### 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.

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.

### Example

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)

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.

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.

#### Related articles

• Ahmed Imtiaz Ferdous
2022-12-16T08:15:27.901663+00:00

Thanks for the explanation. I had only one question. Is there anyway to know the direction of the shear stress on the plane? In the method you described, only the magnitudes can be known.

• Mehmet Zor
2022-04-03T11:05:08.631876+00:00

Thank you for your explanions. But I have question: How can we know that the arc having center C3 pass to two points A and A'? and same quesitions for other arcs and points?

• Pantelis Liolios
2022-04-05T07:43:55+00:00

Hi Mehmet, thank you for your comment. Point A corresponds to a plane parallel to σ3, i.e. σ3 does not have any contribution to the components of stress on this plane. On the other hand, point A' corresponds to a plane parallel to σ2, i.e. no contribution of σ2. These are the two limit cases. Both of these limit cases form angle α with σ1.The arc AA' represents the transition from first limit case to the second. Arcs BB' and ΓΓ' have similar behavior. I hope that I've helped you.

• Isai
2021-05-15T19:23:10.972160+00:00

Thank you for your explanation, it is so useful. Can you cite the literature where the Mohr's Circle method comes from? I only found that Ugural and Fenster, 2003, explain some graphical by trial and error; but other references don't explain the graphical method, anymore. I hope you have other references.

• Pantelis Liolios
2021-06-07T09:52:39+00:00

Hi Isai, I apologize for the late reply. It seems that somehow I missed your comment! I cannot remember if I used a particular book or if this article is a result of multiple sources. However, a good source is the following: Malvern, E.E. Introduction to the mechanics of a continuous medium. Prentice-Hall, Englewood Cliffs, New Jersey, 1969.

• N4S!RODDIN3
2021-02-21T13:54:38.834815+00:00

Sub: How someone get notified? Hi again, Pantelis Liolios Is there any way to subscribe here for new topics and comments? I've searched for RSS feed, but ended up with no results.

• Pantelis Liolios
2021-02-22T09:49:59+00:00

Hi Nas, the blog doesn't have RSS feed yet.

• Joseph
2021-02-14T12:03:02.195137+00:00

For L1 and L2 you have taken alpha and beta anticlockwise but for L3 you have take gamma clockwise. Any particular reason?

• Pantelis Liolios
2021-02-16T11:35:38+00:00

Hi Joseph. Short answer to your question: it doesn't matter. You can take the angle either clockwise or anticlockwise. However, I prefer to measure the angles as it is shown in Figure 1, so that I will end up drawing only a small arc of the circles to find the intersection point (the pole). If I take γ anticlockwise, then ΓΓ' will lie on the lower part (below σ axis) and I'll have to draw the complete circle (with center C1) to find the intersection point. In that case the Figure will start to become messy.

• Jun!
2021-02-14T02:33:28.281047+00:00

Hi, nice to meet you and thanks for kind explanation. I have some question about 3D Mohr's circle. Can I know your e-mail address?

• Pantelis Liolios
2021-02-16T08:52:50+00:00