# Deriving normal and shear stresses

1. Dec 19, 2013

### Niles

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?