Drained and CU tests on two samples

For normally consolidated clay, $c' = 0$ and $c_{cu} = 0$, so $\sin\phi = \dfrac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3}$.

(i) $\phi'$ from the drained test

$\sigma_3' = 300$, $\sigma_1' = 300+650 = 950$ kPa.

$$\sin\phi' = \frac{650}{300+950} = \frac{650}{1250} = 0.52 \;\Rightarrow\; \phi' = 31.3^\circ$$

(ii) $\phi$ from the CU test (total stress)

$\sigma_3 = 200$, $\sigma_1 = 200+250 = 450$ kPa.

$$\sin\phi_{cu} = \frac{250}{200+450} = \frac{250}{650} = 0.385 \;\Rightarrow\; \phi_{cu} = 22.6^\circ$$

(iii) Pore pressure at failure in the CU test

Assume the same effective friction angle holds, $\phi' = 31.3^\circ$, and $c'=0$. At failure the effective Mohr circle has the same deviator stress, 250 kPa:

$$\sin\phi' = \frac{\sigma_1'-\sigma_3'}{\sigma_1'+\sigma_3'} = \frac{250}{2\sigma_3' + 250} = 0.52$$ $$2\sigma_3' + 250 = \frac{250}{0.52} = 480.8 \;\Rightarrow\; \sigma_3' = 115.4\ \text{kPa}$$ $$u_f = \sigma_3 - \sigma_3' = 200 - 115.4 = 84.6\ \text{kPa}$$

Answer: (i) $\phi' = 31.3^\circ$; (ii) $\phi_{cu} = 22.6^\circ$; (iii) $u_f \approx 84.6$ kPa.