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At any point inside a structural element under deformation, we can always find three planes such that when we define a coordinate system with basis vectors lying on those planes, the shear stresses at that point are zero and the stress tensor is diagonal.

The principal stresses are the eigenvalues of the stress tensor, and are invariant of the rotation of the coordinate system.

This last feature on bold is recited everywhere.But my question is, so what? Why is it so useful to find these stresses whose components don't change with rotation of the coordinate system? And why would I even want to rotate my coordinate system?

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  • $\begingroup$ I wouldn't have stated three planes but rather four dimensions. I say this because any stresses are operational in time and are dimensional vectors $\endgroup$ – Rhodie Jun 8 at 12:46
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I'll take your word for it that your quote is "recited everywhere". The point of it is not that you can "rotate your coordinate system" just for fun, but that the principal stresses are independent of whatever coordinate system you choose to work in.

The physical structure doesn't know or care what coordinate systems you are using, so any conclusions you draw from your analysis about failure criteria, crack propagation directions, the onset of plasticity, etc, etc, must be independent of your choice of coordinate system - otherwise they are guaranteed to be wrong!

The principal stress values, and the direction of the coordinate system corresponding to them, at every point in the structure, are a complete description of the state of stress which is independent of the analyst's coordinate system(s), and also very easy to work with since the stress tensor is diagonal.

To give a couple of examples, in brittle materials cracks propagate in the direction of the largest (tensile) principal stress. In ductile materials, a useful failure criterion is the maximum shear stress acting in any direction at a point, and that is simply the difference between the maximum and minimum principal stresses.

Of course it is possible to devise formulas that reproduce those results in a different coordinate system, but they are much more complicated, and don't give much insight into what is going on in the structure.

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  • $\begingroup$ It seems I have some misunderstanding beyond the scope of this question, so I'm going to post another question on Physics SE perhaps. Thank you for the answer though, I will accept it as it has pointed me in the right direction. $\endgroup$ – S. Rotos Jun 8 at 21:26
  • $\begingroup$ Upvoting the answer once wasn't enough, this is the crux of the answer. $\endgroup$ – Sam Farjamirad Jun 20 at 15:59

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