Basic Structural Geology: Stress, Mohr's Circle, Strain, Rock Rheology, Folds and Faults

This chapter answers the Part A heading “Basic structural geology” of the Geology and Geophysics (GG) paper: “Mohr's circle, stress, strain, and material response; Brittle and ductile deformation; Nomenclature and classification of folds and faults.” It teaches stress on a plane and Mohr's circle (described in words), the Coulomb failure criterion, strain from lengths and angles, elastic, plastic and viscous behaviour with the brittle–ductile transition, and the parts, types and classification of folds and faults. The numericals are normal and shear stress on a plane, the failure stress from a Coulomb envelope, shortening and stretch, apparent dip, and the throw and heave of a fault from its net slip.

1. Stress: traction, normal and shear components, principal stresses

Stress is force per unit area (pascal; 1 MPa = 10⁶ Pa). The force on a plane (its traction) splits into a normal stress σ_n, perpendicular to the plane, and a shear stress τ, along it. At every point three mutually perpendicular planes carry no shear stress; the normal stresses on them are the principal stresses σ1 ≥ σ2 ≥ σ3. In geology compression is taken as positive. The mean stress is (σ1 + σ2 + σ3)/3 and the differential stress σ1 − σ3 is what distorts rock. Under a rock column the lithostatic stress is ρ g z, about 26.5 MPa per kilometre for ρ = 2700 kg/m³. Pore fluid at pressure P_f carries part of the load, so the effective stress is σ − P_f.

For a plane whose normal makes an angle θ with σ1 (two-dimensional case):
σ_n = ½(σ1 + σ3) + ½(σ1 − σ3) cos 2θ and τ = ½(σ1 − σ3) sin 2θ.
At θ = 0 the plane is perpendicular to σ1, σ_n = σ1 and τ = 0; at θ = 45° the shear stress is largest, ½(σ1 − σ3). Worked example: σ1 = 80 MPa, σ3 = 20 MPa, θ = 30°: σ_n = 50 + 30 cos 60° = 65 MPa and τ = 30 sin 60° = 26.0 MPa. The commonest slip is to take θ as the angle between the plane itself (not its normal) and σ1: the normal then makes 90° − θ with σ1, so the cosine term changes sign (here σ_n would be 35 MPa, not 65 MPa).

Near the surface one principal stress is vertical, which gives Anderson's three regimes: normal faulting (σ1 vertical), thrust faulting (σ3 vertical, so σ1 is horizontal) and strike-slip faulting (σ2 vertical).

2. Mohr's circle and failure criteria

Mohr's circle displays σ_n and τ for every plane at once. Picture a graph with σ_n along the horizontal axis (compression to the right) and τ upward. The circle is centred on the horizontal axis at ½(σ1 + σ3) and has radius ½(σ1 − σ3), so it cuts the axis at σ3 on the left and σ1 on the right. A plane whose normal makes angle θ with σ1 is the point reached by turning through 2θ about the centre from the σ1 point. The top of the circle is the maximum shear stress, equal to the radius, on planes at 45° to σ1. A larger differential stress makes a bigger circle; raising pore pressure by ΔP shifts the whole circle, in effective stress, leftward by ΔP without changing its size.

The Coulomb criterion for shear failure is a straight line on the diagram: τ = c + σ_n tan φ, with cohesion c and angle of internal friction φ. The rock fails when the circle just touches this envelope. The failure planes are a conjugate pair that contain σ2 and make an angle of 45° − φ/2 with σ1 (their normals make 45° + φ/2), so σ1 bisects the acute angle between them, which is 90° − φ. Touching the envelope gives ½(σ1 − σ3) = sin φ [½(σ1 + σ3) + c cot φ], or
σ1 = σ3 (1 + sin φ)/(1 − sin φ) + 2c cos φ/(1 − sin φ).
Worked example: c = 10 MPa, φ = 30°, σ3 = 20 MPa: σ1 = 20 × 3 + 2 × 10 × 0.866/0.5 = 60 + 34.6 = 94.6 MPa. For φ = 30° the planes lie at 30° to σ1: a new normal fault (σ1 vertical) dips about 60°, a new thrust (σ1 horizontal) about 30°, and a strike-slip fault is near vertical. Where σ3 is tensile and reaches the tensile strength, extension fractures (joints) open perpendicular to σ3.

⚠️ Which angle?
The failure plane makes 45° − φ/2 with σ1, but its normal makes 45° + φ/2, and on Mohr's circle the plane sits at twice the normal's angle, 90° + φ, from the σ1 point. Always state which line the angle is measured from.

3. Strain: stretch, shear strain and strain ellipse

A deformation combines translation, rotation, distortion (change of shape) and dilation (change of volume); strain covers the last two. In homogeneous strain straight lines stay straight and parallel lines stay parallel, so a circle becomes an ellipse, the strain ellipse. For a line of original length l₀ and final length l:
extension e = (l − l₀)/l₀ (negative for shortening), stretch S = l/l₀ = 1 + e, quadratic elongation λ = S².
Shear strain γ = tan ψ, where ψ is the angular change between lines that were originally perpendicular. Worked example: a bed 10 km long is folded and thrust to 7.5 km. e = −0.25, so the shortening is 25 %, S = 0.75 and λ = 0.5625.

  • The strain ellipsoid has principal stretches S1 ≥ S2 ≥ S3 (axes X, Y, Z). Flattening (oblate, pancake-like) has S1 ≈ S2 > S3, constriction (prolate, cigar-like) has S1 > S2 ≈ S3, and plane strain has S2 = 1. Without volume change, S1 S2 S3 = 1.
  • Pure shear is coaxial: the principal axes keep their directions throughout, with no net rotation. Simple shear is non-coaxial: material lines rotate and the strain axes turn progressively toward the shear plane as the strain grows.
ℹ️ Strain you can measure
Strain is read from objects of known original shape: deformed fossils, reduction spots (originally circles) in slate, ooids and pebbles, and boudins. Finite strain is the total, while the strain rate (per second) is what controls whether rock responds in a brittle or ductile way.

4. Material response: elastic, plastic, viscous; brittle and ductile deformation

  • Elastic: strain is fully recovered on unloading and, for small strain, proportional to stress: σ = E ε (Young's modulus E), τ = G γ (shear modulus G). Poisson's ratio ν (about 0.25 for many crustal rocks) is the lateral contraction ÷ the axial extension, and E = 2G(1 + ν). Elastic strain stored around a locked fault is released as an earthquake (elastic rebound).
  • Plastic: above a yield stress the strain is permanent and grows with no further rise of stress (ideal plasticity); it does not recover.
  • Viscous: stress is proportional to strain rate, σ = η (dε/dt) for a Newtonian fluid, with viscosity η in Pa·s (mantle about 10²⁰ to 10²¹ Pa·s). A viscoelastic body (Maxwell) is elastic for fast loading and flows for slow loading, the changeover being the relaxation time η/G.

Rocks show both extremes. Brittle deformation loses cohesion by fracturing and frictional sliding; its strength rises with confining pressure (depth) and barely depends on temperature. Ductile deformation is distributed flow by crystal-plastic processes (dislocation creep, diffusion creep, pressure solution) with no loss of cohesion; its strength falls steeply with temperature and rises with strain rate. Ductility is favoured by high temperature, slow strain rate and the presence of water; brittleness by low temperature, fast strain rate and dry rock. In continental crust a strength profile rises with depth through the brittle field, peaks at the brittle–ductile transition (typically 10 to 15 km, near 300 °C for quartz-rich rock) and then falls; feldspar flows at higher temperature (about 450 °C) and olivine at higher still. Most crustal earthquakes nucleate above the transition.

Brittle and ductile deformation compared
FeatureBrittleDuctile
MechanismFracture, frictional sliding, cataclasisCrystal-plastic flow, pressure solution
Strength depends onConfining pressureTemperature and strain rate
StructuresJoints, veins, faults, breccia, gougeFolds, foliation, boudins, mylonite
BehaviourLocalised, often seismicDistributed, aseismic

5. Folds: nomenclature, classification and apparent dip

A fold is a bend in layering. The hinge is the zone of greatest curvature, and the hinge line joins the hinge points of one surface. The limbs are the flanks. The axial surface contains the hinge lines of all the layers. The plunge is the angle the hinge line makes with the horizontal. The interlimb angle is the angle between the limbs, and amplitude and wavelength describe size. By shape, an antiform closes upward and a synform closes downward. By age, an anticline has the oldest rocks in the core and a syncline the youngest. The two coincide for normal sequences; for overturned ones an antiformal syncline (arch-shaped, youngest in the core) can occur.

Fold classification
BasisTypes
Interlimb angle (limb tightness)Gentle, open, close, tight, isoclinal (limbs parallel, angle about 0°); class limits differ slightly between texts
Axial surface and hingeUpright, inclined, overturned, recumbent; horizontal or plunging
Layer thicknessParallel (concentric): constant thickness at right angles to bedding; similar: constant thickness along the axial surface, thick hinge and thin limbs
ShapeChevron and kink (straight limbs, sharp hinge), box, conjugate, monocline, dome and basin

Outcrop pattern: an eroded anticline shows older beds in the core, a syncline younger ones. A plunging anticline closes in the direction of plunge, a plunging syncline closes against it. A dome shows older beds at the centre of circular outcrops, a basin younger ones. Folds form by buckling of competent layers under layer-parallel shortening (parallel folds, flexural slip) or by passive shearing (similar folds).

Apparent dip: a section that makes an angle β with the strike of a bed of true dip δ shows the dip δ_a given by tan δ_a = tan δ sin β. It is zero along strike, equals δ at β = 90°, and never exceeds δ. Worked example: δ = 60°, β = 30°: tan δ_a = 1.732 × 0.5 = 0.866, so δ_a = 40.9°.

6. Faults: nomenclature, classification and displacement

A fault is a fracture with displacement. The fault plane has a strike and dip; the block above an inclined fault is the hanging wall and the one below is the footwall. The net slip is the true displacement vector. Its rake (pitch) is measured in the fault plane from the strike line. The dip-slip component is s sin(rake) and the strike-slip component is s cos(rake). Dip-slip resolves further into throw (vertical) = dip-slip × sin δ and heave (horizontal) = dip-slip × cos δ, where δ is the fault dip. Worked example: s = 10 m, rake 30°, dip 60°: dip-slip = 5.00 m, strike-slip = 8.66 m, throw = 4.33 m, heave = 2.50 m.

  • Normal fault: hanging wall moves down relative to the footwall; extension; dip typically near 60°. A vertical borehole across it omits part of the sequence. A graben is a block dropped between two normal faults and a horst one left raised.
  • Reverse fault: hanging wall moves up; compression. A thrust is a low-angle reverse fault (dip below about 45°) that can carry rock far over younger rock. A borehole across it repeats part of the sequence.
  • Strike-slip fault: motion along the strike on a near-vertical plane. Dextral (right-lateral): the opposite block moves to the right as you look across the fault; sinistral: to the left. Oblique-slip faults combine strike-slip and dip-slip.
  • Evidence: slickensides and slickenlines on the plane, drag folds, offset markers, and fault rocks: breccia, gouge and cataclasite in brittle faults, mylonite in ductile shear zones, pseudotachylyte where frictional melting occurred.
⚠️ Throw is not slip, and heave is horizontal
Throw and heave are components of the dip-slip, not of the net slip, and the dip-slip is recovered as the square root of throw² + heave². Separation measured on a marker bed is not slip: it depends on the cut through the bed. Right-lateral and left-lateral are judged from either side of the fault and give the same answer.

Key takeaways

  • On a plane whose normal makes θ with σ1: σ_n = ½(σ1 + σ3) + ½(σ1 − σ3) cos 2θ and τ = ½(σ1 − σ3) sin 2θ. Principal planes carry no shear; shear peaks at 45°. Effective stress is σ − P_f, and Anderson's regimes follow the vertical principal stress.
  • Mohr's circle has centre ½(σ1 + σ3) and radius ½(σ1 − σ3), and a plane at θ plots at 2θ. Coulomb failure τ = c + σ_n tan φ gives planes at 45° − φ/2 to σ1 and σ1 = σ3 (1 + sin φ)/(1 − sin φ) + 2c cos φ/(1 − sin φ); raising pore pressure shifts the circle left.
  • Strain: e = (l − l₀)/l₀, S = 1 + e, λ = S², γ = tan ψ. Pure shear is coaxial and simple shear non-coaxial. Flattening has S1 ≈ S2 > S3 and constriction S1 > S2 ≈ S3; without volume change S1 S2 S3 = 1.
  • Elastic strain recovers (σ = E ε), plastic strain is permanent above yield, viscous flow follows σ = η dε/dt. Brittle strength rises with pressure and depth, ductile strength falls with temperature, so the strength peaks at the brittle–ductile transition, near 10 to 15 km in continental crust.
  • Fold parts: hinge, limbs, axial surface, plunge, interlimb angle. Anticline means oldest in the core, antiform means arch shape; parallel folds keep orthogonal thickness, similar folds keep axial-surface thickness. Apparent dip: tan δ_a = tan δ sin β.
  • Normal faults drop the hanging wall and omit section, reverse faults and thrusts raise it and repeat section, strike-slip faults are dextral or sinistral. Dip-slip = s sin(rake), strike-slip = s cos(rake), throw = dip-slip × sin δ, heave = dip-slip × cos δ.

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. What is a principal stress?

    1. The normal stress on a plane that carries no shear stress
    2. The shear stress on the plane where it reaches its largest value
    3. The average of all normal stresses acting on every plane through a point
    4. The difference between the largest and the smallest normal stress at a point
    Show answer

    Answer: A — The normal stress on a plane that carries no shear stress

    Principal stresses are the normal stresses on the three mutually perpendicular planes that have zero shear stress. The difference σ1 − σ3 is the differential stress and the average is the mean stress, which are the tempting but different quantities. Shear stress is largest on planes at 45° to σ1, which carry a normal stress equal to the mean.

  2. At a point the principal stresses (compression positive) are σ1 = 120 MPa and σ3 = 40 MPa in a two-dimensional state. Using σ_n = ½(σ1 + σ3) + ½(σ1 − σ3) cos 2θ, find the normal stress, in MPa, on a plane whose normal makes an angle θ = 60° with the σ1 direction. Give the number only.

    Numerical answer — type the value.

    Show answer

    Answer: 60

    ½(σ1 + σ3) = 80 MPa and ½(σ1 − σ3) = 40 MPa. With 2θ = 120°, cos 120° = −0.5, so σ_n = 80 + 40 × (−0.5) = 60 MPa. Using θ = 60° as if it were 2θ gives 100 MPa, the usual slip.

  3. The total stresses on a block of rock stay constant while the pore-fluid pressure rises by 20 MPa. What happens to the Mohr circle drawn in effective stress?

    1. It grows larger by 20 MPa in diameter, which makes shear failure more likely
    2. It shifts 20 MPa to the left without changing size, which moves it toward the failure envelope
    3. It shifts 20 MPa to the right without changing size, which moves it away from the failure envelope
    4. It shrinks by 20 MPa in diameter because the differential stress is reduced by the fluid
    Show answer

    Answer: B — It shifts 20 MPa to the left without changing size, which moves it toward the failure envelope

    Effective stress is σ − P_f for every principal stress, so each of σ1 and σ3 falls by 20 MPa and the difference σ1 − σ3 is unchanged: the circle keeps its size and moves left toward the Coulomb envelope, which is how high fluid pressure triggers faulting. Enlarging the circle is the tempting answer, but pore pressure does not change the differential stress.

  4. Take the angle of internal friction as 30°. A new fault forms in a stress field with σ1 horizontal and σ3 vertical. Which statement gives the fault type and its approximate dip?

    1. A normal fault dipping at about 60°, because the plane lies 30° from the horizontal σ1
    2. A thrust dipping at about 60°, because the plane lies 60° from the horizontal σ1
    3. A strike-slip fault dipping at about 90°, because the plane contains the vertical σ3
    4. A thrust dipping at about 30°, because the plane lies 30° from the horizontal σ1
    Show answer

    Answer: D — A thrust dipping at about 30°, because the plane lies 30° from the horizontal σ1

    The shear fracture contains σ2 and makes 45° − φ/2 = 30° with σ1. With σ1 horizontal the plane therefore dips about 30°, and horizontal compression with a vertical σ3 gives thrust faulting. A normal fault is the tempting answer only if σ1 is vertical, in which case the plane would dip about 60°. A strike-slip fault needs σ2 vertical.

  5. A rock obeys the Coulomb criterion τ = c + σ_n tan φ with cohesion c = 10 MPa and angle of internal friction φ = 30°. The least principal stress is σ3 = 20 MPa. At what value of σ1, in MPa, to the nearest MPa, does it fail in shear? Use σ1 = σ3 (1 + sin φ)/(1 − sin φ) + 2c cos φ/(1 − sin φ). Give the number only.

    Numerical answer — type the value.

    Show answer

    Answer: 95

    sin 30° = 0.5 and cos 30° = 0.8660, so (1 + sin φ)/(1 − sin φ) = 3 and 2c cos φ/(1 − sin φ) = 2 × 10 × 0.8660/0.5 = 34.64 MPa. Then σ1 = 20 × 3 + 34.64 = 94.64 MPa, which rounds to 95. Dropping the cohesion term gives 60 MPa, the cohesionless result.

  6. Which statement correctly distinguishes simple shear from pure shear?

    1. In pure shear the strain axes rotate steadily toward the shear plane, while in simple shear they keep fixed directions
    2. Simple shear is non-coaxial, so the principal strain axes turn toward the shear plane as strain accumulates
    3. Pure shear changes volume while simple shear never does, so only pure shear produces dilation
    4. Simple shear can only occur in brittle rocks, whereas pure shear is restricted to ductile flow
    Show answer

    Answer: B — Simple shear is non-coaxial, so the principal strain axes turn toward the shear plane as strain accumulates

    Simple shear is non-coaxial: material lines rotate and the principal strain axes turn progressively toward the shear plane. Pure shear is coaxial, with the axes fixed. The claim that pure shear rotates the axes reverses the two, which is the common mistake. Neither type fixes the volume change, since both can be constant-volume, and both can occur in ductile as well as brittle settings.

  7. A bed originally 10.0 km long is shortened by folding and thrusting to a present length of 7.5 km. What is the percentage shortening? Give the number only.

    Numerical answer — type the value.

    Show answer

    Answer: 25

    Extension e = (l − l₀)/l₀ = (7.5 − 10.0)/10.0 = −0.25, so the shortening is 25 %. Quoting the stretch 0.75 or the quadratic elongation 0.5625 instead of the percentage is the usual slip.

  8. Which combination of conditions most favours ductile rather than brittle deformation of quartz-rich rock?

    1. Low temperature, fast loading and dry rock at shallow depth
    2. High temperature, very slow strain rate and water in the rock
    3. High temperature, very fast loading and a total absence of fluid
    4. Low temperature, slow strain rate and a very high pore-fluid pressure
    Show answer

    Answer: B — High temperature, very slow strain rate and water in the rock

    Ductile flow is thermally activated and rate-dependent: high temperature, slow strain rate and water (which weakens quartz) all promote it. The shallow, cold, fast, dry combination is the brittle extreme. High temperature with very fast loading can still produce fracture, and high pore pressure at low temperature promotes brittle failure by reducing effective stress.

  9. Which statements about the strength of continental crust and the brittle–ductile transition are correct?

    1. In the brittle field the strength increases roughly linearly with depth because frictional resistance grows with normal stress
    2. In the ductile field the strength increases with depth because the rising temperature stiffens the rock
    3. Most crustal earthquakes nucleate in the brittle part above the transition
    4. The ductile strength of a rock does not depend on the strain rate, only on temperature
    Show answer

    Answer: A — In the brittle field the strength increases roughly linearly with depth because frictional resistance grows with normal stress; C — Most crustal earthquakes nucleate in the brittle part above the transition

    Frictional strength rises with normal stress, hence with depth, and earthquakes need brittle stick-slip, so they nucleate above the transition. Ductile strength falls with depth because thermally activated creep speeds up as temperature rises, and it also depends on strain rate: a faster rate needs a larger stress for the same flow, so the claims that ductile strength rises with depth and ignores strain rate are wrong.

  10. A rock core is loaded to 50 MPa and unloaded, and returns exactly to its original length. It is then loaded further: once the stress reaches about 120 MPa the strain keeps increasing at that constant stress, and after unloading a permanent strain of 0.3 % remains. Which description fits?

    1. Linear viscous (Newtonian) behaviour at all stresses, since strain increases with time
    2. Brittle failure at 120 MPa, since the sample loses cohesion and breaks into pieces
    3. Elastic up to a yield stress of about 120 MPa, then ideally plastic flow at constant stress
    4. Elastic at all stresses, since the 0.3 % permanent strain is only a measurement error
    Show answer

    Answer: C — Elastic up to a yield stress of about 120 MPa, then ideally plastic flow at constant stress

    Full recovery after 50 MPa shows elastic response there. Strain that grows at constant stress and leaves a permanent residue shows yielding with no strain hardening, which is ideal plasticity, so the yield stress is about 120 MPa. Viscous flow would make strain rate, not strain, proportional to stress, and brittle failure would end in fracture, not steady flow.

  11. A fold is arch-shaped, closing upward, but the beds in its core are the youngest in the sequence. What is this fold called?

    1. An anticline, because every arch-shaped fold is an anticline
    2. A synformal anticline, which is trough-shaped with the oldest beds in the core
    3. An antiformal syncline, which is arch-shaped with the youngest beds in the core
    4. A monocline, because only one limb of the fold is steeply dipping
    Show answer

    Answer: C — An antiformal syncline, which is arch-shaped with the youngest beds in the core

    Shape and age are separate criteria. Closing upward makes it an antiform; youngest beds in the core make it a syncline; together the name is antiformal syncline, found in overturned sequences. Calling it an anticline mixes up the two criteria, which is the tempting error, since an anticline needs the oldest beds in the core. The synformal anticline is the opposite case.

  12. A bed dips at a true dip of 60°. A vertical section is cut in a direction that makes an angle of 30° with the strike of the bed. Using tan δ_a = tan δ sin β, what is the apparent dip in the section, in degrees, to the nearest degree? Give the number only.

    Numerical answer — type the value.

    Show answer

    Answer: 41

    tan δ_a = tan 60° × sin 30° = 1.7321 × 0.5 = 0.8660, so δ_a = arctan 0.8660 = 40.9°, which rounds to 41°. Using the cosine of 30° gives 56°, and taking the apparent dip as 60° × 0.5 = 30° treats the angles as proportional, which they are not.

  13. On a geological map a set of beds forms a V-shaped outcrop closing toward the north. The oldest beds occupy the core of the V. Which fold is this?

    1. An anticline plunging to the north, since an anticline closes in the direction of plunge
    2. A syncline plunging to the north, since a syncline closes in the direction of plunge
    3. An anticline plunging to the south, since an anticline closes against the plunge direction
    4. A syncline plunging to the south, since oldest beds in the core indicate a syncline
    Show answer

    Answer: A — An anticline plunging to the north, since an anticline closes in the direction of plunge

    The oldest rocks in the core mean an anticline. A plunging anticline closes in the direction of its plunge, so the V closing north means it plunges north. A syncline plunging south would also give a V closing north, but it fails on the age criterion, because a syncline has the youngest beds in the core.

  14. Which of the following statements about folds are correct?

    1. In an isoclinal fold the limbs are parallel, so the interlimb angle is about zero
    2. The plunge of a fold is the angle between the hinge line and the vertical
    3. In a parallel (concentric) fold the layer thickness at right angles to the bedding is constant
    4. The axial surface of every fold is vertical, so an inclined axial surface cannot occur
    Show answer

    Answer: A — In an isoclinal fold the limbs are parallel, so the interlimb angle is about zero; C — In a parallel (concentric) fold the layer thickness at right angles to the bedding is constant

    Parallel limbs mean an interlimb angle near 0° (isoclinal), and parallel folds keep a constant orthogonal thickness. Plunge is measured from the horizontal, not the vertical, and axial surfaces can be upright, inclined, overturned or recumbent.

  15. In a normal fault, how does the hanging wall move relative to the footwall?

    1. It moves up, which shortens and thickens the crust
    2. It moves sideways along the strike with no vertical component
    3. It moves down, which extends and thins the crust
    4. It stays fixed while the footwall rotates about a horizontal axis
    Show answer

    Answer: C — It moves down, which extends and thins the crust

    In a normal fault the hanging wall moves down relative to the footwall, which stretches and thins the crust. Upward motion of the hanging wall defines a reverse fault, which shortens and thickens the crust, and pure strike-parallel motion defines a strike-slip fault.

  16. A vertical borehole drilled through horizontal strata crosses a dipping fault and meets the same marker bed twice. Which fault type does this indicate?

    1. A normal fault, since a normal fault repeats part of the sequence
    2. A reverse fault, since a reverse fault repeats part of the sequence
    3. A vertical strike-slip fault, since it moves beds sideways across the borehole
    4. A normal fault, since the hanging wall has moved down along the dipping plane
    Show answer

    Answer: B — A reverse fault, since a reverse fault repeats part of the sequence

    A reverse (or thrust) fault places older rock over younger, so a vertical hole passes through the same bed twice and the sequence is repeated. A normal fault does the opposite and omits a section, which is the tempting confusion. Near-vertical strike-slip faults displace beds horizontally and so neither repeat nor omit beds in a vertical hole.

  17. Which of the following statements about faults are correct?

    1. A thrust is a low-angle reverse fault, so it produces horizontal shortening of the crust
    2. The heave of a fault is the vertical component of its dip-slip displacement
    3. On a dextral strike-slip fault the block across the fault appears to move to the left
    4. A graben is a block that has dropped between two normal faults dipping toward each other
    Show answer

    Answer: A — A thrust is a low-angle reverse fault, so it produces horizontal shortening of the crust; D — A graben is a block that has dropped between two normal faults dipping toward each other

    A thrust is a low-angle reverse fault and shortens the crust horizontally, and a graben is the down-dropped block between inward-dipping normal faults. Heave is the horizontal, not the vertical, component of dip-slip (throw is the vertical one). On a dextral fault the opposite block moves to the right.

  18. A fault dipping at 60° has a net slip of 10.0 m with a rake of 30° measured in the fault plane from the strike line. Taking throw = dip-slip × sin(dip), what is the throw, in metres, to two decimal places? Give the number only.

    Numerical answer — type the value.

    Show answer

    Answer: 4.33

    Dip-slip = 10.0 × sin 30° = 5.00 m (the strike-slip part is 10.0 × cos 30° = 8.66 m). Throw = 5.00 × sin 60° = 5.00 × 0.8660 = 4.33 m, and heave = 5.00 × cos 60° = 2.50 m. Using cos 30° for the dip-slip gives a throw of 7.50 m, which swaps the strike-slip and dip-slip parts.