Mohr’s circle in 3 dimensions

Mohr’s diagram is a useful graphical representation of the stress state at a point. In this graphical representation the state of stress at a point is represented by the Mohr circle diagram, in which the abscissa \( \sigma \) and \( \tau \) give the normal and shear stress acting on a particular cut plane with a fixed normal direction. In the general 3 dimensional case, for a given state of stress at a point, the Mohr circle diagram has three circles as shown in Fig. 1. Mohr’s circle diagram is used frequently in conjunction with failure criteria like the Mohr-Coulomb failure criterion.

3D Mohr's circles
Figure 1: In the three dimensional case there are three Mohr’s circles. The diameters and the centers of the circles are calculated from the three principal stresses. The point representing the normal and shear stress on a plane does not necessarily lie on the perimeter of a circle.

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

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

The center of each circle in Mohr’s diagram lies on \( \sigma \) axis and is given by:

\[ \begin{array}{c}C_{1}=\frac{1}{2}\left(\sigma_{11}+\sigma_{22}\right)\\C_{2}=\frac{1}{2}\left(\sigma_{11}+\sigma_{33}\right)\\C_{3}=\frac{1}{2}\left(\sigma_{22}+\sigma_{33}\right) \end{array} \]
(2)

while the radii of the circles are calculated by:

\[ \begin{array}{c}R_{1}=\frac{1}{2}\sqrt{\left(\sigma_{11}-\sigma_{22}\right)^{2}+\left(\sigma_{12}+\sigma_{21}\right)^{2}}\\R_{2}=\frac{1}{2}\sqrt{\left(\sigma_{11}-\sigma_{33}\right)^{2}+\left(\sigma_{13}+\sigma_{31}\right)^{2}}\\R_{3}=\frac{1}{2}\sqrt{\left(\sigma_{22}-\sigma_{33}\right)^{2}+\left(\sigma_{23}+\sigma_{32}\right)^{2}}\end{array} \]
(3)

for centers \( C_{1} \), \( C_{2} \) and \( C_{3} \), respectively. If the principal stresses are known (may be calculated by the stress tensor as shown in Principal stresses and stress invariants) then the above equations (2) and (3) take the form (for the case \( \sigma_{1}\ge\sigma_{2}\ge\sigma_{3} \) ):

\[ \begin{array}{c} C_{1}=\frac{1}{2}\left(\sigma_{1}+\sigma_{2}\right)\\C_{2}=\frac{1}{2}\left(\sigma_{1}+\sigma_{3}\right)\\C_{3}=\frac{1}{2}\left(\sigma_{2}+\sigma_{3}\right) \end{array} \]
(4)

and

\[ \begin{array}{c}R_{1}=\frac{1}{2}\left(\sigma_{1}-\sigma_{2}\right)\\R_{2}=\frac{1}{2}\left(\sigma_{1}-\sigma_{3}\right)\\R_{3}=\frac{1}{2}\left(\sigma_{2}-\sigma_{3}\right)\end{array} \]
(5)

Consider an arbitrary cut plane that passes through the considered point. All the admissible values of \( \sigma \) and \( \tau \) for this plane lie inside or on the boundaries of the region bounded by the circles \( C_{1} \), \( C_{2} \) and \( C_{3} \) (see Fig.1). The proof, however, will not be given in this article but it can be found in many related books.

In order to calculate the normal and shear stresses acting on any plane, through Mohr’s circle diagram, it is necessary to know the direction cosines of the normal unit vector of the plane with respect to the principal directions. Assume that \( n_{1} \), \( n_{2} \) and \( n_{3} \) are the direction cosines of the plane with respect to the principal directions of \( \sigma_{1} \), \( \sigma_{2} \) and \( \sigma_{3} \), respectively. For a given value of \( n_{1} \) the point \( (\sigma,\tau) \) lies on the arc \( AA’ \) as shown in Fig. 1. To construct this arc we draw line \( L_{1} \) that passes through \( \sigma_{1} \) and is parallel to \( \tau \) axis. Then we measure angle \( \alpha=\cos^{-1}n_{1} \) from that line. This line intersects the circle at points \( A \) and \( A’ \). By using center \( C_{3} \) as center (the only center that does not depend on \( \sigma_{1} \) ) we draw the arc \( AA’ \). Similarly, for direction cosine \( n_{2} \) the point \( (\sigma,\tau) \) lies on the arc \( BB’ \). We draw line \( L_{2} \) and measure angle \( \beta=\cos^{-1}n_{2} \). The intersection points are \( B \) and \( B’ \). Using center \( C_{2} \) we draw the arc \( BB’ \). Finally, we can do the same for direction cosine \( n_{3} \). We measure angle \( \gamma=\cos^{-1}n_{3} \) from \( L_{3} \) and using center \( C_{1} \) we draw the arc \( CC’ \). Since, only two values of \( n_{1} \),\( n_{2} \) and \( n_{3} \) are independent, it is adequate to use only two direction cosines in order to determine the values \( (\sigma,\tau) \). The normal and shear stress is given by the coordinates of intersection point \( P \). All arcs pass through that point, hence, one can use for example \( n_{1} \) and \( n_{3} \) to calculate point \( (\sigma,\tau) \) and use \( n_{2} \) to verify the procedure.

The Mohr’s circle diagram may be used to calculate graphically the normal and shear stresses on a plane. Otherwise, the method described in Calculation of normal and shear stress on a plane may be used.

Exercise

Consider the following stress state acting on a point:

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

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

\[ n=\left(\frac{1}{2},\frac{1}{2},\frac{\sqrt{2}}{2}\right) \]
(7)

Show solution…

From equations (4) and (5) we calculate the centers \( C_{1} \), \( C_{2} \) and \( C_{3} \) and the radii \( R_{1} \), \( R_{2} \) and \( R_{3} \):

\[ \begin{array}{l}C_{1}=\frac{1}{2}\left(5+2\right)=3.5\\C_{2}=\frac{1}{2}\left(5+1\right)=3\\C_{3}=\frac{1}{2}\left(2+1\right)=1.5\\R_{1}=\frac{1}{2}\left(5-2\right)=1.5\\R_{2}=\frac{1}{2}\left(5-1\right)=2\\R_{3}=\frac{1}{2}\left(2-1\right)=0.5\end{array} \]
(8)

Next we draw Mohr’s circle diagram as shown in Fig. 2.

Example of a 3d Mohr's circle construction
Figure 2: Demonstration of the calculation of the normal and shear stress on a plane in the general 3d case by virtue of the 3d Mohr diagram.

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

\[ \begin{array}{l}\alpha=\cos^{-1}\frac{1}{2}=60^{o}\\ \beta=\cos^{-1}\frac{1}{2}=60^{o}\\ \gamma=\cos^{-1}\frac{\sqrt{2}}{2}=45^{o}\end{array} \]
(9)

Using the above angles (we need only two, for example \( \alpha \) and \( \gamma \) ) we draw the arcs and we find the normal and shear stress on the plane:

\[ \begin{array}{c}\sigma=2.25\\ \tau=1.64\end{array} \]
(10)

We can also confirm the solution by using the methodology described in the article: Calculation of normal and shear stress on a plane.