Reservoir Geophysics: Rock Physics, Porosity, Saturation and Permeability, Failure and Fracture Mechanics, and Borehole Applications

This chapter answers the Geology and Geophysics (GG) paper's Part B2 heading “Reservoir Geophysics”: “Rock Physics and Petrophysics, Failure Criteria, stress concentration, Linear elasticity fracture mechanics, Elastic-Plastic Fracture mechanics; Griffith's criterion, dynamic fracture mechanics; Lithology and Porosity Estimation; Saturation and Permeability Estimation; application of borehole geophysics in groundwater, mineral, and oil exploration.” It teaches mixing laws and Gassmann fluid substitution, porosity and lithology from logs, Archie saturation, permeability laws, Mohr-Coulomb and Griffith failure, the stress concentration around a hole, linear-elastic, elastic-plastic and dynamic fracture mechanics, and what boreholes tell us in the three kinds of exploration. The numericals are fluid-mixture modulus, Gassmann bulk modulus, Wyllie porosity, Archie saturation, permeability ratios, Mohr-Coulomb strength, hoop stress and Griffith critical stress.

1. Rock physics: mixing laws and Gassmann fluid substitution

Rock physics links the elastic and electrical properties that logs and seismic measure to porosity, mineral content and fluid. A rock is a mineral frame with pores of volume fraction φ (porosity) filled by fluid, so the bulk density is ρ = (1 − φ)·ρma + φ·ρfl. For a mixture of constituents with volume fractions fi and moduli Mi, the Voigt average (arithmetic, equal strain), M = Σ fi·Mi, is the upper bound and the Reuss average (harmonic, equal stress), 1/M = Σ fi/Mi, is the lower bound; the Voigt-Reuss-Hill value is their mean. Real rocks lie between the bounds, and the Reuss form is exact for a fine-scale mixture of fluids (Wood's law).

That last point has a large consequence. Take brine K = 2.25 GPa and gas K = 0.05 GPa (an illustrative value). A pore fluid of 90% brine and 10% finely mixed gas has 1/K = 0.9/2.25 + 0.1/0.05 = 2.4, so K = 0.42 GPa, far below the volume-weighted 2.03 GPa. A little gas therefore takes the fluid modulus most of the way down towards the gas value.

Gassmann's equation gives the saturated bulk modulus from the dry-frame modulus Kdry, the mineral modulus Kma and the fluid modulus Kfl: Ksat = Kdry + (1 − Kdry/Kma)² / [φ/Kfl + (1 − φ)/Kma − Kdry/Kma²], and the shear modulus is unchanged, μsat = μdry, since a fluid has no rigidity. It assumes an isotropic frame of one mineral, connected pores in pressure equilibrium (so low frequency, as in seismic and log-scale work, not ultrasonic laboratory work) and no chemical effect of the fluid. Worked example: Kdry = 9 GPa, Kma = 36 GPa, Kfl = 2.25 GPa (brine), φ = 0.20 give (1 − 0.25)² = 0.5625 and a denominator 0.0889 + 0.0222 − 0.0069 = 0.1042, so Ksat = 9 + 5.40 = 14.40 GPa. With gas (Kfl = 0.05 GPa) in the same frame, Ksat falls to 9.14 GPa.

⚠️ Gas cuts Vp a lot, but nudges Vs the other way
Vp = √[(Ksat + 4μ/3)/ρ] and Vs = √(μ/ρ). Replacing brine by gas lowers Ksat and the density, so Vp falls clearly, while μ is unchanged and the lighter rock makes Vs rise slightly. With μ = 9 GPa and the example above (taking ρma = 2.65, brine 1.00 and gas 0.10 g/cm³, so that ρ falls from 2.32 to 2.14 g/cm³), Vp goes from about 3370 to 3140 m/s and Vs from about 1970 to 2050 m/s. Higher effective pressure raises both speeds, and high pore pressure lowers them.

2. Porosity and lithology estimation

Three logs give porosity, each with its own matrix assumption. Sonic (Wyllie time-average): the travel time through the rock is the volume-weighted sum of the times in the mineral and the fluid, so φ = (Δt − Δtma)/(Δtfl − Δtma), with Δt in µs/ft; unconsolidated sands read too high and need a compaction factor. Density: φ = (ρma − ρb)/(ρma − ρfl). Neutron: responds to the hydrogen index, which is high in water- or oil-filled pores and low in gas. Worked example: Δt = 81, Δtma = 55 and Δtfl = 185 µs/ft give φ = 26/130 = 0.20; with ρma = 2.65, ρb = 2.31 and ρfl = 1.00 g/cm³ the density log gives 0.34/1.65 = 0.21. A wrong matrix is the commonest error: the same ρb in limestone (2.71) gives 0.23.

Lithology and fluid clues from a standard log suite
RockGamma rayOther clues
Clean sandstoneLowMatrix density about 2.65 g/cm³; photoelectric factor about 1.8
LimestoneLowAbout 2.71 g/cm³; factor about 5.1; density and neutron porosity overlie on a limestone scale
DolomiteLowAbout 2.87 g/cm³; factor about 3.1
ShaleHighHigh neutron porosity from bound water; low resistivity
Gas sandLowNeutron porosity reads low and density porosity reads high, so the curves cross over; high resistivity
CoalLowVery low density, long sonic transit time, high neutron porosity

The photoelectric factor separates sandstone, limestone and dolomite almost independently of porosity, and the sandstone and dolomite curves separate in opposite directions on a neutron-density overlay while limestone overlies. The gamma ray gives a shale volume, Vsh ≈ (GR − GRclean)/(GRshale − GRclean), but radioactive sands (feldspathic, micaceous, uraniferous) read falsely shaly, and a spectral gamma log (K, U, Th) separates them.

3. Saturation and permeability estimation

Archie's law treats a clean rock as an insulating frame with conductive brine in its pores. The formation factor is F = R0/Rw = a/φ^m, where R0 is the resistivity of the rock fully saturated with water of resistivity Rw. With hydrocarbons present the resistivity index is I = Rt/R0 = Sw^(−n), so Sw = [a·Rw/(φ^m·Rt)]^(1/n), with a ≈ 1, cementation exponent m ≈ 2 and saturation exponent n ≈ 2 unless stated. Hydrocarbon saturation is 1 − Sw. Worked example: Rw = 0.05 Ω·m, φ = 0.20, Rt = 20 Ω·m give Sw = [0.05/(0.04 × 20)]^(1/2) = √0.0625 = 0.25, so Sh = 0.75. On a Pickett plot (log Rt against log φ) water-bearing points fall on a line of slope −m, and its intercept at φ = 1 gives a·Rw.

Archie fails in shaly sands, because clay adds its own conduction, so the measured Rt is too low and Archie overestimates Sw; clay-corrected forms (Simandoux, Indonesian) add a term in Vsh. Rw comes from the SP log, water samples or the Pickett plot.

Permeability k, the ability to transmit fluid, follows Darcy's law Q = k·A·ΔP/(μ·L), and 1 darcy is about 0.987 × 10⁻¹² m². It is not measured by logs directly but estimated from porosity and grain or pore size. The Kozeny-Carman relation for a pack of spheres of diameter d is k = d²·φ³/[180·(1 − φ)²], so k rises with the square of the grain size and steeply with porosity: doubling φ from 0.10 to 0.20 at fixed d multiplies k by 2³ × (0.9/0.8)² ≈ 10.1. The empirical Timur relation, k = 0.136·φ^4.4/Swirr² (φ and Swirr in percent, k in mD), says that for the same porosity a rock with more irreducible water (finer pores, more clay) is less permeable: φ = 15 and Swirr = 20 give about 51 mD. NMR logs estimate k from the T2 distribution.

4. Failure criteria and stress concentration

Rock fails under effective stress, σ′ = σ − α·p, where p is the pore pressure and α is the Biot coefficient (at most 1; Terzaghi's form has α = 1). Raising p moves the Mohr circle towards the failure envelope without changing the total stress. The Mohr-Coulomb criterion, τ = c + σ′n·tanφ (cohesion c, friction angle φ), puts the failure plane at 45° − φ/2 to the σ1 direction, and in principal stresses reads σ1′ = C0 + σ3′·tan²(45° + φ/2), where C0 = 2c·cosφ/(1 − sinφ) is the unconfined compressive strength. Worked example: C0 = 30 MPa, φ = 30° (so tan²60° = 3) and σ3′ = 10 MPa give σ1′ = 30 + 3 × 10 = 60 MPa. Strength rises with confinement, which is why Mohr-Coulomb suits brittle rock.

Tresca (maximum shear stress) and von Mises (distortion energy) are yield criteria for ductile material and do not depend on mean stress. The empirical Hoek-Brown envelope fits jointed rock masses. Griffith's locus, from microcracks, is (σ1 − σ3)² = 8T0·(σ1 + σ3) with tensile strength T0 (tension cut-off at σ3 = −T0); it gives C0 = 8·T0, so rocks are far stronger in compression than in tension.

Stress concentration (compression taken as positive). For a circular hole in a large plate (Kirsch solution) with far-field principal stresses S1 and S2, the hoop stress at the wall, at angle θ from the S1 direction, is σθ = S1 + S2 − 2(S1 − S2)·cos2θ. Its maximum, at θ = 90° from S1, is 3·S1 − S2 and its minimum, at θ = 0, is 3·S2 − S1. For a single uniaxial stress S the concentration factor is 3 (with −S at the sides), and for an equal biaxial field it is 2. For an elliptical hole with semi-axis a across the load and b along it, σmax = S·(1 + 2a/b): 3S for a circle, growing without limit as the hole becomes a sharp crack, which is why fracture mechanics is needed. Worked example: S1 = 30 MPa and S2 = 10 MPa give a maximum 3 × 30 − 10 = 80 MPa and a minimum 3 × 10 − 30 = 0.

ℹ️ Why a borehole breaks out along the minimum horizontal stress
For a vertical well with horizontal stresses SH > Sh and mud overpressure ΔP, σθ = SH + Sh − 2(SH − Sh)·cos2θ − ΔP, with θ from the SH azimuth. The compressive hoop stress peaks at the Sh azimuth, 3·SH − Sh − ΔP, so breakouts form on the Sh side. It is smallest at the SH azimuth, 3·Sh − SH − ΔP, where drilling-induced tensile fractures form, and the breakdown pressure in hydraulic fracturing of an impermeable wall is Pb = 3·Sh − SH + T0 − p0.

5. Fracture mechanics: Griffith, LEFM, EPFM and dynamic fracture

A flaw-free elastic solid would break near its atomic bonding strength, yet real rock fails at a small fraction of that because flaws concentrate stress. Griffith's criterion is an energy balance: a crack of half-length a in a thin plate under tension σ extends when the elastic energy released equals the surface energy of the new faces. The energy release rate is G = π·σ²·a/E (plane stress), and growth needs G ≥ 2γ, where γ is the surface energy per unit area (two faces), so the critical stress is σf = √[2E·γ/(π·a)]; for plane strain E becomes E/(1 − ν²). Strength falls as 1/√a, so the largest flaw controls failure. Worked example: E = 50 GPa, γ = 1 J/m² and a = 1 mm give σf = √[2 × 50 × 10⁹ × 1/(π × 10⁻³)] = 5.6 MPa. Real materials need more energy than 2γ because of plastic work or microcracking, which Irwin and Orowan put into a larger effective G.

Linear elastic fracture mechanics (LEFM) writes the stress near the tip as σij = K·fij(θ)/√(2π·r), where the stress intensity factor K = Y·σ·√(π·a) combines load, crack size and geometry (Y = 1 for a central crack in a large plate). The three modes are I (opening), II (in-plane shear, sliding) and III (anti-plane shear, tearing). Fracture occurs when KI ≥ KIc, the fracture toughness, a material property that is lowest in plane strain (thick sections). The two approaches agree: G = K²/E′, with E′ = E in plane stress and E/(1 − ν²) in plane strain. LEFM needs small-scale yielding: the plastic zone, about (1/2π)(K/σy)² in plane stress, must be much smaller than the crack length and the remaining ligament.

  • Elastic-plastic fracture mechanics (EPFM): when yielding at the tip is large, LEFM fails and one uses the J-integral, a path-independent contour integral equal to the energy release rate for non-linear elastic behaviour, which reduces to G in the elastic case, with fracture at J ≥ Jc. Another measure is the crack-tip opening displacement (CTOD). Resistance that rises as the crack grows is described by an R-curve.
  • Dynamic fracture mechanics: for a fast crack, the kinetic energy of the material near the tip enters the energy balance, and the dynamic stress intensity factor equals k(v) times the static value for the same load, with k(v) falling to zero as the crack speed v approaches the Rayleigh wave speed, the theoretical limit for mode I. Observed mode-I speeds are only a fraction of it, since cracks branch well below the limit, and the dynamic toughness depends on speed. Mode II shear ruptures can exceed the shear-wave speed (supershear), as inferred for some earthquakes.
  • Reservoir link: a hydraulic fracture propagates when KI at its tip reaches KIc, and the fracture follows the path of least resistance set by the in situ stresses above.
⚠️ K, G and KIc are not interchangeable
K is the driving force of a given crack under a given load, and KIc is the material's resistance; fracture needs K to reach KIc. Doubling the crack length does not double K: K rises as √a, so the critical stress falls to 1/√2 of its value. Griffith's 2γ is the surface energy only, so it understates the toughness of any material that deforms plastically.

6. Borehole geophysics in groundwater, mineral and oil exploration

Borehole geophysics measures rock and fluid properties in place, along the well, and ties surface geophysics to the rocks. The same quantities (porosity, saturation, lithology) are asked in all three settings, but the question differs.

What borehole logs are used for in the three kinds of exploration
TargetLogs usedWhat they give
Oil and gasGamma ray, deep and shallow resistivity, neutron, density, sonic, photoelectric factor, NMR, formation testingNet pay, porosity, Sw, permeability, lithology, fluid contacts, and the sonic-density tie to seismic
GroundwaterSP, normal and induction resistivity, natural gamma, caliper, fluid temperature and conductivity, flowmeterAquifer top and base, clay beds, salinity from Rw, fracture and producing zones, screen placement
Mineral and coalSpectral gamma, density, resistivity and induced polarisation, magnetic susceptibility, caliper, sonicSeam thickness and grade (coal has low density and high resistivity), potash by K, uranium, sulphides by high chargeability (and low resistivity where connected), magnetite by susceptibility

Three working rules apply everywhere. Depth-match all logs and calibrate them to core. Take Rt from a deep resistivity reading beyond the mud-filtrate invasion, since a shallow reading sees only the flushed zone. In a clean water-bearing formation Rt = F·Rw, so at known porosity a falling resistivity reads rising salinity, which is how a resistivity log screens groundwater quality.

Key takeaways

  • Voigt (arithmetic) and Reuss (harmonic) averages bound the moduli of a mixture, and Reuss is exact for fluids: 10% gas in brine gives Kfl = 0.42 GPa, far below both the volume-weighted 2.03 GPa and the 2.25 GPa of brine alone. Gassmann gives Ksat with μsat = μdry; gas lowers Vp strongly and raises Vs slightly.
  • Porosity: Wyllie φ = (Δt − Δtma)/(Δtfl − Δtma) and density φ = (ρma − ρb)/(ρma − ρfl); a wrong matrix is the commonest error. Gas gives neutron-density crossover, shale reads high on neutron, and the photoelectric factor separates the carbonates and sandstone.
  • Archie: F = a/φ^m, Sw = [a·Rw/(φ^m·Rt)]^(1/n); it holds for clean rock and overestimates Sw in shaly sand. Permeability: Kozeny-Carman k ∝ d²φ³/(1 − φ)² and Timur k = 0.136·φ^4.4/Swirr².
  • Mohr-Coulomb: σ1′ = C0 + σ3′·tan²(45° + φ/2) under effective stress σ − αp. Around a hole the hoop stress is S1 + S2 − 2(S1 − S2)·cos2θ, maximum 3S1 − S2 (factor 3 for a single stress); a borehole breaks out along the Sh azimuth.
  • Griffith: σf = √[2Eγ/(πa)] and G = πσ²a/E; LEFM: K = Yσ√(πa), fracture at KI = KIc, G = K²/E′. EPFM uses the J-integral and CTOD; in dynamic fracture k(v) → 0 at the Rayleigh speed.
  • Logs serve three questions: net pay and contacts in oil, aquifer boundaries and salinity in groundwater, seam and grade in minerals (gamma, density, IP). Use a deep reading for Rt, depth-match and calibrate to core.

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.

  1. A reservoir sandstone has a pore fluid of 90% brine (K = 2.25 GPa) and 10% gas (K = 0.05 GPa) by volume, mixed at a fine scale so that the fluids share the same pressure. Using Wood's (Reuss) law, the effective bulk modulus of the pore fluid is closest to

    1. 2.03 GPa, the volume-weighted arithmetic mean of the two moduli
    2. 0.42 GPa, from the volume-weighted mean of the compliances 1/K
    3. 1.15 GPa, the simple midpoint between the brine and gas moduli
    4. 0.05 GPa, because the softest component always sets the mixture modulus
    Show answer

    Answer: B — 0.42 GPa, from the volume-weighted mean of the compliances 1/K

    For fluids at one pressure the compliances average by volume: 1/K = 0.9/2.25 + 0.1/0.05 = 0.4 + 2.0 = 2.4 GPa⁻¹, so K = 0.417 GPa. The arithmetic mean (2.03 GPa) is the Voigt upper bound and overstates the stiffness enormously. The softest component does not set the value alone, because the brine still contributes 0.4 to the sum.

  2. Which of the following statements about Gassmann fluid substitution are correct?

    1. The shear modulus of the rock is unchanged when the pore fluid is replaced
    2. Replacing brine by gas lowers Vp strongly, while Vs rises slightly because the density falls
    3. The equation holds equally well at ultrasonic laboratory frequencies and at seismic frequencies
    4. Replacing brine by gas raises the saturated bulk modulus because gas is lighter
    Show answer

    Answer: A — The shear modulus of the rock is unchanged when the pore fluid is replaced; B — Replacing brine by gas lowers Vp strongly, while Vs rises slightly because the density falls

    A fluid has no rigidity, so μsat = μdry and Vs = √(μ/ρ) changes only through the density, which falls with gas, so Vs rises a little; the lower Ksat of the gas-filled rock lowers Vp markedly. The equation assumes pressure equilibrium in the pores, which holds at low frequencies only, so ultrasonic data need dispersion corrections. Gas has a much lower bulk modulus than brine, so Ksat falls, whatever the density.

  3. A sandstone has dry-frame bulk modulus Kdry = 12 GPa, mineral modulus Kma = 40 GPa and porosity 0.20. It is saturated with a fluid of Kfl = 2.0 GPa. Using Gassmann's equation, the saturated bulk modulus Ksat, to two decimal places in GPa, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 16.36

    (1 − Kdry/Kma)² = (1 − 0.3)² = 0.49. Denominator = φ/Kfl + (1 − φ)/Kma − Kdry/Kma² = 0.10 + 0.02 − 0.0075 = 0.1125. So Ksat = 12 + 0.49/0.1125 = 12 + 4.356 = 16.36 GPa. Forgetting the squared bracket in the numerator, or the last term of the denominator, gives wrong values.

  4. A clean sandstone has a sonic transit time of 88 µs/ft. Take the matrix transit time as 52 µs/ft and the fluid transit time as 172 µs/ft. By the Wyllie time-average equation the porosity is ____ percent.

    Numerical answer — type the value.

    Show answer

    Answer: 30

    φ = (Δt − Δtma)/(Δtfl − Δtma) = (88 − 52)/(172 − 52) = 36/120 = 0.30, that is 30 percent. Dividing by Δtfl alone (88/172) or inverting the numerator and denominator are the usual slips.

  5. A clean limestone (matrix density 2.71 g/cm³) has a log bulk density of 2.40 g/cm³ and is filled with water of density 1.00 g/cm³. What is the density porosity?

    1. 15%, which results from using the sandstone matrix density of 2.65 g/cm³
    2. 25%, which results from using the dolomite matrix density of 2.87 g/cm³
    3. 18%, from (2.71 − 2.40)/(2.71 − 1.00) with the limestone matrix
    4. 31%, from the difference between the matrix and bulk densities
    Show answer

    Answer: C — 18%, from (2.71 − 2.40)/(2.71 − 1.00) with the limestone matrix

    φ = (ρma − ρb)/(ρma − ρfl) = (2.71 − 2.40)/(2.71 − 1.00) = 0.31/1.71 = 0.181, about 18%. The sandstone matrix would give 15% and the dolomite matrix 25%, which shows how far a wrong matrix shifts the answer; 31% omits the division by the matrix-fluid density contrast.

  6. In a clean sandstone interval with low gamma ray and high resistivity, the neutron and density porosities are plotted in sandstone matrix units, so that a water-filled clean sandstone would give equal values. Here the neutron porosity reads 10% and the density porosity 24%. What is the best interpretation?

    1. Gas-bearing sandstone, because gas lowers the neutron reading and raises the density porosity
    2. Shaly sandstone, because bound water in clay raises the neutron reading above the density porosity
    3. Tight water-bearing sandstone, because both porosities are low and the gap is a calibration error
    4. Limestone, because neutron and density porosity always read different values in a carbonate
    Show answer

    Answer: A — Gas-bearing sandstone, because gas lowers the neutron reading and raises the density porosity

    Gas has a low hydrogen index, so the neutron log under-reads porosity, and its low density makes the density log over-read it, which opens the neutron-density crossover; with low gamma ray and high resistivity this indicates gas. Shale would push the neutron reading up rather than down, a calibration error would not coincide with high resistivity, and a limestone would not produce this pattern: it is not true that the two always disagree (on a limestone scale the curves overlie), and on the sandstone scale of the stem it would, if anything, show the neutron reading above the density porosity.

  7. A clean water-wet sandstone has porosity 0.25, formation-water resistivity Rw = 0.04 Ω·m and true resistivity Rt = 16 Ω·m. Take a = 1, m = 2 and n = 2 in Archie's equation. The water saturation is ____ percent.

    Numerical answer — type the value.

    Show answer

    Answer: 20

    Sw = [a·Rw/(φ^m·Rt)]^(1/n) = [0.04/(0.0625 × 16)]^(1/2) = √0.04 = 0.20, that is 20 percent (the hydrocarbon saturation is 80 percent). Forgetting the square root (giving 4 percent) is the usual slip.

  8. Archie's equation, with clean-sand constants, is applied to a hydrocarbon-bearing shaly sandstone. How is the water saturation affected, and why?

    1. It is underestimated, because clay-bound water raises Rt above that of a clean rock with the same fluids
    2. It is unaffected, because the exponents m and n absorb the clay effect automatically
    3. It is unaffected, because the gamma-ray log corrects Rt before the equation is used
    4. It is overestimated, because clay adds conduction so the measured Rt is lower than that of a clean rock
    Show answer

    Answer: D — It is overestimated, because clay adds conduction so the measured Rt is lower than that of a clean rock

    Clay minerals conduct current through bound water and exchange cations, so Rt is lower than it would be in a clean rock with the same hydrocarbon content. Archie reads the low Rt as more water, so it overestimates Sw and understates the hydrocarbon in place. The exponents are not designed to absorb a clay term, and a gamma-ray log gives Vsh for a clay-corrected equation (Simandoux, Indonesian) but does not correct Rt by itself.

  9. For a pack of equal spheres the Kozeny-Carman relation gives k proportional to d²·φ³/(1 − φ)². At a fixed grain diameter d, by what factor does the permeability change when the porosity increases from 0.10 to 0.20?

    1. About 8, from the φ³ term alone, ignoring the denominator
    2. About 10, from both the φ³ term and the (1 − φ)² term
    3. About 2, because k is taken as simply proportional to φ
    4. About 4, because k is taken as proportional to φ²
    Show answer

    Answer: B — About 10, from both the φ³ term and the (1 − φ)² term

    The ratio is (0.20/0.10)³ × [(1 − 0.10)/(1 − 0.20)]² = 8 × (0.9/0.8)² = 8 × 1.266 = 10.1. The φ³ term alone gives 8, but the (1 − φ)² denominator also falls as φ rises, which raises k further. The choices of 2 and 4 use the wrong exponent.

  10. A rock has unconfined compressive strength C0 = 40 MPa and an internal friction angle of 30°. Using the Mohr-Coulomb criterion in effective stresses, the major effective principal stress at failure when the minor effective principal stress is 5 MPa is

    1. About 42 MPa, using tan²φ as the multiplier
    2. About 49 MPa, using tan(45° + φ/2) without squaring it
    3. 55 MPa, from C0 + σ3′·tan²(45° + φ/2) with a multiplier of 3
    4. 125 MPa, by multiplying C0 by 3 and then adding σ3′
    Show answer

    Answer: C — 55 MPa, from C0 + σ3′·tan²(45° + φ/2) with a multiplier of 3

    For φ = 30°, tan²(45° + 15°) = tan²60° = 3, so σ1′ = 40 + 3 × 5 = 55 MPa. The multiplier belongs to σ3′, not to C0, which rules out 125 MPa, and it is the squared tangent of 45° + φ/2, so 42 MPa (tan²φ) and 49 MPa (unsquared) are both wrong.

  11. A circular hole is drilled in a large elastic plate under far-field principal compressive stresses of 40 MPa and 15 MPa. Using compression as positive, the maximum compressive hoop stress on the wall of the hole is ____ MPa.

    Numerical answer — type the value.

    Show answer

    Answer: 105

    The Kirsch solution gives the maximum hoop stress 3·S1 − S2 at 90° from the larger stress: 3 × 40 − 15 = 105 MPa (the minimum is 3 × 15 − 40 = 5 MPa at the larger-stress azimuth). Using the single-stress factor 3 × 40 = 120 MPa ignores the second stress.

  12. In a vertical well the maximum horizontal stress SH acts north-south and the minimum horizontal stress Sh acts east-west. Where on the borehole wall do compressive breakouts and drilling-induced tensile fractures form?

    1. Breakouts at the east and west sides, and tensile fractures at the north and south sides
    2. Breakouts at the north and south sides, and tensile fractures at the east and west sides
    3. Both at the north and south sides, since SH is the largest stress
    4. Both at the east and west sides, since Sh is the smallest stress
    Show answer

    Answer: A — Breakouts at the east and west sides, and tensile fractures at the north and south sides

    The hoop stress σθ = SH + Sh − 2(SH − Sh)·cos2θ − ΔP is greatest, 3·SH − Sh − ΔP, at 90° from the SH direction, which is the east and west sides, so compressive breakouts form there. It is smallest, 3·Sh − SH − ΔP, along the SH direction (north and south), where tensile fractures form. The reversed assignment confuses where the stress concentrates.

  13. A thin brittle plate has Young's modulus 60 GPa and surface energy γ = 1.5 J/m² per face. It contains a central through-crack of total length 6 mm. Using Griffith's criterion for plane stress, σf = √[2Eγ/(πa)] with a the half-length of the crack, the critical tensile stress is ____ MPa (answer to one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 4.4

    The half-length is a = 3 mm = 0.003 m. σf = √[2 × 60 × 10⁹ × 1.5/(π × 0.003)] = √(1.8 × 10¹¹/0.009425) = √(1.91 × 10¹³) = 4.37 × 10⁶ Pa = 4.4 MPa. Using the full length 6 mm as a gives 3.1 MPa, the usual slip.

  14. A large plate with a central crack carries a constant remote tensile stress. If the half-length of the crack becomes four times as large, what happens to the stress intensity factor KI and the energy release rate G (plane stress)?

    1. KI quadruples and G quadruples, since both are proportional to the crack length
    2. KI doubles and G doubles, since both are proportional to the square root of the crack length
    3. KI quadruples and G rises sixteen-fold, since G is the square of KI
    4. KI doubles and G quadruples, since KI ∝ √a and G ∝ a
    Show answer

    Answer: D — KI doubles and G quadruples, since KI ∝ √a and G ∝ a

    KI = σ√(πa) grows as √a, so four times the half-length doubles it. G = KI²/E′ = π·σ²·a/E is proportional to a, so it quadruples. Treating both as proportional to a, or both as √a, ignores that G is the square of KI divided by a modulus, and the sixteen-fold option wrongly takes KI to quadruple.

  15. When should the J-integral or crack-tip opening displacement be used in place of the stress intensity factor K?

    1. When the crack runs at a speed close to the Rayleigh wave speed, so that inertia matters
    2. When the loading is pure mode II, because K is defined only for mode I
    3. When the plastic zone at the tip is too large for small-scale yielding to hold
    4. When the material is a perfectly brittle glass whose fracture energy equals 2γ
    Show answer

    Answer: C — When the plastic zone at the tip is too large for small-scale yielding to hold

    LEFM and K rest on small-scale yielding. When yielding spreads over a large part of the ligament, the J-integral (or CTOD) replaces K. Fast cracks call for the dynamic stress intensity factor, not J; K is defined for all three modes; and a perfectly brittle glass is the ideal case for LEFM and Griffith's 2γ.

  16. Which of the following statements about fracture mechanics are correct?

    1. Griffith's energy balance gives a critical stress that falls as the inverse square root of the crack half-length
    2. In LEFM fracture is predicted when KI reaches the fracture toughness KIc, which is lowest under plane strain
    3. The J-integral of a non-linear elastic body depends strongly on the contour chosen to evaluate it
    4. The theoretical speed limit of a mode I crack is the P-wave speed, and real cracks reach it
    Show answer

    Answer: A — Griffith's energy balance gives a critical stress that falls as the inverse square root of the crack half-length; B — In LEFM fracture is predicted when KI reaches the fracture toughness KIc, which is lowest under plane strain

    Setting G = π·σ²·a/E equal to 2γ gives σf ∝ 1/√a, and the K criterion KI ≥ KIc with a thickness-dependent toughness that bottoms out in plane strain is the core of LEFM. The J-integral is path-independent, which is what makes it useful. The speed limit for mode I is the Rayleigh wave speed, and observed cracks reach only a fraction of it before branching.

  17. In a clean sandstone aquifer of nearly constant porosity, a resistivity log shows one interval with much lower resistivity than the rest, while the gamma-ray log shows no change. What is the most likely cause?

    1. A clay bed, because clay always lowers resistivity and is invisible to the gamma ray
    2. More saline pore water in that interval, because Rt = F·Rw falls as salinity rises
    3. A zone of lower porosity, because resistivity falls as the formation factor increases
    4. An unsaturated gas-filled zone, because gas lowers the resistivity of the pore space
    Show answer

    Answer: B — More saline pore water in that interval, because Rt = F·Rw falls as salinity rises

    With porosity (hence F) constant and no clay signal on the gamma ray, a drop in Rt = F·Rw must come from a lower Rw, that is, more saline water, which is how resistivity logs map groundwater quality. Clay would raise the gamma ray, lower porosity would raise F and so the resistivity, and gas or air would raise it too.

  18. Which of the following statements about borehole logging in exploration are correct?

    1. A coal seam typically shows low bulk density and a long sonic transit time
    2. Sulphide mineralisation typically shows high induced-polarisation chargeability, and low resistivity where the grains are connected
    3. A shallow resistivity reading is the best measure of the true formation resistivity in an invaded zone
    4. Potassium-rich evaporites such as potash give a very low gamma-ray response
    Show answer

    Answer: A — A coal seam typically shows low bulk density and a long sonic transit time; B — Sulphide mineralisation typically shows high induced-polarisation chargeability, and low resistivity where the grains are connected

    Coal has a very low density and a slow sound speed, and sulphides conduct electronically and polarise under an applied current, so they show high chargeability and, where the grains are connected, low resistivity. A shallow reading mainly sees the flushed zone, so true resistivity is taken from the deep reading, and potash is rich in radioactive potassium-40, so it gives a high gamma-ray response.