A sample of clay (Sample A) is isotropically normally consolidated to cell pressure of 400kPa and

ν0=2.052 (i.e. ν0 is specific volume). A second sample (Sample B) is istropically consolidated to

863kPa and allowed to swell to a mean effective stress of 40kPa when ν0=2.052. Both samples

were then subjected to standard undrained compression tests. The value of the soil constants for

clay are:

M=0.95, =3.15 and =0.19

Find pore pressure at failure of each sample.

To find the pore pressure at failure for both samples, we'll need to utilize concepts from critical state soil mechanics. The pore pressure at failure in undrained conditions can be related to the mean effective stress and deviatoric stress at failure using the following steps:

### Given:
- For both samples:  
  - \(\nu_0 = 2.052\)
  - \(M = 0.95\)
  - \(\Gamma = 3.15\)
  - \(\lambda = 0.19\)

- **Sample A**:
  - Initial isotropic consolidation pressure (\(p_c\)) = 400 kPa

- **Sample B**:
  - Initial isotropic consolidation pressure (\(p_c\)) = 863 kPa
  - Final mean effective stress after swelling (\(p_0'\)) = 40 kPa

### Step 1: Determine the critical state line (CSL) equation.
The critical state line in the \(\nu\) (specific volume) vs \(\ln p'\) (logarithm of mean effective stress) space is given by:

\[
\nu = \Gamma - \lambda \ln p'
\]

### Step 2: Determine specific volumes at the beginning of undrained tests.
Both samples have the same specific volume at the start of the undrained compression test:
\[
\nu_0 = 2.052
\]

### Step 3: Calculate the mean effective stress at critical state (\(p_f'\)) for each sample.
We use the CSL equation for each sample:

\[
p_f' = \exp \left(\frac{\Gamma - \nu_0}{\lambda}\right)
\]

Substituting the given values:

\[
p_f' = \exp \left(\frac{3.15 - 2.052}{0.19}\right) = \exp(5.7632) \approx 319 kPa
\]

This value is the same for both samples as they have the same specific volume at the start of the undrained test.

### Step 4: Calculate the pore pressure at failure (\(u_f\)).
The undrained pore pressure at failure is given by the difference between the initial mean effective stress and the mean effective stress at failure:

\[
u_f = p_c - p_f'
\]

#### For Sample A:
\[
u_{fA} = 400 \, \text{kPa} - 319 \, \text{kPa} = 81 \, \text{kPa}
\]

#### For Sample B:
Here, we take into account the swelling to a mean effective stress of 40 kPa:

\[
u_{fB} = 40 \, \text{kPa} - 319 \, \text{kPa} = -279 \, \text{kPa}
\]

Negative pore pressure (i.e., suction) could occur, but since pore pressure cannot be negative in practice, we interpret this as a state where the soil would be under significant tension and might cavitate, or it might indicate that failure occurs before this condition is reached.

### Final Results:
- **Pore pressure at failure for Sample A**: \( u_{fA} = 81 \, \text{kPa} \)
- **Pore pressure at failure for Sample B**: \( u_{fB} = -279 \, \text{kPa} \) (indicative of cavitation or tension)

### Would you like more details or have any questions?
Here are five related questions to deepen understanding:
1. How does the critical state line influence soil behavior in undrained conditions?
2. What is the physical interpretation of negative pore pressure in soils?
3. Why is the same specific volume used for calculating \(p_f'\) for both samples?
4. How would the results change if Sample B had not been allowed to swell?
5. What factors could affect the value of \(M\) in different types of soils?

**Tip**: In soil mechanics, understanding the relationship between effective stress, specific volume, and the critical state line is crucial for predicting soil behavior under different loading conditions.

Explore how to determine the pore pressure at failure for two isotropically consolidated clay samples (Sample A and Sample B) using critical state soil mechanics principles. Includes detailed calculations and interpretations of the results.

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