Question 3 of 6: Triaxial Testing of Intact Granite

Question 3: Triaxial Testing of Intact Granite (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.

3.1 — Mohr-Coulomb fit

Given. Five (σ₁,σ₃) triaxial pairs on intact granite (table above).

Find. Cohesion c, friction angle φ, and the uniaxial compressive strength implied by the linear Mohr-Coulomb fit $\sigma_1=\sigma_c+\sigma_3\!\left(\dfrac{1+\sin\phi}{1-\sin\phi}\right)$.

Approach. Linear-regress σ₁ against σ₃; the intercept is σ_c and the slope N is the passive-pressure coefficient, from which φ and then c follow.

1. Least-squares fit. $\sigma_1 = 207.8 + 8.267\,\sigma_3\ \text{(MPa)}$, so slope $N=8.267$ and intercept (UCS) $\sigma_c=\boxed{207.8\ \text{MPa}}$.
2. Friction angle. From $N=\dfrac{1+\sin\phi}{1-\sin\phi}$: $\sin\phi=\dfrac{N-1}{N+1}=\dfrac{7.267}{9.267}=0.7842 \Rightarrow \phi=\sin^{-1}(0.7842)=\boxed{51.7^\circ}$.
3. Cohesion. $c=\dfrac{\sigma_c(1-\sin\phi)}{2\cos\phi}=\dfrac{207.8(1-0.7842)}{2(0.6205)}=\dfrac{44.8}{1.241}=\boxed{36.1\ \text{MPa}}$.

Applicability. Both σ_c≈208 MPa and φ≈52° are plausible for a granite (typical UCS 100–250 MPa, φ 45–60°), and the straight line fits the five points closely over the tested confinement range (σ₃ = 0–25 MPa). The linear Mohr-Coulomb form is nonetheless known to be an approximation: the true intact-rock envelope curves downward in slope as confinement increases, so extrapolating this fit well beyond σ₃≈25 MPa, or into the low/negative (tensile) confinement range, will over-predict strength. Within the tested range it is an adequate, simple design tool.

3.2 — Hoek-Brown fit

Given. The same five (σ₁,σ₃) pairs.

Find. The intact-rock constants σ_ci and m_i in $\sigma_1=\sigma_3+\sqrt{m_i\sigma_{ci}\sigma_3+\sigma_{ci}^2}$, and comment on applicability.

Approach. Rearranging, $(\sigma_1-\sigma_3)^2=m_i\sigma_{ci}\,\sigma_3+\sigma_{ci}^2$ is linear in σ₃; regress $(\sigma_1-\sigma_3)^2$ against σ₃ to get the intercept ($\sigma_{ci}^2$) and slope ($m_i\sigma_{ci}$).

1. Transform the data. $(\sigma_1-\sigma_3)$ = 200, 250, 290, 340, 395 MPa for the five tests, so $(\sigma_1-\sigma_3)^2$ = 40,000, 62,500, 84,100, 115,600, 156,025 MPa².
2. Linear fit vs. σ₃. $(\sigma_1-\sigma_3)^2 = 39{,}531 + 4342.8\,\sigma_3$, so intercept $\sigma_{ci}^2=39{,}531 \Rightarrow \sigma_{ci}=\sqrt{39{,}531}=\boxed{198.8\ \text{MPa}}$.
3. Solve for m_i. $m_i=\dfrac{\text{slope}}{\sigma_{ci}}=\dfrac{4342.8}{198.8}=\boxed{21.8}$.

Applicability. σ_ci≈199 MPa is close to the Mohr-Coulomb intercept from 3.1 (as expected, both are UCS estimates from the same data) and m_i≈22 is of the right order for granite (published Hoek-Brown tables give m_i≈28–33 for granite, so this fit is somewhat low but the correct order of magnitude given only five points spanning a modest 0–25 MPa confining range). Hoek-Brown is generally the more defensible criterion for intact rock because it reproduces the genuine curvature of the failure envelope, particularly important near σ₃=0 where Mohr-Coulomb's straight-line extrapolation is weakest; a broader confining-stress range and more test points would tighten the m_i estimate.