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 at least three mutually perpendicular 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). The normal stresses acting on these planes are called principal stresses. From the mathematical point of view, the derivation of the principal stresses and their directions is a problem of determining the eigenvalues and the corresponding eigenvectors of a symmetric matrix. Fig. 1 illustrates the principal stresses and their directions for a point inside the body, compared to the initial coordinate system and the stress tensor.

On a principal plane with unit normal $n_{i}$, the stress vector is parallel to the normal, i.e. $T_{i}=\sigma n_{i}$, where $\sigma$ is the normal stress on the plane. Combined with Cauchy’s formula, $T_{i}=\sigma_{ij}n_{j}$, this leads to the homogeneous system:

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

where $\sigma_{ij}$ is the stress tensor, $\sigma$ is the principal stress (eigenvalue), $\delta_{ij}$ is the Kronecker delta (with $\delta_{ij}=1$ if $i=j$ and $\delta_{ij}=0$ if $i\neq j$) and $n_{j}$ are the components of the unknown eigenvector. To obtain a non-trivial solution for the above system (1), the determinant of the coefficients must vanish:

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

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

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

with (the expanded forms take into account the symmetry of the stress tensor, $\sigma_{ij}=\sigma_{ji}$):

The coefficients $I_{1}$, $I_{2}$ and $I_{3}$, called the first, second and third stress invariants, respectively, are constant and do not depend on the orientation of the coordinate system. Since the stress tensor is symmetric, Eq. (3) has three real roots, $\sigma_{1}$, $\sigma_{2}$ and $\sigma_{3}$, which are the principal stresses or eigenvalues; they are usually ordered so that $\sigma_{1}\ge\sigma_{2}\ge\sigma_{3}$. For each principal stress there is a non-trivial solution for $n_{j}$ in Eq. (1), which is the eigenvector, i.e. the direction cosines of the principal plane on which the principal stress acts. Each eigenvector is defined only up to its sign ($\pm\vec{n}$). If the three principal stresses are distinct, the three principal directions are mutually perpendicular; if two of them are equal, every direction in the plane normal to the third principal direction is a principal direction, so there are infinitely many principal planes.

In the coordinate system of the principal directions the stress tensor is diagonal, and the invariants take the simple form:

$$\begin{array}{l}I_{1}=\sigma_{1}+\sigma_{2}+\sigma_{3}\\I_{2}=\sigma_{1}\sigma_{2}+\sigma_{2}\sigma_{3}+\sigma_{3}\sigma_{1}\\I_{3}=\sigma_{1}\sigma_{2}\sigma_{3}\end{array} \tag{5}$$

Example

Calculate the principal stresses and their directions for the following stress tensor:

$$\sigma_{ij}=\left[\begin{array}{ccc}5 & 3 & 1 \\ 3 & 2 & 0 \\ 1 & 0 & 4\end{array}\right] \tag{6}$$

Solution

From Eqs. (4) we evaluate the stress invariants:

$$\begin{array}{l}I_{1}=11\\I_{2}=28\\I_{3}=2\end{array} \tag{7}$$

and we obtain the characteristic equation:

$$\sigma^{3}-11\sigma^{2}+28\sigma-2=0 \tag{8}$$

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

$$\sigma_{i}=\left(7.091,\ 3.835,\ 0.074\right) \tag{9}$$

As a check, Eqs. (5) give $\sigma_{1}+\sigma_{2}+\sigma_{3}=11$, $\sigma_{1}\sigma_{2}+\sigma_{2}\sigma_{3}+\sigma_{3}\sigma_{1}\approx 28$ and $\sigma_{1}\sigma_{2}\sigma_{3}\approx 2$, in agreement with the invariants calculated above.

For each principal stress we calculate the direction cosines of its principal plane from Eq. (1). For the principal stress $\sigma_{1}=7.091$ we have the following system of equations:

$$\begin{array}{l}(5-7.091)n_{1}^{(1)}+3n_{2}^{(1)}+n_{3}^{(1)}=0\\3n_{1}^{(1)}+(2-7.091)n_{2}^{(1)}=0\\n_{1}^{(1)}+(4-7.091)n_{3}^{(1)}=0\end{array} \tag{10}$$

where the superscript in $n_{i}^{(1)}$ indicates that these direction 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 distinct, only two of the three equations (10) are independent. To solve the system, we also need the condition that $\vec{n}^{(1)}$ is a unit vector:

$$n_{i}^{(1)}n_{i}^{(1)}=\left(n_{1}^{(1)}\right)^{2}+\left(n_{2}^{(1)}\right)^{2}+\left(n_{3}^{(1)}\right)^{2}=1 \tag{11}$$

Then, choosing for example the first two equations of (10) together with Eq. (11), the system of equations takes the form:

$$\begin{array}{l}(5-7.091)n_{1}^{(1)}+3n_{2}^{(1)}+n_{3}^{(1)}=0\\3n_{1}^{(1)}+(2-7.091)n_{2}^{(1)}=0\\\left(n_{1}^{(1)}\right)^{2}+\left(n_{2}^{(1)}\right)^{2}+\left(n_{3}^{(1)}\right)^{2}=1\end{array} \tag{12}$$

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

$$n_{i}^{(1)}=(0.83,\ 0.49,\ 0.27) \tag{13}$$

Alternatively, and more easily, the eigenvector can be found without solving a nonlinear system. Since $\vec{n}^{(1)}$ is perpendicular to every row of the matrix $(\sigma_{ij}-\sigma_{1}\delta_{ij})$, it is parallel to the cross product of any two independent rows. Using the permutation symbol $\varepsilon_{ijk}$, the cross product of the first two rows is written in index notation as:

For the first two rows of Eqs. (10), this gives:

and normalizing this vector to unit length (its magnitude is $\approx 6.134$) gives again $n_{i}^{(1)}=(0.83,\ 0.49,\ 0.27)$.

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

$$\begin{array}{l}n_{i}^{(2)}=(0.16,\ 0.26,\ -0.95)\\n_{i}^{(3)}=(0.54,\ -0.83,\ -0.14)\end{array} \tag{16}$$

Finally, we can verify that the three principal directions are mutually perpendicular: the dot products $n_{i}^{(1)}n_{i}^{(2)}$, $n_{i}^{(1)}n_{i}^{(3)}$ and $n_{i}^{(2)}n_{i}^{(3)}$ are all zero (up to rounding). The three unit vectors therefore form a new coordinate system, and the matrix whose rows are $n_{i}^{(1)}$, $n_{i}^{(2)}$ and $n_{i}^{(3)}$ is the rotation matrix that transforms the stress tensor into this system (see Transformation of a tensor to a new coordinate system), where it becomes diagonal with the principal stresses on its diagonal.