07-Str-A3 · December 2014

Question 6 of 6: CU triaxial test on a saturated clay

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 6: CU triaxial test on a saturated clay (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. Consolidated-drained triaxial tests on the saturated clay gave the effective strength parameters $c' = 10$ kPa and $\phi' = 28^\circ$. A consolidated-undrained test on an identical specimen of the same clay is run at a total cell pressure $\sigma_3 = 100$ kPa, and at failure the measured pore-water pressure is $u_w = 40$ kPa. The specimen is fully saturated, so the effective stress principle $\sigma' = \sigma - u_w$ applies without a suction correction.

Find. The total vertical stress $\sigma_1$ acting on the specimen at failure; then, in prose, what the CU test tells the senior engineer, and when CD and CU tests are each appropriate.

Approach. Failure is governed by effective stress, and the effective-stress envelope is the same one the CD tests produced. So convert the cell pressure to an effective stress, apply the Mohr–Coulomb relation on the formula sheet to obtain $\sigma'_1$, and convert back to total stress by adding the pore pressure.

1. Convert the confining stress to effective stress. The pore pressure at failure acts equally in all directions, so $$\sigma'_3 \;=\; \sigma_3 - u_w \;=\; 100 - 40 \;=\; 60\ \text{kPa}$$
2. Apply the effective-stress failure criterion. The relation given on the formula sheet is $$\sigma'_1 \;=\; \sigma'_3\tan^{2}\!\left(45^\circ + \frac{\phi'}{2}\right) + 2c'\tan\!\left(45^\circ + \frac{\phi'}{2}\right)$$ With $\phi' = 28^\circ$ the half-angle term is $45^\circ + 14^\circ = 59^\circ$, so $\tan 59^\circ = 1.6643$ and $\tan^{2}59^\circ = N_\phi = 2.7698$. This is the step where the CD result earns its keep: the envelope belongs to the soil, not to the test, so the same $c'$ and $\phi'$ apply to the undrained test provided pore pressure is measured.
3. Substitute and solve for the effective major principal stress. $$\sigma'_1 \;=\; 60(2.7698) + 2(10)(1.6643) \;=\; 166.19 + 33.29$$ $$\sigma'_1 \;=\; \boxed{199.5\ \text{kPa}}$$
4. Convert back to total stress. Adding the same pore pressure to the effective vertical stress, $$\sigma_1 \;=\; \sigma'_1 + u_w \;=\; 199.48 + 40 \;=\; \boxed{239.5\ \text{kPa}}$$ which is the applied vertical total stress the question asks for. Note that the deviator stress is unchanged by the conversion: $\sigma_1 - \sigma_3 = 239.5 - 100 = 139.5$ kPa is identical to $\sigma'_1 - \sigma'_3 = 199.5 - 60 = 139.5$ kPa, because pore pressure is isotropic and cancels in the difference. The total-stress circle is simply the effective-stress circle slid 40 kPa to the right, as Figure 6.1 shows.
5. Extract the two derived quantities the test exists to provide. The radius of the circle is the undrained shear strength at this consolidation pressure, $$c_u \;=\; \frac{\sigma_1 - \sigma_3}{2} \;=\; \frac{139.5}{2} \;=\; 69.7\ \text{kPa}$$ and the pore-pressure response is summarised by Skempton’s coefficient at failure, $$A_f \;=\; \frac{\Delta u}{\sigma_1 - \sigma_3} \;=\; \frac{40}{139.5} \;=\; 0.287$$ Both are used in part (i) below.

(i) What the senior engineer is really asking for

The senior engineer already has $c'$ and $\phi'$ from the drained tests, so the CU test is not being run to repeat them. It is being run because a consolidated-undrained test with pore-pressure measurement yields four things a CD test cannot, and yields them in days rather than weeks.

First, the undrained shear strength $c_u = 69.7$ kPa at a consolidation pressure of 100 kPa. This is the parameter for a total-stress ($\phi = 0$) analysis of any short-term, end-of-construction condition — the bearing capacity of a footing placed on this clay and loaded quickly, or the stability of a cut the day it is excavated. The CD test gives no access to it at all.

Second, the pore-pressure response. The measurement $\Delta u = 40$ kPa at failure gives $A_f = 0.287$: positive, so the clay is contractive and generates excess pressure when sheared, but well below the value near 1.0 typical of a soft normally consolidated clay. That places the material in the lightly over-consolidated range and tells the designer what will happen in the field when a load is applied faster than the ground can drain. It is also the number required to predict pore pressures during staged construction and to decide how long a rest period each stage needs.

Third, a normalised strength ratio. Here $c_u/\sigma'_c = 69.7/100 = 0.70$, against the 0.22–0.25 that SHANSEP-type correlations give for a normally consolidated clay. A ratio three times that value is strong independent evidence that the specimen is over-consolidated or lightly cemented, and it confirms the same conclusion the $A_f$ value points to. Two independent indicators agreeing is exactly the sort of cross-check a senior engineer looks for.

Fourth, confirmation and quality control of the drained parameters. Because the effective-stress envelope is unique to the soil, the CU circle plotted in effective stress must be tangent to the CD envelope — and it is, to within rounding. If it were not, either the specimens are not identical, the back pressure has not saturated them ($B < 0.95$), or the shearing rate was too fast for pore pressure to equalise across the specimen. A single CU test is therefore a cheap and fast audit of an expensive drained test programme.

(ii) When to run a CD test and when to run a CU test

Run a CD test when drainage in the field is fast relative to the rate of loading, so that no excess pore pressure ever exists, and when the governing condition is the long-term, fully drained one. That covers all sands and gravels under normal construction rates, and it covers clays whenever the critical case is years after construction: the long-term stability of a permanent cut slope, the long-term earth pressure on a retaining wall, or a slope in which pore pressures have come into equilibrium with a steady groundwater regime. The Canadian example is a permanent cut slope in the sensitive Champlain Sea clay of the Ottawa Valley, where the long-term drained condition — strength $c'$, $\phi'$, with pore pressures set by the steady-state water table — is what governs decades after the cut is made. The price is time: a CD test on a clay must be sheared slowly enough for pore pressure to stay at zero throughout, which can mean weeks per specimen.

Run a CU test when the soil has consolidated under its existing overburden but the design loading is applied faster than it can drain, so that excess pore pressures govern. Typical cases are end-of-construction stability of an embankment built on soft clay, rapid drawdown of a reservoir on the upstream slope of a dam, and the immediate bearing capacity of a footing on saturated clay. The Canadian example is the staged construction of a highway embankment over the soft deltaic silts and clays of the Fraser River lowland at Richmond, British Columbia: each lift is placed faster than the foundation can drain, so the end-of-construction case is analysed in total stress using $c_u$, while the pore pressures measured in the CU tests, together with field piezometers, set how long the fill must rest before the next lift and how much strength gain to credit.

In practice the CU test with pore-pressure measurement (ASTM D4767) is the workhorse of a commercial laboratory precisely because it delivers both sets of parameters — $c_u$ for the short term and $c'$, $\phi'$ for the long term — from one specimen at a fraction of the time a CD test (ASTM D7181) would take on the same clay.