.
    [itex]Δ(dx) = dx \alpha (t(0) - t(x)) \\
    =- \alpha T_0 \sin \dfrac{\pi x}{L} dx \\
    [/itex]

    Integrating both sides

    [itex] ΔL = \dfrac{- L \alpha T_0}{\pi} \left( \cos \dfrac{\pi x}{L} \right)_0^L [/itex]

    Substitute the values to get the answer.
     
  9. May 3, 2014 #8
    Thanks a lot :)
     
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