Bimetal expansion (First year undergrad level I think?)

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  1. Astronomical's Avatar
    1 21-07-2012  18:40
    • Overlord in Training
    • Location: England
    • Posts: 2,144
    Bimetal expansion (First year undergrad level I think?)
    So we have a bimetal and we hold one end fixed and heat it uniformly. One metal wants to expand more than the other, so it curls into a circular arc. I am trying to find h, shown below, which is the vertical height above the fixed level.

    I can't seem to get an equation for h independent of R or theta, but I presume that it should be. Any ideas?

    Click image for larger version.  Name: Photo on 2012-07-21 at 18.30.jpg  Views: 23  Size: 73.3 KB  ID: 164185

    Oops, just to clarify, the question ought to be: \text{Find } h \text{ in terms of } L,\ \Delta T, \ \alpha_1, \ \alpha_2, \ d, \ \delta_L.
    Last edited by Astronomical; 21-07-2012 at 18:44.
  2. Astronomical's Avatar
    2 21-07-2012  19:08
    • Overlord in Training
    • Location: England
    • Posts: 2,144
    Re: Bimetal expansion (First year undergrad level I think?)
    Actually, I just did manage to get an answer.

    I got the following:

    \theta = \dfrac{\delta_L}{d} = \dfrac{(\alpha_2 - \alpha_1) L \Delta T}{d}

    R = \dfrac{Ld(1 + \alpha_1 \Delta T)}{\delta_L} = \dfrac{Ld(1 + \alpha_1 \Delta T)}{L \Delta T (\alpha_2 - \alpha_1) } = \dfrac{d(1 + \alpha_1 \Delta T)}{\Delta T (\alpha_2 - \alpha_1)}

    Leading to:

    h = (R + 2d)(1 - \cos \theta) = \left(\dfrac{d(1 + \alpha_1 \Delta T)}{\Delta T (\alpha_2 - \alpha_1)} + 2d\right)\left(1 - \cos \dfrac{\delta_L}{d}\right) = \left(\dfrac{d(1 + \alpha_1 \Delta T)}{\Delta T (\alpha_2 - \alpha_1)} + 2d\right)\left(1 - \cos \dfrac{(\alpha_2 - \alpha_1) L \Delta T}{d}\right)

    If anyone can be bothered to check this it would be appreciated; I do realise it looks a bit complicated for how simple the problem seems!
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