# Deriving normal and shear stresses

by Niles
Tags: deriving, normal, shear, stresses
 P: 1,863 Hi 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?