Question 6 of 6: Shear Strength Parameters from a CU Triaxial Series

Question 6: Shear Strength Parameters from a CU Triaxial Series (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-undrained triaxial tests on saturated specimens of one clay, each with the pore-water pressure measured at failure.

Find. The effective-stress parameters $c'$ and $\phi'$ from a modified (stress-point) failure envelope, and then answers to the three interpretation questions on why that plot is preferred, on the stress history of the clay, and on which stability case these parameters may be used for.

Part (a) — determining $c'$ and $\phi'$ from the modified failure envelope

Approach. Convert each total-stress test result into effective principal stresses by subtracting the measured pore pressure, reduce each test to the single stress point $\left(p', q\right) = \left(\tfrac12(\sigma_1'+\sigma_3'),\ \tfrac12(\sigma_1'-\sigma_3')\right)$, fit the straight $K_f$ line through the three points, and convert its slope and intercept into $\phi'$ and $c'$ using the transformation printed on the formula sheet.

1. Recover the total major principal stress. The triaxial cell applies $\sigma_3$ all round and the ram adds the deviator stress, so $$\sigma_1 = \sigma_3 + (\sigma_1 - \sigma_3)$$ giving $\sigma_1 = 250$, $500$ and $1000\ \text{kPa}$ for the three tests.
2. Convert to effective stresses. Terzaghi’s principle, $\sigma' = \sigma - u$, applies to both principal stresses with the same measured pore pressure, since $u$ is uniform through a saturated specimen at failure: $$\sigma_3' = \sigma_3 - u, \qquad \sigma_1' = \sigma_1 - u$$

   Effective principal stresses at failure

   Test	$\sigma_3'$ (kPa)	$\sigma_1'$ (kPa)
   1	70	170
   2	140	340
   3	280	680

   Note that the pore pressure is exactly 80 % of the deviator stress in all three tests, a point returned to in part (ii).
3. Reduce each test to one stress point. The modified plot replaces each Mohr circle by the single point at the top of it, using the coordinates given on the formula sheet: $$p' = \tfrac{1}{2}\left(\sigma_1' + \sigma_3'\right), \qquad q = \tfrac{1}{2}\left(\sigma_1' - \sigma_3'\right)$$

   Stress points for the modified envelope

   Test	$p'$ (kPa)	$q$ (kPa)
   1	120	50
   2	240	100
   3	480	200

4. Plot and fit the $K_f$ line. The three points are collinear to the precision of the data and the line passes through the origin, so the intercept is zero: $$\tan\alpha' = \frac{q}{p'} = \frac{50}{120} = \frac{100}{240} = \frac{200}{480} = 0.4167, \qquad a = 0$$ $$\alpha' = \arctan(0.4167) = 22.62^\circ$$
   Figure 6.1 — The modified failure envelope. Three tests, three points, one unambiguous straight line through the origin.
5. Transform to the Mohr–Coulomb parameters. The $K_f$ line is not the failure envelope; it is the locus of circle tops, and the formula sheet supplies the conversion: $$\phi' = \sin^{-1}\!\left(\tan\alpha'\right), \qquad c' = \frac{a}{\cos\phi'}$$ Substituting the fitted values, $$\phi' = \sin^{-1}(0.4167) = 24.62^\circ$$ $$c' = \frac{0}{\cos 24.62^\circ} = 0$$ $$\boxed{\,c' = 0, \qquad \phi' = 24.6^\circ\,}$$
6. Verify against the Mohr circles directly. With $c' = 0$ the tangency condition for every circle reduces to $\sin\phi' = (\sigma_1'-\sigma_3')/(\sigma_1'+\sigma_3')$. Testing all three: $$\frac{170-70}{170+70} = \frac{340-140}{340+140} = \frac{680-280}{680+280} = 0.4167 \ \Rightarrow \ \phi' = 24.62^\circ$$ All three circles are tangent to one envelope through the origin, so the fit is exact rather than a best compromise.

(i) Why plot the modified envelope rather than Mohr circles

Drawing a common tangent to three circles by hand is a genuinely awkward and inaccurate operation. The tangent point on each circle is not marked, the eye has to judge tangency at three places at once, and a small rotation of the straightedge produces a large change in the intercept $c'$ — on a typical plot the same three circles can be fitted with $c'$ anywhere over a range of 10 or 15 kPa depending on who draws the line. When the circles do not have a perfectly consistent envelope, as real data never quite do, the construction has no defined answer at all.

The modified plot removes the ambiguity by representing each test as a single point, the top of its Mohr circle, on axes of $\tfrac12(\sigma_1'-\sigma_3')$ against $\tfrac12(\sigma_1'+\sigma_3')$. Fitting a straight line to a set of points is unambiguous, can be done by least squares rather than by eye, and immediately reveals scatter, curvature and outliers — none of which is visible in a nest of circles. The parameters then follow from the exact transformation $\phi' = \sin^{-1}(\tan\alpha')$ and $c' = a/\cos\phi'$. The same plot also carries the whole stress path of each test, not just its failure state, so it can be used to examine how the specimen approached failure, which the Mohr construction cannot do. This is why stress-path plotting has become the standard laboratory presentation.

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

The clay is normally consolidated, and three independent features of the data say so.

First, the envelope passes through the origin: $c' = 0$ and $a = 0$. A normally consolidated clay has no bonding or locked-in structure to give it an effective cohesion intercept; its strength is purely frictional. An over-consolidated clay, which has been unloaded from a higher stress and retains some of the resulting density and structure, invariably shows a positive $c'$ intercept over the range of stresses below its preconsolidation pressure.

Second, the pore pressure at failure is positive in every test and equal to 80 % of the deviator stress. Skempton’s pore-pressure parameter at failure,

lies squarely in the 0.7 to 1.3 band characteristic of normally consolidated clay, which contracts on shearing and therefore generates positive excess pore pressure. A heavily over-consolidated clay dilates on shearing and produces $A_f$ that is small or negative, typically between $-0.5$ and $0.0$.

Third, the results scale perfectly: doubling $\sigma_3$ doubles the deviator stress and doubles the pore pressure. That proportionality is the signature of a normally consolidated soil, whose behaviour is governed only by the current effective stress and so is self-similar at every stress level. An over-consolidated clay would show a falling strength ratio as the confining pressure rose past its preconsolidation pressure, producing a bilinear envelope rather than the single straight line obtained here.

(iii) Can these parameters be used for short-term or long-term stability?

They apply to the long-term (drained) stability of the earth dam, and to that case only.

The parameters $c'$ and $\phi'$ are effective-stress parameters. They were obtained by subtracting the measured pore pressure from the total stresses, so they describe the frictional strength available at the grain contacts once the pore pressure at a point is known. Using them requires knowing that pore pressure. In the long term, after construction has ended and the excess pore pressures generated by placing the fill have dissipated, the pore pressures in the dam are the steady-state seepage pressures given by the flow net — known, computable and stable. Effective-stress analysis with $c' = 0$ and $\phi' = 24.6^\circ$ is then exactly the right tool, and this is the standard long-term analysis for the downstream slope of an earth dam and for the upstream slope under steady seepage.

For the short-term case the same parameters cannot be applied directly. Immediately after rapid construction the clay fill is loaded far faster than it can drain, and the excess pore pressures at that instant are not known independently — they depend on the fill’s placement water content, its degree of saturation and its compressibility. Short-term stability is therefore normally checked in total stress, using the undrained shear strength $c_u$ (with $\phi_u = 0$ for a saturated clay) from unconsolidated-undrained tests, or in effective stress but only if the construction pore pressures are measured with piezometers or predicted through a pore-pressure ratio $r_u$. The consolidated-undrained series above does provide the raw material for a total-stress envelope — plotting the same three circles on total-stress axes would give $c_{cu}$ and $\phi_{cu}$ — but $c'$ and $\phi'$ themselves are the wrong parameters for an undrained analysis.

The critical short-term case for the upstream slope is rapid drawdown, where the reservoir falls faster than the fill can drain; that too is an undrained or partially drained problem and is analysed with the appropriate undrained strengths or with measured pore pressures, not with $c'$ and $\phi'$ applied to pre-drawdown seepage pressures.