# Why are the principal planes where principal stresses occur perpendicular to each other?

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 $$90^\circ$$?

• Please provide the source of your equation. – Bob D Jul 24 '20 at 17:43

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 $$\sigma_{xy}$$.