Mohr's circle

Figure 1. Mohr's circles for a three-dimensional state of stress

Mohr's circle, named after Christian Otto Mohr, is a two-dimensional graphical representation of the state of stress at a point. The abscissa, $\sigma_\mathrm{n}\,\!$, and ordinate, $\tau_\mathrm{n}\,\!$, of each point on the circle's circumference are the normal stress and shear stress components, respectively, acting on a particular cut plane (of the object being analyzed) with a unit vector $\mathbf n\,\!$ with components $\left(n_1, n_2, n_3 \right)\,\!$. In other words, the circumference of the circle is the locus of points that represent the state of stress on individual planes at all their orientations, where the axes represent the principal axes of the stress element.

Karl Culmann was the first to conceive a graphical representation for stresses while considering longitudinal and vertical stresses in horizontal beams during bending. Mohr's contribution extended the use of this representation for both two- and three-dimensional stresses and developed a failure criterion based on the stress circle.

There is also a similar Mohr's circle for strain where x-axis depicts strain and the y-axis represents half of shear strain which can be found out by Generalised Hooke's Law.

Other graphical methods for the representation of the stress state at a point include the Lame's stress ellipsoid and Cauchy's stress quadric.

Motivation for the Mohr's Circle

Figure 2 Stress in a loaded deformable material body assumed as a continuum.

Figure 3 - Stress transformation at a point in a continuum under plane stress conditions.

When a physical deformable object, assumed as a continuum, is acted upon by external forces, either surface forces or body forces, internal reactive forces are produced between its particles. This reaction follows from Euler's laws of motion for a continuum, which are the equivalent to Newton's laws of motion for a particle. A measure of the intensity of these internal forces is called stress. Because the object is assumed as a continuum, these internal forces are distributed continuously within the volume of the object.

It is of interest for engineering, e.g., structural, mechanical, or geotechnical engineering, to perform a stress analysis to determine the stress distribution within objects, i.e. stresses in a rock mass around a tunnel, airplane wings, or building columns. Calculating the stress distribution implies the determination of stresses at every point (material particle) in the object. According to Cauchy, the stress at any point in an object (Figure 2), assumed as a continuum, is completely defined by obtaining the nine components $\sigma_{ij}\,\!$ of a second order tensor of type (2,0) known as the Cauchy stress tensor, $\boldsymbol\sigma\,\!$:

$\boldsymbol{\sigma}= \left[{\begin{matrix} \sigma _{11} & \sigma _{12} & \sigma _{13} \\ \sigma _{21} & \sigma _{22} & \sigma _{23} \\ \sigma _{31} & \sigma _{32} & \sigma _{33} \\ \end{matrix}}\right] \equiv \left[{\begin{matrix} \sigma _{xx} & \sigma _{xy} & \sigma _{xz} \\ \sigma _{yx} & \sigma _{yy} & \sigma _{yz} \\ \sigma _{zx} & \sigma _{zy} & \sigma _{zz} \\ \end{matrix}}\right] \equiv \left[{\begin{matrix} \sigma _x & \tau _{xy} & \tau _{xz} \\ \tau _{yx} & \sigma _y & \tau _{yz} \\ \tau _{zx} & \tau _{zy} & \sigma _z \\ \end{matrix}}\right] \,\!$

After the stress distribution has been determined within the object, with respect to a coordinate system, it is sometimes necessary to calculate the components of the stress tensor at a point with respect to a different coordinate system, i.e., the stresses on a plane passing through that point of interest with a different orientation —forming an angle with the coordinate system (Figure 3). For example, it is of interest to find the maximum normal stress and maximum shear stress, as well as the orientation of the plane where they act upon. To achieve this, it is necessary to perform a tensor transformation under a change in the system of coordinates. From the definition of tensor, the Cauchy stress tensor obeys the tensor transformation law. A graphical representation of this transformation law for the Cauchy stress tensor is the Mohr's circle for stress.

Stress transformation in two-dimensions (plane stress and plane strain)

Figure 4 - Stress components at a plane passing through a point in a continuum under plane stress conditions.

To understand where the equation of Mohr's circle comes from, consider a point $P\,\!$ in a two dimensional object under a state of plane stress, or plane strain, with an associated stress tensor with known components $(\sigma_x, \sigma_y, \tau_{xy})\,\!$ and all other stress components equal to zero (Figure 3):

$\boldsymbol{\sigma}= \left[{\begin{matrix} \sigma _x & \tau _{xy} & 0 \\ \tau _{yx} & \sigma _y & 0 \\ 0 & 0 & 0 \\ \end{matrix}}\right] \equiv \left[{\begin{matrix} \sigma _x & \tau _{xy} \\ \tau _{yx} & \sigma _y \\ \end{matrix}}\right] \,\!$

Knowing the stress components $(\sigma_x, \sigma_y, \tau_{xy})\,\!$ on any two perpendicular directions at $P\,\!$, the objective is to find the stress components $\sigma_\mathrm{n}\,\!$ and $\tau_\mathrm{n}\,\!$ on any other plane passing through that point as a function of the angle $\theta\,\!$ between that plane and the original coordinate systems. It is important to remember that we are considering a unit area of the infinitesimal element in the direction parallel to the $y\,\!$-$z\,\!$ plane.

From equilibrium of forces on an infinitesimal material element at $P\,\!$ (Figure 4), the normal stress $\sigma_\mathrm{n}\,\!$ and the shear stress $\tau_\mathrm{n}\,\!$ on any plane perpendicular to the $x\,\!$-$y\,\!$ plane passing through $P\,\!$ with a unit vector $\mathbf n\,\!$ making an angle of $\theta\,\!$ with the horizontal, i.e. $\cos \theta\,\!$ is the direction cosine in the $x\,\!$ direction, are given by:

$\sigma_\mathrm{n} = \frac{1}{2} ( \sigma_x + \sigma_y ) + \frac{1}{2} ( \sigma_x - \sigma_y )\cos 2\theta + \tau_{xy} \sin 2\theta\,\!$

$\tau_\mathrm{n} = -\frac{1}{2}(\sigma_x - \sigma_y )\sin 2\theta + \tau_{xy}\cos 2\theta\,\!$

Derivation of Mohr's circle parametric equations - Equilibrium of forces

From equilibrium of forces in the direction of $\sigma_\mathrm{n}\,\!$ ($x'\,\!$-axis) (Figure 4), and knowing that the area of the plane where $\sigma_\mathrm{n}\,\!$ acts is $dA\,\!$, we have: $\ \begin{align} \sum F_{x'} &= \sigma_\mathrm{n} dA - \sigma_x dA \cos ^2 \theta - \sigma_y dA \sin ^2 \theta - \tau_{xy} dA \cos \theta \sin \theta - \tau_{xy} dA \sin \theta \cos \theta = 0 \\ \sigma_\mathrm{n} &= \sigma_x \cos ^2 \theta + \sigma_y \sin ^2 \theta + 2\tau_{xy} \sin \theta \cos \theta \\ \end{align}$ However, knowing that $\cos ^2 \theta = \frac{1+\cos 2\theta}{2}, \qquad \sin ^2 \theta = \frac{1-\cos 2\theta}{2} \qquad \text{and} \qquad \sin 2\theta= 2\sin\theta\cos\theta\,\!$ we obtain $\sigma_\mathrm{n} = \frac{1}{2} ( \sigma_x + \sigma_y ) + \frac{1}{2} ( \sigma_x - \sigma_y )\cos 2\theta + \tau_{xy} \sin 2\theta\,\!$ Now, from equilibrium of forces in the direction of $\tau_\mathrm{n}\,\!$ ($y'\,\!$-axis) (Figure 4), and knowing that the area of the plane where $\tau_\mathrm{n}\,\!$ acts is $dA\,\!$, we have: $\ \begin{align} \sum F_{y'} &= \tau_\mathrm{n} dA + \sigma_x dA \cos \theta \sin \theta - \sigma_y dA \sin \theta \cos \theta - \tau_{xy} dA \cos ^2 \theta + \tau_{xy} dA \sin ^2 \theta = 0 \\ \tau_\mathrm{n} &= -(\sigma_x-\sigma_y) \sin\theta\cos\theta + \tau_{xy} \left( \cos^2 \theta -\sin^2 \theta \right) \\ \end{align}$ However, knowing that $\cos ^2 \theta - \sin^2\theta=\cos 2\theta \qquad \text{and} \qquad \sin 2\theta= 2\sin\theta\cos\theta\,\!$ we obtain $\tau_\mathrm{n} = -\frac{1}{2}(\sigma_x - \sigma_y )\sin 2\theta + \tau_{xy}\cos 2\theta\,\!$

Both equations can also be obtained by applying the tensor transformation law on the known Cauchy stress tensor, which is equivalent to performing the static equilibrium of forces in the direction of $\sigma_\mathrm{n}\,\!$ and $\tau_\mathrm{n}\,\!$.

Derivation of Mohr's circle parametric equations - Tensor transformation

The stress tensor transformation law can be stated as $\begin{align} \boldsymbol{\sigma}' &= \mathbf A \boldsymbol{\sigma} \mathbf A^T \\ \left[{\begin{matrix} \sigma'_{x} & \sigma'_{xy} \\ \sigma'_{yx} & \sigma'_{y} \\ \end{matrix}}\right] &=\left[{\begin{matrix} a_{x} & a_{xy} \\ a_{yx} & a_{y} \\ \end{matrix}}\right] \left[{\begin{matrix} \sigma_{x} & \sigma_{xy} \\ \sigma_{yx} & \sigma_{y} \\ \end{matrix}}\right]\left[{\begin{matrix} a_{x} & a_{yx} \\ a_{xy} & a_{y} \\ \end{matrix}}\right] \\ &= \left[{\begin{matrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{matrix}}\right]\left[{\begin{matrix} \sigma_{x} & \tau_{xy} \\ \tau_{yx} & \sigma_{y} \\ \end{matrix}}\right] \left[{\begin{matrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{matrix}}\right] \end{align}$ Expanding the right hand side, and knowing that $\sigma'_{x}=\sigma_\mathrm{n}\,\!$ and $\tau'_{xy}=\tau_\mathrm{n}\,\!$, we have: $\sigma_\mathrm{n} = \sigma_x \cos ^2 \theta + \sigma_y \sin ^2 \theta + 2\tau_{xy} \sin \theta \cos \theta\,\!$ However, knowing that $\cos ^2 \theta = \frac{1+\cos 2\theta}{2}, \qquad \sin ^2 \theta = \frac{1-\cos 2\theta}{2} \qquad \text{and} \qquad \sin 2\theta= 2\sin\theta\cos\theta\,\!$ we obtain $\sigma_\mathrm{n} = \frac{1}{2} ( \sigma_x + \sigma_y ) + \frac{1}{2} ( \sigma_x - \sigma_y )\cos 2\theta + \tau_{xy} \sin 2\theta\,\!$ $\tau_\mathrm{n} = -(\sigma_x-\sigma_y) \sin\theta\cos\theta + \tau_{xy} \left( \cos^2 \theta -\sin^2 \theta \right)\,\!$ However, knowing that $\cos ^2 \theta - \sin^2\theta=\cos 2\theta \qquad \text{and} \qquad \sin 2\theta= 2\sin\theta\cos\theta\,\!$ we obtain $\tau_\mathrm{n} = -\frac{1}{2}(\sigma_x - \sigma_y )\sin 2\theta + \tau_{xy}\cos 2\theta\,\!$ It is not necessary at this moment to calculate the stress component $\sigma'_{y}\,\!$ acting on the plane perpendicular to the plane of action of $\sigma'_{x}\,\!$ as it is not required for deriving the equation for the Mohr circle.

These two equations are the parametric equations of the Mohr circle. This means that by choosing a coordinate system with abscissa $\sigma_\mathrm{n}\,\!$ and ordinate $\tau_\mathrm{n}\,\!$, giving values to the parameter $\theta\,\!$ will place the points obtained lying on a circle. Eliminating the parameter $\theta\,\!$ from these parametric equations will yield the non-parametric equation of the Mohr circle. This can be achieved by rearranging the equations for $\sigma_\mathrm{n}\,\!$ and $\tau_\mathrm{n}\,\!$, first transposing the first term in the first equation and squaring both sides of each of the equations then adding them. Thus we have

$\begin{align} \left[ \sigma_\mathrm{n} - \tfrac{1}{2} ( \sigma_x + \sigma_y )\right]^2 + \tau_\mathrm{n}^2 &= \left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2 \\ (\sigma_\mathrm{n} - \sigma_\mathrm{avg})^2 + \tau_\mathrm{n}^2 &= R^2 \end{align}\,\!$

where

$R = \sqrt{\left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2} \quad \text{and} \quad \sigma_\mathrm{avg} = \tfrac{1}{2} ( \sigma_x + \sigma_y )\,\!$

This is the equation of a circle (the Mohr circle) of the form

$(x-a)^2+(y-b)^2=r^2\,\!$

with radius $r=R\,\!$ centered at a point with coordinates $(a,b)=(\sigma_\mathrm{avg}, 0)\,\!$ in the $\sigma_\mathrm{n}:\tau_\mathrm{n}\,\!$ coordinate system.

Mohr's circle for two-dimensional stress states

A two-dimensional Mohr's circle can be constructed if we know the normal stresses $\sigma_x$, $\sigma_y$, and the shear stress $\tau_{xy}$. The following sign conventions are usually used:

1. Tensile stresses (positive) are to the right.
2. Compressive stresses (negative) are to the left.
3. Clockwise shear stresses are plotted upward.
4. Counterclockwise shear stresses are plotted downward.

The reason for the above sign convention is that, in engineering mechanics, the normal stresses are positive if they are outward to the plane of action (tension), and shear stresses are positive if they rotate clockwise about the point in consideration. In geomechanics, i.e. soil mechanics and rock mechanics, however, normal stresses are considered positive when they are inward to the plane of action (compression), and shear stresses are positive if they rotate counterclockwise about the point in consideration.

To construct the Mohr circle of stress for a state of plane stress, or plane strain, first we plot two points in the $\sigma_\mathrm{n}:\tau_\mathrm{n}\,\!$ space corresponding to the known stress components on both perpendicular planes, i.e. $A(\sigma_y, \tau_{xy})\,\!$ and $B(\sigma_x, -\tau_{yx})\,\!$ (Figure 5 and 2). We then connect points $A\,\!$ and $B\,\!$ by a straight line and find the midpoint $O\,\!$ which corresponds to the intersection of this line with the $\sigma_\mathrm{n}\,\!$ axis. Finally, we draw a circle with diameter $\overline{AB}\,\!$ and centre at $O\,\!$.

The radius $R\,\!$ of the circle is $R = \sqrt{\left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2}\,\!$, and the coordinates of its centre are $\left[\tfrac{1}{2}(\sigma_x + \sigma_y), 0\right]\,\!$.

The principal stresses are then the abscissa of the points of intersection of the circle with the $\sigma_\mathrm{n}\,\!$ axis (note that the shear stresses are zero for the principal stresses).

Drawing a Mohr's circle

Figure 5. Mohr's circle for plane stress and plane strain conditions (double angle approach).

The following procedure is used to draw a Mohr's circle and to find the magnitude and direction of maximum stresses from it.

• First, the $x$- and $y$-axes of a Cartesian coordinate system are identified as the $\sigma_n$-axis and $\tau_n$-axis, respectively.
• Next, two points of the Mohr's circle are plotted. These are the points B ($\sigma_x$, $-\tau_{yx}$) and A ($\sigma_y$, $\tau_{xy}$). The line connecting these two points is a diameter of the Mohr's circle.
• The center of the Mohr's circle, O, is located where the diameter, AB, intersects the σ-axis. This point gives the average normal stress (σ_avg). The average normal stress can be read directly from a plot of the Mohr's circle. Alternatively, it can be calculated using

$\sigma_\text{avg}=\tfrac{1}{2}(\sigma_x+ \sigma_y)$.

• The Mohr's circle intersects the $\sigma_n$ axis at two points, C and E. The stresses at these two end points of the horizontal diameter are $\sigma_1$ and $\sigma_2$, the principal stresses. The point $\sigma_1$ represents the maximum normal stress (σ_max) and the point $\sigma_2$ is the minimum normal stress (σ_min). The equations for finding these values are

$\sigma_1 = \sigma_\max = \tfrac{1}{2}(\sigma_x + \sigma_y) + \sqrt{\left[\tfrac{1}{2}(\sigma_x- \sigma_y)\right]^2+ \tau_{xy}^2}$

$\sigma_2 = \sigma_\min = \tfrac{1}{2}(\sigma_x + \sigma_y) - \sqrt{\left[\tfrac{1}{2}(\sigma_x- \sigma_y)\right]^2+ \tau_{xy}^2}$

• Next we examine the points where the circle intersects the line parallel to $\tau_n$-axis passing through the center of the circle, O. The vertical diameter of the circle passes through O (σ_avg) and goes up to positive $\tau_\max$ and down to negative $\tau_\min$. The magnitudes of extreme values are equal to the radius of the Mohr's circle, but with different signs. The equation to find these extreme values of the shear stress is

$\tau_{\max,\min}= \pm \sqrt{\left[\tfrac{1}{2}(\sigma_x- \sigma_y)\right]^2+ {\tau_{xy}}^2}$ .

• The next value to determine is the angle that the plane of maximum normal stress makes with the $x$-axis. Let us create a new $X$-axis by drawing a line from the center of the Mohr circle, O, through point A. Let the angle between the $X$-axis and the $\sigma$-axis be $\phi$. If $\theta$ is the angle between the maximum normal stress and the $x$-axis, then it can be shown that $\phi$ = 2$\theta_{p1}$. The angle $\phi$ is found by:

$\phi = 2\theta_{p1} = \arctan\left[2\tau_{xy}/(\sigma_x- \sigma_y)\right]$.

• To find the angle that the direction that the plane of maximum shear stress makes with the $x$-axis, we use the relation

$2\theta_s = \arctan\left[-(\sigma_x - \sigma_y)/(2\tau_{xy})\right]$. It is important to pay attention to the use of these two equations as they look similar.

• Often, the final step of the process is to also draw a square stress element indicating the orientations of the maximum normal and shear stresses; the normal stress element at an angle $\theta$ and the maximum shear stress element at an angle of $\theta_s$.

The previous discussion assumes, implicitly, that there are two orthogonal directions $x$ and $y$ that define a plane in which the stress components $\sigma_x\,\!$. $\sigma_y\,\!$, and $\tau_{xy}\,\!$ are known. It is also implicit that these stresses are known at a point $P\,\!$ in a continuum body under plane stress or plane strain. The Mohr circle, once drawn, can be used to find the components of the stress tensor for any other choice of orthogonal directions in the plane.

Stress components on an arbitrary plane

Figure 6. Mohr's circle for plane stress and plane strain conditions (Pole approach). Any straight line drawn from the pole will intersect the Mohr circle at a point that represents the state of stress on a plane inclined at the same orientation (parallel) in space as that line.

Using the Mohr circle one can find the stress components $(\sigma_\mathrm{n}, \tau_\mathrm{n})\,\!$ on any other plane with a different orientation $\theta\,\!$ that passes through point $P\,\!$. For this, two approaches can be used:

• The first approach relies on the fact that the angle $\theta\,\!$ between two planes passing through $P\,\!$ is half the angle between the lines joining their corresponding stress points $(\sigma_\mathrm{n}, \tau_\mathrm{n})\,\!$ on the Mohr circle and the centre of the circle (Figure 5). In other words, the stresses $(\sigma_\mathrm{n}, \tau_\mathrm{n})\,\!$ acting on a plane at an angle $\theta\,\!$ counterclockwise to the plane on which $\sigma_x\,\!$ acts is determined by traveling counterclockwise around the circle from the known stress point $(\sigma_x, \tau_{xy})\,\!$ a distance subtending an angle $2\theta\,\!$ at the centre of the circle (Figure 5).
• The second approach involves the determination of a point on the Mohr circle called the pole or the origin of planes. Any straight line drawn from the pole will intersect the Mohr circle at a point that represents the state of stress on a plane inclined at the same orientation (parallel) in space as that line. Therefore, knowing the stress components $\sigma\,\!$ and $\tau\,\!$ on any particular plane, one can draw a line parallel to that plane through the particular coordinates $\sigma_\mathrm{n}\,\!$ and $\tau_\mathrm{n}\,\!$ on the Mohr circle and find the pole as the intersection of such line with the Mohr circle. As an example, let's assume we have a state of stress with stress components $\sigma_x,\!$, $\sigma_y,\!$, and $\tau_{xy},\!$, as shown on Figure 6. First, we can draw a line from point $B\,\!$ parallel to the plane of action of $\sigma_x\,\!$, or, if we choose otherwise, a line from point $A\,\!$ parallel to the plane of action of $\sigma_y\,\!$. The intersection of any of these two lines with the Mohr circle is the pole. Once the pole has been determined, to find the state of stress on a plane making an angle $\theta\,\!$ with the vertical, or in other words a plane having its normal vector forming an angle $\theta\,\!$ with the horizontal plane, then we can draw a line from the pole parallel to that plane (See Figure 6). The normal and shear stresses on that plane are then the coordinates of the point of intersection between the line and the Mohr circle.

Mohr's circle for a general three-dimensional state of stresses

Figure 7. Mohr's circle for a three-dimensional state of stress

To construct the Mohr's circle for a general three-dimensional case of stresses at a point, the values of the principal stresses $\left(\sigma_1, \sigma_2, \sigma_3 \right)\,\!$ and their principal directions $\left(n_1, n_2, n_3 \right)\,\!$ must be first evaluated.

Considering the principal axes as the coordinate system, instead of the general $x_1\,\!$, $x_2\,\!$, $x_3\,\!$ coordinate system, and assuming that $\sigma_1 > \sigma_2 > \sigma_3\,\!$, then the normal and shear components of the stress vector $\mathbf T^{(\mathbf n)}\,\!$, for a given plane with unit vector $\mathbf n\,\!$, satisfy the following equations

$\begin{align} \left( T^{(n)} \right)^2 &= \sigma_{ij}\sigma_{ik}n_jn_k \\ \sigma_\mathrm{n}^2 + \tau_\mathrm{n}^2 &= \sigma_1^2 n_1^2 + \sigma_2^2 n_2^2 + \sigma_3^2 n_3^2 \end{align}\,\!$

$\sigma_\mathrm{n} = \sigma_1 n_1^2 + \sigma_2 n_2^2 + \sigma_3 n_3^2.\,\!$

Knowing that $n_i n_i = n_1^2+n_2^2+n_3^2 = 1\,\!$, we can solve for $n_1^2\,\!$, $n_2^2\,\!$, $n_3^2\,\!$, using the Gauss elimination method which yields

$\begin{align} n_1^2 &= \frac{\tau_\mathrm{n}^2+(\sigma_\mathrm{n} - \sigma_2)(\sigma_\mathrm{n} - \sigma_3)}{(\sigma_1 - \sigma_2)(\sigma_1 - \sigma_3)} \ge 0\\ n_2^2 &= \frac{\tau_\mathrm{n}^2+(\sigma_\mathrm{n} - \sigma_3)(\sigma_\mathrm{n} - \sigma_1)}{(\sigma_2 - \sigma_3)(\sigma_2 - \sigma_1)} \ge 0\\ n_3^2 &= \frac{\tau_\mathrm{n}^2+(\sigma_\mathrm{n} - \sigma_1)(\sigma_\mathrm{n} - \sigma_2)}{(\sigma_3 - \sigma_1)(\sigma_3 - \sigma_2)} \ge 0. \end{align}\,\!$

Since $\sigma_1 > \sigma_2 > \sigma_3\,\!$, and $(n_i)^2\,\!$ is non-negative, the numerators from the these equations satisfy

$\tau_\mathrm{n}^2+(\sigma_\mathrm{n} - \sigma_2)(\sigma_\mathrm{n} - \sigma_3) \ge 0\,\!$ as the denominator $\sigma_1 - \sigma_2 > 0\,\!$ and $\sigma_1 - \sigma_3 > 0\,\!$

$\tau_\mathrm{n}^2+(\sigma_\mathrm{n} - \sigma_3)(\sigma_\mathrm{n} - \sigma_1) \le 0\,\!$ as the denominator $\sigma_2 - \sigma_3 > 0\,\!$ and $\sigma_2 - \sigma_1 < 0\,\!$

$\tau_\mathrm{n}^2+(\sigma_\mathrm{n} - \sigma_1)(\sigma_\mathrm{n} - \sigma_2) \ge 0\,\!$ as the denominator $\sigma_3 - \sigma_1 < 0\,\!$ and $\sigma_3 - \sigma_2 < 0.\,\!$

These expressions can be rewritten as

$\begin{align} \tau_\mathrm{n}^2 + \left[ \sigma_\mathrm{n}- \tfrac{1}{2} (\sigma_2 + \sigma_3) \right]^2 \ge \left( \tfrac{1}{2}(\sigma_2 - \sigma_3) \right)^2 \\ \tau_\mathrm{n}^2 + \left[ \sigma_\mathrm{n}- \tfrac{1}{2} (\sigma_1 + \sigma_3) \right]^2 \le \left( \tfrac{1}{2}(\sigma_1 - \sigma_3) \right)^2 \\ \tau_\mathrm{n}^2 + \left[ \sigma_\mathrm{n}- \tfrac{1}{2} (\sigma_1 + \sigma_2) \right]^2 \ge \left( \tfrac{1}{2}(\sigma_1 - \sigma_2) \right)^2 \\ \end{align}\,\!$

which are the equations of the three Mohr's circles for stress $C_1\,\!$, $C_2\,\!$, and $C_3\,\!$, with radii $R_1=\tfrac{1}{2}(\sigma_2 - \sigma_3)\,\!$, $R_2=\tfrac{1}{2}(\sigma_1 - \sigma_3)\,\!$, and $R_3=\tfrac{1}{2}(\sigma_1 - \sigma_2)\,\!$, and their centres with coordinates $\left[\tfrac{1}{2}(\sigma_2 + \sigma_3), 0\right]\,\!$, $\left[\tfrac{1}{2}(\sigma_1 + \sigma_3), 0\right]\,\!$, $\left[\tfrac{1}{2}(\sigma_1 + \sigma_2), 0\right]\,\!$, respectively.

These equations for the Mohr's circles show that all admissible stress points $(\sigma_\mathrm{n}, \tau_\mathrm{n})\,\!$ lie on these circles or within the shaded area enclosed by them (see Figure 7). Stress points $(\sigma_\mathrm{n}, \tau_\mathrm{n})\,\!$ satisfying the equation for circle $C_1\,\!$ lie on, or outside circle $C_1\,\!$. Stress points $(\sigma_\mathrm{n}, \tau_\mathrm{n})\,\!$ satisfying the equation for circle $C_2\,\!$ lie on, or inside circle $C_2\,\!$. And finally, stress points $(\sigma_\mathrm{n}, \tau_\mathrm{n})\,\!$ satisfying the equation for circle $C_3\,\!$ lie on, or outside circle $C_3\,\!$.

Bibliography

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• Jumikis, Alfreds R. (1969). Theoretical soil mechanics: with practical applications to soil mechanics and foundation engineering. Van Nostrand Reinhold Co. ISBN 0-442-04199-3. 
• Parry, Richard Hawley Grey (2004). Mohr circles, stress paths and geotechnics (2 ed.). Taylor & Francis. pp. 1–30. ISBN 0-415-27297-1. 
• Timoshenko, Stephen P.; James Norman Goodier (1970). Theory of Elasticity (Third ed.). McGraw-Hill International Editions. ISBN 0-07-085805-5. 
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External links

• Mohr's Circle and more circles by Rebecca Brannon
• DoITPoMS Teaching and Learning Package- "Stress Analysis and Mohr's Circle"
• Online Mohr's Circle Calculator Online Mohr's Circle Calculator
• TAA_Mohrs_Circle_2D Free Downloadable Mohr's Circle Desktop Application