1
$\begingroup$

I am trying to get an intuitive understanding of the normal components of the stress tensor for a Newtonian Fluid. The stress on the x face in the x direction is $$\sigma_{xx} = 2*\mu\frac{\partial u}{\partial x} $$ Where u is the velocity in the x direction.
I am trying to figure why this is the case. In solid mechanics, the shear stress for an element under pure tension is maximum at 45$^o$ (this is evident from Mohr's circle. The resulting shear stress is $\frac{1}{2}\sigma$ with $\sigma$ being the normal stress applied.
So, coming back to the fluid "realm",I would then think (With Newton's Viscosity Law in mind) that the normal stress would be related to the shear stress by: $$\sigma_{xx} = \frac{1}{2}*\mu\frac{\partial u}{\partial x} $$ Where $\frac{1}{2}*\frac{\partial u}{\partial x}$ is the shear rate along the 45$^o$ plane.
It would help if someone could explain where my logic is off in terms of basic Newtonian mechanics and not tensor math.

$\endgroup$
3

1 Answer 1

2
$\begingroup$

$\tau_{xx}$ is the $x$-component of the stress on a small area element whose normal is in the $x$ direction. Likewise, $\tau_{xy}$ is the $y$-component of the same stress. When Newton's law of viscosity is first introduced the flow is uniform along the $x$-direction but is shown to vary along the $y$-direction. That is why, we see an expression like \begin{equation} \mu\frac{du}{dy}. \end{equation} This force acts in the $y$ direction. When $u$ varies in $x$ direction as well there will be a contribution from the partial derivative with respect to $x$ as well. In general, the velocity gradient is \begin{equation} \gamma_{ij} = \frac{\partial u_j}{\partial x_i}. \end{equation} We decompose it as \begin{equation} \gamma_{ij} = \frac{1}{2}\left(\frac{\partial u_j}{\partial x_i} + \frac{\partial u_i}{\partial x_j}\right) + \frac{1}{2}\left(\frac{\partial u_j}{\partial x_i} - \frac{\partial u_i}{\partial x_{ij}}\right) = e_{ij} + \Omega_{ij}. \end{equation} The second term is the vorticity tensor and it represents the "rigid rotation" of a fluid element. The first terms is the rate of strain tensor and is related to viscous stress.

Your next question is about symmetry of the stress tensor. A mathematical answer could be that the symmetry follows from $\tau_{ij} = \mu e_{ij}$. Since $e_{ij}=e_{ji}$ so is $\tau_{ij}=\tau_{ji}$. However, the stress tensor is always symmetric even though it is not related to the rate of strain tensor in this way. Its symmetry follows from the law of conservation of angular momentum. You may refer to G. K. Batchelor's Fluid Dynamics or any other book on continuum mechanics for more details. There is also a Wikipedia page on the topic.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.