Question 4 of 5: Mohr–Coulomb parameters from triaxial compression data

Question 4: Mohr–Coulomb parameters from triaxial compression data (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. Five paired (confining stress, failure axial stress) triaxial test results.

Find. The Mohr-Coulomb failure-plane angle $\Psi$, friction angle $\phi$, cohesion $C$, UCS $S_c$ and tensile strength $S_T$; then predicted failure axial stresses at $\sigma_3=5,15,22.5$ MPa; then data-quality issues and a verification method.

Approach. The exam's own Mohr-Coulomb sheet gives $\sigma_1=\sigma_3\tan^2\Psi+2C\tan\Psi=\sigma_3\tan^2\Psi+S_c$, which is linear in $\sigma_3$ — fit a least-squares line to the five data points to get the slope ($\tan^2\Psi$) and intercept ($S_c$) directly, then back out $\Psi$, $\phi=2(\Psi-45^\circ)$, $C=S_c/(2\tan\Psi)$, and $S_T=C/\tan\phi$.

1. (a) Least-squares fit of $\sigma_1=\sigma_3\tan^2\Psi+S_c$. Regressing the five pairs gives slope $\tan^2\Psi\approx3.50$ and intercept $S_c\approx104.6$ MPa ($R^2\approx0.94$): $$\tan^2\Psi=3.50\ \Rightarrow\ \Psi=\arctan\sqrt{3.50}\approx61.9^\circ$$ $$\phi=2(\Psi-45^\circ)=2(61.9-45)\approx33.8^\circ$$ $$C=\frac{S_c}{2\tan\Psi}=\frac{104.6}{2(1.871)}\approx28.0\ \text{MPa},\qquad S_T=\frac{C}{\tan\phi}=\frac{28.0}{\tan33.8^\circ}\approx41.8\ \text{MPa}$$ $$\boxed{\phi\approx33.8^\circ,\ \ \Psi\approx61.9^\circ,\ \ C\approx28.0\ \text{MPa},\ \ S_c\approx104.6\ \text{MPa}}$$
2. (b) Problems evident from the data. All five confining stresses tested fall in a narrow mid-range band (9.8–25.1 MPa) — no test was run near $\sigma_3=0$ (unconfined compression), so the reported UCS ($S_c\approx104.6$ MPa) and tensile strength ($S_T\approx41.8$ MPa) are both extrapolations of the fitted line well outside the tested range, not directly measured quantities. The fit itself is reasonably good ($R^2\approx0.94$) but not perfect — residuals of several MPa on individual points (largest at the highest-confinement test, $\sigma_3=25.1$ MPa) show real scatter consistent with normal specimen-to-specimen variability, but with only one specimen per confining stress there is no way to separate genuine material scatter from test-to-test error.
3. (c) Failure axial stresses at $\sigma_3=5,15,22.5$ MPa. Applying the fitted line directly (no extrapolation problem here — 5 and 15 MPa sit inside the tested range, 22.5 MPa is close to the tested maximum of 25.1 MPa): $$\sigma_1=\sigma_3(3.50)+104.6$$ $$\sigma_1(5)=5(3.50)+104.6\approx122.1\ \text{MPa}$$ $$\sigma_1(15)=15(3.50)+104.6\approx157.1\ \text{MPa}$$ $$\sigma_1(22.5)=22.5(3.50)+104.6\approx183.4\ \text{MPa}$$ $$\boxed{\sigma_1\approx122.1,\ 157.1,\ 183.4\ \text{MPa at }\sigma_3=5,15,22.5\ \text{MPa}}$$
4. (d) Verifying the results. Run additional triaxial tests at (or near) $\sigma_3=0$ to measure UCS directly rather than extrapolating it, run replicate specimens at one or two of the existing confining stresses to quantify genuine scatter versus test error, and independently cross-check the fitted $\phi$/$C$ against a direct-shear test on the same rock (a different loading path testing the same failure criterion) — agreement between two independent test methods is the standard way to confirm a Mohr-Coulomb envelope rather than trusting a single regression.