Question 6 of 6: Consolidated-undrained triaxial series — effective and total strength parameters

Question 6: Consolidated-undrained triaxial series — effective and total strength parameters (30 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. (i) $c'$ and $\phi'$ from the modified (stress-point) envelope; (ii) $c_{cu}$ and $\phi_{cu}$ analytically; (iii) the deviator stress at failure for a third specimen consolidated to $\sigma'_3 = 250$ kPa; (iv) whether the clay is normally consolidated or overconsolidated, with reasons.

Approach. Reduce each test to its total and effective principal stresses, plot both sets as stress points $(s, t)$, fit the two straight $K_f$ lines, convert their slopes and intercepts to $\phi$ and $c$, then use the effective envelope to predict the third test and the pore-pressure parameter to classify the clay.

1. Reduce the test data. The deviator stress is the same in total and in effective terms because subtracting $u$ from both principal stresses leaves their difference unchanged. For each specimen $\sigma_1 = \sigma_3 + (\sigma_1-\sigma_3)_f$, $\sigma'_3 = \sigma_3 - u_f$ and $\sigma'_1 = \sigma_1 - u_f$:

   Specimen	$\sigma_1$	$\sigma'_3$	$\sigma'_1$	$s = \tfrac{\sigma_1+\sigma_3}{2}$	$s' = \tfrac{\sigma'_1+\sigma'_3}{2}$	$t = \tfrac{\sigma_1-\sigma_3}{2}$
   A	253	68	171	201.5	119.5	51.5
   B	502	131	333	401.0	232.0	101.0

   All values in kPa. Note that $t$ is common to both columns — only the abscissa shifts left by $u_f$ when moving from total to effective stress, which is why the two envelopes below share the same ordinates.
2. (i) Plot the modified failure envelope in effective stress. The modified (or “stress-point”, or $K_f$) plot replaces each Mohr circle by the single point at its crown, so a straight line can be fitted by eye or by algebra instead of drawing a common tangent. Through the two effective stress points, $$\tan\alpha' = \frac{t_B - t_A}{s'_B - s'_A} = \frac{101.0 - 51.5}{232.0 - 119.5} = \frac{49.5}{112.5} = 0.4400 ,$$ $$d' = t_A - s'_A \tan\alpha' = 51.5 - 119.5(0.4400) = -1.08\ \text{kPa}.$$
3. Convert the modified envelope to $c'$ and $\phi'$. The formula sheet gives the two conversions $\sin\phi' = \tan\alpha'$ and $c' = d'/\cos\phi'$: $$\phi' = \sin^{-1}(0.4400) = \boxed{26.1^{\circ}}, \qquad c' = \frac{-1.08}{\cos 26.1^{\circ}} = -1.2\ \text{kPa} \approx \boxed{0}$$ A cohesion intercept of −1.2 kPa on stresses of 70–330 kPa is zero to the precision of the test, so the effective envelope passes through the origin: $c' = 0$, $\phi' = 26.1^{\circ}$. That is the signature of a normally consolidated clay, and part (iv) returns to it.
4. (ii) Repeat in total stress for the analytical solution. The same two specimens plotted as total stress points give $$\tan\alpha = \frac{101.0 - 51.5}{401.0 - 201.5} = \frac{49.5}{199.5} = 0.2481, \qquad d = 51.5 - 201.5(0.2481) = 1.50\ \text{kPa},$$ so $$\phi_{cu} = \sin^{-1}(0.2481) = \boxed{14.4^{\circ}}, \qquad c_{cu} = \frac{1.50}{\cos 14.4^{\circ}} = \boxed{1.6\ \text{kPa}}$$ This is a genuinely analytical route — no tangent has to be drawn to two circles, only two points joined — which is what part (ii) asks for. The total-stress friction angle is little more than half the effective one because the pore pressure generated during undrained shear grows with the confining stress and eats most of the strength gain that the extra confinement would otherwise deliver.
   The modified failure envelopes. Circles are the effective stress points $(s', t)$ and squares the total stress points $(s, t)$; each pair shares its ordinate $t$ and is displaced horizontally by the pore pressure at failure. The slopes give $\phi'$ and $\phi_{cu}$ through $\sin\phi = \tan\alpha$.
5. (iii) Predict the third test at $\sigma'_3 = 250$ kPa. A specimen consolidated to $\sigma'_3 = 250$ kPa fails on the effective envelope, so writing $t$ for the unknown half-deviator and noting $s' = \sigma'_3 + t$, $$t = \tan\alpha'\,(\sigma'_3 + t) + d' \quad\Longrightarrow\quad t = \frac{\tan\alpha'\,\sigma'_3 + d'}{1 - \tan\alpha'} = \frac{0.4400(250) - 1.08}{1 - 0.4400} = 194.5\ \text{kPa},$$ and the principal stress difference at failure is twice that: $$(\sigma_1 - \sigma_3)_f = 2t = \boxed{389\ \text{kPa}}$$ Treating the clay as purely frictional ($c' = 0$) gives the equivalent closed form $(\sigma_1-\sigma_3)_f = 2\sigma'_3\sin\phi'/(1-\sin\phi') = 2(250)(0.44)/0.56 = 393$ kPa, or through the formula sheet’s Rankine relation $\sigma'_1 = \sigma'_3\tan^{2}(45^{\circ}+\phi'/2) = 250(2.571) = 643$ kPa, so $\sigma'_1 - \sigma'_3 = 393$ kPa. The two routes differ by 1 %, which is the weight of the tiny negative cohesion intercept; 389 kPa is the value the fitted envelope gives, and either is defensible provided the assumption is stated.
6. (iv) Classify the clay. Three independent pieces of evidence, all pointing the same way. First, the effective envelope found in Step 3 has $c' \approx 0$ — an overconsolidated clay carries a real cohesion intercept, a normally consolidated one does not. Second, the pore-water pressure at failure is large and positive in both tests, which means both specimens wanted to contract during shear: contractive behaviour is normally consolidated behaviour, whereas an overconsolidated clay dilates and generates negative excess pore pressure. Third, and quantitatively, Skempton’s pore-pressure parameter at failure $$A_f = \frac{u_f}{(\sigma_1-\sigma_3)_f}: \qquad A_f^{(A)} = \frac{82}{103} = 0.80, \qquad A_f^{(B)} = \frac{169}{202} = 0.84,$$ and entering Figure 3 with $A_f \approx 0.8$ lands at the extreme left of the curve, at $OCR \approx 1$. The clay is normally consolidated.