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 (Value: 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.

Find. The pore-water pressure at failure, Skempton’s pore-pressure coefficients $A$ and $B$ for this specimen, and a reasoned classification of the clay as normally or over consolidated.

Approach. The effective-stress failure criterion with $c' = 0$ ties the two effective principal stresses together through a single constant, so imposing it on the measured total stresses leaves one unknown — the pore pressure — which is then fed into Skempton’s equation to recover $A$, with $B$ following from saturation.

1. Write down the total stresses at failure. The cell pressure is the minor principal stress and the deviator adds to it, $$\sigma_3 = 100\ \text{kPa}, \qquad \sigma_1 = 100 + 60 = 160\ \text{kPa} .$$ As a first consistency check on the data, the total-stress circle these define has $\sin\phi = (\sigma_1 - \sigma_3)/(\sigma_1 + \sigma_3) = 60/260 = 0.231$, giving $\phi = 13.3^{\circ}$ — exactly the CU angle quoted in the question, so the printed parameters and the printed test result describe the same specimen.
2. Impose the effective-stress failure criterion. With $c' = 0$ the Mohr–Coulomb relation on the formula sheet reduces to a proportionality between the effective principal stresses, $$\sigma_1' = \sigma_3' \tan^{2}\left(45^{\circ} + \frac{\phi'}{2}\right) = \sigma_3' \tan^{2}60^{\circ} = 3\,\sigma_3' .$$ Since both effective stresses differ from their total counterparts by the same pore pressure, $\sigma_1' = \sigma_1 - u$ and $\sigma_3' = \sigma_3 - u$.
3. Solve for the pore-water pressure. Substituting the measured totals, $$160 - u = 3\,(100 - u) \;\Longrightarrow\; 2u = 140 \;\Longrightarrow\; \boxed{u = 70\ \text{kPa}} .$$ The effective stresses at failure are therefore $\sigma_3' = 30\ \text{kPa}$ and $\sigma_1' = 90\ \text{kPa}$. Checking back, $\sin\phi' = (90-30)/(90+30) = 0.500$, so $\phi' = 30^{\circ}$ as required — the effective circle is tangent to the drained envelope, which is the whole point of the construction.
4. State the coefficient $B$. Skempton’s coefficient $B$ measures how much of an all-round pressure increment is taken by the pore water, $\Delta u = B\,\Delta\sigma_3$ under isotropic loading. For a fully saturated soil the pore fluid is far more compressible than nothing and far less compressible than the skeleton, so the water carries the entire increment and $$\boxed{B = 1.0} .$$ The specimen is stated to be a saturated clay, and the fact that $B = 1$ would in practice be confirmed at the start of the test by a saturation check before shearing begins.
5. Extract the coefficient $A$ at failure. Skempton’s general expression is $$\Delta u = B\left[\Delta\sigma_3 + A\left(\Delta\sigma_1 - \Delta\sigma_3\right)\right].$$ During the shearing stage of a triaxial test the cell pressure is held constant, so $\Delta\sigma_3 = 0$ and the bracket collapses to the deviator term. With $B = 1$ this leaves $$A_f = \frac{\Delta u}{\Delta\sigma_d} = \frac{70}{60} = \boxed{1.17} .$$ The subscript matters: $A_f$ is the value at failure, and $A$ varies through the test, starting near 0.3 to 0.5 at small strain and climbing as the specimen contracts.
6. The Mohr circles make the result visual. The total-stress circle runs from 100 to 160 kPa and is tangent to the flat $13.3^{\circ}$ total envelope; the effective circle has the same 30 kPa radius but is shifted 70 kPa to the left, from 30 to 90 kPa, where it is tangent to the steeper $30^{\circ}$ effective envelope. The horizontal shift between the two circles is the pore pressure, which is the geometric statement of $\sigma' = \sigma - u$.
   Total-stress (blue, dashed) and effective-stress (red) Mohr circles at failure. The circles have the same radius; the 70 kPa leftward shift of the effective circle is the measured pore-water pressure. Stresses in kPa.
7. Classify the clay: it is normally consolidated. Three independent lines of evidence, as the question requires, and they agree. First, the magnitude of $A_f$. The value 1.17 exceeds unity. Normally consolidated clays typically fall in the range 0.5 to 1.0, with sensitive clays running above 1.0, whereas lightly over-consolidated clays give roughly 0 to 0.5 and heavily over-consolidated clays give negative values. A value above 1 belongs unambiguously to the normally consolidated, and probably sensitive, end of that scale. Second, the sign and size of the pore pressure. The specimen generated 70 kPa of positive excess pore pressure, fully seventy per cent of the cell pressure, and it did so all the way to failure. That is contractive behaviour: the skeleton is trying to reduce its volume as it shears and, being prevented, hands the load to the water. An over-consolidated clay behaves the opposite way, dilating on shear and generating pore pressures that fall and often go negative near failure. Third, the failure envelopes pass through the origin. Both the drained envelope, $c' = 0$ with $\phi' = 30^{\circ}$, and the undrained envelope, $c = 0$ with $\phi = 13.3^{\circ}$, have zero intercept. A normally consolidated clay has never carried a higher effective stress, so it has no locked-in structure to give it a cohesion intercept and its envelope must pass through the origin; an over-consolidated clay shows a curved envelope with an apparent intercept $c' \gt 0$ over the stress range below its preconsolidation pressure.
8. A fourth check, available without further data. The stress path confirms the same conclusion. The effective stress at failure, $\sigma_3' = 30\ \text{kPa}$, is well below the 100 kPa to which the specimen was consolidated: the mean effective stress has fallen during undrained shear, which is the signature of a contractive, normally consolidated soil. A heavily over-consolidated clay moves the other way, its mean effective stress rising as it dilates. Had the clay been over consolidated, the test would also have shown a distinct peak followed by strain softening rather than the ductile, plateau-type response expected here.