# Viscous stress tensor for an incompressible Newtonian fluid

I am reading the cengel book and found that the viscous stress tensor for an incompressible Newtonian fluid with constant properties is given by: $$\tau_{ij} = 2 \mu \epsilon_{ij}$$ where $$\epsilon_{ij}$$ is the strain rate tensor. My question is if there is a derivation of this property or any reference that can demonstrate it

• It is the generalized form of a Newtonian fluid description. The shear stress is proportional to the shear strain rate.
– Fire
Commented Jul 8, 2022 at 3:26

It's convenient when modeling viscosity to relate the shear stress $$\tau$$ to the rate of change $$\frac{d\gamma}{dt}$$ in a corner angle that was originally 90°, as this parameter is easily accessible in experiments:

$$\tau\sim\frac{d\gamma}{dt}=\dot\gamma;$$

$$\tau=\mu\dot\gamma.$$

We call the constant of proportionality the viscosity $$\mu$$.

It's also convenient to model strains as

$$\varepsilon_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)$$

because this compact expression works for both normal and shear strains:

$$\varepsilon_{11}=\frac{1}{2}\left(\frac{\partial u_1}{\partial x_1}+\frac{\partial u_1}{\partial x_1}\right)=\frac{\partial u_1}{\partial x_1};$$

$$\varepsilon_{22}=\frac{1}{2}\left(\frac{\partial u_2}{\partial x_2}+\frac{\partial u_2}{\partial x_2}\right)=\frac{\partial u_2}{\partial x_2};$$

$$\varepsilon_{12}=\frac{1}{2}\left(\frac{\partial u_1}{\partial x_2}+\frac{\partial u_2}{\partial x_1}\right);$$

for example. We actually get all nine components of the strain tensor this way.

You can prove to yourself through diagrams and trigonometry that the tensorial shear strain $$\varepsilon_{12}$$ ends up being one-half the so-called engineering shear strain $$\gamma$$. For this reason, we need to insert a factor of two in the original equation to obtain

$$\tau_{ij}=2\mu\dot\varepsilon_{ij}.$$

Does this make sense?