Derive Mohr-Coulomb principal stress relation

Mohr's failure theory

Mohr's theory states that a material fails when the shear stress on a plane reaches a limiting value that depends on the normal stress on that plane. Failure is therefore governed by a combination of normal and shear stress, not by the maximum stress alone:

$$\tau_f = f(\sigma)$$

The curve of $\tau_f$ against $\sigma$ is the failure envelope. Coulomb approximated it for soils by a straight line, giving the Mohr-Coulomb criterion:

$$\tau_f = c + \sigma\tan\phi \qquad \text{(effective stress: } \tau_f = c' + \sigma'\tan\phi'\text{)}$$

Here $c$ is cohesion and $\phi$ is the angle of internal friction. A stress state is safe if its Mohr circle lies below the line, and the soil fails when the circle touches it.

Derivation of the relation between $\sigma_1$ and $\sigma_3$

  tau |          . failure line
      |        .   tau = c + sigma tan(phi)
      |   _.-'-._.
      | .'   |   '.
      |/  phi|  C   \
 -----+-----+-+---+---+------ sigma
      c cot(phi) O   (s3)  (s1)

Consider the Mohr circle at failure, touching the failure line at point $T$. Its centre $C$ is at $\frac{\sigma_1+\sigma_3}{2}$ and its radius is $\frac{\sigma_1-\sigma_3}{2}$. The failure line meets the $\sigma$-axis at $O'$, a distance $c\cot\phi$ to the left of the origin. In the right-angled triangle $O'TC$, the angle at $O'$ is $\phi$ and $CT \perp O'T$:

$$\sin\phi = \frac{CT}{O'C} = \frac{\dfrac{\sigma_1-\sigma_3}{2}}{c\cot\phi + \dfrac{\sigma_1+\sigma_3}{2}}$$ $$\sigma_1 - \sigma_3 = (\sigma_1+\sigma_3)\sin\phi + 2c\cos\phi$$ $$\sigma_1(1-\sin\phi) = \sigma_3(1+\sin\phi) + 2c\cos\phi$$ $$\boxed{\sigma_1 = \sigma_3\,\frac{1+\sin\phi}{1-\sin\phi} + 2c\,\frac{\cos\phi}{1-\sin\phi}}$$

Using $\frac{1+\sin\phi}{1-\sin\phi} = \tan^2\left(45^\circ+\frac{\phi}{2}\right)$ and $\frac{\cos\phi}{1-\sin\phi} = \tan\left(45^\circ+\frac{\phi}{2}\right)$:

$$\sigma_1 = \sigma_3\tan^2\left(45^\circ+\frac{\phi}{2}\right) + 2c\tan\left(45^\circ+\frac{\phi}{2}\right)$$

Equivalent form: $\sigma_3 = \sigma_1\tan^2\left(45^\circ-\frac{\phi}{2}\right) - 2c\tan\left(45^\circ-\frac{\phi}{2}\right)$.

Failure plane angle: in triangle $O'TC$ the angle $\angle TCO'$ is $90^\circ-\phi$. The radius $CT$ therefore makes an angle $2\theta_f = 180^\circ - (90^\circ-\phi) = 90^\circ+\phi$ with the radius to the $\sigma_1$ point, so the failure plane is inclined at $\theta_f = 45^\circ + \frac{\phi}{2}$ to the major principal plane.

For cohesionless soil ($c=0$): $\sigma_1 = \sigma_3\tan^2(45^\circ+\phi/2)$. For a saturated clay in undrained loading ($\phi_u=0$): $\sigma_1 - \sigma_3 = 2c_u$.