Geomechanics and Ground Control: Rock Properties, Rock Mass Classification, In-Situ Stress, Failure Theories, Openings, Subsidence, Slopes, Pillars, Supports and Filling
1. Geotechnical properties of rocks and rock mass classification
The index and strength properties of intact rock are measured on cores. Uniaxial compressive strength (UCS, σc) comes from loading a cylinder with length about 2–2.5 times its diameter to failure. Tensile strength is measured indirectly by the Brazilian test, a disc of diameter D and thickness t loaded across a diameter to failure at load P: σt = 2P/(π D t). The point load test gives an index Is = P/De², corrected to a 50 mm core as Is(50), and UCS is estimated from it by a site-calibrated factor (commonly of the order of 20–25). The triaxial test supplies σ1 at several confining pressures σ3 and so the failure envelope. Deformability is given by Young’s modulus and Poisson’s ratio, and durability of weak rocks by the slake durability index.
A rock mass is intact rock cut by discontinuities, and rock mass classification puts a number on how much the joints weaken it. RQD (Deere) is the percentage of a core run made up of intact pieces 10 cm or longer: RQD = Σ(lengths of pieces ≥ 10 cm)/total run × 100, with 0–25 very poor, 25–50 poor, 50–75 fair, 75–90 good and 90–100 excellent. The RMR (Bieniawski) adds ratings for six parameters — intact strength, RQD, discontinuity spacing, discontinuity condition, groundwater, and an adjustment for joint orientation — to give 0–100, in five classes of twenty points from very poor (below 21) to very good (81–100). The Q-system (Barton) multiplies three ratios: Q = (RQD/Jn)(Jr/Ja)(Jw/SRF) — block size, inter-block shear strength, and active stress — on a logarithmic scale from 0.001 to 1000. The GSI carries the classification into the Hoek–Brown criterion.
2. In-situ stress, instrumentation, and theories of rock failure
Before any opening is made the ground carries a virgin (in-situ) stress. The vertical component is the weight of the overburden, σv = γH, about 0.025–0.027 MPa per metre of depth for typical rock unit weights. The horizontal stress is expressed as k = σh/σv; elastic confinement alone would give k = ν/(1 − ν), but tectonic stress often makes k greater than 1 at shallow depth. Stress is measured by:
| Method or instrument | What it gives |
|---|---|
| Hydraulic fracturing | in a sealed borehole section, the shut-in pressure Ps equals the minimum horizontal stress σh; with breakdown pressure Pb, pore pressure P0 and tensile strength T, σH = 3σh − Pb − P0 + T |
| Overcoring (CSIRO hollow-inclusion cell, USBM gauge) | strain relief when the gauge is cored free; the HI cell gives the full 3-D stress tensor from one hole, the USBM gauge the stresses in the plane normal to the hole |
| Flat jack | the stress normal to a slot, as the pressure that restores the pins across it to their original spacing |
| Borehole extensometer, convergence indicator | displacement of anchors at depth; closure of an opening |
| Load cell, stress cell (stressmeter), piezometer | load on a support or bolt; stress change in the rock; pore-water pressure |
The Mohr–Coulomb criterion writes shear strength on a plane as τ = c + σn tan φ. In principal stresses it becomes σ1 = σc + σ3 tan²(45° + φ/2), with the uniaxial strength σc = 2c cos φ/(1 − sin φ); the plane of failure makes 45° + φ/2 with the minor principal stress direction. It over-predicts tensile strength, so a tension cut-off is added. The empirical Hoek–Brown criterion, σ1 = σ3 + σc(m σ3/σc + s)^a with a = 0.5 for intact and good-quality rock, captures the curvature of real envelopes, and its constants m and s fall as the mass becomes more jointed (s = 1 for intact rock). Griffith’s theory attributes failure to the growth of flaws from their tips and predicts a uniaxial compressive strength of 8 times the tensile strength.
3. Stress around mine openings, and subsidence
An opening diverts the virgin stress round itself. For a circular opening in an elastic medium (the Kirsch solution) with vertical stress σv and horizontal stress kσv, the tangential stress on the boundary is σθ = σv(3k − 1) at the crown and floor and σθ = σv(3 − k) at the sidewalls. Under hydrostatic stress (k = 1) it is 2σv everywhere; when k < 1/3 the crown goes into tension, which is why roofs fail first in low-horizontal-stress ground. The concentration dies out within about three radii. Rectangular and elliptical openings concentrate stress more at their corners and at the ends of the long axis; an ellipse elongated in the direction of the major principal stress is the most favourable shape.
Subsidence is the lowering of the surface above an extraction. Its trough extends beyond the edges of the working, bounded by the angle of draw measured from the vertical at the panel edge. The subsidence at the centre grows with panel width until the panel reaches the critical width, when the maximum possible subsidence — a fraction of the extracted thickness called the subsidence factor, higher for caving than for stowing — is reached; narrower panels are sub-critical and wider ones super-critical, with a flat-bottomed trough. The surface suffers tilt, curvature and horizontal strain (tensile over the edges, compressive over the goaf), which damage structures. Control is by leaving pillars, partial extraction, backfilling or stowing, harmonic mining and support pillars under important structures.
4. Slope and dump stability
Rock slopes fail along structures, so the first step is kinematic analysis on a stereonet: which failure modes can happen at all, given the joint sets and the face. Plane failure needs a plane that strikes within about ±20° of the face, daylights (dips less steeply than the face), and dips more steeply than its friction angle. Wedge failure slides on the line of intersection of two planes, with the same conditions on that line. Toppling occurs where steep joints dip into the face; circular failure occurs in heavily jointed rock, soil and waste dumps, where no single structure controls.
The factor of safety is resisting over driving force along the slip surface. For a dry infinite slope of cohesionless material at angle β, FoS = tan φ/tan β. For a dry plane failure of a block of weight W on a plane of area A dipping at ψ, FoS = (cA + W cos ψ tan φ)/(W sin ψ); water pressure in a tension crack and on the plane reduces the normal force and adds a driving force, and is the commonest trigger of failure. Circular failures are analysed by the method of slices (Fellenius, Bishop). Waste dumps fail by base failure on a weak foundation, by circular failure through the dump, or by sliding on the dump–floor contact, and their stability depends on dump height, overall angle, the strength of loose spoil, the foundation and drainage.
5. Pillar design, supports, mine filling and strata control
The tributary area method assumes each pillar carries the full overburden over its share of the area. For square pillars of width w on centres w + B (B the gallery width), the average pillar stress is σp = σv (w + B)²/w², equivalently σp = σv/(1 − e) with the extraction ratio e = 1 − (w/(w + B))². Pillar strength falls with size and rises with width-to-height ratio; the empirical formulas include Obert–Duvall σs = σ1(0.778 + 0.222 w/h), Bieniawski σs = σ1(0.64 + 0.36 w/h), both with σ1 the strength of a cubical specimen of critical size, and Salamon–Munro σs = 7.176 w^0.46/h^0.66 MPa (w, h in m), from South African coal. The factor of safety is σs/σp. Pillars with a width-to-height ratio well above 5 to 10 develop a confined core and behave as practically indestructible.
- Supports. Passive supports — timber and steel props, chocks, steel arches and sets — carry load only as the ground moves onto them. Rock bolts are active or semi-active reinforcement: mechanical (expansion shell) bolts, full-column resin or cement-grouted bolts, friction bolts (split sets, Swellex), and long cable bolts for wide spans. In the suspension design a bolt hangs a loosened roof layer of thickness t from competent rock above, so the load per bolt is γ t × (spacing)² and FoS = bolt capacity/load; in the beam-building design bolts clamp thin beds into a thicker composite beam. Shotcrete and mesh hold the rock between bolts. Longwall faces use powered supports (chock, shield, chock-shield) of a rated yield load.
- Mine filling. Voids are filled to support the ground, limit subsidence, raise extraction of pillars and dispose of waste: hydraulic sand stowing (sand carried in water, drained in place — long used under built-up areas in coalfields), pneumatic stowing, paste fill (dewatered tailings with a little cement, high solids and little drainage), cemented hydraulic fill and cemented rock fill in metal mines.
- Strata control manages the behaviour of the roof over a working face: the immediate roof caves behind the supports, the main fall occurs when the overhanging main roof first breaks, and periodic weighting follows at regular intervals of advance as each new cantilever breaks. The ground reaction curve shows the support pressure needed against the convergence allowed — a stiffer support installed early takes more load, a yielding one installed later takes less.
Key takeaways
- RQD counts core pieces of 10 cm or more; RMR adds six ratings (81–100 very good) and Q multiplies block size, joint shear strength and active stress.
- Brazilian σt = 2P/(πDt); Mohr–Coulomb σc = 2c cos φ/(1 − sin φ) and σ1 = σc + σ3 tan²(45° + φ/2); Griffith predicts σc = 8σt.
- Kirsch: σθ = σv(3k − 1) at the crown and σv(3 − k) at the walls; hydraulic fracturing gives σh = Ps and σH = 3σh − Pb − P0 + T.
- Infinite slope FoS = tan φ/tan β; plane failure FoS = (cA + W cos ψ tan φ)/(W sin ψ); plane failure needs strike within ±20°, daylighting and ψ > φ.
- Tributary area: σp = σv(w + B)²/w² = σv/(1 − e); FoS = pillar strength/pillar stress; a suspension bolt carries γ t s².
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
In a 150 cm core run, the intact pieces 10 cm or longer add up to 110 cm. The RQD, in %, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 73.33
RQD = 110/150 × 100 = 73.33%, which is “fair” (50–75). Dividing by the recovered length rather than the drilled run, or counting pieces shorter than 10 cm, inflates it; 110/150 = 0.7333 is the fraction, not the percentage.In Barton’s Q-system, Q = (RQD/Jn)(Jr/Ja)(Jw/SRF). The quotient Jr/Ja represents:
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Answer: B — the inter-block shear strength
Joint roughness over joint alteration measures how rough and how clay-filled the joints are — the shear strength between blocks. RQD/Jn is the block size and Jw/SRF the active stress; intact strength does not appear in Q at all, which is one difference from RMR.A Brazilian test on a disc of 54 mm diameter and 27 mm thickness fails at 15 kN. The tensile strength, in MPa, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 6.55
σt = 2P/(πDt) = 2 × 15 000/(π × 0.054 × 0.027) = 30 000/0.0045804 = 6.55 × 10⁶ Pa = 6.55 MPa. Omitting the 2 gives 3.27 MPa; P/(Dt) without π gives 10.3 MPa.A rock has cohesion 10 MPa and friction angle 30°. By the Mohr–Coulomb criterion, its strength σ1 at a confining pressure of 5 MPa, in MPa, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 49.64
σc = 2c cos φ/(1 − sin φ) = 20 × 0.86603/0.5 = 34.64 MPa, and tan²(45° + 15°) = tan² 60° = 3, so σ1 = 34.64 + 3 × 5 = 49.64 MPa. Using tan²(45° − φ/2) = 1/3 gives 36.31, and adding σ3 once instead of three times gives 39.64.Griffith’s theory of brittle fracture predicts that the uniaxial compressive strength of a rock is about:
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Answer: C — eight times its tensile strength
The original two-dimensional Griffith criterion gives σc = 8σt from the stress concentration at the tips of randomly oriented flat flaws. Measured ratios are often larger, 10 to 20, which is one reason empirical criteria such as Hoek–Brown replaced it in design.A circular tunnel is driven where the vertical stress is 10 MPa and the horizontal stress is 5 MPa. By the Kirsch solution, the tangential stress at the sidewall, in MPa, is ____.
Numerical answer — type the value.
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Answer: 25
k = 0.5, so σθ(wall) = σv(3 − k) = 10 × 2.5 = 25 MPa, while the crown carries σv(3k − 1) = 10 × 0.5 = 5 MPa. Swapping the two formulae puts 5 MPa at the wall; the hydrostatic value 2σv = 20 MPa applies only when k = 1.A hydraulic fracturing test gives a breakdown pressure of 12 MPa and a shut-in pressure of 8 MPa. The pore pressure is 2 MPa and the tensile strength of the rock is 5 MPa. The maximum horizontal stress, in MPa, is ____.
Numerical answer — type the value.
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Answer: 15
σh = Ps = 8 MPa, and σH = 3σh − Pb − P0 + T = 24 − 12 − 2 + 5 = 15 MPa. Treating the breakdown pressure as σH gives 12, and dropping the tensile strength (the re-opening form) gives 10.Square coal pillars 15 m wide on 20 m centres lie at a depth of 300 m, where the overburden unit weight is 25 kN/m³. The pillar strength is 18.88 MPa. The factor of safety by the tributary area method, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 1.42
σv = 25 × 300 = 7500 kPa = 7.5 MPa; σp = 7.5 × (20/15)² = 7.5 × 1.7778 = 13.33 MPa; FoS = 18.88/13.33 = 1.42. Using the ratio 20/15 unsquared gives σp = 10 MPa and FoS 1.89, forgetting that area, not width, carries the load.For square pillars 15 m wide on 20 m centres, the extraction ratio, to four decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.4375
e = 1 − (w/(w + B))² = 1 − (15/20)² = 1 − 0.5625 = 0.4375, and 1/(1 − e) = 1.778 is the same stress multiplier as (20/15)². Answering 0.25 uses the linear ratio 1 − 15/20, which would be right only for long rib pillars.Roof bolts on a 1.2 m × 1.2 m pattern suspend a loosened roof layer 1.5 m thick, of unit weight 25 kN/m³, from competent strata. Each bolt has a capacity of 120 kN. The factor of safety of the bolting, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 2.22
Load per bolt = γ t s² = 25 × 1.5 × 1.44 = 54 kN, so FoS = 120/54 = 2.22. Using the spacing unsquared (25 × 1.5 × 1.2 = 45 kN) gives 2.67, and forgetting the thickness gives a load in kN/m, not kN.A dry, cohesionless dump material with a friction angle of 35° stands as an infinite slope at 30°. The factor of safety, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 1.21
FoS = tan φ/tan β = tan 35°/tan 30° = 0.70021/0.57735 = 1.21. The ratio of the angles, 35/30 = 1.17, is not the factor of safety, and if seepage parallel to the slope were present the value would fall to roughly half.A rock block of weight 1000 kN per metre run rests on a plane dipping at 35° with a contact area of 20 m² per metre run. The plane has cohesion 20 kPa and friction angle 30°, and is dry. The factor of safety against plane failure, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 1.52
FoS = (cA + W cos ψ tan φ)/(W sin ψ) = (20 × 20 + 1000 × 0.81915 × 0.57735)/(1000 × 0.57358) = (400 + 472.93)/573.58 = 1.52. Leaving out cohesion gives tan 30°/tan 35° = 0.82 — cohesion is what keeps this block in place.For plane failure to be kinematically possible on a joint in a rock slope, which of the following conditions must hold? (Select all that apply.)
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Answer: B — The joint strikes within about ±20° of the slope face; C — The joint daylights in the face, dipping less steeply than the face; D — The joint dips more steeply than its angle of friction
The block must be able to slide out of the face (daylight, ψp < ψf), roughly along the dip of the face (strike within ±20°), and on a plane steep enough to overcome friction (ψp > φ). A joint steeper than the face does not outcrop in it, so the block has no free surface to slide onto.The maximum possible subsidence over an extracted panel is reached only when:
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Answer: B — the panel width equals or exceeds the critical width
Below the critical width the overburden arches across the panel and the centre subsides less than the maximum; at critical width the full subsidence factor is reached, and super-critical panels only widen the flat bottom of the trough. Stowing reduces the subsidence factor, and the angle of draw is never zero in practice.Which of the following statements about rock instrumentation and stress measurement are correct? (Select all that apply.)
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Answer: A — A CSIRO hollow-inclusion cell can give the complete 3-D stress tensor from a single overcored borehole; B — A flat jack measures the stress component normal to its slot; C — A borehole extensometer measures displacement of the rock at depth
The HI cell carries strain gauges in several directions, enough for all six components; the flat jack’s cancellation pressure is the stress across its slot; the extensometer reads anchor displacement. The USBM gauge measures diametral changes only, so it gives the stresses in the plane normal to the hole, and three non-parallel holes are needed for the full tensor.The method of filling voids in which sand carried in water is placed in the goaf and drained in place is called:
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Answer: B — Hydraulic sand stowing
Hydraulic stowing slurries sand to the goaf through pipes and drains the water away, leaving a dense fill that limits subsidence. Pneumatic stowing blows dry material with compressed air, cemented rock fill uses cement-bound waste rock, and caving leaves the goaf unfilled.