The correct answer is: $\frac{{{{\text{p}}_1} + {{\text{p}}_2}}}{2} + \frac{{{{\text{p}}_1} – {{\text{p}}_2}}}{2}\sin \,2\theta$
The normal stress on any plane inclined at angle $\theta$ to the principal plane carrying the principal stress $p_1$ is given by the following equation:
$$\sigma = \frac{{{{\text{p}}_1} + {{\text{p}}_2}}}{2} + \frac{{{{\text{p}}_1} – {{\text{p}}_2}}}{2}\sin \,2\theta$$
where:
- $\sigma$ is the normal stress on the inclined plane
- $p_1$ is the major principal stress
- $p_2$ is the minor principal stress
- $\theta$ is the angle between the inclined plane and the principal plane carrying the principal stress $p_1$
The equation can be derived using the Mohr-Coulomb failure criterion. The Mohr-Coulomb failure criterion states that failure occurs when the shear stress on a plane reaches a critical value, which is dependent on the normal stress on the plane and the cohesion and angle of internal friction of the soil.
The shear stress on an inclined plane can be calculated using the following equation:
$$\tau = \frac{{{{\text{p}}_1} – {{\text{p}}_2}}}{2}\tan \,\phi$$
where:
- $\tau$ is the shear stress on the inclined plane
- $\phi$ is the angle of internal friction of the soil
Substituting this equation into the Mohr-Coulomb failure criterion gives the following equation:
$$\frac{{{{\text{p}}_1} – {{\text{p}}_2}}}{2}\tan \,\phi = \frac{{{{\text{p}}_1} + {{\text{p}}_2}}}{2} + \frac{{{{\text{p}}_1} – {{\text{p}}_2}}}{2}\sin \,2\theta$$
Solving for $\sigma$ gives the following equation:
$$\sigma = \frac{{{{\text{p}}_1} + {{\text{p}}_2}}}{2} + \frac{{{{\text{p}}_1} – {{\text{p}}_2}}}{2}\sin \,2\theta$$