The soil friction angle $\phi$ can be determined using the results of the drained triaxial test. For a normally consolidated clay, the friction angle is derived using the relationship between the confining pressure, deviator stress, and effective stress at failure.

In a drained triaxial test, the deviator stress $\sigma_d$ is the difference between the axial stress $\sigma_1$ (major principal stress) and the confining pressure $\sigma_3$ (minor principal stress):

$\sigma_d = \sigma_1 - \sigma_3$

Given:

• Confining pressure ($\sigma_3$) = 145 kN/m²
• Deviator stress ($\sigma_d$) = 198 kN/m²

To find the major principal stress $\sigma_1$:

$\sigma_1 = \sigma_3 + \sigma_d = 145 + 198 = 343 \, \text{kN/m²}$

Now, the relationship between the major and minor principal stresses and the friction angle $\phi$ is given by:

$\frac{\sigma_1}{\sigma_3} = \frac{1 + \sin\phi}{1 - \sin\phi}$

Substituting the values of $\sigma_1$ and $\sigma_3$:

$\frac{343}{145} = \frac{1 + \sin\phi}{1 - \sin\phi}$

First, simplify the ratio:

$\frac{343}{145} = 2.3655$

Now, solve for $\sin\phi$:

$2.3655 = \frac{1 + \sin\phi}{1 - \sin\phi}$

Multiplying both sides by $1 - \sin\phi$:

$2.3655(1 - \sin\phi) = 1 + \sin\phi$

Expanding the equation:

$2.3655 - 2.3655\sin\phi = 1 + \sin\phi$

Combine like terms:

$2.3655 - 1 = \sin\phi + 2.3655\sin\phi$

$1.3655 = (1 + 2.3655)\sin\phi$

$1.3655 = 3.3655\sin\phi$

Solving for $\sin\phi$:

$\sin\phi = \frac{1.3655}{3.3655} = 0.4058$

Now, to find $\phi$:

$\phi = \arcsin(0.4058) \approx 23.94^\circ$

Therefore, the soil friction angle $\phi$ is approximately $23.94^\circ$.

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Would you like further details or have any questions?

Here are 5 related questions:

1. What is the significance of the soil friction angle in geotechnical design?
2. How would the results change for an undrained triaxial test on the same specimen?
3. What assumptions are made when using the Mohr-Coulomb failure criterion in triaxial tests?
4. How does the effective stress principle apply in a drained triaxial test?
5. Can the presence of cohesion affect the calculation of the friction angle?

Tip: In normally consolidated clays, the friction angle tends to be constant, while the cohesion is often assumed to be zero.