A rock specimen has a porosity of 0.20. Its void ratio, defined as the volume of voids divided by the volume of solids, is

0.25, because the void ratio equals n/(1 − n) for the stated porosity

0.20, because void ratio and porosity are the same quantity expressed differently

0.25, because the void ratio equals n/(1 − n) for the stated porosity

0.80, because the void ratio equals the fraction of the volume that is solid

0.40, because the void ratio is always twice the porosity of the rock

In a slake-durability test, a mudstone sample retains only 25 per cent of its dry mass after two cycles, whereas a granite sample retains 99 per cent. Which conclusion is best supported?

The mudstone will disintegrate readily on wetting and drying, so exposed faces in it need protection

The mudstone has the higher tensile strength, because its fragments were reduced to fines by the drum

The test result is meaningless, because slake durability applies only to rock with an RQD above 90 per cent

The granite has the lower porosity of the two, because slake durability is a direct measure of porosity

A rock obeys the Mohr-Coulomb criterion with cohesion c = 20 MPa and friction angle φ = 35° (compression positive; the intermediate principal stress is ignored). Its axial strength at a confining stress σ₃ = 10 MPa is closest to

113.7 MPa, from σ_c = 2c cos φ/(1 − sin φ) plus σ₃ tan²(45° + φ/2)

56.9 MPa, taking the cohesion itself as the unconfined strength

79.5 MPa, using the factor tan²(45° − φ/2) on the confining stress

86.8 MPa, adding the confining stress without any friction multiplier

113.7 MPa, from σ_c = 2c cos φ/(1 − sin φ) plus σ₃ tan²(45° + φ/2)

Which statement correctly distinguishes the Griffith, Mohr-Coulomb and Hoek-Brown failure criteria?

Hoek-Brown has a curved envelope fitted to triaxial data and scaled by GSI, whereas Mohr-Coulomb is a straight line with constant c and φ

Mohr-Coulomb accounts for the intermediate principal stress through its cohesion term, whereas the other two criteria ignore it

Griffith predicts a compressive strength equal to the tensile strength times tan²(45° + φ/2), whereas Mohr-Coulomb predicts a fixed ratio of eight

Hoek-Brown has a curved envelope fitted to triaxial data and scaled by GSI, whereas Mohr-Coulomb is a straight line with constant c and φ

Griffith was fitted empirically to triaxial tests on rock masses with GSI, whereas Hoek-Brown follows from the theory of elliptical cracks

In the Q system, a joint set that was clean and rough becomes coated with soft clay and begins to carry flowing water. With the other parameters unchanged, which description of the effect on Q is correct?

J_a rises and J_w falls, so the shear strength and active stress quotients both drop and Q decreases

J_r rises because the clay coat fills the joint and J_w rises because water flows through it, so Q increases

Only RQD can change, because Q depends on block size alone and joint wall condition is not part of Q

J_a rises and J_w falls, so the shear strength and active stress quotients both drop and Q decreases

J_a falls because the clay fills the aperture, so inter-block strength rises and Q increases

At a depth of 800 m in rock of unit weight 27 kN/m³, the horizontal stress is estimated by treating the rock as a laterally confined elastic mass with Poisson’s ratio 0.20. The horizontal stress is closest to

5.4 MPa, equal to ν/(1 − ν) times the vertical stress

4.3 MPa, taking ν itself as the ratio of horizontal to vertical stress

10.8 MPa, taking k = 0.5 as the usual ratio for elastic rock

21.6 MPa, taken equal to the vertical stress (k = 1)

A long circular tunnel is driven in elastic rock where the vertical stress is 12 MPa and the horizontal stress is 3 MPa (both compressive). Using the Kirsch solution, with compression taken as positive, the tangential stresses on the tunnel wall at the crown and at the sidewall are respectively

−3 MPa (tension) and 33 MPa, with the crown in tension and the sidewall heavily compressed

24 MPa and 24 MPa, since the wall stress is twice the vertical stress everywhere

−3 MPa (tension) and 33 MPa, with the crown in tension and the sidewall heavily compressed

3 MPa and 33 MPa, with the crown stress compressive because it equals the horizontal stress

33 MPa and −3 MPa, with the compressive peak at the crown and tension in the sidewall

A rock cut has a face dipping at 70° and the friction angle of the joints is 30°. Which discontinuity makes planar sliding kinematically possible?

A plane dipping 50° out of the face and striking within 10° of it, as it daylights and exceeds the friction angle

A plane dipping 50° into the slope and striking parallel to the face, because it is steeper than the friction angle

A plane dipping 50° out of the face and striking at 70° to the face, because it is steeper than the friction angle

A plane dipping 20° out of the face and striking parallel to it, because a low dip releases the block easily

A cohesionless infinite slope has a factor of safety of 1.40 when dry. Steady seepage then flows parallel to the slope with the water table at the ground surface. Take γ_sat = 20 kN/m³ and γ_w = 10 kN/m³. The factor of safety becomes

0.70, since FS is multiplied by γ′/γ_sat = 0.5

1.40, because the strength of a cohesionless slope does not depend on water

0.90, obtained by subtracting a fixed 0.5 for the effect of seepage

0.70, since FS is multiplied by γ′/γ_sat = 0.5

1.05, since only a quarter of the dry resistance is lost

Which of the following statements about rock index tests are correct?

I_s(50) is obtained by applying the size correction F = (D_e/50)^0.45 to the uncorrected point-load index

The Brazilian test gives the tensile strength of a rock disc as σ_t = 2P/(πDt), although the applied load is compressive

I_s(50) is obtained by applying the size correction F = (D_e/50)^0.45 to the uncorrected point-load index

The Brazilian test gives the tensile strength of a rock disc as σ_t = 2P/(πDt), although the applied load is compressive

A high Schmidt rebound number on a smooth dry surface proves that the rock mass behind it has a high rock-mass rating

A low slake-durability index indicates a rock that resists breakdown well under wetting and drying

Which of the following statements about the shear strength of rock discontinuities are correct?

In the Barton equation the roughness contribution JRC log₁₀(JCS/σ_n) decreases as the normal stress rises

When a clay infill is thicker than the asperity amplitude, the shear strength is governed mainly by the infill

A laboratory shear strength measured on a small joint sample usually underestimates the strength of a long field joint

In the Barton equation the roughness contribution JRC log₁₀(JCS/σ_n) decreases as the normal stress rises

When a clay infill is thicker than the asperity amplitude, the shear strength is governed mainly by the infill

Water pressure u in a joint raises the effective normal stress to σ_n + u and so raises the strength

Which of the following statements about geological factors in dams and tunnels are correct?

Uplift under a concrete dam reduces the effective normal stress on its base and so its sliding resistance

A water take of 1 Lugeon is 1 litre per minute per metre of test length at a test pressure of 1 MPa

Squeezing ground is the sudden violent failure of strong brittle rock under high stress

Uplift under a concrete dam reduces the effective normal stress on its base and so its sliding resistance

A tunnel is best aligned along a fault zone and close to it, so that its whole length lies in the same uniform weak rock

A water take of 1 Lugeon is 1 litre per minute per metre of test length at a test pressure of 1 MPa

Engineering Geology: Rock and Soil Properties, Failure Criteria, Rock-Mass Classification, In-situ Stress, Structures and Slopes

This chapter answers the Part B1 heading “Engineering Geology”: “Physico-mechanical properties of rocks and soils; rock index tests; Rock failure criteria (Mohr-Coulomb, Griffith, and Hoek-Brown criteria); Shear strength of rock discontinuities; Rock mass classifications (RMR and Q Systems); in-situ stresses; Geological factors in the construction of engineering structures including dams, tunnels, and excavation sites; Analysis of slope stability.” It is the first subject of Part B1 in the Geology and Geophysics (GG) paper. Its numericals are strength criteria, RQD, RMR and Q, in-situ stress and slope factors of safety.

1. Physico-mechanical properties of rocks and soils

Physical (index) properties say what a material is; mechanical properties say how it answers a load. For rock the basics are density, unit weight, water content and porosity n = V_v/V. For soil the phase relations tie solid, water and air together. With void ratio e = V_v/V_s: n = e/(1 + e), e = n/(1 − n), S·e = w·G (S saturation, w water content, G specific gravity of solids), γ_d = Gγ_w/(1 + e) and γ_sat = (G + e)γ_w/(1 + e). Worked example: G = 2.70, e = 0.80, γ_w = 9.81 kN/m³ give γ_sat = 3.50 × 9.81/1.80 = 19.1 kN/m³ and γ_d = 14.7 kN/m³.

Fine soils are described by the Atterberg limits: liquid limit w_L, plastic limit w_P and plasticity index PI = w_L − w_P (high PI means clay-rich, swelling-prone). Strength follows Terzaghi’s effective stress σ′ = σ − u through τ_f = c′ + σ′ tan φ′, so pore-water pressure u weakens soil and rock alike without changing the total load.

UCS σ_c: peak axial stress of an unconfined cylinder (length : diameter about 2.5 to 3). Tensile strength σ_t is far smaller, of the order of a tenth of σ_c.
Young’s modulus E is stress over strain in the near-linear part; Poisson’s ratio ν is minus lateral strain over axial strain.
Foliated rock is weakest when loaded along the foliation, shale loses strength on wetting, and a jointed mass is weaker than its intact core.

⚠️ Core strength is not rock-mass strength

Using the UCS of an intact core directly in a tunnel or slope calculation overstates the strength of a jointed mass. The criteria, discontinuity strength and classifications below bridge that gap.

2. Rock index tests

Index tests are quick, cheap and run on irregular pieces; they correlate with costly parameters (UCS, durability) but never replace them.

Point-load test: a core or lump is crushed between conical platens at load P. I_s = P/D_e², with D_e² = D² for a diametral test on a core of diameter D and D_e² = 4A/π for axial or block tests (A the minimum cross-section through the platen points). Size-corrected: I_s(50) = F·I_s, F = (D_e/50)^0.45. UCS is roughly 20 to 25 times I_s(50), a rule of thumb to be calibrated for each rock.
Brazilian test (indirect tension): a disc of diameter D and thickness t, loaded across a diameter to failure load P, gives σ_t = 2P/(πDt); the disc splits in tension although the load is compressive.
Schmidt rebound hammer: rebound number R measures surface hardness and correlates with UCS, but orientation, roughness, moisture and weathering affect it.
Slake-durability test: oven-dried lumps rotate in a 2 mm mesh drum partly immersed in water for two 10-minute cycles; I_d2 = (dry mass retained ÷ initial dry mass) × 100. A low value (shale, mudstone, weathered rock) warns of breakdown on wetting and drying.
Aggregate tests (Los Angeles abrasion, impact, crushing) give the percentage broken down: lower means tougher aggregate.

Worked example. A diametral test on a 50 mm core fails at P = 10 kN: I_s = 10 000/2500 = 4.0 MPa; F = 1 at D = 50 mm, so I_s(50) = 4.0 MPa and, with a calibration factor of 24, UCS ≈ 96 MPa. For a 60 mm core F = (60/50)^0.45 = 1.086.

⚠️ An index is not the strength

I_s(50) is not the UCS, and 20 to 25 is only a rough average. A high Schmidt number on a smooth dry surface says nothing about the joints behind it.

3. Rock failure criteria: Mohr-Coulomb, Griffith and Hoek-Brown

A failure criterion is the stress state (σ₁ ≥ σ₂ ≥ σ₃, compression positive) at which rock fails. Mohr-Coulomb: shear fails on a plane when τ = c + σ_n tan φ (cohesion c, friction angle φ). In principal stresses σ₁ = σ_c + σ₃ tan²(45° + φ/2), with σ_c = 2c cos φ/(1 − sin φ). The shear plane contains σ₂ and makes 45° − φ/2 with σ₁. The envelope is straight, σ₂ is ignored, and a tension cut-off is added because the line overestimates tensile strength. Example: c = 20 MPa, φ = 35° give σ_c = 76.8 MPa and tan²(62.5°) = 3.69, so at σ₃ = 10 MPa, σ₁ = 76.8 + 36.9 = 113.7 MPa.

Griffith argued from physics: rock holds microscopic elliptical flaws, tensile stress concentrates at their tips, and failure starts when the tip stress reaches the material strength even under overall compression. For σ₁ + 3σ₃ ≥ 0: (σ₁ − σ₃)² = 8σ_t(σ₁ + σ₃); otherwise σ₃ = −σ_t. Putting σ₃ = 0 gives σ_c = 8σ_t; in shear form τ² = 4σ_t(σ_t + σ_n), a parabola. Real rocks show a larger ratio because compression closes cracks and friction acts across them (modified Griffith).

Hoek-Brown is empirical, fitted to triaxial tests on intact rock and scaled to rock masses: σ₁ = σ₃ + σ_ci (m_b σ₃/σ_ci + s)^a. Intact rock has m_b = m_i, s = 1, a = 0.5. For a mass, with the Geological Strength Index GSI (0 to 100) and disturbance factor D (0 undisturbed, 1 heavily blast-damaged): m_b = m_i exp[(GSI − 100)/(28 − 14D)], s = exp[(GSI − 100)/(9 − 3D)], a = ½ + (e^(−GSI/15) − e^(−20/3))/6. Then σ_c = σ_ci s^a and σ_t = −s σ_ci/m_b. Example: σ_ci = 80 MPa, m_b = 2, s = 0.01, a = 0.5, σ₃ = 5 MPa give σ₁ = 5 + 80 √(0.125 + 0.01) = 34.4 MPa.

⚠️ Do not mix the three

σ_c = 8σ_t is Griffith only; in Hoek-Brown σ_c = σ_ci s^a. None of the three uses σ₂.

4. Shear strength of rock discontinuities

A discontinuity (joint, bedding plane, foliation, fault) has almost no tensile strength, so at low stress the strength of the mass is the strength of sliding on these surfaces. It depends on roughness, wall strength, infill, water and normal stress σ_n. A clean smooth surface follows τ = σ_n tan φ_b (basic friction angle φ_b, c ≈ 0).

Patton (saw-tooth asperities of inclination i): at low σ_n the block rides up the teeth, τ = σ_n tan(φ_b + i); at high σ_n the teeth shear off and strength becomes c_a + σ_n tan φ_r (a bilinear envelope).
Barton and Choubey: τ = σ_n tan[φ_r + JRC log₁₀(JCS/σ_n)]. JRC is the joint roughness coefficient (0 smooth planar to about 20 rough undulating), JCS the wall compressive strength, φ_r the residual friction angle. The roughness term shrinks as σ_n rises and vanishes when σ_n = JCS.
Infill and water: an infill thicker than the asperity amplitude controls the strength (a clay seam can be as weak as clay itself), and water pressure u replaces σ_n by σ_n − u.

Worked example. JRC = 10, JCS = 100 MPa, σ_n = 1 MPa, φ_r = 30°: the roughness term is 10 × log₁₀(100) = 20°, so τ = tan 50° = 1.19 MPa. At σ_n = 10 MPa the term is 10°, τ = 10 tan 40° = 8.39 MPa, and τ/σ_n has fallen from 1.19 to 0.84.

🎯 Scale effect

JRC and JCS fall as a joint gets longer, since large-scale waviness is gentler and large walls are weaker, so small-sample tests overestimate the strength of a long joint.

5. Rock-mass classification: RQD, RMR and the Q system

A classification rates the mass from a few observable parameters and links the rating to stand-up time, strength and support. RQD (Deere) is the sum of sound core pieces of at least 10 cm divided by the core run, in percent (drilling breaks excluded). Classes: below 25 very poor, 25 to 50 poor, 50 to 75 fair, 75 to 90 good, 90 to 100 excellent. From joints per cubic metre, RQD ≈ 115 − 3.3 J_v (100 if J_v is below about 4.5).

RMR (Bieniawski, 1989) adds the ratings of five parameters: intact UCS (maximum 15), RQD (20), joint spacing (20), joint condition (30) and groundwater (15), a basic total of at most 100. An orientation adjustment is then deducted (zero if favourable). Classes of the adjusted RMR are in the table; class I stands about 20 years unsupported over a 15 m span, class V only about half an hour over 1 m.

RMR classes and the rock-mass strength they suggest
Class	RMR	Description	Cohesion (kPa), friction angle
I	81–100	Very good	above 400, above 45°
II	61–80	Good	300–400, 35°–45°
III	41–60	Fair	200–300, 25°–35°
IV	21–40	Poor	100–200, 15°–25°
V	below 21	Very poor	below 100, below 15°

The Q system (Barton, Lien and Lunde, 1974) is Q = (RQD/J_n)·(J_r/J_a)·(J_w/SRF): relative block size (J_n joint-set number), inter-block shear strength (J_r roughness, J_a alteration) and active stress (J_w water reduction, 1 if dry; SRF stress reduction factor). RQD below 10 is taken as 10. Q is logarithmic, 0.001 to 1000: below 0.01 exceptionally poor, 0.01–0.1 extremely poor, 0.1–1 very poor, 1–4 poor, 4–10 fair, 10–40 good, 40–100 very good, 100–400 extremely good, above 400 exceptionally good. Support is read from Q and D_e = span ÷ ESR, where the excavation support ratio ESR is larger for temporary openings and smaller for openings that must be very safe.

Worked examples. Q: RQD = 70, J_n = 9, J_r = 1.5, J_a = 2, J_w = 1, SRF = 1 give Q = (70/9)(0.75)(1) = 5.83, fair. RMR: ratings 7, 13, 10, 20, 10 total 60; an orientation adjustment of −5 gives 55, class III. The systems are loosely related by RMR ≈ 9 ln Q + 44, and GSI ≈ RMR(1989) − 5 when the groundwater rating is taken as 15 (dry) and no orientation adjustment is made.

⚠️ Adjusted RMR sets the class; Q is a product

The orientation deduction can drop a mass across a class boundary (62 to 57 moves class II to III). In Q, a wet or altered joint lowers J_w or raises J_a and Q falls sharply, because Q is a product, not a sum.

6. In-situ stresses

Rock at depth is stressed before anything is excavated. The vertical stress is the overburden weight, σ_v = γz (about 0.027z MPa, z in metres, for γ = 27 kN/m³). The horizontal stress is σ_h = kσ_v. A laterally confined elastic mass gives k = ν/(1 − ν) (0.33 for ν = 0.25), while weak or creeping rock and great depth tend to k = 1. Near the surface k is often larger, because tectonic compression and erosional unloading keep horizontal stress high; measured values lie within 100/z + 0.3 ≤ k ≤ 1500/z + 0.5 (z in metres), a wide band that narrows with depth.

Measurement: hydraulic fracturing (pressurise a sealed borehole interval until the wall cracks; breakdown and shut-in pressures give the horizontal stresses), overcoring (strain relief of a gauge glued in a borehole) and the flat-jack test. Example: z = 600 m, γ = 27 kN/m³ give σ_v = 16.2 MPa; for ν = 0.25, σ_h = 0.333 × 16.2 = 5.4 MPa.

An opening concentrates stress. For a circular tunnel (Kirsch solution) the tangential wall stress is (3k − 1)σ_v at the crown and floor and (3 − k)σ_v at the sidewalls; for k = 1 it is 2σ_v all round. Tension appears at the crown when k < 1/3. High wall stress against σ_c causes spalling and rockburst in strong brittle rock and squeezing in weak rock. Example: σ_v = 10 MPa, k = 0.5 give 5 MPa at the crown and 25 MPa at the sidewalls.

⚠️ The crown can go into tension

“2σ_v at the wall” holds only for k = 1. With low k the sidewalls are heavily loaded and the roof is lightly loaded or in tension; with high k the roof is critical. Compression is positive, so a negative result is tension.

7. Geological factors in dams, tunnels and excavation sites

Dams. Concrete gravity dams need sound rock; arch dams also need strong abutments to take the thrust; embankment dams tolerate weaker foundations if seepage and settlement are controlled. Water under a dam creates uplift, which lowers effective normal stress and sliding resistance. Permeability is tested by the Lugeon test: 1 Lugeon is a water take of 1 litre per minute per metre of test length at 1 MPa, of the order of 10⁻⁷ m/s. Curtain grouting cuts seepage and consolidation grouting stiffens the foundation. Reservoirs may leak through karst, slide on their banks, silt up and rarely induce earthquakes.
Tunnels. Rock is classified (RMR, Q) to choose support: rock bolts, shotcrete, steel ribs or lining. Avoid driving along and close to a major fault or shear zone; cross it at a large angle. In steeply dipping layered rock an axis across the strike is preferred to one along it. Portals belong in sound rock, not loose colluvium on a steep slope.
Cuts and excavation sites. The stable angle depends on joints daylighting in the face, the depth of weathering (residual soil over rock), groundwater (dewatering, pore pressure) and blast damage. Foundations on weathered rock face variable depth to sound rock; solution cavities in limestone and expansive clay must be proved before building.

ℹ️ Ground behaviour in tunnels

Squeezing is slow inward flow of weak rock under high stress; swelling comes from clay minerals and anhydrite taking up water; rockburst is sudden brittle failure of strong rock under high stress. Identify the mechanism before choosing support.

8. Analysis of slope stability

Factor of safety FS = resisting ÷ driving, with FS = 1 at limit equilibrium. In rock, the failure mode is set first by discontinuity geometry against the face (kinematic analysis, done on a stereonet by plotting poles and a friction circle). Planar failure needs a plane that strikes within about 20° of the face, daylights (plane dip ψ_p less than face dip ψ_f) and is steeper than the friction angle (ψ_p > φ). Wedge failure needs two planes whose intersection line plunges out of the face at more than φ. Toppling needs steep layers dipping into the slope. Circular failure occurs in soil, waste or heavily fractured weak rock.

For a dry block of weight W on a plane of area A dipping at ψ_p: FS = (cA + W cos ψ_p tan φ)/(W sin ψ_p), which is tan φ/tan ψ_p if c = 0. Water pressure U on the plane lowers the normal force; water thrust V in a tension crack adds to the driving force. Example per metre of slope: W = 4000 kN, ψ_p = 35°, φ = 30°, c = 25 kPa, A = 20 m²: FS = (500 + 4000 cos 35° tan 30°)/(4000 sin 35°) = (500 + 1892)/2294 = 1.04.

For an infinite slope of inclination β with the failure plane at depth z: FS = [c + (γ z cos²β − u) tan φ]/(γ z sin β cos β). Dry and cohesionless it reduces to tan φ/tan β, independent of depth. With seepage parallel to the slope and the water table at the surface, FS = (γ′/γ_sat) tan φ/tan β, about half the dry value. Circular failure is analysed by the method of slices as the ratio of resisting to driving moments about a trial centre.

⚠️ Stable against sliding is not stable

A plane dipping into the slope cannot slide planarly, yet steep layers dipping into the face can topple; a plane flatter than φ is safe only while dry. Check every mode. Remedies follow the cause: flatten or bench the cut, drain, bolt or anchor, shotcrete, scale loose blocks, or support the toe.

Key takeaways

Soil phase relations: n = e/(1 + e), S·e = wG, γ_sat = (G + e)γ_w/(1 + e); strength follows σ′ = σ − u. Core strength overstates a jointed mass.
Index tests: I_s(50) = (D_e/50)^0.45 P/D_e², UCS ≈ 20–25 I_s(50); Brazilian σ_t = 2P/(πDt); slake durability I_d2 measures breakdown on wetting and drying.
Mohr-Coulomb is linear, σ₁ = σ_c + σ₃ tan²(45° + φ/2); Griffith has σ_c = 8σ_t; Hoek-Brown is empirical and curved, scaled by GSI and D. None uses σ₂.
Joint strength: Patton τ = σ_n tan(φ_b + i); Barton τ = σ_n tan[φ_r + JRC log₁₀(JCS/σ_n)]. Lab values overstate long joints; infill and water cut strength.
RQD = Σ sound pieces ≥ 10 cm ÷ core run. RMR sums five ratings (15, 20, 20, 30, 15) then deducts orientation. Q = (RQD/J_n)(J_r/J_a)(J_w/SRF) is a product; RMR ≈ 9 ln Q + 44.
σ_h = kσ_v, k = ν/(1 − ν) when confined; tunnel wall stress is (3k − 1)σ_v at the crown, (3 − k)σ_v at the sidewalls. Planar FS = (cA + W cos ψ tan φ)/(W sin ψ); dry cohesionless infinite slope FS = tan φ/tan β.

Practice questions (18)

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.

A rock specimen has a porosity of 0.20. Its void ratio, defined as the volume of voids divided by the volume of solids, is
1. 0.20, because void ratio and porosity are the same quantity expressed differently
2. 0.25, because the void ratio equals n/(1 − n) for the stated porosity
3. 0.80, because the void ratio equals the fraction of the volume that is solid
4. 0.40, because the void ratio is always twice the porosity of the rock
Show answer
Answer: B — 0.25, because the void ratio equals n/(1 − n) for the stated porosity
Porosity is n = V_v/V and void ratio is e = V_v/V_s, so e = n/(1 − n) = 0.20/0.80 = 0.25. Equating e with n is the tempting slip, but the denominators differ (total volume against solid volume), and e exceeds n whenever n is non-zero. The solid fraction is 0.80, not the void ratio.
In a slake-durability test, a mudstone sample retains only 25 per cent of its dry mass after two cycles, whereas a granite sample retains 99 per cent. Which conclusion is best supported?
1. The mudstone will disintegrate readily on wetting and drying, so exposed faces in it need protection
2. The mudstone has the higher tensile strength, because its fragments were reduced to fines by the drum
3. The test result is meaningless, because slake durability applies only to rock with an RQD above 90 per cent
4. The granite has the lower porosity of the two, because slake durability is a direct measure of porosity
Show answer
Answer: A — The mudstone will disintegrate readily on wetting and drying, so exposed faces in it need protection
The slake-durability index measures resistance to breakdown under wetting, drying and tumbling, so 25 per cent against 99 per cent says the mudstone disintegrates on exposure and needs shotcrete, sealing or a protective cover. It is not a porosity or a tensile-strength measure, and a mass loss in the drum does not mean higher tensile strength (the opposite is expected). The test is run on lumps and has no RQD condition.
A rock obeys the Mohr-Coulomb criterion with cohesion c = 20 MPa and friction angle φ = 35° (compression positive; the intermediate principal stress is ignored). Its axial strength at a confining stress σ₃ = 10 MPa is closest to
1. 56.9 MPa, taking the cohesion itself as the unconfined strength
2. 79.5 MPa, using the factor tan²(45° − φ/2) on the confining stress
3. 86.8 MPa, adding the confining stress without any friction multiplier
4. 113.7 MPa, from σ_c = 2c cos φ/(1 − sin φ) plus σ₃ tan²(45° + φ/2)
Show answer
Answer: D — 113.7 MPa, from σ_c = 2c cos φ/(1 − sin φ) plus σ₃ tan²(45° + φ/2)
The unconfined strength is σ_c = 2 × 20 × cos 35°/(1 − sin 35°) = 76.8 MPa and the multiplier is tan²(62.5°) = 3.69, so σ₁ = 76.8 + 10 × 3.69 = 113.7 MPa. Using 45° − φ/2 inverts the multiplier (0.27, giving 79.5), omitting it adds only 10 MPa (86.8), and treating c as σ_c ignores the friction term (56.9).
Which statement correctly distinguishes the Griffith, Mohr-Coulomb and Hoek-Brown failure criteria?
1. Mohr-Coulomb accounts for the intermediate principal stress through its cohesion term, whereas the other two criteria ignore it
2. Griffith predicts a compressive strength equal to the tensile strength times tan²(45° + φ/2), whereas Mohr-Coulomb predicts a fixed ratio of eight
3. Hoek-Brown has a curved envelope fitted to triaxial data and scaled by GSI, whereas Mohr-Coulomb is a straight line with constant c and φ
4. Griffith was fitted empirically to triaxial tests on rock masses with GSI, whereas Hoek-Brown follows from the theory of elliptical cracks
Show answer
Answer: C — Hoek-Brown has a curved envelope fitted to triaxial data and scaled by GSI, whereas Mohr-Coulomb is a straight line with constant c and φ
Hoek-Brown is an empirical, curved envelope whose parameters m_b, s and a are obtained from GSI, whereas Mohr-Coulomb is linear in τ–σ_n with constant c and φ. The two alternatives that swap origins are reversed: Griffith comes from crack theory and Hoek-Brown is empirical. The σ₂ claim is false, since none of the three uses σ₂. The fixed ratio of eight belongs to Griffith, not Mohr-Coulomb.
A 150 cm core run contains sound pieces of lengths 28, 8, 15, 12, 40, 6 and 22 cm; the rest of the run was lost. Using the standard definition of RQD (pieces of at least 10 cm) and the classes below 25 very poor, 25 to 50 poor, 50 to 75 fair, 75 to 90 good, 90 to 100 excellent, the RQD and class are
1. 60 per cent, fair
2. 78 per cent, good
3. 87 per cent, good
4. 78 per cent, fair
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Answer: B — 78 per cent, good
Only pieces of at least 10 cm count: 28 + 15 + 12 + 40 + 22 = 117 cm, so RQD = 117/150 × 100 = 78 per cent, which lies in the 75 to 90 band, good. Adding the 8 cm and 6 cm pieces gives 131 cm and 87 per cent, which is core recovery and not RQD. Counting only pieces over 20 cm gives 60 per cent. 78 per cent is not fair, because the fair band ends at 75.
In the Q system, a joint set that was clean and rough becomes coated with soft clay and begins to carry flowing water. With the other parameters unchanged, which description of the effect on Q is correct?
1. J_r rises because the clay coat fills the joint and J_w rises because water flows through it, so Q increases
2. Only RQD can change, because Q depends on block size alone and joint wall condition is not part of Q
3. J_a rises and J_w falls, so the shear strength and active stress quotients both drop and Q decreases
4. J_a falls because the clay fills the aperture, so inter-block strength rises and Q increases
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Answer: C — J_a rises and J_w falls, so the shear strength and active stress quotients both drop and Q decreases
Clay coating raises the alteration number J_a, lowering J_r/J_a, and flowing water lowers the water reduction factor J_w (1 for dry), lowering J_w/SRF. Since Q is a product, both changes reduce it. Rougher joints would raise J_r, but a soft clay coat does not make the joint rougher; Q is not a function of RQD alone, and a clay coat weakens the joint rather than strengthening it.
At a depth of 800 m in rock of unit weight 27 kN/m³, the horizontal stress is estimated by treating the rock as a laterally confined elastic mass with Poisson’s ratio 0.20. The horizontal stress is closest to
1. 5.4 MPa, equal to ν/(1 − ν) times the vertical stress
2. 4.3 MPa, taking ν itself as the ratio of horizontal to vertical stress
3. 10.8 MPa, taking k = 0.5 as the usual ratio for elastic rock
4. 21.6 MPa, taken equal to the vertical stress (k = 1)
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Answer: A — 5.4 MPa, equal to ν/(1 − ν) times the vertical stress
The vertical stress is 27 × 800 = 21.6 MPa. For lateral confinement k = ν/(1 − ν) = 0.20/0.80 = 0.25, so σ_h = 0.25 × 21.6 = 5.4 MPa. Using ν alone gives 4.3 MPa, which omits the (1 − ν) in the denominator; k = 1 (equal stress) holds for weak, creeping rock or very great depth, not for this elastic case.
A long circular tunnel is driven in elastic rock where the vertical stress is 12 MPa and the horizontal stress is 3 MPa (both compressive). Using the Kirsch solution, with compression taken as positive, the tangential stresses on the tunnel wall at the crown and at the sidewall are respectively
1. 24 MPa and 24 MPa, since the wall stress is twice the vertical stress everywhere
2. −3 MPa (tension) and 33 MPa, with the crown in tension and the sidewall heavily compressed
3. 3 MPa and 33 MPa, with the crown stress compressive because it equals the horizontal stress
4. 33 MPa and −3 MPa, with the compressive peak at the crown and tension in the sidewall
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Answer: B — −3 MPa (tension) and 33 MPa, with the crown in tension and the sidewall heavily compressed
With k = 3/12 = 0.25, the crown stress is (3k − 1)σ_v = (0.75 − 1) × 12 = −3 MPa, i.e. tension, and the sidewall stress is (3 − k)σ_v = 2.75 × 12 = 33 MPa. The twice-σ_v rule holds only for k = 1. Swapping the two locations puts the tension in the wrong place, and giving the crown a positive 3 MPa ignores the sign of 3k − 1, which is negative when k is below 1/3.
A rock cut has a face dipping at 70° and the friction angle of the joints is 30°. Which discontinuity makes planar sliding kinematically possible?
1. A plane dipping 50° out of the face and striking within 10° of it, as it daylights and exceeds the friction angle
2. A plane dipping 50° into the slope and striking parallel to the face, because it is steeper than the friction angle
3. A plane dipping 50° out of the face and striking at 70° to the face, because it is steeper than the friction angle
4. A plane dipping 20° out of the face and striking parallel to it, because a low dip releases the block easily
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Answer: A — A plane dipping 50° out of the face and striking within 10° of it, as it daylights and exceeds the friction angle
Planar sliding needs a plane that daylights in the face (dips out of it, at less than the face dip of 70°), is steeper than φ = 30°, and strikes within about 20° of the face so that the block can move without constraint. Only a plane dipping 50° out of the face and striking within 10° meets all three. A 20° plane is flatter than φ and holds by friction, a plane dipping into the slope cannot daylight, and a plane striking 70° from the face is held in by its own sides.
A cohesionless infinite slope has a factor of safety of 1.40 when dry. Steady seepage then flows parallel to the slope with the water table at the ground surface. Take γ_sat = 20 kN/m³ and γ_w = 10 kN/m³. The factor of safety becomes
1. 1.40, because the strength of a cohesionless slope does not depend on water
2. 0.90, obtained by subtracting a fixed 0.5 for the effect of seepage
3. 0.70, since FS is multiplied by γ′/γ_sat = 0.5
4. 1.05, since only a quarter of the dry resistance is lost
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Answer: C — 0.70, since FS is multiplied by γ′/γ_sat = 0.5
For seepage parallel to the slope with the water table at the surface, FS = (γ′/γ_sat) tan φ/tan β. Here γ′ = 20 − 10 = 10 kN/m³, so γ′/γ_sat = 0.5 and FS = 0.5 × 1.40 = 0.70, which is below 1, so the slope fails. Water does reduce the effective stress on the sliding plane, so the dry value is not retained, and the fall is a multiplicative factor, not a fixed subtraction.
Which of the following statements about rock index tests are correct?
1. I_s(50) is obtained by applying the size correction F = (D_e/50)^0.45 to the uncorrected point-load index
2. The Brazilian test gives the tensile strength of a rock disc as σ_t = 2P/(πDt), although the applied load is compressive
3. A high Schmidt rebound number on a smooth dry surface proves that the rock mass behind it has a high rock-mass rating
4. A low slake-durability index indicates a rock that resists breakdown well under wetting and drying
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Answer: A — I_s(50) is obtained by applying the size correction F = (D_e/50)^0.45 to the uncorrected point-load index; B — The Brazilian test gives the tensile strength of a rock disc as σ_t = 2P/(πDt), although the applied load is compressive
The point-load index is size-corrected to a 50 mm reference with the exponent 0.45, and the Brazilian test converts a compressive diametral load into a tensile strength through 2P/(πDt). A rebound number reflects only the near-surface hardness and says nothing reliable about the joints in the mass. A low slake-durability index means the opposite of good resistance: the rock breaks down readily.
Which of the following statements about the shear strength of rock discontinuities are correct?
1. A laboratory shear strength measured on a small joint sample usually underestimates the strength of a long field joint
2. In the Barton equation the roughness contribution JRC log₁₀(JCS/σ_n) decreases as the normal stress rises
3. When a clay infill is thicker than the asperity amplitude, the shear strength is governed mainly by the infill
4. Water pressure u in a joint raises the effective normal stress to σ_n + u and so raises the strength
Show answer
Answer: B — In the Barton equation the roughness contribution JRC log₁₀(JCS/σ_n) decreases as the normal stress rises; C — When a clay infill is thicker than the asperity amplitude, the shear strength is governed mainly by the infill
Because JCS/σ_n shrinks as σ_n grows, the logarithmic roughness term falls (and vanishes at σ_n = JCS), and an infill thicker than the asperities separates the walls so the seam, not the rock, controls sliding. Small samples are rougher relative to their size and so overestimate the strength of long joints (the scale effect), and water pressure lowers, not raises, the effective normal stress to σ_n − u.
Which of the following statements about geological factors in dams and tunnels are correct?
1. Squeezing ground is the sudden violent failure of strong brittle rock under high stress
2. Uplift under a concrete dam reduces the effective normal stress on its base and so its sliding resistance
3. A tunnel is best aligned along a fault zone and close to it, so that its whole length lies in the same uniform weak rock
4. A water take of 1 Lugeon is 1 litre per minute per metre of test length at a test pressure of 1 MPa
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Answer: B — Uplift under a concrete dam reduces the effective normal stress on its base and so its sliding resistance; D — A water take of 1 Lugeon is 1 litre per minute per metre of test length at a test pressure of 1 MPa
Uplift lowers effective normal stress and therefore friction on the base, and the Lugeon is defined as 1 L/min per metre at 1 MPa. Squeezing is the slow inward flow of weak rock; the sudden violent failure of strong rock is rockburst. Driving a tunnel along and near a fault zone exposes its length to the worst ground, so the alignment should cross the zone at a large angle.
A diametral point-load test on a 60 mm diameter core fails at P = 18 kN. Take I_s = P/D² (D_e = D for a diametral test), F = (D_e/50)^0.45 and UCS = 22 × I_s(50). The UCS, to the nearest whole MPa, is ____ MPa.
Numerical answer — type the value.
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Answer: 119
I_s = 18 000 N / 3600 mm² = 5.00 MPa. F = (60/50)^0.45 = 1.2^0.45 = 1.0855. I_s(50) = 5.00 × 1.0855 = 5.43 MPa. UCS = 22 × 5.43 = 119.4, which is 119 MPa to the nearest whole number. Forgetting F gives 110 MPa.
A rock mass follows the Hoek-Brown criterion σ₁ = σ₃ + σ_ci (m_b σ₃/σ_ci + s)^a with σ_ci = 80 MPa, m_b = 2, s = 0.01 and a = 0.5. At a confining stress σ₃ = 5 MPa its axial strength σ₁, to one decimal place, is ____ MPa.
Numerical answer — type the value.
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Answer: 34.4
m_b σ₃/σ_ci = 2 × 5/80 = 0.125. Adding s gives 0.135, and 0.135^0.5 = 0.3674. σ₁ = 5 + 80 × 0.3674 = 5 + 29.39 = 34.39, i.e. 34.4 MPa. The unconfined strength would be only σ_ci s^a = 8 MPa, so confinement of 5 MPa raises the strength more than fourfold.
A rock joint has JRC = 12, JCS = 60 MPa and residual friction angle φ_r = 28°. Use the Barton equation τ = σ_n tan[φ_r + JRC log₁₀(JCS/σ_n)]. At a normal stress σ_n = 0.6 MPa the peak shear strength, to two decimal places, is ____ MPa.
Numerical answer — type the value.
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Answer: 0.77
JCS/σ_n = 60/0.6 = 100, so log₁₀ = 2 and the roughness term is 12 × 2 = 24°. The effective angle is 28° + 24° = 52°. τ = 0.6 × tan 52° = 0.6 × 1.2799 = 0.768, i.e. 0.77 MPa. Using φ_r alone would give only 0.32 MPa.
Barton’s Q is computed from RQD = 55, J_n = 9, J_r = 2, J_a = 1.5, J_w = 0.5 and SRF = 2.5. Q, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 1.6
Q = (RQD/J_n)(J_r/J_a)(J_w/SRF) = (55/9)(2/1.5)(0.5/2.5) = 6.111 × 1.333 × 0.2 = 1.63, i.e. 1.6, which falls in the poor class (1 to 4). The three quotients are block size 6.11, inter-block shear strength 1.33 and active stress 0.2.
A dry rock block of weight W = 3000 kN per metre of slope length rests on a plane dipping at ψ = 40°. The plane has area A = 15 m² per metre, cohesion c = 20 kPa and friction angle φ = 35°. There is no water and no tension crack. The factor of safety FS = (cA + W cos ψ tan φ)/(W sin ψ), to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.99
Resisting force = cA + W cos ψ tan φ = 20 × 15 + 3000 × 0.7660 × 0.7002 = 300 + 1609 = 1909 kN. Driving force = W sin ψ = 3000 × 0.6428 = 1928 kN. FS = 1909/1928 = 0.99, so the block is at the point of sliding. Dropping the cohesion would give 0.83.