Question 6 of 6: Shear strength parameters from CD triaxial tests

Question 6: Shear strength parameters from CD triaxial tests (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. Three consolidated-drained (CD) triaxial tests on a saturated clay, initial specimen $38\ \text{mm}$ diameter $\times\ 76\ \text{mm}$ long ($A_0 = 1134\ \text{mm}^2$, $V_0 = 86.2\ \text{cm}^3$).

Find. $c'$ and $\phi'$ from the modified ($K_f$) envelope; the advantage of that plot; whether the clay is NC or OC; and whether the parameters suit long-term dam stability.

Approach. Correct the specimen area for axial and volumetric strain (drained, so total = effective), form the deviator stress and hence $\sigma_1'$, plot the stress points $\big(p',q\big)=\big(\tfrac12(\sigma_1'+\sigma_3'),\tfrac12(\sigma_1'-\sigma_3')\big)$, fit the $K_f$ line and convert its slope and intercept to $\phi'$ and $c'$.

1. Corrected area. With axial strain $\varepsilon_a = \Delta L/L_0$ and volumetric strain $\varepsilon_v = \Delta V/V_0$, $A_c = A_0\dfrac{1-\varepsilon_v}{1-\varepsilon_a}$. For Test 1, $\varepsilon_a = 7.22/76 = 0.0950$, $\varepsilon_v = 5.25/86.2 = 0.0609$, giving $A_c = 1134\times\dfrac{0.939}{0.905} = 1177\ \text{mm}^2$.
2. Deviator and major principal stress. $\sigma_d = P/A_c$ and $\sigma_1' = \sigma_3' + \sigma_d$ (drained: $\sigma' = \sigma$). Test 1: $\sigma_d = 480/1177 = 0.408\ \text{MPa} = 408\ \text{kPa}$, so $\sigma_1' = 200+408 = 608\ \text{kPa}$. Repeating for all three:

   Test	$A_c$ (mm$^2$)	$\sigma_d$ (kPa)	$\sigma_1'$ (kPa)	$p'$ (kPa)	$q$ (kPa)
   1	1177	408	608	404	204
   2	1165	768	1168	784	384
   3	1155	1126	1726	1163	563

3. Fit the $K_f$ (stress-point) line. A least-squares fit of $q$ on $p'$ gives slope $\tan\alpha' = 0.473$ and intercept $a = 13\ \text{kPa}$: $$q = 13 + 0.473\,p'.$$
4. Convert to Mohr–Coulomb parameters. $\phi' = \sin^{-1}(\tan\alpha') = \sin^{-1}(0.473) = \boxed{28.2^\circ}$ and $c' = \dfrac{a}{\cos\phi'} = \dfrac{13}{\cos 28.2^\circ} = \boxed{14.8\ \text{kPa}}$.

Advantage of the modified ($K_f$) envelope. Instead of drawing a common tangent to several Mohr circles — a construction that is awkward and subjective when the circles overlap or when scatter means no single line touches them all — the $K_f$ plot reduces each test to a single stress point at the top of its circle. A best-fit straight line (here by least squares) through those points is objective, repeatable, and lets scatter be averaged rather than forcing the envelope through the worst circle. The slope and intercept then convert to $\phi'$ and $c'$ by the standard relations.

(a) Normally consolidated or over consolidated?

The three stress points fall on a straight line that passes very close to the origin: the cohesion intercept $c'\approx 15\ \text{kPa}$ is small compared with the 200–600 kPa stress range. A linear effective-stress envelope through (near) the origin is the signature of a normally consolidated clay; the tiny positive intercept indicates at most very light over-consolidation and is within the scatter expected from area correction and reading of a hand-drawn failure. Reasons: (i) the envelope is essentially straight (no strength curvature or distinct pre-consolidation break), and (ii) it has negligible true cohesion, both of which are characteristic of NC behaviour, where strength derives almost entirely from friction.

Can these parameters be used for long-term dam stability?

Yes. The tests are consolidated-drained, so $c'$ and $\phi'$ are effective-stress parameters measured with no excess pore pressure. Long-term (steady-seepage) stability of an earth dam is precisely a drained condition: over the life of the structure any construction pore pressures dissipate and the pore pressures are set by the steady seepage regime. An effective-stress analysis using these drained $c'$ and $\phi'$, with pore pressures from the long-term flow net, is therefore the correct approach. (The short-term, end-of-construction case would instead need undrained strength.)