### Mohr Coulomb failure criterion

The Mohr Coulomb failure criterion is one of the most used failure models for quasi-brittle materials like rocks. It is very popular due to its simplicity and its parameters are direct physical properties of the material. Its calibration can be done with typical uniaxial and triaxial compression tests or in some cases by using only uniaxial compression tests; however with some uncertainty. In the analysis that follows, the compressive stresses will be assumed positive.

Consider a box lying on a flat horizontal surface as illustrated in Fig. 1. A force $$F_n$$, normal (perpendicular) to the horizontal surface, is acting on the box. That can be, for example, the gravitational force or some external force or a combination of them. Consider, also, the horizontal force $$F_s$$ that is acting on the box and is parallel to the contact surface. The minimum force $$F_s$$ that is required to slide the box is given by Amontons' law of friction:

$F_s=\mu F_n$
(1)

where $$\mu$$ is called the coefficient of friction and it is a property of the interface where the two materials (box and horizontal surface) are in contact. By dividing both sides of the above Eq. (1) by the contact area $$A$$, the relation between the normal and the shear stress acting on the interface plane is derived:

$\tau=\mu \sigma$
(2)

where $$\tau$$, $$\sigma$$ are the shear and normal stresses, respectively. Equation (2) depicts a linear relationship between the normal and the shear stress as illustrated in Fig. (2). The slope of the line is the friction coefficient $$\mu$$ or equivalently:

$\mu = \tan\phi$
(3)

where $$\phi$$ is the angle between the straight line in Fig. (2) and the horizontal axis and it is called the friction angle.

Consider, now, that the box and the horizontal surface are glued together. In this case, the shear stress $$\tau$$ must overcome both the friction resistance and the glue strength to slide the box. The latter does not depend on the normal stress at all; hence, it is a constant quantity that bonds the two pieces together. Including this constant strength into the friction model of Eq. (2) leads to:

$\tau=c + \sigma\tan\phi$
(4)

The above Eq. (4) is known as the Coulomb failure criterion in honor of Charles-Augustin de Coulomb who verified Amontons' friction laws. The quantity $$c$$ is called cohesion, it is a shearing strength and it is a property of the glue at the interface. The addition of the cohesion to the model translates the straight line upwards in the $$\sigma - \tau$$ diagram as depicted if Fig. 3. The cohesion represents the point where the graphical representation crosses the vertical axis.

According to the Coulomb failure criterion (Eq. 4), the orientation of the sliding surface does not matter. The failure will occur at any arbritrary plane whenever the combination of the normal and shear stresses that act on it satisfy Eq. (4). In other words, if the shear stress $$\tau$$ that acts on the plane is equal (or greater) to the maximum shear stress $$\sigma\tan\phi$$ that this plane can withstand, siliding will occur. Another observation is that it is not necessary for the sliding plane to be visible aforehand. The failure surface will develop within the material as soon as the criterion is satisfied. In this case, the cohesion is the stress (glue) that keeps the particles of the material bonded together and the friction develops between the contacting grains. For this reason, angle $$\phi$$ is called internal friction angle.

The Coulomb failure criterion can be nicely combined with the Mohr's circle and for that reason this failure criterion is widely known as the Mohr-Coulomb failure criterion. Mohr's circle is the graphical representation of the stress state within a body in static equilibrium. The circle is drawn in the $$\sigma-\tau$$ plane, like the Coulomb criterion, and it represents all the normal and shear stress pairs that act in the internal planes of the material (cf. Fig. 4). During loading the center of the circle translates horizontally and the size of the radius changes while the position of the criterion remains constant. If the circle touches the line of the criterion then the material will fail. The limiting case is when the line is tangent to the circle, i.e. the line intersects the circle at a single point, as illustrated in Fig. 4.

At the intersection point, the pair of the normal and shear stresses that act on an internal plane of the material satisfy Eq. (4). The angle of this plane with respect to the major principal stress $$\sigma_1$$ can be found geometrically. The theory of Mohr's circle dictates that the plane whose normal forms angle $$\beta$$ with respect to the axis of $$\sigma_1$$ (Fig. 5) is represented by the central angle $$2\beta$$ in the Mohr's diagram (Fig. 4). Furthermore, in the right triangle $$ABC$$, as shown in Fig. 4, the angle at the apex $$A$$ is equal to the friction angle $$\phi$$ (two angles with mutually perpendicular sides). The sum of all the angles in the triangle must be equal to $$\pi$$:

$\phi+\frac{\pi}{2}+(\pi-2\beta)=\pi$
(5)

$\beta=\frac{\phi}{2}+\frac{\pi}{4}$
(6)

The above Eq. (6) directly relates the geometrical angle of the failure plane with the internal friction angle.

The expression of the Coulomb failure criterion in Eq. (4) is very simple but difficult to calibrate from experimental data. A much easier to calibrate expression can be derived with the help of Mohr's circle and Fig. 4. Let the normal and shear stresses at the point $$A$$ in Fig. 4 to be $$\sigma_n$$ and $$\tau_n$$, respectively. The right triangle $$ABC$$ represents the projections of the radius in the horizontal and vertical directions:

$BC=AC\sin\phi=\frac{1}{2}(\sigma_1-\sigma_3)\sin\phi$
(7)

and

$\tau_n=AB=AC\cos\phi=\frac{1}{2}(\sigma_1-\sigma_3)\cos\phi$
(8)

where $$\tau_n$$ coincides with the vertical projection. The normal stress $$\sigma_n$$ can be calculated as follows:

$\sigma_n=OC-BC=\frac{1}{2}(\sigma_1+\sigma_3)-\frac{1}{2}(\sigma_1-\sigma_3)\sin\phi$
(9)

On the other hand, at the point $$A$$, the pair $$(\sigma_n, \tau_n)$$ must satisfy Eq. (4) which leads to:

$\frac{1}{2}(\sigma_1-\sigma_3)\cos\phi=c+\left[\frac{1}{2}(\sigma_1+\sigma_3)-\frac{1}{2}(\sigma_1-\sigma_3)\sin\phi\right]\tan\phi$
(10)

Collecting all the $$\sigma_1$$ terms to the left hand side and all the $$\sigma_3$$ and constant terms to the right hand side and after some manipulations leads to:

$\sigma_1=\frac{2c\cos\phi}{1-\sin\phi}+\frac{1+\sin\phi}{1-\sin\phi}\sigma_3$
(11)

The above Eq. (11) is the Mohr-Coulomb failure criterion expressed in terms of the major $$\sigma_1$$ and the minor $$\sigma_3$$ principal stresses. As illustrated in Fig. 6, it represents a linear relationship between $$\sigma_1$$ and $$\sigma_3$$. The point of intersection of the criterion and the vertical axis is the uniaxial compression strength of the material:

$\sigma_c=\frac{2c\cos\phi}{1-\sin\phi}$
(12)

The point of intersection of the criterion and the horizontal line represents the uniaxial tensile strength of the material. However, the Mohr-Coulomb criterion overestimates the tensile strength $$\sigma_t$$ of the quasi-brittle materials and therefore it is usually combined with a tension cut-off as illustrated in Fig. 6.

The above Eq. (11) is much easier to calibrate from experimental data compared to Eq. (4). Typically, a laboratory will conduct a series of uniaxial and triaxial compression tests to calibrate the Mohr-Coulomb criterion, i.e. to estimate the cohesion and the internal friction of the material. In a uniaxial compression test, the specimen, usually cylindrical, is loaded by a normal stress along its axis. The applied normal stress is considered to be the major principal stress $$\sigma_1$$; however, friction forces are also present and proper lubricant should be applied to reduce them. The other two principal stresses are equal to zero ($$\sigma_2=\sigma_3=0$$). The loading curve exhibits a peak maximum strength where the material fails. This peak stress is called the Uniaxial Compression Strength. The failure may develop either in axial splitting mode, which indicates tensile fracture, or in shear mode where inclined sliding planes develop. In the latter case, Eqs. (6) and (12) may be used to approximately estimate the internal friction angle and the cohesion of the material. In a typical triaxial compression test, the specimen is, additionally, loaded with a constant lateral pressure throughout the experiment (cf. Fig. 5). The remaining procedure remains the same with the uniaxial compression test. In this case the minor and the intermediate principal stresses are both equal to the lateral pressure.

The peak strengths of the experimental data will be a set of $$(\sigma_1, \sigma_3)$$ pairs that all satisfy Eq. (11). To estimate the cohesion and the internal friction angle of the material, a least squares approach must be followed. The overdetermined system of $$n$$ equations with $$2$$ unknowns may be nicely written by virtue of matrices:

$\mathrm{A} \mathrm{X}=\mathrm{B}$
(13)

where:

$\mathrm{A}=\begin{bmatrix} 1 & \sigma_3^{(1)}\\ 1 & \sigma_3^{(2)}\\ \vdots & \vdots\\ 1 & \sigma_3^{(n)} \end{bmatrix}$
(14)
$\mathrm{X}=\begin{bmatrix} \sigma_c\\ k \end{bmatrix}$
(15)

and

$\mathrm{B}=\begin{bmatrix} \sigma_1^{(1)}\\ \sigma_1^{(2)}\\ \vdots\\ \sigma_1^{(n)} \end{bmatrix}$
(16)

The term $$k$$ is the slope of the linear failure criterion, $$\sigma_c$$ is the uniaxial compression strength and $$(\sigma_3^{(i)},\sigma_1^{(i)})$$ represent the $$i$$-th experimental point. The superscript $$T$$ in the above equations denotes the transpose of the matrix. The solution of the above system with the least squares method is given by:

$\mathrm{X}=\left( \mathrm{A^T}\mathrm{A}\right)^{-1}\mathrm{A^T B}$
(17)

where the superscript $$-1$$ indicates the inverse of the matrix. After the solution of the system, the internal friction angle can be computed from the expression:

$k=\frac{1+\sin\phi}{1-\sin\phi}\Leftrightarrow\sin\phi=\frac{k-1}{k+1}$
(18)

Consequently, the cohesion is calculated from Eq. (12).

The Mohr-Coulomb failure criterion is a simple, yet widely used failure model for brittle materials like rocks and concrete. It works well in the compression regime for low confinements and its two parameters are direct physical properties of the material. Its disadvantage is that it overestimates the tensile strength and a tension cut-off limit must be considered.

### Example

A quasi-brittle material subjected to uniaxial and triaxial compression tests. In the following table, the peak strength of each test is given. Calculate the cohesion and the internal friction angle of the material.

Table 1: Example of peak strengths of uniaxial and triaxial compression tests on a quasi-brittle material.
$$\sigma_3$$ (MPa) $$\sigma_1$$ (MPa)
0 74.9
2 81.3
4 87.3
6 92.9

The overdetermined system of equations is written in the form of Eq. (13). Matrices $$\mathrm{A}$$ and $$\mathrm{B}$$ are the following:

$\mathrm{A}=\begin{bmatrix} 1 & 0\\ 1 & 2\\ 1 & 4\\ 1 & 6 \end{bmatrix}$
(19)

and

$\mathrm{B}=\begin{bmatrix} 74.9\\ 81.3\\ 87.3\\ 92.9 \end{bmatrix}$
(20)

Next, the normal matrix $$\mathrm{A^T A}$$ and the moment matrix $$\mathrm{A^T B}$$ are computed:

$\mathrm{A^T A}=\begin{bmatrix} 4 & 12\\ 12 & 56 \end{bmatrix}$
(21)
$\mathrm{A^T B}=\begin{bmatrix} 336.4\\ 1069.2 \end{bmatrix}$
(22)

The inverse of $$\mathrm{A^T A}$$ is:

$\left(\mathrm{A^T A}\right)^{-1}=\begin{bmatrix} 0.7 & -0.15\\ -0.15 & 0.05 \end{bmatrix}$
(23)

and the solution of the system is:

$\mathrm{X}=\begin{bmatrix} 75.1\\ 3 \end{bmatrix}$
(24)

The uniaxial compression strength is $$\sigma_c=75.1$$ MPa. The internal friction angle is calculated from the slope $$k=3$$ and Eq. (18):

$\sin\phi=\frac{1}{2}\Leftrightarrow\phi=30^o$
(25)

Finally, the cohesion is calculated from Eq. (12):

$c=\frac{1-\sin\phi}{2\cos\phi}\sigma_c=21.7\mbox{ MPa}$
(26)

Altenatively, the overdetermined system of equations may, also, be solved with the help of Python and the math library numpy as follows.

import numpy as np

# define the A, B matrices
A = np.array([[1, 0],[1,2], [1,4],[1,6]])
B = np.array([74.9,81.3,87.3,92.9])

# calculate the normal matrix
ATA=np.dot(A.T, A)

# calculate the moment matrix
ATB=np.dot(A.T, B)

# calculate the inverse of the normal matrix
ATAinv=np.linalg.inv(AAT)

# finally, calculate the least squares solution
X=np.dot(ATAinv, ATB)