0
$\begingroup$

Let's define :

$$ \sigma_{ij}=\tilde{\sigma_{ij}} - P \delta_{ij} $$ with $\sigma$ the stress tensor, $P$ the pressure and $\tilde{\sigma}$ the anisotropic stress tensor.

The balance of force imposes : $$ \partial_{i}\tilde{\sigma}_{ij} - \partial_jP=0 $$.

My question is do anisotropic stresses cancel at the center because of the symmetries. For example, for a fluid in spherical coordinates and radial symmetry we have :

$$ \tilde{\sigma}_{rr}\sim\tilde{\sigma}_{\theta\theta}\sim \tilde{\sigma}_{\phi\phi} \sim \partial_rv-v/r$$

Do we have at the center of the sphere : $\partial_rv-v/r=0$ ?

I'm asking the question because of some intuition that anisotropic stresses have some directions, and that's not consistant with the center of symmetry, I mean we will have a singularity at the center. Is it right ? And if not does anisotropic stress resect certain conditions at the center ?

$\endgroup$
2
  • $\begingroup$ What is the "constraints tensor"? $\endgroup$
    – nicoguaro
    Jul 11 '19 at 15:01
  • $\begingroup$ Sorry I used the French word by mistake :) $\endgroup$
    – J.A
    Jul 11 '19 at 15:03
1
$\begingroup$

If you consider a problem with spherical symmetry the balance of forces is given by

$$\frac{\partial \sigma_{rr}}{\partial r} + \frac{2}{r}[\sigma_{rr} - \sigma_{\phi\phi}] + f_r = 0\, ,$$

when $\sigma_{\phi\phi} = \sigma_{\theta\theta}$ have been used.

We also have the following (non-trivial) deformation components

\begin{align} &\epsilon_{rr} =\frac{\mathrm{d} u_r}{\mathrm{d} r}\, \\ &\epsilon_{\phi\phi} = \epsilon_{\theta\theta} =\frac{ u_r}{r}\, . \end{align}

This leads to the following

$$ \sigma_{rr} = \frac{2 \lambda u_r}{r} + \left(\lambda + 2 \mu\right) \frac{d}{d r} u_r\, ,\\ \sigma_{\phi\phi} = \sigma_{\theta\theta} = \lambda \frac{d}{d r} u_r + \frac{\lambda u_r}{r} + \frac{\left(\lambda + 2 \mu\right) u_r}{r}\, . $$

I suppose that what you call "anisotropic stress" is the deviatoric part of the stress tensor, i.e.,

$$\tilde{\sigma} = \frac{2}{3} \mu \left(\frac{d}{d r} u_r - \frac{u_r}{r}\right) \begin{pmatrix} 2 &0 &0\\ 0 &-1 &0\\ 0 &0 &-1\end{pmatrix}\, .$$

And I don't see that clear what you suggest.

$\endgroup$
5
  • $\begingroup$ What if u = kr ? $\endgroup$ Jul 12 '19 at 1:19
  • 1
    $\begingroup$ @ChetMiller, in that case, the deviatoric would be zero everywhere. That case corresponds to a sphere with uniform pressure applied in the exterior. $\endgroup$
    – nicoguaro
    Jul 12 '19 at 15:34
  • $\begingroup$ I think you're writing the stress for an elastic solid and I wrote it for a fluid, don't you ? $\endgroup$
    – J.A
    Nov 23 '19 at 22:31
  • 1
    $\begingroup$ @J.A, yes I wrote for an elastic solid. In your question it was never implied that you wanted it for a fluid (Newtonian?). $\endgroup$
    – nicoguaro
    Nov 23 '19 at 22:54
  • 1
    $\begingroup$ I wrote the stress in function of a velocity, so I had in mind more a fluid, but actually I think the answer should be the same. So my question is $\tilde{\sigma}(r=0)=0$ ? I mean does the anisotropy implies something concerning the anisotropic stress at the center ? $\endgroup$
    – J.A
    Nov 23 '19 at 22:58

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.