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Deriving normal and shear stresses

  1. Dec 19, 2013 #1

    When we talk about shear stresses in a fluid, we find that the shear stress is given by
    \tau_{xy} = \mu(\partial_y u + \partial_x v) = \tau_{yx}
    This relation we get when only looking at one side of our fluid-"cube". Now, in order to take into account the opposite side we assume that the fluid element is so small that the shear stress is constant, leading to the average
    \tau_{xy} = \frac{1}{2}2\mu(\partial_y u + \partial_x v) = \mu(\partial_y u + \partial_x v) = \tau_{yx}
    Applying the same logic to the normal stresses gives me
    \tau_{xx} = \frac{1}{2}\mu(\partial_x u + \partial_x u) = \mu(\partial_x u)
    However, in my textbook (White) it is given as
    \tau_{xx} = 2\mu(\partial_x u)
    Where does this extra factor of 2 come from in the normal stress?
  2. jcsd
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