Principal Stress Calculation for Sandy Soil

This problem involves calculating the minor principal stress ($\sigma_3$) at failure for a sandy soil, given the major principal stress ($\sigma_1$) and the angle of internal friction ($\phi$). Sandy soil is typically considered cohesionless, meaning it has no inherent shear strength due to cohesion.

Mohr-Coulomb Failure Criterion

The relationship between the major and minor principal stresses at failure for a cohesionless soil (where cohesion, $c=0$) is described by the Mohr-Coulomb failure criterion. The formula relating $\sigma_1$ and $\sigma_3$ is:

$$ \sigma_1 = \sigma_3 \left( \frac{1 + \sin\phi}{1 - \sin\phi} \right) $$

In this formula:

• $\sigma_1$ represents the major principal stress at failure.
• $\sigma_3$ represents the minor principal stress at failure.
• $\phi$ is the angle of internal friction.

Applying Values and Calculating Minor Principal Stress

We are given the following values:

• Major principal stress, $\sigma_1 = 50$ kN/m$^2$.
• Angle of internal friction, $\phi = 30^\circ$.

First, let's calculate the value of $\sin\phi$:

$$ \sin(30^\circ) = 0.5 $$

Now, substitute this value into the formula:

$$ \sigma_1 = \sigma_3 \left( \frac{1 + 0.5}{1 - 0.5} \right) $$

$$ \sigma_1 = \sigma_3 \left( \frac{1.5}{0.5} \right) $$

$$ \sigma_1 = \sigma_3 (3) $$

We need to find the minor principal stress, $\sigma_3$. Rearranging the formula:

$$ \sigma_3 = \frac{\sigma_1}{3} $$

Now, substitute the given value for $\sigma_1$:

$$ \sigma_3 = \frac{50 \text{ kN/m}^2}{3} $$

$$ \sigma_3 \approx 16.67 \text{ kN/m}^2 $$

Result Confirmation

The calculated minor principal stress is approximately $16.67$ kN/m$^2$. Comparing this with the given options, $16.66$ kN/m$^2$ is the closest value.