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Deriving normal and shear stresses 
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#1
Dec1913, 07:12 AM

P: 1,863

Hi
When we talk about shear stresses in a fluid, we find that the shear stress is given by [tex] \tau_{xy} = \mu(\partial_y u + \partial_x v) = \tau_{yx} [/tex] 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 [tex] \tau_{xy} = \frac{1}{2}2\mu(\partial_y u + \partial_x v) = \mu(\partial_y u + \partial_x v) = \tau_{yx} [/tex] Applying the same logic to the normal stresses gives me [tex] \tau_{xx} = \frac{1}{2}\mu(\partial_x u + \partial_x u) = \mu(\partial_x u) [/tex] However, in my textbook (White) it is given as [tex] \tau_{xx} = 2\mu(\partial_x u) [/tex] Where does this extra factor of 2 come from in the normal stress? 


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