Dry sand triaxial, φ = 36°, σ3 = 100

For dry cohesionless soil, $c = 0$ and

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

Given: $\phi = 36^\circ$, $\sigma_3 = 100$ kN/m².

$$\tan^2(63^\circ) = (1.9626)^2 = 3.852$$ $$\sigma_1 = 100 \times 3.852 = 385.2\ \text{kN/m}^2$$ $$\sigma_d = \sigma_1 - \sigma_3 = 385.2 - 100 = 285.2\ \text{kN/m}^2$$

Check: $\sin\phi = \frac{285.2}{385.2+100} = 0.588 = \sin36^\circ$.

Answer: deviator stress at failure $\approx 285\ \text{kN/m}^2$.