Stress tensor, hydrostatic stress tensor, deviatoric stress tensor, directional stress tensor, invariants of stress tensor and deviatoric stress tensor, principal stresses and shear stresses, von Mises stress, stress triaxiality, Lode angle, octahedral stresses and Mohr's diagram in Matlab for the case when all components of stress tensor are given.
The script is programmed with extended formulas for the purpose of stress-free export to other languages, if necessary.
$$
\boldsymbol{\sigma} =
\begin{bmatrix}
\sigma_x&\tau_{xy}&\tau_{xz}\\
\tau_{yx}&\sigma_y&\tau_{yz}\\
\tau_{zx}&\tau_{zy}&\sigma_z
\end{bmatrix}.
$$
$$\begin{gather}
\sigma_\text{v} = \text{tr}\boldsymbol{\sigma}, \\\
\sigma_\text{m} = \frac{1}{3}\sigma_\text{v}.
\end{gather}$$
Hydrostatic Stress Tensor $$
\boldsymbol{\sigma}_\text{hyd} = \sigma_\text{m}\mathbf{I}.
$$
$$
\boldsymbol{\sigma}' =\boldsymbol{\sigma} - \boldsymbol{\sigma}_\text{hyd}.
$$
Invariants of the Stress Tensor $$
\begin{bmatrix}
I_1 \\
I_2 \\
I_3
\end{bmatrix} =
\begin{bmatrix}
\text{tr} \boldsymbol{\sigma}\\
\frac{1}{2}\left[\text{tr}^2 \boldsymbol{\sigma} - \text{tr} \left(\boldsymbol{\sigma}^2\right)\right]\\
\det \boldsymbol{\sigma}
\end{bmatrix}.
$$
Invariants of the Deviatoric Stress Tensor $$
\begin{bmatrix}
J_1 \\
J_2 \\
J_3
\end{bmatrix}
=\begin{bmatrix}
\text{tr}\boldsymbol{\sigma}'\\
\frac{1}{2}\text{tr}\left(\boldsymbol{\sigma}'^2\right)\\
\det \boldsymbol{\sigma}'
\end{bmatrix}=\begin{bmatrix}
0\\
\frac{1}{3}I_1^2 - I_2\\
\frac{2}{27}I_1^3 - \frac{1}{3}I_1 I_2 + I_3
\end{bmatrix}.
$$
The principal stresses are eigenvalues of the stress tensor, therefore
$$\begin{gather}
\det\left(\boldsymbol{\sigma}-\sigma\mathbf{I}\right)=0\Rightarrow\sigma^3-I_1\sigma^2+I_2\sigma-I_3=0,\\\
\det\left(\boldsymbol{\sigma}'-\sigma'\mathbf{I}\right)=0\Rightarrow \sigma'^3-J_2\sigma'-J_3=0.
\end{gather}$$
Invariants in Terms of Prinicipal Stresses $$
\begin{bmatrix}
I_1 \\
I_2 \\
I_3
\end{bmatrix} =
\begin{bmatrix}
\sigma_1+\sigma_2+\sigma_3\\
\sigma_1\sigma_2+\sigma_2\sigma_3+\sigma_3\sigma_1\\
\sigma_1\sigma_2\sigma_3
\end{bmatrix},
$$
$$
\begin{bmatrix}
J_1 \\
J_2 \\
J_3
\end{bmatrix} =
\begin{bmatrix}
0\\
-\left(\sigma_1'\sigma_2'+\sigma_2'\sigma_3'+\sigma_3'\sigma_1'\right)\\
\sigma_1'\sigma_2'\sigma_3'
\end{bmatrix}.
$$
Von Mises Equivalent Stress $$
\sigma_\text{vM} = \sqrt{\frac{3}{2}\boldsymbol{\sigma}':\boldsymbol{\sigma}'}= \sqrt{3J_2}= \frac{1}{\sqrt{2}}\sqrt{\left(\sigma_1-\sigma_2\right)^2+\left(\sigma_2-\sigma_3\right)^2+\left(\sigma_3-\sigma_1\right)^2}.
$$
Closed-form solution for ordered eigenvalues (principal stresses)
$$\begin{gather}
\sigma_1\geq\sigma_2\geq\sigma_3,\\\
\sigma_1'\geq\sigma_2'\geq\sigma_3'
\end{gather}$$
can be obtained via Lode angle $\Theta$ (or azimuthal Lode angle $\bar\Theta$ ). Define Lode parameter $\mathcal{L}$ as
$$
\mathcal{L}=\frac{27}{2} \frac{J_3}{\sigma_\text{vM}^3}=\cos 3\Theta=-\sin 3\bar\Theta,
$$
then
$$\begin{gather}
\Theta = \frac{1}{3}\arccos \mathcal{L} \in \left[0,\dfrac{\pi}{3}\right],\\\
\bar\Theta = \Theta - \frac{\pi}{6} = -\frac{1}{3}\arcsin\mathcal{L} \in \left[-\frac{\pi}{6},\frac{\pi}{6}\right],
\end{gather}$$
so principal stresses $\sigma_1,\sigma_2,\sigma_3$
$$
\begin{bmatrix}
\sigma_1\\
\sigma_2\\
\sigma_3
\end{bmatrix} = \begin{bmatrix}
\sigma_\text{m}\\
\sigma_\text{m}\\
\sigma_\text{m}
\end{bmatrix} + \frac{2}{3}\sigma_\text{vM} \cos\begin{bmatrix}
\Theta\\
\Theta-\dfrac{2\pi}{3}\\
\Theta+\dfrac{2\pi}{3}
\end{bmatrix}= \begin{bmatrix}
\sigma_\text{m}\\
\sigma_\text{m}\\
\sigma_\text{m}
\end{bmatrix} + \frac{2}{3}\sigma_\text{vM} \sin\begin{bmatrix}
\bar\Theta+\dfrac{2\pi}{3}\\
\bar\Theta\\
\bar\Theta-\dfrac{2\pi}{3}
\end{bmatrix},
$$
and deviatoric principal stresses $\sigma_1',\sigma_2',\sigma_3'$
$$
\begin{bmatrix}
\sigma_1'\\
\sigma_2'\\
\sigma_3'
\end{bmatrix} = \begin{bmatrix}
\sigma_1\\
\sigma_2\\
\sigma_3
\end{bmatrix} - \begin{bmatrix}
\sigma_\text{m}\\
\sigma_\text{m}\\
\sigma_\text{m}
\end{bmatrix} = \frac{2}{3}\sigma_\text{vM} \cos\begin{bmatrix}
\Theta\\
\Theta-\dfrac{2\pi}{3}\\
\Theta+\dfrac{2\pi}{3}
\end{bmatrix} = \frac{2}{3}\sigma_\text{vM} \sin\begin{bmatrix}
\bar\Theta+\dfrac{2\pi}{3}\\
\bar\Theta\\
\bar\Theta-\dfrac{2\pi}{3}
\end{bmatrix}.
$$
$$
T_X = \frac{\sigma_\text{m}}{\sigma_\text{vM}}.
$$
$$\begin{gather}
\sigma_\text{oct.} = \sigma_\text{m},\\\
\tau_\text{oct.} = \sqrt{\frac{2}{3}J_2} = \frac{1}{3}\sqrt{\left(\sigma_1-\sigma_2\right)^2+\left(\sigma_2-\sigma_3\right)^2+\left(\sigma_3-\sigma_1\right)^2}.
\end{gather}$$
Directional Stress Tensor The directional stress tensor determines only the principal stresses and the relationship between the components of the stress tensor directions of stresses and the relationship between the components of the stress tensor, but doesn't determine their values, since the components of the directional stress tensor are dimensionless.
$$
\overline{\boldsymbol{\sigma}}' = \frac{1}{\tau_\text{oct.}}\boldsymbol{\sigma}'.
$$
$$
\sigma_\text{norm} = \left\lVert\boldsymbol{\sigma}\right\rVert = \sqrt{\boldsymbol{\sigma}:\boldsymbol{\sigma}}=\sqrt{\text{tr}\left(\boldsymbol{\sigma}^2\right)}.
$$
$$
\sigma_\text{tm} = \sqrt{\sigma_1^2+\sigma_2^2+\sigma_3^2}=\sqrt{\frac{1}{3}I_1^2+2J_2}.
$$
$$\begin{gather}
\begin{bmatrix}
\tau_{13}\\\
\tau_{12}\\\
\tau_{23}
\end{bmatrix}
=\frac{1}{2}
\begin{bmatrix}
\sigma_1-\sigma_3\\\
\sigma_1-\sigma_2\\\
\sigma_2-\sigma_3
\end{bmatrix},
\\\
\tau_\text{max} = \tau_{13}.
\end{gather}$$
The corresponding normal stresses $\sigma_{13}, \sigma_{12}, \sigma_{23}$ acting on the sections where
$\tau_{13}, \tau_{12}, \tau_{23}$ are acting, respectively, are
$$\begin{gather}
\begin{bmatrix}
\sigma_{13}\\\
\sigma_{12}\\\
\sigma_{23}
\end{bmatrix}
=\frac{1}{2}
\begin{bmatrix}
\sigma_1+\sigma_3\\\
\sigma_1+\sigma_2\\\
\sigma_2+\sigma_3
\end{bmatrix}
\end{gather}$$
Radii of circles:
$$
\begin{bmatrix}
R_1\\
R_2 \\
R_3
\end{bmatrix}=
\begin{bmatrix}
\tau_{23}\\
\tau_{13}\\
\tau_{12}
\end{bmatrix}.
$$
Centers of circles:
$$\begin{gather}
O_1=\left(\sigma_{23}, 0\right),\\\
O_2= \left(\sigma_{13}, 0\right),\\\
O_3= \left(\sigma_{12}, 0\right).
\end{gather}$$
In the script StressTensor.m the plotting of the Mohr's diagram for the tensor is considered
$$
\boldsymbol{\sigma} =
\begin{bmatrix}
-22.2 & 9.1 & 7.3 \\
9.1 & -16.9 & -4.6 \\
7.3 & -4.6 & 31.8 \\
\end{bmatrix}
$$
Chakrabarty, J.(2016). Theory of Plasticity: Third edition
Brannon, R.M. (2009). KAYENTA: Theory and User's Guide
Yu, M.-H. (2004). Unified Strength Theory and Its Applications. https://doi.org/10.1007/978-3-642-18943-2
Mattiello, A., & Desmorat, R. (2020). Lode angle dependency due to anisotropic damage. International Journal of Damage Mechanics, 105678952094856. https://doi.org/10.1177/1056789520948563
