Question 3 of 6: Shear Strength

Question 3: Shear Strength (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. CD triaxial tests on an over-consolidated (OC) clay: Test 1 — $\sigma_3=20$ kPa, $(\sigma_1-\sigma_3)_f=80$ kPa; Test 2 — $\sigma_3=135$ kPa, $(\sigma_1-\sigma_3)_f=300$ kPa.

Find. (a) qualitative stress-strain/volume-strain sketches; (b) $c'$, $\phi'$; (c) $\sigma_{ff}$, $\tau_{ff}$ for Test 2; (d) failure-plane angle; (e) $(\sigma_1-\sigma_3)_f$ at $\sigma_3=75$ kPa; (f) appropriate lab tests for the two field scenarios.

Approach. Compute $\sigma_1$ at failure for each test, fit the common tangent (Mohr–Coulomb envelope) through both failure circles (an OC clay legitimately carries a real, nonzero $c'$, unlike cohesionless sand), then use the tangent-point stresses and $\theta_f=45^{\circ}+\phi'/2$ for the remaining parts.

1. Failure stresses and circle geometry. Test 1: $\sigma_{1f}=20+80=100$ kPa; centre $C_1=\tfrac{100+20}{2}=60$ kPa, radius $R_1=\tfrac{100-20}{2}=40$ kPa. Test 2: $\sigma_{1f}=135+300=435$ kPa; centre $C_2=\tfrac{435+135}{2}=285$ kPa, radius $R_2=\tfrac{435-135}{2}=150$ kPa.
2. (b) Common-tangent envelope (shear strength parameters). Solving $R=c'\cos\phi'+C\sin\phi'$ simultaneously for both circles: $\sin\phi'=\dfrac{R_2-R_1}{C_2-C_1}=\dfrac{150-40}{285-60}=0.4889\ \Rightarrow\ \phi'=\boxed{29.3^{\circ}}$, and back-substituting, $c'=\dfrac{R_1-C_1\sin\phi'}{\cos\phi'}=\dfrac{40-60(0.4889)}{0.8724}=\boxed{12.2\text{ kPa}}$. A real, nonzero $c'$ is physically expected here (unlike the cohesionless-sand case) — it reflects the clay's over-consolidated, denser and more interlocked fabric.
3. (c) Stresses on the failure plane, Test 2. At the tangent point, $\sigma_{ff}=C_2-R_2\sin\phi'=285-150(0.4889)=\boxed{211.7\text{ kPa}}$; $\tau_{ff}=R_2\cos\phi'=150(0.8724)=\boxed{130.9\text{ kPa}}$.
4. (d) Failure-plane orientation, Test 2. $\theta_f=45^{\circ}+\phi'/2=45+14.63=\boxed{59.6^{\circ}}$ from the plane on which $\sigma_3$ acts (i.e. from horizontal, since $\sigma_3$ is the horizontal cell pressure in a standard triaxial compression test).
5. (e) Test 3 at $\sigma_3=75$ kPa. Using the Mohr–Coulomb failure envelope in principal-stress form, $\sigma_{1f}=\sigma_3 N_\phi+2c'\sqrt{N_\phi}$ with $N_\phi=\tan^2(45^{\circ}+\phi'/2)=\tan^2(59.63^{\circ})=2.913$: $\sigma_{1f}=75(2.913)+2(12.23)\sqrt{2.913}=218.5+41.7=260.2$ kPa, so $\boxed{(\sigma_1-\sigma_3)_f=260.2-75=185.2\text{ kPa}}$ — consistent with lying between Test 1's 80 kPa (at $\sigma_3=20$) and Test 2's 300 kPa (at $\sigma_3=135$), as expected for an intermediate cell pressure.
6. (f) Appropriate strength tests. (i) First-time quick loading of a foundation on clay: load is applied faster than the clay can drain, so the relevant strength is the short-term, total-stress (undrained) strength — an unconsolidated–undrained (UU) triaxial test ($\phi_u\approx0$, $c=s_u$). (ii) Long-term, steady-state slope stability: excess pore pressures have long since dissipated, so the relevant strength is the drained, effective-stress strength — a consolidated–drained (CD) test (or an equivalent CU test with pore-pressure measurement), using $c'$, $\phi'$ exactly as determined in part (b).

Qualitatively, for an over-consolidated clay sheared in CD compression: the deviator stress $(\sigma_1-\sigma_3)$ rises steeply, peaks near a small axial strain (a few percent), then softens slightly toward a critical-state value at larger strain; the volumetric strain first compresses marginally, then dilates (volume increase) past the peak as the denser, over-consolidated fabric is forced to expand to shear.