Question 2 of 5: Pillar RMR, Hoek-Brown strength and factor of safety

Question 2: Pillar RMR, Hoek-Brown strength and factor of safety (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. Room-and-pillar limestone mine at 80 m depth, 7 m×7 m pillars on 6 m clear spacing; horizontally bedded, dry, single bedding discontinuity set of moderate (200–600 mm) spacing; point-load UCS = 100 MPa; one triaxial pair (σ₁,σ₃) = (110, 4) MPa; γ = 28 kN/m³; k = σ_h/σ_v,centre = 0.075.

Find. (a) RMR; (b) Hoek-Brown m, s; (c) maximum face stress; (d) maximum centre stress; (e) tributary-area average stress and factor of safety.

Approach. Rate the six RMR parameters from the field description, use Figure 1 of the exam's own reference material to resolve the UCS rating precisely at the 100 MPa band boundary, back out the intact-rock Hoek-Brown constant m_i from the single triaxial pair, scale it to rock-mass m, s at the computed RMR, evaluate the Hoek-Brown criterion at the unconfined face (σ₃=0) and at the confined centre (σ₃=kσ₁), then compare the confined-core strength against the tributary-area average stress.

1. Part (a) — Rate the six RMR parameters. UCS = 100 MPa sits exactly on the boundary between Table 1's 50–100 MPa (rating 7) and 100–250 MPa (rating 12) bands, so the discrete table cannot resolve it; the exam's own Figure 1 (rating vs. UCS, a smooth calibrated curve) is read instead, interpolating between the plotted points at 80 MPa (rating ≈ 8.3) and 120 MPa (rating ≈ 11.7): at 100 MPa this gives a strength rating of 10.
   Check: the strength rating is a chart-digitization at a genuine table-boundary case; Table 1 alone would give either 7 or 12 depending on which side of 100 MPa is assumed — Figure 1's continuous curve is used here specifically because it exists to resolve this ambiguity.
   The remaining five parameters follow directly from the field description (Bieniawski Table 1, this exam's reference material): a single bedding set of “moderate” (200–600 mm, ISRM terminology) spacing gives a spacing rating of 10, and at any representative spacing in that band the volumetric joint count J_v=1/spacing gives RQD = 115−3.3J_v ≥ 99% (capped at 100%) — robust across the whole 200–600 mm range — so the RQD rating is 20. Condition of discontinuities is scored from Table 1's Section E sub-ratings, since the description uses the exact wording of that finer table: persistence assumed continuous (regional bedding) = 0, separation none (“no visible aperture”) = 6, roughness smooth = 1, infilling none = 6, weathering slightly weathered = 5, summing to a condition rating of 18. Groundwater is completely dry = 15, and no orientation adjustment applies to a global material rating, giving 0.
2. Sum to RMR. $$RMR = 10 + 20 + 10 + 18 + 15 + 0 = 73$$ $$\boxed{RMR = 73 \text{(Class II, Good rock)}}$$
3. Part (b) — Intact-rock constant m_i from the triaxial pair. For intact rock (s = 1) the exam's own Hoek-Brown form gives $$\frac{\sigma_1}{\sigma_c}=\frac{\sigma_3}{\sigma_c}+\sqrt{m_i\frac{\sigma_3}{\sigma_c}+1}$$ Substituting σ₁=110, σ₃=4, σ_c=100 MPa: $$1.10 = 0.04+\sqrt{0.04m_i+1}\ \Rightarrow\ 1.1236=0.04m_i+1\ \Rightarrow\ m_i=3.09$$
   Check: m_i=3.09 is below the literature range usually quoted for limestone (m_i≈7–12, Hoek 2007 Table); it is the mathematically consistent result of the single low-confinement (σ₃=4 MPa) triaxial pair given in the question. Practice would regress m_i from several triaxial pairs across a range of confinement; only one pair is given here, so it is used as given.
4. Scale to rock-mass m, s at RMR = 73. $$m=m_i\exp\!\left(\frac{RMR-100}{28}\right)=3.09\exp\!\left(\frac{-27}{28}\right)=1.18$$ $$s=\exp\!\left(\frac{RMR-100}{9}\right)=\exp\!\left(\frac{-27}{9}\right)=\exp(-3)=0.0498$$ $$\boxed{m = 1.18,\quad s = 0.0498}$$
5. Part (c) — Maximum stress at the (unconfined) pillar face. At the free face the boundary is unconfined, σ₃=0, so the Hoek-Brown criterion reduces to $$\sigma_1=\sigma_3+\sigma_c\sqrt{m\frac{\sigma_3}{\sigma_c}+s}=\sigma_c\sqrt{s}$$ $$\sigma_{1,face}=100\sqrt{0.0498}=22.3\ \text{MPa}$$ $$\boxed{\sigma_{1,face} = 22.3\ \text{MPa}}$$
6. Part (d) — Maximum stress at the confined pillar centre. At the centre the given ratio k = σ₃/σ₁ = 0.075 confines the core, so σ1 and σ3 must be solved simultaneously with the Hoek-Brown criterion: $$\sigma_1=k\sigma_1+\sigma_c\sqrt{m\frac{k\sigma_1}{\sigma_c}+s}$$ Substituting the numbers and solving numerically (single positive root) gives σ_1,centre = 29.8 MPa, with the corresponding confining stress σ₃ = kσ₁ = 2.24 MPa. $$\boxed{\sigma_{1,centre} = 29.8\ \text{MPa}}$$ The confined core is roughly 34% stronger than the unconfined face, which is the expected sense — the small lateral confinement at the pillar centre measurably raises its Hoek-Brown strength above the free-face value.
7. Part (e) — Tributary area theory and factor of safety. Each pillar supports its own footprint plus half the surrounding opening on every side: $$A_{trib}=(W_p+W_o)^2=(7+6)^2=169\ \text{m}^2,\qquad A_{pillar}=7^2=49\ \text{m}^2$$ $$\sigma_{v,avg}=\gamma Z\frac{A_{trib}}{A_{pillar}}=(28)(80)\frac{169}{49}=7.73\ \text{MPa}$$ This corresponds to an extraction ratio r = 1−A_pillar/A_trib = 0.71 (71% of the ore removed, 29% left as pillars) — consistent with a 7 m pillar on a 6 m opening. The pillar's load-bearing capacity is governed by the confined core (the face may show minor spalling, but that does not by itself collapse the pillar), so the factor of safety is taken against the centre strength from part (d): $$FoS=\frac{\sigma_{1,centre}}{\sigma_{v,avg}}=\frac{29.8}{7.73}=3.86$$ $$\boxed{FoS_{centre} = 3.86}$$ Even judged against the more conservative, unconfined face strength, FoS_face = 22.3/7.73 = 2.89 — still comfortably above 1.0. Both checks indicate the old room-and-pillar workings are structurally adequate to be repurposed as an underground storage facility.