# shear and normal stresses

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?

$$\tau_{ij} = \mu \left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right)$$
So, if you look at $\tau_{ii}$, you get $2\frac{\partial u_i}{\partial x_i}$. That's really where the factors come from, not from "averaging" over any fluid element or anything like that. It's just due to the symmetry of the tensor.
• So you are adding the angular deformation for adjacent sides in $\tau_{ij}$, but in $\tau_{ii}$ you are adding opposite sides. Why is that? IMO, seeing as there are no adjacent sides to $\tau_{ii}$, it should merely be $\partial u_i/\partial x_i$ – BillyJean Dec 20 '13 at 8:30
• I'm not adding any sides in anything. This is the definition of the stress tensor on each face. So if you wanted to take the average for some reason, you would get $1/2 (2*\tau_{ij})$ for all $i,j$ which gives you the correct answer also. You're confusing two different things -- the balance of the stress tensor across a differential fluid element, and the definition of the stress tensor. – tpg2114 Dec 20 '13 at 14:48