## Principal stresses and stress invariants

In this article we will discuss the derivation of the principal stresses and the stress invariants from the Cauchy stress tensor. The principal stresses and the stress invariants are important parameters that are used in failure criteria, plasticity, Mohr’s circle etc.

For every point inside a body under static equilibrium there are three planes, called the principal planes, where the stress vector is normal to the plane and there is no shear component (see also: Calculation of normal and shear stress on a plane). These normal stress vectors are called principal stresses. From the mathematical point of view, the derivation of the principal stresses and their direction is known as a problem of determining the eigenvalues and their corresponding eigenvectors from a square matrix. Fig. 1 illustrates the principal stresses and their direction for a point inside the body compared to the initial system of coordinates and the stress tensor.

The eigenvalues and the eigenvectors of the stress tensor are derived by solving the homogenous system:

$(\sigma_{ij}-\sigma\delta_{ij})n_{j}=0$
(1)

where $$\sigma_{ij}$$ is the stress tensor, $$\sigma$$ is a constant of proportionality and in this case corresponds to the principal stresses (eigenvalues), $$\delta_{ij}$$ is the Kronecker delta (with $$\delta_{ij}=1$$ if $$i=j$$ and $$\delta_{ij}=0$$ if $$i\neq j$$ ) and $$n_{j}$$ are the unknown eigenvectors. To obtain a non-trivial solution for the above system (1) the determinant of the coefficients must vanish:

$|\sigma_{ij}-\sigma\delta_{ij}|=\left|\begin{array}{ccc}\sigma_{11}-\sigma & \sigma_{12} & \sigma_{13}\\ \sigma_{21} & \sigma_{22}-\sigma & \sigma_{23}\\ \sigma_{31} & \sigma_{32} & \sigma_{33}-\sigma\\ \end{array}\right|=0$
(2)

The above determinant leads to the so-called characteristic equation:

$\sigma^{3}-I_{1}\sigma^{2}+I_{2}\sigma-I_{3}=0$
(3)

with

$\begin{array}{rl}I_{1}=&\sigma_{11}+\sigma_{22}+\sigma_{33}\\=&\sigma_{kk}\\I_{2}=&\left|\begin{array}{cc}\sigma_{22} & \sigma_{23}\\ \sigma_{32} & \sigma_{33}\end{array}\right|+\left|\begin{array}{cc}\sigma_{11} & \sigma_{13}\\ \sigma_{31} & \sigma_{33}\end{array}\right|+\left|\begin{array}{cc}\sigma_{11} & \sigma_{12}\\ \sigma_{21} & \sigma_{22}\end{array}\right|\\=&\sigma_{11}\sigma_{22}+\sigma_{22}\sigma_{33}+\sigma_{11}\sigma_{33}-\sigma_{12}^2-\sigma_{23}^2-\sigma_{13}^2\\=&\frac{1}{2}\left(\sigma_{ii}\sigma_{jj}-\sigma_{ij}\sigma_{ji}\right)\\I_{3}=&\left|\begin{array}{ccc}\sigma_{11} & \sigma_{12} & \sigma_{13}\\\sigma_{21} & \sigma_{22} & \sigma_{23}\\ \sigma_{31} & \sigma_{32} & \sigma_{33}\end{array}\right|\\=&\sigma_{11}\sigma_{22}\sigma_{33}+2\sigma_{12}\sigma_{23}\sigma_{31}-\sigma_{12}^{2}\sigma_{33}-\sigma_{23}^{2}\sigma_{11}-\sigma_{13}^{2}\sigma_{22}\end{array}$
(4)

Coefficients $$I_{1}$$, $$I_{2}$$ and $$I_{3}$$, called first, second and third stress invariants, respectively, are constant and don’t depend on the orientation of the coordinate system. Equation (3) has 3 real roots, $$\sigma_{1}$$, $$\sigma_{2}$$ and $$\sigma_{3}$$ which are the principal stresses or eigenvalues. For each principal stress there is a non-trivial solution for $$n_{j}$$ in equation (1) which is the eigenvector or the direction (direction cosines) of the principal plane where the principal stress acts.

### Example

Calculate the principal stresses and their direction from the following stress tensor:

$\sigma_{ij}=\left[\begin{array}{ccc}5 & 3 & 1 \\ 3 & 2 & 0 \\ 1 & 0 & 4\end{array}\right]$
(5)

From equations (4) we evaluate the stress invariants:

$\begin{array}{l}I_{1}=11\\I_{2}=28\\I_{3}=2\end{array}$
(6)

and we calculate the characteristic equation:

$\sigma^{3}-11\sigma^{2}+28\sigma-2=0$
(7)

The principal stresses are the roots of the polynomial (approximate solution):

$\sigma_{i}=\left(7.09, 3.84, 0.07 \right)$
(8)

For each principal stress we calculate the direction cosines of its principal plane by virtue of equation (1). For principal stress $$\sigma_{1}=7.09$$ we have the following system of equations:

$\begin{array}{l}(5-7.09)n_{1}^{(1)}+3n_{2}^{(1)}+n_{3}^{(1)}=0\\3n_{1}^{(1)}+(2-7.09)n_{2}^{(1)}=0\\n_{1}^{(1)}+(4-7.09)n_{3}^{(1)}=0\end{array}$
(9)

where superscript $$n_{i}^{(1)}$$ indicates that these directional cosines correspond to $$\sigma_{1}$$ (similarly we will have $$n_{i}^{(2)}$$ for $$\sigma_{2}$$ and $$n_{i}^{(3)}$$ for $$\sigma_{3}$$ ). According to linear algebra, if the three principal stresses are different between them, then only the two of three equations (9) are independent. In order to solve the system we will need also the identity:

$n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=1$
(10)

Then by choosing for example the first two equations from (9) and equation (10) the system of equations takes the form:

$\begin{array}{l}(5-7.09)n_{1}^{(1)}+3n_{2}^{(1)}+n_{3}^{(1)}=0\\3n_{1}^{(1)}+(2-7.09)n_{2}^{(1)}=0\\n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=1\end{array}$
(11)

By solving the above system we get the three direction cosines of $$\sigma_{1}$$:

$n_{i}^{(1)}=(0.83, 0.49, 0.27)$
(12)

Similarly, we derive the direction cosines for $$\sigma_{2}$$ and $$\sigma_{3}$$:

$\begin{array}{l}n_{i}^{(2)}=(0.16, 0.26, -0.95)\\n_{i}^{(3)}=(0.53, -0.83, -0.14)\end{array}$
(13)