Tlie following information was obtained from laboratory tests on specimens from a complete...

7272-13-25P AID: 1825 | 28/06/2013

RID: 3213 | 03/07/2013

Consider the test result of the drained test conducted under the direct shear procedure in a

completely saturated clay sample.

Soil is normally consolidated as the preconsolidation pressure \(\sigma_{p}^{\prime}\) is less than the normal stress

\(\sigma\) in the direct shear test.

Find the angle of internal friction \(\phi^{\prime}\) for effective condition using the formula:

$$ \tan \phi^{\prime}=\frac{\tau}{\sigma} $$

Denote shear strength of the soil as \(\tau\).

Substitute the values of \(350 \mathrm{kPa}\) for \(\tau\) and \(600 \mathrm{kPa}\) for \(\sigma\) to get:

$$ \begin{aligned} \tan \phi^{\prime} &=\frac{350}{600} \\ \phi^{\prime} &=30.26^{\circ} \end{aligned} $$

Therefore, the angle of internal friction (effective) of normally consolidated clay is \(\sqrt{30.26^{\circ}}\)

Consider the test result of the undrained test conducted under direct shear procedure in a

completely saturated clay sample.

Find the angle of internal friction \(\phi^{\prime}\) for total stress condition using the formula:

$$ \tan \phi^{\prime}=\frac{\tau}{\sigma} $$

Denote shear strength of the soil as \(\tau\).

Substitute the values of \(175 \mathrm{kPa}\) for \(\tau\) and \(600 \mathrm{kPa}\) for \(\sigma\) to get:

$$ \begin{aligned} \tan \phi^{\prime} &=\frac{175}{600} \\ \phi^{\prime} &=16.26^{\circ} \end{aligned} $$

Therefore, the angle of internal friction (total) of normally consolidated clay is \(16.26^{\circ}\).

Show the Mohr failure envelope for a series of direct shear tests performed under drained and

undrained conditions as shown in Fiqure (1).