Question

Undisturbed samples from a normally consolidated clay layer were collected during a subsoil exploration. Drained triaxial tests showed that the effective friction angle was φ’ = 25°. The unconfined compressive strength, qu, of a similar specimen was found to be 133 kN/m2. Find the pore pressure at failure for the unconfined compression test.

Answer 1

Answer:

pre water pressure $= \sigma_1 = 208.76 kN/m2$

Explanation:

given data:

frictional angle  25°

unconfined xompressive strength  qu = 133 kN/m2

for clay, unconfined compressive strenth is cu

$cu =\frac{qu}{2}$

cu = 133/2 =66.5 kN/m2

cell pressure $\sigma_3$ is zero in uncomfined compressive strenth test

from principle effective stress we have

$\sigma_1 = \sigma_2 tan^{2}(45+ \frac{\theta}{2}) +2cu tan(45+ \frac{\theta}{2})$

$\sigma_1 = 2cu tan(45+ \frac{\theta}{2})$

$\sigma_1 = 2*66.5 tan(45+ \frac{25}{2})$

$\sigma_1 = 208.76 kN/m2$

as this test is quick test i.e. undrained test therefore

pre water pressure $= \sigma_1 = 208.76 kN/m2$