The Mohr–Coulomb failure criterion represents the linear envelope that is obtained from a plot of the shear strength of a material versus the applied normal stress. This relation is expressed as : \tau = \sigma~\tan(\phi) + c where \tau is the shear strength, \sigma is the normal stress, c is the intercept of the failure envelope with the \tau axis, and \tan(\phi) is the slope of the failure envelope. The quantity c is often called the
cohesion and the angle \phi is called the
angle of internal friction. Compression is assumed to be positive in the following discussion. If compression is assumed to be negative then \sigma should be replaced with -\sigma. If \phi = 0, the Mohr–Coulomb criterion reduces to the
Tresca criterion. On the other hand, if \phi = 90^\circ the Mohr–Coulomb model is equivalent to the Rankine model. Higher values of \phi are not allowed. From
Mohr's circle we have \sigma = \sigma_m - \tau_m \sin\phi ~;~~ \tau = \tau_m \cos\phi where \tau_m = \cfrac{\sigma_1-\sigma_3}{2} ~;~~ \sigma_m = \cfrac{\sigma_1+\sigma_3}{2} and \sigma_1 is the maximum principal stress and \sigma_3 is the minimum principal stress. Therefore, the Mohr–Coulomb criterion may also be expressed as \tau_m = \sigma_m \sin\phi + c \cos\phi ~. This form of the Mohr–Coulomb criterion is applicable to failure on a plane that is parallel to the \sigma_2 direction.
Mohr–Coulomb failure criterion in three dimensions The Mohr–Coulomb criterion in three dimensions is often expressed as : \left\{\begin{align} \pm\cfrac{\sigma_1 - \sigma_2}{2} & = \left[\cfrac{\sigma_1 + \sigma_2}{2}\right]\sin(\phi) + c\cos(\phi) \\ \pm\cfrac{\sigma_2 - \sigma_3}{2} & = \left[\cfrac{\sigma_2 + \sigma_3}{2}\right]\sin(\phi) + c\cos(\phi)\\ \pm\cfrac{\sigma_3 - \sigma_1}{2} & = \left[\cfrac{\sigma_3 + \sigma_1}{2}\right]\sin(\phi) + c\cos(\phi). \end{align}\right. The
Mohr–Coulomb failure surface is a cone with a hexagonal cross section in deviatoric stress space. The expressions for \tau and \sigma can be generalized to three dimensions by developing expressions for the normal stress and the resolved shear stress on a plane of arbitrary orientation with respect to the coordinate axes (basis vectors). If the unit normal to the plane of interest is : \mathbf{n} = n_1~\mathbf{e}_1 + n_2~\mathbf{e}_2 + n_3~\mathbf{e}_3 where \mathbf{e}_i,~~ i=1,2,3 are three orthonormal unit basis vectors, and if the principal stresses \sigma_1, \sigma_2, \sigma_3 are aligned with the basis vectors \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3, then the expressions for \sigma,\tau are : \begin{align} \sigma & = n_1^2 \sigma_{1} + n_2^2 \sigma_{2} + n_3^2 \sigma_{3} \\ \tau & = \sqrt{(n_1\sigma_{1})^2 + (n_2\sigma_{2})^2 + (n_3\sigma_{3})^2 - \sigma^2} \\ & = \sqrt{n_1^2 n_2^2 (\sigma_1-\sigma_2)^2 + n_2^2 n_3^2 (\sigma_2-\sigma_3)^2 + n_3^2 n_1^2 (\sigma_3 - \sigma_1)^2}. \end{align} The Mohr–Coulomb failure criterion can then be evaluated using the usual expression \tau = \sigma~\tan(\phi) + c for the six planes of maximum shear stress. : == Mohr–Coulomb failure surface in Haigh–Westergaard space ==