Octahedral stresses

The normal and shear stresses acting on certain planes of a stressed body, the octahedral planes, are called octahedral stresses. If we take the principal directions as the coordinate axes (see also the article: Principal stresses and stress invariants), then a plane whose normal vector makes equal angles with the coordinate axes is called an octahedral plane. There are eight such planes, which form a regular octahedron, as illustrated in Fig. 1.

Since the normal of an octahedral plane makes equal angles with the coordinate axes and $n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=1$, its direction cosines are $n_{1}=n_{2}=n_{3}=\pm1/\sqrt{3}$. The $2^{3}=8$ combinations of signs give the unit normals $\vec{n}=(\pm1,\pm1,\pm1)/\sqrt{3}$ of the eight faces of the octahedron. In this coordinate system the stress tensor at point $O$ (the origin) is diagonal:

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

where $\sigma_{1}$, $\sigma_{2}$ and $\sigma_{3}$ are the principal stresses. The components of the stress vector acting on an octahedral plane are given by Cauchy’s formula (for more details see: Calculation of normal and shear stress on a plane):

$$T_{i}^{(oct)}=\sigma_{ij}n_{j} \tag{2}$$

which, for the plane with $\vec{n}=(1,1,1)/\sqrt{3}$, gives simply $T_{i}^{(oct)}=\left(\sigma_{1},\sigma_{2},\sigma_{3}\right)/\sqrt{3}$. Then, the octahedral normal stress $\sigma_{oct}$ is:

$$\begin{array}{rl}\sigma_{oct}=&T_{i}^{(oct)}n_{i}=\sigma_{ij}n_{i}n_{j}\\= & \sigma_{1}n_{1}n_{1}+\sigma_{2}n_{2}n_{2}+\sigma_{3}n_{3}n_{3}\\=&\frac{1}{3}\left(\sigma_{1}+\sigma_{2}+\sigma_{3}\right)=\frac{1}{3}I_{1}=p\end{array} \tag{3}$$

where $I_{1}$ is the first invariant of the stress tensor and $p$ is the mean or hydrostatic stress (see the article: Deviatoric stress and invariants). Next, the magnitude of the stress vector squared is:

$$T_{i}^{(oct)}T_{i}^{(oct)}=\frac{1}{3}\left(\sigma_{1}^{2}+\sigma_{2}^{2}+\sigma_{3}^{2}\right) \tag{4}$$

and the octahedral shear stress $\tau_{oct}$ follows from the Pythagorean theorem:

$$\begin{array}{rl}\tau_{oct}=&\sqrt{T_{i}^{(oct)}T_{i}^{(oct)}-\sigma_{oct}^{2}}\\=&\frac{1}{3}\sqrt{\left(\sigma_{1}-\sigma_{2}\right)^2+\left(\sigma_{2}-\sigma_{3}\right)^2+\left(\sigma_{3}-\sigma_{1}\right)^2}\\=&\frac{1}{3}\sqrt{2I_{1}^{2}-6I_{2}}=\sqrt{\frac{2}{3}J_{2}}\end{array} \tag{5}$$

where $I_{2}$ is the second invariant of the stress tensor and $J_{2}$ is the second invariant of the deviatoric stress tensor. In an arbitrary (non-principal) coordinate system, the square of the octahedral shear stress can be written in terms of the stress components as:

Since $\sigma_{oct}$ and $\tau_{oct}$ depend only on the stress invariants, they are the same on all eight octahedral planes and can be calculated without determining the principal stresses. Moreover, they reflect the decomposition of the stress tensor into its hydrostatic and deviatoric parts: $\sigma_{oct}$ is equal to the mean stress $p$, while $\tau_{oct}$ depends only on the deviatoric stress tensor, through $J_{2}$.

Example

Calculate the octahedral stresses for the following stress tensor:

$$\sigma_{ij}=\left[\begin{array}{ccc}4&-2&-1\\-2&3&4\\-1&4&-2\end{array}\right] \tag{7}$$

Solution

The given tensor is not expressed in the principal coordinate system. However, according to Eqs. (3) and (5), the octahedral stresses depend only on the invariants, so we only have to calculate the first and second invariants of the stress tensor. Using the expressions from the article Principal stresses and stress invariants we get:

$$\begin{array}{rl}I_{1}= & \sigma_{11}+\sigma_{22}+\sigma_{33}\\=&5\\I_{2}=&\sigma_{11}\sigma_{22}+\sigma_{22}\sigma_{33}+\sigma_{11}\sigma_{33}-\sigma_{12}^2-\sigma_{23}^2-\sigma_{13}^2\\=&-23\end{array} \tag{8}$$

Thus the octahedral stresses are:

$$\sigma_{oct}=\frac{1}{3}I_{1}=\frac{5}{3}\approx 1.667 \tag{9}$$

and

$$\tau_{oct}=\frac{1}{3}\sqrt{2I_{1}^{2}-6I_{2}}=\frac{\sqrt{188}}{3}=\frac{2\sqrt{47}}{3}\approx 4.570 \tag{10}$$

As a check, the second deviatoric invariant is $J_{2}=\frac{1}{3}I_{1}^{2}-I_{2}=\frac{25}{3}+23=\frac{94}{3}$, so $\sqrt{\frac{2}{3}J_{2}}=\frac{\sqrt{188}}{3}$. The same result follows directly from the stress components with Eq. (6):