16-Civ-A4 Geotechnical Materials and Analysis · December 2018
Question 6 of 6: CU Triaxial — Effective Strength Parameters
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Examinations — December 2018 · 16-Civ-A4 Geotechnical Materials and Analysis · closed book, 3 hours, 100 marks · answer ALL six questions.
Reference texts: R.F. Craig & J. Knappett, Craig’s Soil Mechanics (8th ed.); B.M. Das, Principles of Geotechnical Engineering; R.D. Holtz, W.D. Kovacs & T.C. Sheahan, An Introduction to Geotechnical Engineering (2nd ed.); M. Budhu, Soil Mechanics and Foundations. Take $\gamma_w = 9.81\ \text{kN/m}^3$ throughout.
Question 6: CU Triaxial — Effective Strength Parameters (20 marks)
Find. $c'$ and $\phi'$; the expected deviator stress at failure for $\sigma_3'=250$ kPa; and the NC/OC state and short-term usability of the parameters.
Figure Q6 — modified failure (K₣) envelope of effective stress points. The best-fit line passes through the origin (c'≈0), so the clay is normally consolidated.
Approach. Convert each test to effective stresses, plot the failure stress points $(p',q)$, fit the modified ($K_f$) line, and convert its slope and intercept to $c'$ and $\phi'$; then apply the effective envelope to the new confining pressure using the measured pore-pressure response.
Effective stresses at failure. $\sigma_3'=\sigma_3-u$ and $\sigma_1'=\sigma_3'+(\sigma_1-\sigma_3)$:
$$\sigma_3' = 68,\ 131,\ 198\ \text{kPa};\qquad \sigma_1' = 171,\ 333,\ 503\ \text{kPa}.$$
Stress points. With $p'=\tfrac12(\sigma_1'+\sigma_3')$ and $q=\tfrac12(\sigma_1'-\sigma_3')$ (note $q$ equals half the deviator, so the pore pressure cancels):
$$(p',q) = (119.5,\,51.5),\ (232,\,101),\ (350.5,\,152.5)\ \text{kPa}.$$
Fit the $K_f$ line. A least-squares fit gives slope $\tan\alpha' = 0.437$ and an intercept $a\approx 0$ (the line passes through the origin).
Effective parameters. Converting the modified-line constants,
$$\phi' = \sin^{-1}(\tan\alpha') = \sin^{-1}(0.437) = \boxed{25.9^\circ},\qquad c' = \frac{a}{\cos\phi'} \approx \boxed{0\ \text{kPa}}.$$
Deviator at failure for $\sigma_3'=250$ kPa. The tests give a pore-pressure parameter $A_f = u/(\sigma_1-\sigma_3) \approx 0.82$. With the envelope through the origin, $\sigma_1'/\sigma_3' = N = \tan^2(45^\circ+\phi'/2)=2.55$; solving the undrained-shear balance $ (\sigma_1-\sigma_3)\big[(1-A_f)+N A_f\big] = \sigma_3'(N-1)$ with $\sigma_3'=250$:
$$(\sigma_1-\sigma_3)_f = \frac{250\,(2.55-1)}{(1-0.82)+2.55(0.82)} = \boxed{171\ \text{kPa}}.$$
A direct interpolation of the measured deviator against $\sigma_3$ gives 170 kPa — the same answer.
(i) NC or OC? The effective envelope passes through the origin ($c'\approx0$) and every test develops a large positive pore pressure ($A_f\approx0.82\gt0.5$, i.e. contractive). Both are signatures of a normally consolidated clay.
(ii) Short-term dam stability? No. $c'$ and $\phi'$ are effective (drained) parameters and describe the long-term stability of the dam, once construction pore pressures have dissipated. Short-term (end-of-construction) stability is an undrained problem governed by the undrained strength $s_u$ (a total-stress $\phi_u=0$ analysis), or by the effective parameters only if the construction excess pore pressures are separately predicted. The effective parameters alone are therefore not sufficient for the short-term case.
