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

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

1 Answer 1


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:



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


Does this make sense?


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