Question 7 of 7: Apparent vs. Effective Shear-Strength Parameters from Two Consolidated-Undrained Triaxial Tests

Question 7: Apparent vs. Effective Shear-Strength Parameters from Two Consolidated-Undrained Triaxial Tests (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.

a) What "undrained" means in a CU test. In a consolidated-undrained (CU) test the specimen is first allowed to fully consolidate under an all-round cell pressure (so the effective stresses are known before shearing begins), then the drainage valve is closed and the specimen is sheared to failure with no further volume change permitted — any tendency of the soil skeleton to compress or dilate during shear instead generates excess pore water pressure, which is measured (in a CU test with pore-pressure measurement) so that both total- and effective-stress strength parameters can be obtained from the same pair of tests, without waiting for the very long drainage times a fully drained (CD) test on a clay would require.

Approach. Two Mohr circles fix a unique common tangent (the Mohr–Coulomb envelope). Rather than a graphical fit, use the equivalent p–q (Kf-line) method: with $p=(\sigma_1+\sigma_3)/2$, $q=(\sigma_1-\sigma_3)/2$, the Kf line $q = a + p\sin\phi$ passes through both test points, and $c = a/\cos\phi$. Subtracting u from both principal stresses before repeating the same fit gives the effective parameters (q is unchanged by u, since it cancels in $\sigma_1'-\sigma_3' = \sigma_1-\sigma_3$; only p shifts).

1. b) Apparent (total-stress) parameters. $p_1=127.5,\ q_1=62.5$; $p_2=235.0,\ q_2=105.0$ (kPa). The Kf-line slope and intercept: $$\sin\phi = \frac{q_2-q_1}{p_2-p_1} = \frac{42.5}{107.5} = 0.3953 \ \Rightarrow\ \phi = \boxed{23.3^\circ}$$ $$a = q_1 - p_1\sin\phi = 62.5 - 127.5(0.3953) = 12.09, \qquad c = \frac{a}{\cos\phi} = \frac{12.09}{\cos 23.3^\circ} = \boxed{13.2\ \text{kPa}}$$
2. c) Effective-stress parameters. Subtracting u from both principal stresses: Test 1 gives $\sigma_1'=155,\ \sigma_3'=30$ ($p_1'=92.5,\ q_1'=62.5$); Test 2 gives $\sigma_1'=280,\ \sigma_3'=70$ ($p_2'=175.0,\ q_2'=105.0$). Note q is unchanged by u, exactly as expected (u cancels in $\sigma_1'-\sigma_3'$), so only the shift in p changes the fitted line: $$\sin\phi' = \frac{105.0-62.5}{175.0-92.5} = \frac{42.5}{82.5} = 0.5152 \ \Rightarrow\ \phi' = \boxed{31.0^\circ}$$ $$a' = q_1' - p_1'\sin\phi' = 62.5 - 92.5(0.5152) = 14.85, \qquad c' = \frac{a'}{\cos\phi'} = \frac{14.85}{\cos 31.0^\circ} = \boxed{17.3\ \text{kPa}}$$

d) Other means of determining shear strength. Besides the triaxial test: the direct shear box test (simple, fast, but forces failure onto a pre-defined horizontal plane and cannot control drainage as cleanly); the unconfined compression test, a fast special case (σ₃ = 0) giving undrained shear strength for saturated clays directly from $s_u = q_u/2$; the field or laboratory vane shear test, widely used for soft, sensitive clays where sampling disturbance is a concern; cone penetration testing (CPT) with empirical correlations to undrained strength; the standard penetration test (SPT) N-value, correlated empirically to strength/consistency; and the ring shear test, used specifically to obtain residual (large-displacement) strength parameters for slope-stability back-analysis.