Question

A conventional drained triaxial compression test was conducted on a normally consolidated clay sample under an effective confining pressure of 200 kPa. The deviator stress at failure was found to be 400 kPa. An identical specimen of the same clay sample is isotropically consolidated to a confining pressure of 200 kPa and subjected to standard undrained triaxial compression test. If the deviator stress at failure is 150 kPa, the pore pressure developed (in kPa, upto one decimal place) is .

• A. 125

Answer 1

The correct option is A 125
For normally consolidated clay C = 0

NOTE: Here ${\sigma }_a={\sigma }_d$

CD Test :

${\sigma }_d=400KPa,{\overline{\sigma }}_3=200KPa$

${\overline{\sigma }}_1={\overline{\sigma }}_d+{\overline{\sigma }}_3=400+200=600KPa$

${\overline{\sigma }}_1=\overline{{\sigma }_3}ta{n}^2(45+\frac{{\varphi }^{\prime }}{2})+2Dtan(45+\frac{{\varphi }^{\prime }}{2})$

$600=200ta{n}^2(45+\frac{\varphi }{2})$

${\varphi }^{\prime }={30}^{\circ }$

Now,

CU Test

Now,
${\overline{\sigma }}_1={\overline{\sigma }}_3(45+\frac{{\varphi }^{\prime }}{2})+2Ctan(45+\frac{{\varphi }^{\prime }}{2})$

${\sigma }_c+\Delta {\sigma }_a-U_0=[{\sigma }_c-U_0]ta{n}^2{60}^{\circ }+0$

[200 + 150 - $U_0$] = [200 - $U_0$] $\times$ 3

$2U_0=600-350$

$U_0=125KPa$