1.21 For a sand given: D_(85)=0.21mm D_(50)=0.13mm D_(15)=0.09mm Uniformity coefficient. C_(n)=2.1 Void ratio. e=0.68 Relative density =53% Estimate the soil friction angle using Eq.(1.83) b Eq.(1.84) 1.22 A consolidated -drained triaxial test on a normally consolidated clay yields the fol- lowing results. All around confining pressure. sigma _(3)'=138kN/m^2 Added axial stress at failure. Delta sigma =276kN/m^2 Determine the shear-strength parameters.

1.21 Estimating the Soil Friction Angle

Given:

- \( D_{85} = 0.21 \, \text{mm} \)
- \( D_{50} = 0.13 \, \text{mm} \)
- \( D_{15} = 0.09 \, \text{mm} \)
- Uniformity coefficient, \( C_u = 2.1 \)
- Void ratio, \( e = 0.68 \)
- Relative density = 53%

We need to estimate the soil friction angle using the given equations.

Using Eq. (1.83):

The formula for the soil friction angle (\(\phi\)) using the uniformity coefficient is:
\[ \tan \phi = \frac{\log_{10}(C_u)}{\log_{10}(e^2) + 3 \log_{10}(e) + 3} \]

Substitute the given values:
\[ \tan \phi = \frac{\log_{10}(2.1)}{\log_{10}(0.68^2) + 3 \log_{10}(0.68) + 3} \]

Calculate each term:
\[ \log_{10}(2.1) \approx 0.322 \]
\[ \log_{10}(0.68^2) \approx \log_{10}(0.4624) \approx -0.336 \]
\[ 3 \log_{10}(0.68) \approx 3 \times (-0.169) \approx -0.507 \]

Now, sum up the terms:
\[ \tan \phi = \frac{0.322}{-0.336 - 0.507 + 3} \approx \frac{0.322}{2.157} \approx 0.149 \]

Finally, find the friction angle:
\[ \phi = \tan^{-1}(0.149) \approx 8.6^\circ \]

Using Eq. (1.84):

The formula for the soil friction angle (\(\phi\)) using the relative density (\(D_r\)) is:
\[ \tan \phi = \frac{D_r}{e^2 + 0.35 e + 0.37} \]

Substitute the given values:
\[ \tan \phi = \frac{0.53}{0.68^2 + 0.35 \times 0.68 + 0.37} \]

Calculate each term:
\[ 0.68^2 \approx 0.4624 \]
\[ 0.35 \times 0.68 \approx 0.238 \]

Now, sum up the terms:
\[ \tan \phi = \frac{0.53}{0.4624 + 0.238 + 0.37} \approx \frac{0.53}{1.07} \approx 0.494 \]

Finally, find the friction angle:
\[ \phi = \tan^{-1}(0.494) \approx 26.6^\circ \]

1.22 Determining the Shear-Strength Parameters

Given:

- Confining pressure, \(\sigma_3' = 138 \, \text{kN/m}^2\)
- Added axial stress at failure, \(\Delta \sigma = 276 \, \text{kN/m}^2\)

The total stress at failure (\(\sigma_1'\)) is:
\[ \sigma_1' = \sigma_3' + \Delta \sigma = 138 + 276 = 414 \, \text{kN/m}^2 \]

The major principal stress (\(\sigma_1\)) is:
\[ \sigma_1 = \sigma_1' = 414 \, \text{kN/m}^2 \]

The minor principal stress (\(\sigma_3\)) is:
\[ \sigma_3 = \sigma_3' = 138 \, \text{kN/m}^2 \]

The intermediate principal stress (\(\sigma_2\)) is:
\[ \sigma_2 = \frac{\sigma_1 + \sigma_3}{2} = \frac{414 + 138}{2} = 276 \, \text{kN/m2 \]

The shear stress (\(\tau\)) is:
\[ \tau = \frac{\sigma_1 - \sigma_3}{2} = \frac{414 - 138}{2} = 238 \, \text{kN/m}^2 \]

The Mohr-Coulomb failure criterion is given by:
\[ \tau = c + \sigma_n \tan \phi \]

Where:
- \(c\) is the cohesion intercept.
- \(\sigma_n\) is the normal stress.
- \(\phi\) is the friction angle.

Assuming a typical clay behavior, let's