Major principal stress=8mN/m$^{2}$

Minor principal stress=2mN/m$^{2}$

Angle of inclination of rupture plane is 60$^{\circ}$ to the horizontal. Present the above data by means of a Mohr's circle of stress diagram. Find the Cohesion and angle of internal friction.

• The minor and major principal stresses are plotted as OA and OB to a convenient scale on the σ-axis. The mid-point of AB is located as C. With C as centre and CA or CB as radius, the Mohr’s stress circle is drawn.
• Angle BCD is plotted as 2θcr or 2 × 60° = 120° to cut the circle in D.
• A tangent to the circle drawn at D (perpendicular to CD) gives the strength envelope. The intercept of this envelope, on the τ-axix gives the cohesion, c, and the inclination of the envelope with σ-axis gives the angle of internal friction,30° and Cohesion ,C =.575MN/m^2