Question 3 of 5: Limestone pillar design — RMR, Hoek–Brown strength, tributary-area FoS

Question 3: Limestone pillar design — RMR, Hoek–Brown strength, tributary-area FoS (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. Square pillars in a room-and-pillar limestone mine, with strength data from point-load and triaxial testing.

Find. RMR; Hoek-Brown $m$, $s$; the unconfined (face) and confined (centre) pillar strengths; the tributary-area average pillar stress; and the resulting factor of safety.

Approach. Build RMR from the five Bieniawski parameters, calibrate the intact-rock Hoek–Brown constant $m_i$ from the one triaxial data point, scale to the rock mass via RMR, then evaluate the Hoek–Brown strength at $\sigma_3=0$ (pillar face) and at the pillar's own confined core ($\sigma_3=0.09\sigma_1$, per the given ratio), and finally compare the confined strength against the tributary-area average vertical stress for the factor of safety.

1. (a) RMR from the five Bieniawski parameters. Strength (UCS 120 MPa, the 100–250 MPa band) rates 12. RQD is not stated directly; “moderate spacing” bedding (200–600 mm band, taken at 0.4 m) gives a single-set volumetric joint count $J_v=1/0.4=2.5\ \text{m}^{-1}$, so $RQD=115-3.3(2.5)=106.8\%\to$ capped at 100% — rating 20. Discontinuity spacing (moderate, 200–600 mm) rates 10. Condition is read from Table E (Guidelines for Classification of Discontinuity Conditions), since the stated description does not match a single Table A band exactly: persistence (continuous bedding) 0 + separation (none visible) 6 + roughness (smooth) 1 + infilling (none) 6 + weathering (slightly weathered) 5 = 18. Groundwater (dry) rates 15. No orientation adjustment applies to a pillar-strength problem (it is a Table B correction for tunnel/foundation/ slope orientation relative to a specific discontinuity, not to pillar loading). $$RMR=12+20+10+18+15=75$$ $$\boxed{RMR=75\ \text{(Class II, Good rock)}}$$
2. (b) Intact-rock $m_i$ from the triaxial point, then Hoek–Brown $m,s$ for the rock mass. For intact rock ($s=1$, $m=m_i$), the given criterion $\sigma_1/\sigma_c=\sigma_3/\sigma_c+\sqrt{m\,\sigma_3/\sigma_c+s}$ rearranges to $$m_i=\frac{\left[\left(\dfrac{\sigma_1-\sigma_3}{\sigma_{ci}}\right)^2-1\right]\sigma_{ci}}{\sigma_3} =\frac{\left[\left(\dfrac{140-6}{120}\right)^2-1\right](120)}{6}\approx4.94$$ Scaling to the rock mass at $RMR=75$: the exam's printed sheet gives $s=(RMR-100)/9=(75-100)/9=-2.78$, which is negative and not a valid Hoek–Brown $s$ (physically $0\le s\le1$) — this is a typo in the formula sheet, missing the exponential that the standard Hoek–Brown (1980) correlation requires. Using the correct form $s=\exp\!\left(\frac{RMR-100}{9}\right)$: $$m=m_i\exp\!\left(\frac{75-100}{28}\right)=4.94\exp(-0.893)=2.02$$ $$s=\exp\!\left(\frac{75-100}{9}\right)=\exp(-2.778)=0.0622$$ $$\boxed{m\approx2.02,\ \ s\approx0.0622}$$
3. (c) Pillar face strength (unconfined, $\sigma_3=0$). At the free face $\sigma_3=0$, so $$\sigma_{1,\text{face}}=\sigma_3+\sigma_{ci}\sqrt{m\,\sigma_3/\sigma_{ci}+s}=\sigma_{ci}\sqrt{s}=120\sqrt{0.0622}$$ $$\boxed{\sigma_{1,\text{face}}\approx29.9\ \text{MPa}}$$
4. (d) Pillar centre strength (confined, $\sigma_3=0.09\sigma_1$). Substituting the given ratio into the Hoek–Brown criterion and solving the resulting quadratic in $\sigma_1$: $$\sigma_1=0.09\sigma_1+\sigma_{ci}\sqrt{\frac{m(0.09\sigma_1)}{\sigma_{ci}}+s} \ \Rightarrow\ (0.91)^2\sigma_1^2-(0.09\,m\,\sigma_{ci})\sigma_1-s\,\sigma_{ci}^2=0$$ Solving: $\sigma_{1,\text{centre}}\approx48.6$ MPa, so $\sigma_{3,\text{centre}}=0.09(48.6)\approx4.4$ MPa. $$\boxed{\sigma_{1,\text{centre}}\approx48.6\ \text{MPa}}$$ The centre carries roughly 1.6× the face's unconfined strength, purely from the lateral confinement the pillar's own core provides — the pillar is strongest where it is least accessible to inspect.
5. (e) Tributary-area average stress and factor of safety. Overburden (vertical) stress at 100 m: $\sigma_v=\gamma Z=26.5(100)/1000=2.65$ MPa. The tributary area per pillar is $A_T=(W_p+W_o)^2=(9+8)^2=289\ \text{m}^2$ against a pillar area $A_p=W_p^2=81\ \text{m}^2$, so $$\sigma_{v,\text{avg}}=\sigma_v\frac{A_T}{A_p}=2.65\left(\frac{289}{81}\right)\approx9.45\ \text{MPa}$$ This average, tributary stress applies across the pillar's full (confined) cross-section, so it is compared against the confined centre strength from (d) rather than the unconfined face strength: $$FoS=\frac{\sigma_{1,\text{centre}}}{\sigma_{v,\text{avg}}}=\frac{48.6}{9.45}$$ $$\boxed{FoS\approx5.1}$$