CU test on NC clay, three samples

Step 1: Principal stresses. $\sigma_1 = \sigma_3 + \sigma_d$, effective: $\sigma' = \sigma - u$.

Step 2: Plot. Draw three total stress circles (diameters 200 to 444, 300 to 614, 400 to 784) and three effective stress circles (145 to 389, 193 to 507, 241 to 625). Draw a common tangent for each set.

 tau
  |       _.-'''-._   total-stress envelope
  |   _.-'  _.-''-. `-.     (phi_cu, c_cu)
  |  /   .-'  effective  \   envelope is steeper
  +-+--+----+-----+----+---+---- sigma
   145 200  241   389  444 ...

Step 3: Parameters. Both sets of points fit straight lines of the form $\sigma_1 = N\sigma_3 + 2c\sqrt{N}$.

Total stress: slope $N = \dfrac{614-444}{300-200} = 1.70$, intercept $= 104$ kPa.

$$\phi_{cu} = \sin^{-1}\frac{N-1}{N+1} = \sin^{-1}\frac{0.70}{2.70} = 15.0^\circ, \qquad c_{cu} = \frac{104}{2\sqrt{1.70}} = 39.9\ \text{kPa}$$

Effective stress: slope $N' = \dfrac{507-389}{193-145} = 2.458$, intercept $= 32.5$ kPa.

$$\phi' = \sin^{-1}\frac{1.458}{3.458} = 24.9^\circ, \qquad c' = \frac{32.5}{2\sqrt{2.458}} = 10.4\ \text{kPa}$$

Answer: total stress $c_{cu} \approx 40$ kN/m², $\phi_{cu} \approx 15^\circ$; effective stress $c' \approx 10$ kN/m², $\phi' \approx 25^\circ$. (For an ideal normally consolidated clay the intercepts would be close to zero; the small cohesion shown here comes from the given data.)