Dry sand failure plane 25°, find σ3

Given: dry sand ($c=0$), deviator stress at failure $\sigma_1-\sigma_3 = 500$ kN/m². The failure plane makes $25^\circ$ with the axis of the sample.

Step 1: Angle of the failure plane. The axis of the sample is the direction of the major principal stress $\sigma_1$, so the failure plane makes $25^\circ$ with the direction of $\sigma_1$ and hence $90^\circ - 25^\circ = 65^\circ$ with the major principal plane (the horizontal plane). The theoretical failure angle is

$$\theta = 45^\circ + \frac{\phi}{2} = 65^\circ \;\Rightarrow\; \phi = 40^\circ$$

Step 2: Flow value

$$N_\phi = \tan^2\left(45^\circ+\frac{\phi}{2}\right) = \tan^2 65^\circ = 4.599$$

Step 3: Lateral pressure. For $c=0$:

$$\sigma_1 = N_\phi\sigma_3 \;\Rightarrow\; \sigma_1 - \sigma_3 = \sigma_3(N_\phi - 1)$$ $$\sigma_3 = \frac{500}{4.599 - 1} = \frac{500}{3.599} = 138.9\ \text{kN/m}^2$$

Check: $\sigma_1 = 638.9$, $\sin\phi = \frac{500}{638.9+138.9} = 0.643 = \sin40^\circ$.

Answer: $\phi = 40^\circ$ and the lateral pressure $\sigma_3 \approx 139\ \text{kN/m}^2$.