In a previous tutorial, Mohr's Circle for Stresses, we learned about how we could use 2D transformations to find the maximum shear stress as well as the principal stresses, or maximum normal stresses as well as their directions. Now, when principal stresses are present, there are no shear stresses because they go to zero! Let's learn how to find principal stresses.
Principal stresses and their directions can be found similar to the process of finding eigenvalue and eigenvectors for a system. We can start out by defining our stress field as
[ 1 ]
$$ [\sigma] = \begin{bmatrix} \sigma_x & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_y & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_z \end{bmatrix} $$
and then taking the determinant of our stress field, $ [\sigma] $. The determinant can be taken by subtracting $ [\sigma] $ by $ \bar{\sigma} [I] $, which is a placeholder value for the principal stresses times the identity matrix. Doing this, we get that
$$ \det( [\sigma] - \bar{\sigma} [I] ) = 0 $$
$$ \Rightarrow \det \begin{pmatrix} \begin{bmatrix} \sigma_x - \bar{\sigma} & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_y - \bar{\sigma} & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_z - \bar{\sigma} \end{bmatrix} \end{pmatrix} = 0 $$
From this, we get a very intense looking equation, but we will get more used to it as we see it more! Below, Eq. (2) is called the characteristic equation.
[ 2 ]
$$ (\sigma_x - \bar{\sigma})[(\sigma_y - \bar{\sigma})(\sigma_z - \bar{\sigma}) - \tau_{yz} \tau_{zy}] - \tau_{xy}[\tau_{yx}(\sigma_z - \bar{\sigma}) - \tau_{yz} \tau_{zx}] + \tau_{xz}[\tau_{yx} \tau_{zy} - \tau_{zx} (\sigma_y - \bar{\sigma})] = 0 $$
Eq. (2) can be solved for the three $ \bar{\sigma} $'s; these will then be assigned to $ \sigma_I $, $ \sigma_{II} $, and $ \sigma_{III} $, where $ \sigma_I > \sigma_{II} > \sigma_{III} $. Now we have found the principal stresses! The only thing left is to find their principal directions.
The way this is done is by writing out the original $ [\sigma] - \bar{\sigma} [I] $, but by inserting the value for $ \sigma_I $ for $ \bar{\sigma} $. This gives us
$$ \begin{bmatrix} \sigma_x - \sigma_I & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_y - \sigma_I & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_z - \sigma_I \end{bmatrix} $$
We can then multiply this by a normal vector
[ 3 ]
$$ \vec{n} = \begin{Bmatrix} n_x \\ n_y \\ n_z \end{Bmatrix} $$
and set it equal to a zero vector. Leaving us with
[ 4 ]
$$ \begin{bmatrix} \sigma_x - \sigma_I & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_y - \sigma_I & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_z - \sigma_I \end{bmatrix} \begin{Bmatrix} n_x \\ n_y \\ n_z \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \\ 0 \end{Bmatrix} $$
All that is left is to solve the system of equations for $ n_x $, $ n_y $, and $ n_z $, then the direction for the principal stress, $ \sigma_I $ will be known. This will need to be done for all 3 stresses!