Chapter 10, Problem 10.13P

To determine

Find the pore water pressure at failure for the second specimen.

Find the Skempton’s pore pressure parameter at failure.

Answer to Problem 10.13P

The pore water pressure at failure for the second specimen is $\underset{\_}{83.8{\text{kN/m}}^{\text{2}}}$.

The Skempton’s pore pressure parameter at failure is $\underset{\_}{0.73}$.

Explanation of Solution

Given information:

Specimen 1:

The confining pressure of the clay $\left({{\sigma }^{\prime }}_3\right)$ for consolidated drained triaxial test is $150{\text{kN/m}}^{\text{2}}$.

The deviator stress at failure ${\left(\Delta d\right)}_f$ for consolidated drained triaxial test is $260{\text{kN/m}}^{\text{2}}$.

Specimen 2:

The confining pressure of the clay for consolidated undrained triaxial test is $150{\text{kN/m}}^{\text{2}}$.

The deviator stress at failure ${\left(\Delta d\right)}_f$ for consolidated undrained triaxial test is $115{\text{kN/m}}^{\text{2}}$.

Calculation:

The consolidated drained triaxial test was conducted for normally consolidated clay (specimen 1) and the consolidated undrained triaxial test was conducted for specimen 2.

Consider the consolidated drained triaxial test (specimen 1).

Find the major principal effective stress at failure $\left({{\sigma }^{\prime }}_1\right)$ using the formula:

${{\sigma }^{\prime }}_1={{\sigma }^{\prime }}_3+{\left(\Delta {\sigma }_d\right)}_f$

Here, ${{\sigma }^{\prime }}_3$ is confining pressure or minor principal effective stress and ${\left(\Delta {\sigma }_d\right)}_f$ is deviator stress at failure.

Substitute $150{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_3$ and $260{\text{kN/m}}^{\text{2}}$ for ${\left(\Delta {\sigma }_d\right)}_f$.

$\begin{array}{c}{{\sigma }^{\prime }}_1={{\sigma }^{\prime }}_3+{\left(\Delta {\sigma }_d\right)}_f\\ =150{\text{kN/m}}^{\text{2}}+260{\text{kN/m}}^{\text{2}}\\ =410{\text{kN/m}}^{\text{2}}\end{array}$

Find effective friction angle $\left({\varphi }^{\prime }\right)$ using the using Mohr-Coulomb’s failure criteria.

${{\sigma }^{\prime }}_1={{\sigma }^{\prime }}_3{\tan }^2\left(45°+\frac{{\varphi }^{\prime }}{2}\right)+2c^{\prime }\tan \left(45°+\frac{{\varphi }^{\prime }}{2}\right)$

Here, ${{\sigma }^{\prime }}_1$ is major principal stress and ${{\sigma }^{\prime }}_3$ is confining pressure or minor principal stress

Consider that the specimen as normally consolidated clay. Hence the effective stress cohesion $\left(c^{\prime }\right)$ is 0.

Substitute 0 for $c^{\prime }$.

$\begin{array}{l}{{\sigma }^{\prime }}_1={{\sigma }^{\prime }}_3{\tan }^2\left(45°+\frac{{\varphi }^{\prime }}{2}\right)+0\\ {{\sigma }^{\prime }}_1={{\sigma }^{\prime }}_3{\tan }^2\left(45°+\frac{{\varphi }^{\prime }}{2}\right)\end{array}$ (1)

Rearrange the Equation.

$\begin{array}{l}\frac{{{\sigma }^{\prime }}_1}{{{\sigma }^{\prime }}_3}={\tan }^2\left(45°+\frac{{\varphi }^{\prime }}{2}\right)\\ \tan \left(45°+\frac{{\varphi }^{\prime }}{2}\right)={\left(\frac{{{\sigma }^{\prime }}_1}{{{\sigma }^{\prime }}_3}\right)}^{0.5}\\ 45°+\frac{{\varphi }^{\prime }}{2}={\tan }^{-1}{\left(\frac{{{\sigma }^{\prime }}_1}{{{\sigma }^{\prime }}_3}\right)}^{0.5}\\ {\varphi }^{\prime }=2\left[{\tan }^{-1}{\left(\frac{{{\sigma }^{\prime }}_1}{{{\sigma }^{\prime }}_3}\right)}^{0.5}-45°\right]\end{array}$

Substitute $410{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_1$ and $150{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_3$.

$\begin{array}{l}{\varphi }^{\prime }=2\left[{\tan }^{-1}{\left(\frac{410}{150}\right)}^{0.5}-45°\right]\\ {\varphi }^{\prime }=27.7°\end{array}$

Consider the consolidated undrained triaxial test (specimen 2).

Find the major principal stress $\left({\sigma }_1\right)$ using the formula:

${\sigma }_1={\sigma }_3+{\left(\Delta {\sigma }_d\right)}_f$

Here, ${\sigma }_3$ is minor principal stress.

Substitute $150{\text{kN/m}}^{\text{2}}$ for ${\sigma }_3$ and $115{\text{kN/m}}^{\text{2}}$ for ${\left(\Delta {\sigma }_d\right)}_f$.

$\begin{array}{c}{\sigma }_1={\sigma }_3+{\left(\Delta {\sigma }_d\right)}_f\\ =150{\text{kN/m}}^{\text{2}}+115{\text{kN/m}}^{\text{2}}\\ =265{\text{kN/m}}^{\text{2}}\end{array}$

Show the formula for major principal effective stress.

${{\sigma }^{\prime }}_1={\sigma }_1-{\left(\Delta u_d\right)}_f$

Here, ${\left(\Delta u_d\right)}_f$ is pore water pressure at failure.

Substitute $265{\text{kN/m}}^{\text{2}}$ for ${\sigma }_1$.

${{\sigma }^{\prime }}_1=265-{\left(\Delta u_d\right)}_f$ (2)

Show the formula for minor principal effective stress.

${{\sigma }^{\prime }}_3={\sigma }_3-{\left(\Delta u_d\right)}_f$

Substitute $150{\text{kN/m}}^{\text{2}}$ for ${\sigma }_3$.

${{\sigma }^{\prime }}_3=150-{\left(\Delta u_d\right)}_f$ (3)

Calculate the pore water pressure using the Equation (1).

Substitute Equation (2), (3) in Equation (1) and $27.7°$ for ${\varphi }^{\prime }$.

$\begin{array}{l}{{\sigma }^{\prime }}_1={{\sigma }^{\prime }}_3{\tan }^2\left(45°+\frac{{\varphi }^{\prime }}{2}\right)\\ 265-{\left(\Delta u_d\right)}_f=\left(150-{\left(\Delta u_d\right)}_f\right)\left[{\tan }^2\left(45°+\frac{27.7°}{2}\right)\right]\\ 265-{\left(\Delta u_d\right)}_f=\left(150-{\left(\Delta u_d\right)}_f\right)\left(2.737\right)\\ 265-{\left(\Delta u_d\right)}_f=410.55-2.737{\left(\Delta u_d\right)}_f\end{array}$

$\begin{array}{l}1.737{\left(\Delta u_d\right)}_f=145.55\\ {\left(\Delta u_d\right)}_f=83.8{\text{kN/m}}^{\text{2}}\end{array}$

Therefore, the pore water pressure at failure for specimen 2 is $\underset{\_}{83.8{\text{kN/m}}^{\text{2}}}$.

Find the Skempton’s pore pressure parameter $\left(\overline{A}\right)$ at failure using the formula:

$\overline{A}=\frac{{\left(\Delta u_d\right)}_f}{{\left(\Delta {\sigma }_d\right)}_f}$

Substitute $83.8{\text{kN/m}}^{\text{2}}$ for ${\left(\Delta u_d\right)}_f$ and $115{\text{kN/m}}^{\text{2}}$ for ${\left(\Delta {\sigma }_d\right)}_f$.

$\begin{array}{c}\overline{A}=\frac{83.8{\text{kN/m}}^{\text{2}}}{115{\text{kN/m}}^{\text{2}}}\\ =0.73\end{array}$

Therefore, the Skempton’s pore pressure parameter at failure is $\underset{\_}{0.73}$.