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Equilibrium equation for forces in tangential direction

  1. Oct 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Derive the differential equilibrium equation in the tangential direction [itex]\theta[/itex]


    2. Relevant equations
    [itex]\sum F_\theta = 0[/itex]

    [itex]V = 1/2(2r + \Delta r)\Delta r \Delta \theta[/itex]

    Small angle approximation:
    [itex]
    \cos(\frac{\Delta \theta}{2}) = 1
    [/itex]

    [itex]
    \sin(\frac{\Delta \theta}{2}) = \frac{\Delta \theta}{2}
    [/itex]

    3. The attempt at a solution

    [itex]
    (\sigma_{\theta \theta} + \Delta \sigma_{\theta \theta})\cos(\frac{\Delta \theta}{2})\Delta r(1) - \sigma_{\theta \theta}\cos(\frac{\Delta \theta}{2})\Delta r(1) + (\sigma_{r\theta} + \Delta \sigma_{r\theta})\sin(\frac{\Delta \theta}{2})\Delta r(1) - \sigma_{r\theta}\sin(\frac{\Delta \theta}{2})\Delta r(1) + (\sigma_{r\theta} + \Delta\sigma_{r\theta})(r + \Delta r)\Delta \theta - \sigma_{r\theta}r\Delta \theta + \Theta V = 0
    [/itex]

    Simplifying:
    [itex]
    \Delta \sigma_{\theta \theta} \Delta r + \Delta \sigma_{r \theta} \frac{\Delta \theta}{2} \Delta r + \sigma_{r \theta} \Delta r \Delta \theta + \Delta \sigma_{r \theta} r \Delta \theta + \Delta \sigma_{r \theta}\Delta r \Delta \theta + 1/2\Theta(2r + \Delta r)\Delta r \Delta \theta = 0
    [/itex]


    Divide by [itex]\Delta r \Delta \theta[/itex]

    [itex]
    \frac{\Delta \sigma_{\theta \theta}}{\Delta \theta} + \frac{\Delta \sigma_{r \theta}}{2} + \sigma_{r \theta} + \frac{\Delta \sigma_{r \theta}r}{\Delta r} + \Delta \sigma_{r \theta} + 1/2\Theta(2r + \Delta r) = 0
    [/itex]

    Take the limit as [itex] \Delta r[/itex] approaches 0 and [itex]\Delta \theta[/itex] approaches 0

    [itex]
    \frac{\partial \sigma_{\theta \theta}}{\partial \theta} + (?) + \sigma_{r \theta} + \frac{\partial \sigma_{r \theta}r}{\partial r} + (?) + r\Theta = 0
    [/itex]

    [itex]
    \frac{1}{r} \frac{\partial \sigma_{\theta \theta}}{\partial \theta} + \frac{1}{r}(?) + \frac{1}{r} \sigma_{r \theta} + \frac{\partial \sigma_{r \theta}}{\partial r} + \frac{1}{r}(?) + \Theta = 0
    [/itex]

    I searched online for the correction equation and it is as follows:

    [itex]
    \frac{1}{r} \frac{\partial \sigma_{\theta \theta}}{\partial \theta} + \frac{\partial \sigma_{r \theta}}{\partial r} + \frac{2\sigma_{r \theta}}{r} + \Theta = 0
    [/itex]

    I'm not sure what to do about the terms with no [itex]\Delta[/itex] in the denominator when I take the limit. Attached is the figure I'm working from.
     

    Attached Files:

  2. jcsd
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