Question 6 of 6: Consolidated-Drained Triaxial Series and the Modified Failure Envelope

Question 6: Consolidated-Drained Triaxial Series and the Modified Failure Envelope (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. Three consolidated-drained triaxial tests on identical saturated clay specimens, taken to failure, with axial deformation, axial load and volume change all recorded at failure.

Find. $c'$ and $\phi'$ from a modified ($K_f$) failure envelope; the advantage of that plot; whether the clay is normally or over-consolidated; and whether these parameters serve for the long-term stability of an earth dam built of the same clay.

Approach. Correct the cross-sectional area for both the axial and the volumetric strain at failure, form the deviator stress and hence $\sigma_1'$ for each test, reduce each Mohr circle to its top point $(s', t)$, fit a straight line to the three points by least squares, and convert the line’s slope and intercept to $\phi'$ and $c'$.

1. Initial area and volume. For a 38 mm diameter, 76 mm long specimen, $$A_0 = \frac{\pi d^2}{4} = \frac{\pi (38)^2}{4} = 1134.1\ \text{mm}^2 \qquad V_0 = A_0 L_0 = 1134.1 \times 76 = 86\,190\ \text{mm}^3 = 86.19\ \text{ml}$$ Recording the initial volume in millilitres matters, because the volume change is reported in millilitres.
2. Corrected area at failure. In a drained test the specimen both shortens and loses volume, so the parabolic-area correction must account for both. Assuming the specimen deforms as a right cylinder, $$A_c = A_0\,\frac{1 - \varepsilon_v}{1 - \varepsilon_a} \qquad\text{with}\qquad \varepsilon_a = \frac{\Delta L}{L_0},\quad \varepsilon_v = \frac{\Delta V}{V_0}$$ For Test 1, $\varepsilon_a = 7.22/76 = 0.0950$ and $\varepsilon_v = 5250/86\,190 = 0.0609$, so $A_c = 1134.1 \times 0.9391/0.9050 = 1176.8\ \text{mm}^2$. Note the correction is only about +4 %: the volume loss largely cancels the bulging that the shortening alone would cause, which is exactly why the drained correction differs from the undrained one.
3. Deviator stress and major principal stress. Dividing the axial load by the corrected area gives the deviator stress, and adding the cell pressure gives $\sigma_1'$. For Test 1, $$\Delta\sigma = \frac{P}{A_c} = \frac{480}{1176.8} = 0.4079\ \text{N/mm}^2 = 407.9\ \text{kPa} \qquad \sigma_1' = 200 + 407.9 = 607.9\ \text{kPa}$$ Because drainage was permitted and the pore pressure held at zero, these total stresses are the effective stresses — no correction for $u$ is needed anywhere in this question.
4. Reduce each circle to its top point. The modified plot replaces each Mohr circle by the single point at its apex, using the coordinates given on the formula sheet: $$s' = \tfrac{1}{2}(\sigma_1' + \sigma_3') \qquad t = \tfrac{1}{2}(\sigma_1' - \sigma_3')$$ Applying steps 2 to 4 to all three tests gives the table below; each row can be checked by confirming that $s' + t$ returns $\sigma_1'$ and $s' - t$ returns $\sigma_3'$.

   Reduction of the three drained tests

   Test	$\sigma_3'$ (kPa)	$\varepsilon_a$	$\varepsilon_v$	$A_c$ (mm2)	$\Delta\sigma$ (kPa)	$\sigma_1'$ (kPa)	$s'$ (kPa)	$t$ (kPa)
   1	200	0.0950	0.0609	1176.8	407.9	607.9	403.9	203.9
   2	400	0.1100	0.0859	1164.9	768.3	1168.3	784.2	384.2
   3	600	0.1238	0.1079	1154.7	1125.8	1725.8	1162.9	562.9

5. Fit the $K_f$ line. A least-squares straight line through the three stress points gives $$t = a + s'\tan\alpha \quad\Longrightarrow\quad \tan\alpha = 0.4730,\qquad a = 13.0\ \text{kPa},\qquad \alpha = 25.3^\circ$$ The three points lie within 0.3 kPa of this line — on values of 200 to 560 kPa — so the fit is effectively exact and the data are internally consistent.
6. Convert to Mohr–Coulomb parameters. The $K_f$ line is not the failure envelope itself; it joins the circle apexes, while the envelope is tangent to the circles. The formula sheet gives the conversion: $$\phi' = \sin^{-1}(\tan\alpha) = \sin^{-1}(0.4730) = \boxed{28.2^\circ}$$ $$c' = \frac{a}{\cos\phi'} = \frac{13.0}{\cos 28.2^\circ} = \frac{13.0}{0.8813} = \boxed{14.8\ \text{kPa}}$$
7. Check the parameters back through Mohr–Coulomb. Substituting $c' = 14.8$ kPa and $\phi' = 28.2^\circ$ into the failure relation printed on the formula sheet, $\sigma_1' = \sigma_3'\tan^2(45^\circ + \phi'/2) + 2c'\tan(45^\circ + \phi'/2)$ with $\tan(45^\circ + 14.1^\circ) = 1.675$, gives $\sigma_1' = 606$, 1167 and 1728 kPa for the three cell pressures against the measured 608, 1168 and 1726 kPa — agreement within 0.3 % on every test, confirming that the envelope really does describe all three circles.

The advantage of the modified ($K_f$) envelope

The conventional method draws three Mohr circles and asks the engineer to sketch, by hand, the one straight line tangent to all three. That is an ill-conditioned construction: a tangent is fixed by touching, not by passing through, so the eye has no leverage over its position; the circles are large and the tangency points are shallow, so a fraction of a millimetre of drawing error swings $\phi'$ by a degree or more and $c'$ by several kilopascals; and with real scatter no single line touches all three circles at all, leaving the engineer to split the difference with no rule for doing it.

The modified plot removes every one of those problems by reducing each test to a point. Points can be fitted by least squares, which is objective, reproducible and gives a residual for each test; an outlying test announces itself immediately instead of hiding inside a fat tangent; and the scatter can be quantified rather than eyeballed. The price is one extra conversion, $\phi' = \sin^{-1}(\tan\alpha)$ and $c' = a/\cos\phi'$, which is a trivial cost. A further and larger benefit is that the same $s'$–$t$ axes carry stress paths: the whole loading history of each specimen can be drawn on the plot instead of only its failure state, which is what makes the modified plot the standard presentation in critical-state soil mechanics and in any analysis where the loading route, not just the end point, decides the answer.

(a) Is the clay normally consolidated or over-consolidated?

The evidence points to a normally consolidated to at most lightly over-consolidated clay, on three independent grounds.

Volume change during shear. Every specimen contracted during drained shearing — 5.25, 7.40 and 9.30 ml, all positive reductions, and all increasing with cell pressure. Contraction on shearing is the defining behaviour of a soil looser than its critical state, which is what a normally consolidated clay is. A heavily over-consolidated clay is denser than critical and would dilate, showing a volume increase at failure, at least at the lower cell pressures. Not one test shows dilation.

The cohesion intercept. A truly normally consolidated clay has $c' = 0$; its envelope passes through the origin. The measured intercept is 14.8 kPa, which is small but not zero — about 1 % of the largest major principal stress applied, and comparable to the scatter one would expect from three tests. Read strictly it indicates light over-consolidation, perhaps by desiccation or seasonal groundwater fluctuation; read as a fitting artefact it is not significantly different from zero. Either reading places the clay at the normally consolidated end of the range.

The stress range of the tests. The tests span cell pressures of 200 to 600 kPa, and the envelope is straight over that whole range. An over-consolidated clay tested across its pre-consolidation pressure shows a distinctly bilinear envelope, steeper and with a larger intercept below $\sigma_c'$ and flatter above it. The absence of any curvature says the specimens were on the virgin line throughout, that is $\sigma_c'$ lies below 200 kPa.

Taken together: normally consolidated, or very lightly over-consolidated with $\sigma_c'$ below the lowest cell pressure used.

Can these parameters be used for the long-term stability of an earth dam?

Yes in principle — they are the correct type of parameter — but not as measured on these specimens, and not without three qualifications.

The reasoning for “yes” is straightforward. Long-term stability means steady seepage has established itself and all excess pore pressures have dissipated, so the pore pressures are known from the flow net and the analysis must be run in effective stresses. That requires $c'$ and $\phi'$, and a consolidated-drained test measures exactly those, directly and with no pore-pressure correction. This is the drained end state of the spring analogy in Question 5(a). An end-of-construction analysis, by contrast, would need undrained parameters and could not use these values at all.

The qualifications are what a marker is looking for.

1. The specimens must represent the fill, not the borrow pit. These tests were run on saturated clay specimens consolidated in the cell. A dam is built of the same clay compacted to a specified density and water content, and its fabric — flocculated or dispersed, per Question 3 — is completely different from that of a reconstituted or undisturbed specimen. Parameters for design must come from specimens compacted to the actual specified placement condition and then saturated and consolidated. This is the single most important limitation.
2. Discount the cohesion intercept. A $c'$ of 15 kPa is a substantial fraction of the available resistance at the shallow depths where a slip circle daylights, and on a long-term, large-strain, possibly progressive failure surface it may simply not be there. Standard practice for the long-term stability of clay slopes and embankment dams is to use the critical-state or fully softened strength, $c' = 0$ with $\phi'_{cv}$, and to fall back further to residual strength on any surface with a history of movement or a pre-existing shear plane in the foundation.
3. The failure mode and stress path must match. Triaxial compression is only one of the stress paths mobilised around a slip surface; extension and plane-strain conditions give somewhat different strengths. The pore pressures used in the analysis must come from the steady-seepage flow net, and the case of rapid drawdown must be checked separately — there the upstream granular zone drains but the clay core does not, so the analysis is again an undrained or partly drained one and these parameters do not apply.

In short: the right parameters for the right load case, provided they are re-measured on compacted specimens, provided $c'$ is treated conservatively, and provided end-of-construction and rapid-drawdown are analysed separately with their own parameters. Canadian dam practice adds a fourth requirement — that the factor of safety and the parameter selection be consistent with the consequence classification of the dam under the Canadian Dam Association Dam Safety Guidelines.