Octahedral stresses

Posted by: Pantelis Liolios | Sept. 17, 2020

Octahedral stresses we call the normal and shear stresses that are acting on some specific planes inside the stressed body, the octahedral planes. If we consider the principal directions as the coordinate axes (see also the article: Principal stresses and stress invariants), then the plane whose normal vector forms equal angles with the coordinate system is called octahedral plane. There are eight such planes forming a regular octahedron as it is illustrated in Fig. 1.

Octahedral stresses and their corresponding planes — Figure 1: A plane whose normal vector makes equal angles to the axes of the principal stresses is called octahedral plane. There are eight such planes that form a regular octahedron. The normal and shear stresses that act on these planes are called octahedral stresses.

The direction cosines of the octahedral plane are equal to \( n_{1}=n_{2}=n_{3}=1/\sqrt{3} \) (since the plane forms equal angles with the coordinate axes and \( n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=1 \) ). The stress tensor acting on the point \( O \) (origin) has the form:

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

(1)

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

\[ T_{i(oct)}^{(n)}=\sigma_{ij}n_{j} \]

(2)

Then, the octahedral normal stress \( \sigma_{oct} \) is given by:

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

(3)

where \( I_{1} \) is the first invariant of the stress tensor and \( p \) is the mean or hydrostatic stress (see article: Deviatoric stress and invariants). Next, the octahedral shear stress \( \tau_{oct} \) is given by:

\[ \begin{array}{rl}\tau_{oct}=&\sqrt{T_{i(oct)}^{(n)}T_{i(oct)}^{(n)}-\sigma_{oct}^{2}}\\=&\frac{1}{3}\left[\left(\sigma_{11}-\sigma_{22}\right)^2+\left(\sigma_{22}-\sigma_{33}\right)^2+\left(\sigma_{33}-\sigma_{11}\right)^2\right]^\frac{1}{2}\\=&\frac{1}{3}\sqrt{2I_{1}^{2}-6I_{2}}=\sqrt{\frac{2}{3}J_{2}}\end{array} \]

(4)

where \( I_{2} \) is the second invariant of the stress tensor and \( J_{2}\) is the second invariant of the deviatoric stress tensor.

Example

Calculate the octahedral stresses for the following stress tensor:

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

(5)

Show solution...

According to equations (3) and (4) we only have to calculate the first and second invariant of the stress tensor. From article Principal stresses and stress invariants we get:

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

(6)

Thus the octahedral stresses are:

\[ \sigma_{oct}=\frac{1}{3}I_{1}=\frac{5}{3} \]

(7)

and

\[ \tau_{oct}=\frac{1}{3}\sqrt{2I_{1}^{2}-6I_{2}}=\frac{\sqrt{188}}{3} \]

(8)

Tags: eigenvalues | invariants | mechanics | tensors

3 comments

Abizo

2022-06-03T10:53:55.999467+00:00

Add more diagram
Ahmet Giz

2021-10-10T08:56:46.929901+00:00

Thank you for useful article. 1. equation in line below fig 1. : should be: n1= n2= n3 = 1/sqrt(3)
Pantelis Liolios

2021-10-11T12:17:11+00:00

Thank you Ahmet for your comment. You are correct, it is 1/sqrt(3).