Question 6 of 6: Pore pressure and Skempton coefficients from a CU triaxial test

Question 6: Pore pressure and Skempton coefficients from a CU triaxial test (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. CU test: cell pressure $\sigma_3 = 100\text{ kPa}$, deviator at failure $\sigma_d = 60\text{ kPa}$ (so total $\sigma_1 = 160\text{ kPa}$). Effective (CD) strength: $c'=0,\ \phi'=30^\circ$. Total-stress (CU) strength: $c=0,\ \phi_{cu}=13.3^\circ$.

Find. The pore pressure $u_w$ at failure, the Skempton coefficients $A$ and $B$, and whether the clay is normally or over-consolidated (with three justifications).

Approach. The effective-stress failure envelope is unique to the soil, so apply the CD parameters to the effective stresses in the CU test to back-figure $u_w$; then use Skempton’s definitions for $A$ and $B$.

1. Effective-stress failure ratio (from CD). With $c'=0$, at failure $$\frac{\sigma_1'}{\sigma_3'}=\tan^2\!\left(45^\circ+\tfrac{\phi'}{2}\right)=\tan^2(60^\circ)=3.0.$$
2. Solve for the pore pressure. The effective stresses are $\sigma_1'=160-u_w$ and $\sigma_3'=100-u_w$, so $$160-u_w = 3\,(100-u_w)\ \Rightarrow\ 2u_w = 140\ \Rightarrow\ \boxed{u_w = 70\ \text{kPa}.}$$ Check: $\sigma_3'=30$, $\sigma_1'=90$, ratio $=3.0$ ✓. This also reproduces the CU envelope: $\sigma_1/\sigma_3=1.60=\tan^2(45^\circ+\tfrac{13.3^\circ}{2})$ ✓.
3. Coefficient B. The specimen is saturated, so a change in cell pressure is carried entirely by the pore water: $$B=\frac{\Delta u}{\Delta\sigma_3}=1.0.$$
4. Coefficient A (at failure). During the shear stage the cell pressure is constant ($\Delta\sigma_3=0$), so Skempton’s equation $\Delta u = B[\Delta\sigma_3 + A(\Delta\sigma_1-\Delta\sigma_3)]$ reduces to $\Delta u = A\,\sigma_d$. The failure pore pressure of 70 kPa was all generated by shear, hence $$A_f=\frac{\Delta u}{\sigma_d}=\frac{70}{60}=\boxed{1.17.}$$
5. Normally or over-consolidated? The clay is normally consolidated. Three reasons:
   1. The effective-stress envelope passes through the origin ($c'=0$); an over-consolidated clay would show a cohesion intercept.
   2. The pore-pressure parameter $A_f=1.17$ is large and positive — normally consolidated clays give $A_f$ roughly 0.5–1.0+ (sensitive clays even higher), whereas heavily over-consolidated clays give low or negative $A_f$.
   3. Positive pore pressure was generated throughout shear (the soil is contractive, tending to reduce volume); over-consolidated (dense) clays dilate and develop negative pore pressure at failure.