Mohr's circle in three dimensions

Mohr’s circle is a graphical representation of the state of stress at a point: the abscissa $\sigma$ and the ordinate $\tau$ of a point of the diagram are the normal and shear stress acting on a plane through that point. In the general three-dimensional case, the Mohr diagram consists of three circles, as shown in Fig. 1. The Mohr diagram is often used together with failure criteria such as the Mohr–Coulomb failure criterion.

Assume that the stress state at a point is given by the stress tensor:

$$\sigma_{ij}=\left[\begin{array}{ccc}\sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22} & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33}\end{array}\right] \tag{1}$$

The circles are drawn from the principal stresses $\sigma_{1}\ge\sigma_{2}\ge\sigma_{3}$, which are calculated from the stress tensor as shown in Principal stresses and stress invariants. The circles cannot be drawn directly from the components of the tensor: a circle drawn with $\sigma_{11}$, $\sigma_{22}$ and $\sigma_{12}$, as in the two-dimensional case, ignores the shear stresses $\sigma_{13}$ and $\sigma_{23}$ and is one of Mohr’s circles only if $x_{3}$ is a principal direction. The centers of the circles lie on the $\sigma$ axis and are given by:

$$\begin{array}{c} C_{1}=\frac{1}{2}\left(\sigma_{1}+\sigma_{2}\right)\\C_{2}=\frac{1}{2}\left(\sigma_{1}+\sigma_{3}\right)\\C_{3}=\frac{1}{2}\left(\sigma_{2}+\sigma_{3}\right) \end{array} \tag{2}$$

and

$$\begin{array}{c}R_{1}=\frac{1}{2}\left(\sigma_{1}-\sigma_{2}\right)\\R_{2}=\frac{1}{2}\left(\sigma_{1}-\sigma_{3}\right)\\R_{3}=\frac{1}{2}\left(\sigma_{2}-\sigma_{3}\right)\end{array} \tag{3}$$

Thus, the circle with center $C_{1}$ passes through $\sigma_{1}$ and $\sigma_{2}$, the circle with center $C_{2}$ through $\sigma_{1}$ and $\sigma_{3}$, and the circle with center $C_{3}$ through $\sigma_{2}$ and $\sigma_{3}$.

As in the article on Mohr’s circle, compressive stresses are taken as positive. In three dimensions, however, the shear stress on a plane is a vector in the plane whose direction is not fixed by a single in-plane axis, so $\tau$ is the magnitude of the shear stress, $\tau\ge0$, and only the upper half of the diagram represents stress states. The special sign rule of the two-dimensional Mohr’s circle does not apply here; the parts of the circles below the $\sigma$ axis are used only for the construction.

Normal and shear stress on a plane

Consider an arbitrary plane through the point, and let $n_{1}$, $n_{2}$ and $n_{3}$ be the direction cosines of its unit normal vector with respect to the principal directions of $\sigma_{1}$, $\sigma_{2}$ and $\sigma_{3}$, respectively. The stress vector on the plane has components $\sigma_{1}n_{1}$, $\sigma_{2}n_{2}$ and $\sigma_{3}n_{3}$ along the principal directions (see also Calculation of normal and shear stress on a plane), so the normal and shear stress satisfy:

$$\begin{array}{l} \sigma=\sigma_{1}n_{1}^{2}+\sigma_{2}n_{2}^{2}+\sigma_{3}n_{3}^{2}\\ \sigma^{2}+\tau^{2}=\sigma_{1}^{2}n_{1}^{2}+\sigma_{2}^{2}n_{2}^{2}+\sigma_{3}^{2}n_{3}^{2}\\ 1=n_{1}^{2}+n_{2}^{2}+n_{3}^{2} \end{array} \tag{4}$$

Solving Eqs. (4) for $n_{1}^{2}$, $n_{2}^{2}$ and $n_{3}^{2}$ and using Eqs. (2) and (3), we obtain:

$$\begin{array}{l} \left(\sigma-C_{3}\right)^{2}+\tau^{2}=R_{3}^{2}+n_{1}^{2}\left(\sigma_{1}-\sigma_{2}\right)\left(\sigma_{1}-\sigma_{3}\right)\\ \left(\sigma-C_{2}\right)^{2}+\tau^{2}=R_{2}^{2}-n_{2}^{2}\left(\sigma_{1}-\sigma_{2}\right)\left(\sigma_{2}-\sigma_{3}\right)\\ \left(\sigma-C_{1}\right)^{2}+\tau^{2}=R_{1}^{2}+n_{3}^{2}\left(\sigma_{1}-\sigma_{3}\right)\left(\sigma_{2}-\sigma_{3}\right) \end{array} \tag{5}$$

Each of Eqs. (5) is a circle: for a given $n_{1}$ the point $(\sigma,\tau)$ lies on a circle with center $C_{3}$, for a given $n_{2}$ on a circle with center $C_{2}$, and for a given $n_{3}$ on a circle with center $C_{1}$. Since $n_{i}^{2}\ge0$ and $\sigma_{1}\ge\sigma_{2}\ge\sigma_{3}$, Eqs. (5) also show where the admissible points lie: the first and the third equation give $(\sigma-C_{3})^{2}+\tau^{2}\ge R_{3}^{2}$ and $(\sigma-C_{1})^{2}+\tau^{2}\ge R_{1}^{2}$, and the second gives $(\sigma-C_{2})^{2}+\tau^{2}\le R_{2}^{2}$. Hence, the point $(\sigma,\tau)$ of any plane lies inside or on the large circle (between $\sigma_{1}$ and $\sigma_{3}$) and outside or on the two small circles (between $\sigma_{1}$ and $\sigma_{2}$, and between $\sigma_{2}$ and $\sigma_{3}$). The points on the circles themselves belong to planes that contain one of the principal directions ($n_{1}=0$, $n_{2}=0$ or $n_{3}=0$).

The circles of Eqs. (5) can be drawn without calculation (Fig. 1):

1. For $n_{1}$: draw the line $L_{1}$ through $\sigma_{1}$, parallel to the $\tau$ axis. Draw a line through $\sigma_{1}$ at the angle $\alpha=\cos^{-1}n_{1}$ from $L_{1}$, towards the smaller stresses. It intersects the circle between $\sigma_{1}$ and $\sigma_{2}$ at $A$ and the circle between $\sigma_{1}$ and $\sigma_{3}$ at $A'$. With center $C_{3}$, the only center that does not depend on $\sigma_{1}$, draw the arc $AA'$.
2. For $n_{2}$: draw the line $L_{2}$ through $\sigma_{2}$ and a line through $\sigma_{2}$ at the angle $\beta=\cos^{-1}n_{2}$ from $L_{2}$, towards $\sigma_{3}$. It intersects the circle between $\sigma_{2}$ and $\sigma_{3}$ at $B'$; extended below the $\sigma$ axis, it intersects the circle between $\sigma_{1}$ and $\sigma_{2}$ at $B$. With center $C_{2}$, draw the arc $BB'$.
3. For $n_{3}$: draw the line $L_{3}$ through $\sigma_{3}$ and a line through $\sigma_{3}$ at the angle $\gamma=\cos^{-1}n_{3}$ from $L_{3}$, towards the larger stresses. It intersects the circle between $\sigma_{2}$ and $\sigma_{3}$ at $\Gamma$ and the circle between $\sigma_{1}$ and $\sigma_{3}$ at $\Gamma'$. With center $C_{1}$, draw the arc $\Gamma\Gamma'$.

The normal and shear stress on the plane are the coordinates of the intersection point $P$ of the arcs. Since $n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=1$, only two of the direction cosines are independent, so two arcs are enough to find $P$. For example, we can use $n_{1}$ and $n_{3}$ to find $P$ and $n_{2}$ to check the construction, since all three arcs pass through $P$.

The Mohr diagram can thus be used to calculate the normal and shear stresses on a plane graphically. Alternatively, Eqs. (4) or the method described in Calculation of normal and shear stress on a plane may be used.

Example

Consider the following stress state at a point:

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

Calculate the normal and shear stress on the plane with normal vector:

$$\vec{n}=\left(\frac{1}{2},\frac{1}{2},\frac{\sqrt{2}}{2}\right)$$

Solution

The stress tensor (6) is diagonal, so the coordinate axes are the principal directions and $\sigma_{1}=5$, $\sigma_{2}=2$ and $\sigma_{3}=1$. Hence, the components of $\vec{n}$ are the direction cosines $n_{1}$, $n_{2}$ and $n_{3}$ with respect to the principal directions. From Eqs. (2) and (3) we calculate the centers $C_{1}$, $C_{2}$ and $C_{3}$ and the radii $R_{1}$, $R_{2}$ and $R_{3}$:

$$\begin{array}{ll}C_{1}=\frac{1}{2}\left(5+2\right)=3.5, & R_{1}=\frac{1}{2}\left(5-2\right)=1.5\\C_{2}=\frac{1}{2}\left(5+1\right)=3, & R_{2}=\frac{1}{2}\left(5-1\right)=2\\C_{3}=\frac{1}{2}\left(2+1\right)=1.5, & R_{3}=\frac{1}{2}\left(2-1\right)=0.5\end{array}$$

Next we draw the Mohr diagram, as shown in Fig. 2.

From the direction cosines we calculate the angles $\alpha$, $\beta$ and $\gamma$:

$$\begin{array}{l}\alpha=\cos^{-1}\frac{1}{2}=60^\circ\\ \beta=\cos^{-1}\frac{1}{2}=60^\circ\\ \gamma=\cos^{-1}\frac{\sqrt{2}}{2}=45^\circ\end{array}$$

Using these angles (only two are needed, for example $\alpha$ and $\gamma$) we draw the arcs, and their intersection point $P$ gives the normal and shear stress on the plane:

$$\sigma=2.25,\quad \tau=1.64$$

We can confirm the result with Eqs. (4):

$$\begin{array}{l} \sigma=5\cdot\frac{1}{4}+2\cdot\frac{1}{4}+1\cdot\frac{1}{2}=2.25\\ \sigma^{2}+\tau^{2}=25\cdot\frac{1}{4}+4\cdot\frac{1}{4}+1\cdot\frac{1}{2}=7.75\\ \tau=\sqrt{7.75-2.25^{2}}=\sqrt{2.6875}\approx1.64 \end{array}$$