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Stress tensor, hydrostatic stress tensor, deviatoric stress tensor, invariants of stress tensor and deviatoric stress tensor, principal stresses, von Mises stress and Mohr's diagram in Matlab for the case when all components of stress tensor are given.

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Stress Tensor in Matlab

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.

Table of Contents

Theoretical background

Cauchy Stress Tensor

$$ \boldsymbol{\sigma} = \begin{bmatrix} \sigma_x&\tau_{xy}&\tau_{xz}\\ \tau_{yx}&\sigma_y&\tau_{yz}\\ \tau_{zx}&\tau_{zy}&\sigma_z \end{bmatrix}. $$

Volumetric & Mean Stress

$$\begin{gather} \sigma_\text{v} = \text{tr}\boldsymbol{\sigma}, \\\ \sigma_\text{m} = \frac{1}{3}\sigma_\text{v}. \end{gather}$$

Hydrostatic Stress Tensor

$$ \boldsymbol{\sigma}_\text{hyd} = \sigma_\text{m}\mathbf{I}. $$

Deviatoric Stress Tensor

$$ \boldsymbol{\sigma}' =\boldsymbol{\sigma} - \boldsymbol{\sigma}_\text{hyd}. $$

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

$$ \begin{bmatrix} J_1 \\ J_2 \\ J_3 \end{bmatrix} =\begin{bmatrix} \text{tr}\boldsymbol{\sigma}'\\ \frac{1}{2}\text{tr}\left(\boldsymbol{\sigma}'^2\right)\\ \det \boldsymbol{\sigma}' \end{bmatrix}=\begin{bmatrix} 0\\ \frac{1}{3}I_1^2 - I_2\\ \frac{2}{27}I_1^3 - \frac{1}{3}I_1 I_2 + I_3 \end{bmatrix}. $$

Principal Stresses

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

$$ \sigma_\text{vM} = \sqrt{\frac{3}{2}\boldsymbol{\sigma}':\boldsymbol{\sigma}'}= \sqrt{3J_2}= \frac{1}{\sqrt{2}}\sqrt{\left(\sigma_1-\sigma_2\right)^2+\left(\sigma_2-\sigma_3\right)^2+\left(\sigma_3-\sigma_1\right)^2}. $$

Lode Angle Dependency

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 $\sigma_1',\sigma_2',\sigma_3'$

$$ \begin{bmatrix} \sigma_1'\\ \sigma_2'\\ \sigma_3' \end{bmatrix} = \begin{bmatrix} \sigma_1\\ \sigma_2\\ \sigma_3 \end{bmatrix} - \begin{bmatrix} \sigma_\text{m}\\ \sigma_\text{m}\\ \sigma_\text{m} \end{bmatrix} = \frac{2}{3}\sigma_\text{vM} \cos\begin{bmatrix} \Theta\\ \Theta-\dfrac{2\pi}{3}\\ \Theta+\dfrac{2\pi}{3} \end{bmatrix} = \frac{2}{3}\sigma_\text{vM} \sin\begin{bmatrix} \bar\Theta+\dfrac{2\pi}{3}\\ \bar\Theta\\ \bar\Theta-\dfrac{2\pi}{3} \end{bmatrix}. $$

Stress Triaxiality

$$ T_X = \frac{\sigma_\text{m}}{\sigma_\text{vM}}. $$

Octahedral Stress

$$\begin{gather} \sigma_\text{oct.} = \sigma_\text{m},\\\ \tau_\text{oct.} = \sqrt{\frac{2}{3}J_2} = \frac{1}{3}\sqrt{\left(\sigma_1-\sigma_2\right)^2+\left(\sigma_2-\sigma_3\right)^2+\left(\sigma_3-\sigma_1\right)^2}. \end{gather}$$

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.

$$ \overline{\boldsymbol{\sigma}}' = \frac{1}{\tau_\text{oct.}}\boldsymbol{\sigma}'. $$

Stress Norm

$$ \sigma_\text{norm} = \left\lVert\boldsymbol{\sigma}\right\rVert = \sqrt{\boldsymbol{\sigma}:\boldsymbol{\sigma}}=\sqrt{\text{tr}\left(\boldsymbol{\sigma}^2\right)}. $$

Stress Total Measure

$$ \sigma_\text{tm} = \sqrt{\sigma_1^2+\sigma_2^2+\sigma_3^2}=\sqrt{\frac{1}{3}I_1^2+2J_2}. $$

Principal Shear Stresses

$$\begin{gather} \begin{bmatrix} \tau_{13}\\\ \tau_{12}\\\ \tau_{23} \end{bmatrix} =\frac{1}{2} \begin{bmatrix} \sigma_1-\sigma_3\\\ \sigma_1-\sigma_2\\\ \sigma_2-\sigma_3 \end{bmatrix}, \\\ \tau_\text{max} = \tau_{13}. \end{gather}$$

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

$$\begin{gather} \begin{bmatrix} \sigma_{13}\\\ \sigma_{12}\\\ \sigma_{23} \end{bmatrix} =\frac{1}{2} \begin{bmatrix} \sigma_1+\sigma_3\\\ \sigma_1+\sigma_2\\\ \sigma_2+\sigma_3 \end{bmatrix} \end{gather}$$

Mohr's Diagram

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:

$$\begin{gather} O_1=\left(\sigma_{23}, 0\right),\\\ O_2= \left(\sigma_{13}, 0\right),\\\ O_3= \left(\sigma_{12}, 0\right). \end{gather}$$

Example

In the script StressTensor.m the plotting of the Mohr's diagram for the tensor is considered

$$ \boldsymbol{\sigma} = \begin{bmatrix} -22.2 & 9.1 & 7.3 \\ 9.1 & -16.9 & -4.6 \\ 7.3 & -4.6 & 31.8 \\ \end{bmatrix} $$

The Mohr's Diagram

References

  1. Chakrabarty, J.(2016). Theory of Plasticity: Third edition
  2. Brannon, R.M. (2009). KAYENTA: Theory and User's Guide
  3. Yu, M.-H. (2004). Unified Strength Theory and Its Applications. https://doi.org/10.1007/978-3-642-18943-2
  4. Mattiello, A., & Desmorat, R. (2020). Lode angle dependency due to anisotropic damage. International Journal of Damage Mechanics, 105678952094856. https://doi.org/10.1177/1056789520948563

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Stress tensor, hydrostatic stress tensor, deviatoric stress tensor, invariants of stress tensor and deviatoric stress tensor, principal stresses, von Mises stress and Mohr's diagram in Matlab for the case when all components of stress tensor are given.

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