Deviatoric stress and invariants

Posted by: Pantelis Liolios | Sept. 16, 2020

The stress tensor can be expressed as the sum of two stress tensors, namely: the hydrostatic stress tensor and the deviatoric stress tensor. In this article we will define the hydrostatic and the deviatoric part of the stress tensor and we will calculate the invariants of the stress deviator tensor. The invariants of the deviatoric stress are used frequently in failure criteria.

Consider a stress tensor \( \sigma_{ij} \) acting on a body. The stressed body tends to change both its volume and its shape. The part of the stress tensor that tends to change the volume of the body is called mean hydrostatic stress tensor or volumetric stress tensor. The part that tends to distort the body is called stress deviator tensor. Hence, the stress tensor may expressed as:

\[ \sigma_{ij}=s_{ij}+p\delta_{ij} \]
(1)

where \( \delta_{ij} \) is the Kronecker delta (with \( \delta_{ij}=1 \) if \( i=j \) and \( \delta_{ij}=0 \) if \( i\neq j \) ), \( p \) is the mean stress given by:

\[ p=\frac{1}{3}\sigma_{kk}=\frac{1}{3}\left(\sigma_{11}+\sigma_{22}+\sigma_{33}\right)=\frac{1}{3}I_{1} \]
(2)

where \( I_{1} \) is the first invariant of the stress tensor (see also: Principal stresses and stress invariants). The product \( p\delta_{ij} \) is the hydrostatic stress tensor and contains only normal stresses. The deviatoric stress tensor can be obtained by subtracting the hydrostatic stress tensor from the stress tensor:

\[ \begin{array}{rl}s_{ij}=&\sigma_{ij}-p\delta_{ij}\\=&\left[\begin{array}{ccc}\sigma_{11}-p & \sigma_{12} & \sigma_{13}\\ \sigma_{21} & \sigma_{22}-p & \sigma_{23}\\ \sigma_{31} & \sigma_{32} & \sigma_{33}-p\end{array}\right]\end{array} \]
(3)

In order to calculate the invariants of the stress deviator tensor we will follow the same procedure used in the article Principal stresses and stress invariants. It must be mentioned that the principal directions of the stress deviator tensor coincide with the principal directions of the stress tensor. The characteristic equation for \( s_{ij} \) is:

\[ \left| s_{ij}-s\delta_{ij}\right|=s^{3}-J_{1}s^{2}-J_{2}s-J_{3}=0 \]
(4)

where \( J_{1} \), \( J_{2} \) and \( J_{3} \) are the first, second and third deviatoric stress invariants, respectively. The roots of the polynomial are the three principal deviatoric stresses \( s_{1} \), \( s_{2} \) and \( s_{3} \). \( J_{1} \), \( J_{2} \) and \( J_{3} \) may be calculated by the following expressions:

\[ \begin{array}{rl}J_{1}=&s_{kk}=0\\J_{2}=&\frac{1}{2}s_{ij}s_{ji}\\=&\frac{1}{6}\left[\left(\sigma_{11}-\sigma_{22}\right)^2+\left(\sigma_{22}-\sigma_{33}\right)^2+\left(\sigma_{33}-\sigma_{11}\right)^2\right]\\&+\sigma_{12}^2+\sigma_{23}^2+\sigma_{31}^2\\=&\frac{1}{3}I_{1}^{2}-I_{2}\\J_{3}=&\det(s_{ij})\\=&\frac{1}{3}s_{ij}s_{jk}s_{ki}\\=&\frac{2}{27}I_{1}^{3}-\frac{1}{3}I_{1}I_{2}+I_{3}\end{array} \]
(5)

where \( I_{1} \), \( I_{2} \) and \( I_{3} \) are the three invariants of the stress tensor and \( \det(s_{ij}) \) is the determinant of \( s_{ij} \). It should be mentioned that since \( J_{1}=s_{kk}=0 \), the stress deviator tensor describes a state of pure shear.

Example

Calculate the stress deviator tensor and its invariants for the following stress tensor:

\[ \sigma_{ij}=\left[\begin{array}{ccc}2&-3&4\\-3&-5&1\\4&1&6\end{array}\right] \]
(6)

Show solution...

Firstly we calculate the mean pressure \( p \):

\[ p=\frac{1}{3}\left(2-5+6\right)=1 \]
(7)

From equation (3) we calculate the stress deviator tensor:

\[ s_{ij}=\left[\begin{array}{ccc}1&-3&4\\-3&-6&1\\4&1&5\end{array}\right] \]
(8)

For the stress deviator tensor invariants we will use equations (5) and we get:

\[ \begin{array}{rl}J_{1}=&0\\J_{2}=&\frac{1}{2}s_{ij}s_{ji}\\=&\frac{1}{2}\left(s_{11}^2+s_{22}^2+s_{33}^2+2s_{12}^2+2s_{23}^2+2s_{13}^2\right)\\=&\frac{1}{2}\left(1^2+(-6)^2+5^2+2*(-3)^2+2*1^2+2*4^2\right)\\=&57\\J_{3}=&\frac{1}{3}s_{ij}s_{jk}s_{ki}\\=&s_{11}s_{22}s_{33}+2s_{12}s_{23}s_{13}-s_{11}s_{23}^2-s_{22}s_{13}^2-s_{33}s_{12}^2\\=&1*(-6)*5+2*(-3)*1*4\\&-1*1^2-(-6)*4^2-5*(-3)^2\\=&-4\end{array} \]
(9)

Finally the characteristic equation is:

\[ s^3-57s+4=0 \]
(10)

Tags: algebra |  eigenvalues |  invariants |  mechanics |  tensors

Related articles

2 comments

  • Alex Teesdale
    2021-11-29T07:36:13.147289+00:00

    Dr. Liolios Thank you for your explanation on how to derive the magnitude of the octahedral stress at yield using the description of cylindrical coordinates and deviatoric invariants. This is new learning territory for me, but I found your development of the answer to my question both logical and comprehensible, which is all the more impressive since I actually understood it! It has been my experience that people who are happy to explain complex topics in a coherent manner are typically those not only with an excellent comprehension of the subject matter, but with an obvious devotion as an educator in doing so. Thanks again for your immediate response! Best regards Alex

  • Pantelis Liolios
    2021-12-04T15:44:14+00:00

    Thank you Alex for your kind words.

Leave a comment