Question

A consolidated drained (CD) triaxial test was carried out on a sand sample with the known effective shear strength parameters, \(c' = 0\) and \(\phi' = 30^\circ\). In the test, prior to the failure, when the sample was undergoing axial compression under constant cell pressure, the drainage valve was accidentally closed. At the failure, 360 kPa deviatoric stress was recorded along with 70 kPa pore water pressure. If the test is repeated without such error, and no back pressure is applied in either of the tests, what is the deviatoric stress (in kPa, in integer) at the failure?

Hint:

In triaxial testing, when pore water pressure is included, the total stress at failure consists of both deviatoric stress and pore water pressure. When the drainage valve is closed, pore water pressure must be accounted for, but in the corrected test, with no pore water pressure, the deviatoric stress equals the total stress minus the pore pressure.

Answer 1

We are given the following conditions in the problem: - The deviatoric stress at failure in the initial test with the drainage valve closed is \(360 \, \text{kPa}\). - The pore water pressure at failure is \(70 \, \text{kPa}\). - The test is repeated without the drainage valve error and no back pressure is applied. Now, let's calculate the total stress at failure in the initial test, where the drainage valve was closed: \[ \sigma_{\text{total}} = \sigma_d + u = 360 \, \text{kPa} + 70 \, \text{kPa} = 430 \, \text{kPa} \] where: - \(\sigma_d\) is the deviatoric stress. - \(u\) is the pore water pressure. In the repeated test, no pore water pressure will be present since the drainage valve is open. Therefore, the deviatoric stress will be equal to the total stress measured earlier: \[ \sigma_d = \sigma_{\text{total}} - u = 430 \, \text{kPa} - 70 \, \text{kPa} = 500 \, \text{kPa} \] Thus, the deviatoric stress at failure in the corrected test is \(\boxed{500 \, \text{kPa}}\).