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Equation of principal angles:

$$\tan 2\theta_p=\frac{2\tau_{xy}}{\sigma_x-\sigma_y}$$

Equation of principal stresses: $$\sigma_{max}, \sigma_{min} = {\sigma_{xx} + \sigma_{yy} \over 2} \pm \sqrt{ \left( {\sigma_{xx} - \sigma_{yy} \over 2} \right)^2 + \tau_{xy}^2 }$$

Source of equations: Lectures notes on Mechanics of solids, Course code- BME-203, prepared by Prof. P.R.Dash, page 45 and 46.

Above is the equation used for finding the principal angles corresponding to the two principal planes where principal stresses (maximum and minimum stresses) occur.

In solid Mechanics, the difference between the two values of principal angles is $90^\circ$. Why is it equal to $90^\circ$?

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    $\begingroup$ Please provide the source of your equation. $\endgroup$ – Bob D Jul 24 '20 at 17:43
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The answer is probably because the stress tensor is symmetric $\sigma_{ij}=\sigma_{ji}$, and the principal (not principle!) planes are perpendicular to the eigenvectors, which for a symmetric matrix are always mutually perpendicular. Note that the equation you added for the pricipal stresses is indeed the equation for the eigenvalues of the matrix $$ \left[\matrix{\sigma_{xx}& \sigma_{xy}\cr \sigma_{yx} & \sigma_{yy}}\right] $$ This is assuming that for some reason you have written $\tau_{xy}$ for the shear stress $\sigma_{xy}$.

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  • $\begingroup$ Thank you for your answer. Unfortunately, the textbook I'm learning from has mistakenly used the term principle instead of principal, so thank you for the remark. Please can you explain to me what are the eigenvectors in this case ? $\endgroup$ – user260844 Jul 24 '20 at 21:20
  • $\begingroup$ Thank you for the edit. The condition of symmetry is not enough for the eigenvectors to be perpendicular, but we will need to see that the eigenvalues are of different values. Also, how do we know that the principal stresses are the eigenvectors of the stress tensor? Is it because they contain the elements of the stress tensor ? $\endgroup$ – user260844 Jul 26 '20 at 16:48
  • $\begingroup$ The pricipal stress are by definition the eigenvectors. See en.wikipedia.org/wiki/… $\endgroup$ – mike stone Jul 26 '20 at 17:00

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