4
$\begingroup$

The rate of strain tensor is given as $$e_{ij} = \frac{1}{2}\Big[\frac{\partial v_i}{\partial x_j}+ \frac{\partial v_j}{\partial x_i}\Big]$$ where $v_i$ is the $i$th component of the velocity field and $x_i$ is the $i$th component of the position vector. From what I read, I understand that $e_{ij}$ is the rate of strain tensor or the symmetric part of the deformation tensor i.e $\nabla \bf{v}$. The rate of strain tensor can be decomposed in the following form: $$e_{ij} = [e_{ij} - \frac{1}{3}e_{kk}\delta_{ij}] + \frac{1}{3}e_{kk}\delta_{ij} $$ From what I could gather, $e_{kk}$ can be written as $\nabla \cdot \bf{v}$ which represents the pure volumetric expansion of a fluid element and the first term is some kind of strain which does not encompass volumetric change. Is this correct or is there more to it. What is the correct physical interpretation for it, and why is it useful?

Further more I read that any such symmetric part of tensor can be decomposed into a “isotropic” part and an “anisotropic” part. I am unable to understand Why we can do this and what it represents physically. I would like to have a mathematical as well as a physical understanding for this sort of decomposition. I am very new to tensors and fluid mechanics and would like to have a complete understanding of this. Thank you for the answers.

$\endgroup$
  • 1
    $\begingroup$ Here is a suggestion. Write it out in component form for the case of the principal directions of the rate of strain tensor. This will eliminate the cross terms, and will give you better insight into how the terms play out. $\endgroup$ – Chet Miller Jul 26 '18 at 16:00
4
$\begingroup$

There are many different answers to your question (since usefulness is subjective), but here's what I would consider the "main" one.

Very often we assume fluids are incompressible: that is, that the density $\rho$ is constant, and consequently $\nabla \cdot \mathbf{v} = 0$ from the mass continuity equation. By splitting the strain rate tensor $\bf{D}$ into a sum of an isotropic tensor $\mathbf{P}$ and a trace-less deviatoric tensor $\mathbf{S}$,

$$\mathbf{D} = \mathbf{P} + \mathbf{S} = \frac{1}{3}\text{tr}(\mathbf{D})\mathbf{I} + \left(\mathbf{D} - \frac{1}{3}\text{tr}(\mathbf{D})\mathbf{I}\right) = \frac{1}{3}(\nabla\cdot\mathbf{v})\mathbf{I} + \mathbf{S}$$ we can isolate the source of compressibility effects as $\mathbf{P}$ and ignore it in the case where $\rho$ is constant, simplifying constitutive equations considerably.

This can be useful, for example, to give us a straightforward way to mathematically analyze the behavior of fluids in the regime where they become slightly compressible: we know the effects will show up in the strain rate tensor as an extra diagonal term $\epsilon \mathbf{I}$ where $\epsilon \ll 1$, and we can use perturbation theory to see how compressibility propagates into the mechanics.

From a more general perspective, when formulating constitutive laws involving tensors of arbitrary type in classical mechanics, we seek to formulate such laws so that they satisfy objectivity (Galilean transformation invariance). Such laws can only depend on the invariants of tensors, and as a result it's useful to isolate the terms which depend on each individual invariant, of which the trace is one.

$\endgroup$
  • 1
    $\begingroup$ I think your equation should read: $$\mathbf{D} = \mathbf{P} + \mathbf{S} = \frac{1}{3}\text{tr}(\mathbf{D})\mathbf{I} + \left(\mathbf{D} - \frac{1}{3}\text{tr}(\mathbf{D})\mathbf{I}\right)$$ $\endgroup$ – Chet Miller Jul 26 '18 at 21:42
  • 1
    $\begingroup$ You’re absolutely right! Editing now. $\endgroup$ – aghostinthefigures Jul 26 '18 at 22:01
3
$\begingroup$

In principal component form, $$D_{11}=\frac{1}{3}\left[\frac{\partial v_1}{\partial x_1}+\frac{\partial v_2}{\partial x_2}+\frac{\partial v_3}{\partial x_3}\right]+\left[\frac{1}{3}\left(\frac{\partial v_1}{\partial x_1}-\frac{\partial v_2}{\partial x_2}\right)+\frac{1}{3}\left(\frac{\partial v_1}{\partial x_1}-\frac{\partial v_3}{\partial x_3}\right)\right]$$ The first term in brackets represents the isotropic expansion/compression contribution to the rate of deformation tensor. The two terms in the second brackets can be interpreted as non-isotropic "pure shear" deformation contributions to the rate of deformation tensor. This same type of pure shear kinematics is encountered in the interpretation of solid mechanics deformations. Google "pure shear" in solid mechanics.

https://www.google.com/search?q=pure+shear

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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