Fluid stress tensor in cylindrical coordinates?

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SUMMARY

The discussion focuses on expressing the fluid stress tensor in cylindrical coordinates, specifically for a viscous fluid characterized by viscosity \(\mu\). The stress-strain relationship is defined as \(\sigma_{ij} = -p \delta_{ij} + 2 \mu u_{ij}\). The participants conclude that the cylindrical representation mirrors the rectangular case, with the indices \(i,j\) transitioning to \{r, \theta, z\}, resulting in components \(\sigma_{rr}, \sigma_{\theta\theta}, \sigma_{zz}\) for the pressure term. The discussion also touches on the comparison with the Maxwell stress tensor, noting the radial and tangential components.

PREREQUISITES
  • Understanding of fluid mechanics principles, particularly stress tensors
  • Familiarity with cylindrical coordinate systems
  • Knowledge of viscosity and its role in fluid dynamics
  • Basic understanding of the Maxwell stress tensor in electromagnetism
NEXT STEPS
  • Research the derivation of the stress tensor in cylindrical coordinates
  • Study the relationship between viscosity and stress in fluid mechanics
  • Explore the applications of the Maxwell stress tensor in electromagnetism
  • Examine the implications of non-viscous fluids in stress analysis
USEFUL FOR

Fluid mechanics students, researchers in theoretical physics, and engineers working with stress analysis in cylindrical systems will benefit from this discussion.

Peeter
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For fluid with viscosity \mu our stress strain relationship takes the form

\sigma_{ij} = -p \delta_{ij} + 2 \mu u_{ij}.

I was wondering how to express this in cylindrical coordinates. The strain tensor I can calculate in cylindrical coordinates (what I get matches eq 1.8 in [1]). But how would the \delta_{ij} portion of the stress strain relationship be expressed in cylindrical coordinates?

For example, if we considered a non-viscous fluid, the very simplest stress tensor, we have in rectangular coordinates

\sigma_{ij} = -p \delta_{ij}.

It's not obvious to me how this would be expressed in cylindrical form. I wanted to try some calculations with the traction vector T_i = \sigma_{ij} n_j in a cylindrical coordinate system, but I'm not sure how to express it. I figured the place to start was with the stress tensor.

References:

[1] L.D. Landau, EM Lifgarbagez, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. 1960.
 
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After some thought, I think the answer is that this would be exactly like the rectangular case, where for a non-viscious stress \sigma_{ij} = -p \delta_{ij}, the cylindrical representation would be exactly the same, with the only difference being that i,j have to index over \{r, \theta, z\}. So, we'd have a -p component for \sigma_{rr}, \sigma_{\theta\theta}, \sigma_{zz}.
 
Last edited:
Peeter said:
After some thought, I think the answer is that this would be exactly like the rectangular case, where for a non-viscious stress \sigma_{ij} = -p \delta_{ij}, the cylindrical representation would be exactly the same, with the only difference being that i,j have to index over \{r, \theta, z\}. So, we'd have a -p component for \sigma_{rr}, \sigma_{\theta\theta}, \sigma_{zz}.

I know nothing about fluid mechanics but comparing this tensor with Maxwell stress tensor in electromagnetism, you seem to be right. However in Maxwell stress tensor, only two components of matter, the radial and the tangential. Hence I expect an expression like the following:

\sigma_{rt}=-p\delta_{rt}+2\mu\mu_{rt}

r: radial
t: tangential

see http://en.wikipedia.org/wiki/Maxwell_stress_tensor
 

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