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Deriving normal and shear stresses

by Niles
Tags: deriving, normal, shear, stresses
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Niles
#1
Dec19-13, 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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