Structural Geology I: Stress, Strain, Deformation Mechanisms, Fabrics and Folds
2. Strain: stretch, shear strain and the strain ellipse
Strain is the change of shape and size once translation and rotation are removed. In homogeneous strain straight lines stay straight and a circle becomes an ellipse, the strain ellipse (axes X ≥ Z), or an ellipsoid X ≥ Y ≥ Z in three dimensions. For a line, extension e = (l − l0)/l0, stretch S = l/l0 = 1 + e and quadratic elongation λ = S². Shear strain γ = tan ψ, where ψ is the angular shear. The axial ratio is Rs = X/Z. Flinn's k = (X/Y − 1)/(Y/Z − 1) gives the shape: k > 1 is constriction (cigar, prolate, L-tectonite), k < 1 is flattening (pancake, oblate, S-tectonite), and k = 1 is plane strain. Fabric follows the axes: foliation lies in the XY plane, perpendicular to the shortening axis Z, and lineation is parallel to X.
Pure shear is coaxial: the strain axes keep their orientation in the material. Simple shear is non-coaxial: displacement parallel to one plane with area conserved. In simple shear the long axis of the strain ellipse makes angle θ′ with the shear plane, tan 2θ′ = 2/γ, so it starts near 45° and rotates towards the shear plane as γ grows. The principal stretches are S1 = ½[γ + √(γ² + 4)] and S2 = 1/S1, so Rs = S1². For γ = 1, S1 = 1.618, Rs = 2.62 and θ′ = 31.7°. Natural deformation often combines both (general shear).
Shortening of a folded layer is (l0 − l)/l0 × 100%: a layer 10 km long that now spans 6 km has e = −0.4, a shortening of 40%. Strain is measured from deformed markers of known initial shape, such as reduction spots, ooids, pebbles and fossils, using the Rf/φ method for initially non-spherical grains, and from boudinage. Mean strain rate is strain divided by time, with about 10⁻¹⁴ s⁻¹ a typical tectonic order.
3. Material response and deformation mechanisms
Rocks respond elastically (recoverable, σ = Eε), viscously (flow, σ = η ε̇) or plastically (permanent beyond a yield stress). Brittle deformation loses cohesion by fracture and frictional sliding, and frictional strength rises linearly with depth (Byerlee: τ ≈ 0.85 σn up to about 200 MPa, then τ ≈ 50 MPa + 0.6 σn, nearly independent of rock type). Ductile deformation spreads without loss of cohesion and weakens exponentially with temperature, so lithospheric strength peaks at the brittle–ductile transition, near 300 °C for quartz-rich crust (roughly 10 to 15 km depth) and at higher temperature for feldspar and olivine. Cold and fast favours brittle behaviour, hot and slow favours ductile.
| Mechanism | Conditions | Signature and rate law |
|---|---|---|
| Cataclasis, frictional sliding | Shallow, cold, fast | Breccia, gouge, cataclasite; pressure-dependent strength |
| Pressure solution | Fluids present; low to moderate temperature | Stylolites, spaced cleavage, fibrous pressure shadows |
| Diffusion creep | High temperature, fine grains, low stress | Strain rate ∝ σ (n = 1) and ∝ d⁻² (Nabarro–Herring, through the lattice) or d⁻³ (Coble, along boundaries) |
| Dislocation creep | High temperature, higher stress | ε̇ = A σⁿ exp(−Q/RT), n about 3 to 5, no grain-size term; crystallographic preferred orientation, undulose extinction |
| Twinning, kinking | Low temperature, strong anisotropy | Calcite twins, kinked mica |
In quartz, dynamic recrystallisation records temperature: bulging (BLG) at roughly 300 to 400 °C, subgrain rotation (SGR) at roughly 400 to 500 °C and grain-boundary migration (GBM), with lobate boundaries, above about 500 °C. Because dislocation creep has n of 3 to 5, a small rise of stress raises strain rate steeply: with n = 3, increasing σ from 20 to 30 MPa multiplies ε̇ by 1.5³ = 3.4 at constant temperature and grain size, but only by 1.5 for diffusion creep.
4. Primary and secondary structures, planar and linear fabrics
Primary structures form with the rock: bedding, lamination, cross-bedding, graded bedding, ripple marks, mud cracks and sole marks in sediments; flow banding, vesicles and pillows in lavas. Secondary structures form later by deformation: folds, faults, joints, cleavage, foliation, lineation, boudins and veins. Primary structures give the younging (way-up) direction, which is essential in folded ground.
- Way-up criteria: graded bedding is coarse at the base and fine at the top; cross-bed foresets are truncated at the top and tangential at the base; mud cracks taper downward; flute and load casts lie on the base of a bed; pillows are convex upward with cusps pointing down; vesicles concentrate at the top of a flow.
- Foliation lies in the XY plane, perpendicular to the shortening direction, and is usually parallel to the axial surface of folds (axial planar). Slaty cleavage is microscopic alignment of phyllosilicates in very low to low-grade rock, with perfect splitting; phyllitic foliation has a silky sheen; schistosity shows visible aligned mica at medium to high grade; gneissic banding is compositional layering at high grade. Spaced (disjunctive) cleavage forms by pressure solution along thin domains, and crenulation cleavage is a microfold of an earlier foliation.
- Lineations: stretching lineation (elongate pebbles and grains, parallel to X, the transport direction in shear zones); mineral lineation (aligned prisms such as amphibole); intersection lineation (bedding with cleavage, parallel to fold hinges); crenulation lineation; fold hinges; and slickenlines on faults. A rock with both foliation and lineation is an L–S tectonite, and the pair fixes the XY plane and the X axis.
Bedding–cleavage rule: when axial-planar cleavage and bedding dip in the same direction, cleavage steeper than bedding means the beds are upright, and bedding steeper than cleavage means they are overturned, because the overturned limb of an inclined fold is steeper than its axial surface. Cleavage refraction: in alternating sandstone and shale, cleavage meets the bedding at a larger angle in the competent sandstone than in the shale.
5. Folds: geometry and classification
A fold is a bent surface. The hinge is the line of maximum curvature, the limbs are the flanks, the axial surface joins the hinges of successive layers and the fold axis is the line that generates the fold when moved parallel to itself. Wavelength is the distance between successive hinges of the same kind and amplitude is half the crest-to-trough height. Antiform and synform describe shape alone; anticline (older rocks in the core) and syncline (younger in the core) need age. A plunging anticline closes in the direction of plunge on the map and a plunging syncline closes against it. A dome and a basin are doubly closing folds and a monocline is a single step in dip.
| Property | Classes and ranges |
|---|---|
| Interlimb angle | Gentle 180 to 120, open 120 to 70, close 70 to 30, tight 30 to 0, isoclinal 0 |
| Axial surface dip | Recumbent 0 to 10, gently inclined 10 to 30, moderately 30 to 60, steeply 60 to 80, upright 80 to 90 |
| Hinge plunge | Horizontal 0 to 10, gentle 10 to 30, moderate 30 to 60, steep 60 to 80, vertical 80 to 90 |
Ramsay's classes use dip isogons, the lines joining points of equal dip on the top and bottom surfaces of a layer. In class 1 the isogons converge towards the inner arc; class 1B is the parallel (concentric) fold, with constant orthogonal thickness. In class 2 the isogons are parallel and the thickness measured parallel to the axial surface is constant: this is the similar fold, with thick hinges and thin limbs. In class 3 the isogons diverge towards the inner arc. Parallel folds cannot persist indefinitely downward, because the layers run out of room and must be taken up in a detachment, whereas similar folds can; they typify ductile rocks with axial-planar cleavage. Pure geometry is not mechanism, since a fold's class says how thickness varies, not how it formed.
6. Fold mechanisms, wavelength and asymmetry
In buckling a competent layer shortened along its length folds at a dominant wavelength set by its thickness and the viscosity contrast: λd = 2π h (η1/(6 η2))^(1/3), for a layer of thickness h and viscosity η1 in a matrix of viscosity η2 (Biot–Ramberg theory). Thicker layers therefore fold with longer wavelengths, and λd/h depends only on the viscosity ratio. For η1/η2 = 162, (162/6)^(1/3) = 3, so λd = 6π h ≈ 18.8 h.
- Flexural slip is bed-parallel slip, like bending a pack of cards: it is zero at the hinge, greatest on the limbs, and leaves slickenlines perpendicular to the hinge. Flexural flow spreads that shear through the layer. In tangential longitudinal strain the outer arc is stretched and the inner arc compressed, with a neutral surface between.
- Passive flow (shear folding) lets layers act only as markers and gives similar folds by shear across the axial plane; bending acts across the layering, as over a rising fault block. Angular kink and chevron folds develop in strongly layered or anisotropic rock, and boudinage is necking of a competent layer in extension.
- Parasitic folds are small folds on the limbs of a large fold whose asymmetry reverses across the hinge. Viewed down the plunge, they show S shapes on one limb, Z shapes on the other, and symmetric M shapes at the hinge; the large hinge lies where S changes to Z.
Key takeaways
- Mohr circle: σn = (σ1 + σ3)/2 + (σ1 − σ3)/2 cos 2θ and τ = (σ1 − σ3)/2 sin 2θ with θ from σ1 to the plane's normal. Coulomb failure planes lie at 45° − φ/2 to σ1; pore pressure shifts the effective circle left; the vertical principal stress sets the Anderson fault type.
- Strain: S = 1 + e, λ = S², shortening = (l0 − l)/l0. Flinn's k separates constriction (k > 1) from flattening (k < 1). Simple shear is non-coaxial with tan 2θ′ = 2/γ and Rs = S1² where S1 = ½[γ + √(γ² + 4)]. Foliation lies in XY, lineation along X.
- Brittle strength rises with depth (Byerlee), ductile strength falls with temperature, so strength peaks at the brittle–ductile transition. Diffusion creep has n = 1 and depends on grain size; dislocation creep has n of 3 to 5 and builds a crystallographic fabric; quartz goes BLG, SGR, GBM with rising temperature.
- Primary structures give way-up; secondary structures are deformational. Slaty cleavage, schistosity and gneissic banding mark rising grade, foliation is axial planar, and lineations are stretching (parallel to X), mineral, intersection or slickenlines. Bedding steeper than cleavage in the same dip direction means overturned.
- Folds are described by interlimb angle, axial-surface dip and hinge plunge (Fleuty) and by thickness variation (Ramsay): class 1B parallel, class 2 similar. Anticline and syncline need age; antiform and synform do not.
- Buckling wavelength is λd = 2π h (η1/(6 η2))^(1/3). Flexural slip is greatest on the limbs and zero at the hinge, tangential longitudinal strain stretches the outer arc, and parasitic folds change from S to Z across the hinge of the major fold.
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.
According to Anderson's theory of faulting, which orientation of the principal stresses produces thrust (reverse) faults near the surface?
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Answer: A — σ3 vertical, with σ1 and σ2 both horizontal
In a thrust regime the vertical stress, from the weight of the rock, is the least principal stress, so σ1 and σ2 are horizontal and faults form at about 30° to the horizontal. The case with σ1 vertical produces normal faults and the case with σ2 vertical produces strike-slip faults. A purely lithostatic state has no differential stress, so no shear failure can occur.At a point the principal stresses are σ1 = 150 MPa and σ3 = 50 MPa (compression positive). A plane has its normal at 25° to the σ1 direction. The normal stress on this plane, in MPa to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 132.1
With θ the angle from σ1 to the plane's normal, σn = (σ1 + σ3)/2 + (σ1 − σ3)/2 · cos 2θ = 100 + 50 cos 50° = 100 + 32.14 = 132.1 MPa. Treating 25° as the angle between σ1 and the plane itself reverses the sign of the cosine term and gives 67.9 MPa, which is the usual slip.A rock under σ1 = 120 MPa and σ3 = 60 MPa is stable, its Mohr circle lying just below a Coulomb failure envelope. Pore fluid pressure then rises by 20 MPa while the total σ1 and σ3 stay unchanged. What happens on the effective-stress Mohr diagram?
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Answer: B — The circle keeps its diameter of 60 MPa and shifts 20 MPa to the left, so it may now touch the envelope and the rock can fail
Effective stress is σ′ = σ − Pf, so both effective principal stresses fall by 20 MPa (to 100 and 40 MPa). The differential stress, and so the circle's diameter of 60 MPa, is unchanged and the circle moves left towards the envelope, which is how fluid pressure triggers failure without any change in tectonic load. Shifting right or shrinking the circle misreads the sign of the pore-pressure term, and fluid pressure does not change the differential stress.For a rock with an angle of internal friction φ = 30°, a new shear fracture forms in the Coulomb criterion at what angle to the σ1 direction?
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Answer: B — 30°
The shear fracture forms at 45° − φ/2 to σ1, which is 45° − 15° = 30°. The conjugate fracture lies on the other side of σ1, so the acute angle between the pair is 60°, and 60° is the tempting wrong answer because it is that angle between the fractures, not the angle to σ1. A value of 45° would hold only for φ = 0.Which of the following statements about strain are correct?
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Answer: A — When Flinn's k is greater than 1 the strain ellipsoid is prolate, and a lineation develops more strongly than a foliation; B — In simple shear the long axis of the strain ellipse rotates towards the shear plane as the shear strain increases
Constriction (k greater than 1) gives cigar-shaped ellipsoids and lineation-dominated fabrics, and in simple shear the long axis starts near 45° and rotates towards the shear plane with growing γ. Pure shear is coaxial, not non-coaxial, so its axes keep their orientation in the material. Foliation lies in the XY plane, perpendicular to the shortening axis Z, and not in the XZ plane.A competent bed originally 12 km long is now folded so that it spans 7.5 km. The shortening, in percent to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 37.5
Shortening = (l0 − l)/l0 × 100 = (12 − 7.5)/12 × 100 = 4.5/12 × 100 = 37.5%, that is e = −0.375. Dividing by the final length instead of the original length gives 60%, which is the usual slip.A rock undergoes simple shear with shear strain γ = 2. The principal stretches are S1 = ½[γ + √(γ² + 4)] and S2 = 1/S1. The axial ratio Rs = X/Z of the strain ellipse, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 5.83
S1 = ½[2 + √8] = ½[2 + 2.828] = 2.414 and S2 = 1/2.414 = 0.414, so Rs = S1/S2 = S1² = 5.83. Reporting S1 itself (2.41) as the axial ratio, which is the ratio of the longest stretch to unity rather than to the shortest stretch, is the usual slip.Which deformation mechanism has a strain rate that is linearly proportional to differential stress (n = 1) and depends strongly on grain size?
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Answer: C — Diffusion creep, in which atoms move through the lattice or along grain boundaries and the rate falls as grain size rises
Diffusion creep (Nabarro–Herring or Coble) is Newtonian, with n = 1, and its rate scales as d⁻² or d⁻³, so fine-grained rocks creep faster. Dislocation creep is the tempting alternative but has n of 3 to 5 and no grain-size term. Cataclasis and twinning are not governed by such a flow law.A quartzite deforms by dislocation creep, ε̇ ∝ σⁿ, with stress exponent n = 4 at constant temperature and grain size. The differential stress rises from 20 MPa to 50 MPa. The strain rate increases by a factor of ____ (to the nearest whole number).
Numerical answer — type the value.
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Answer: 39
The ratio is (σ2/σ1)ⁿ = (50/20)⁴ = 2.5⁴ = 39.06, so the strain rate rises about 39 times. Multiplying the stress ratio by n (2.5 × 4 = 10) instead of raising it to the power n is the usual slip.Two quartz mylonites are compared. One shows small new grains formed by progressive rotation of subgrains. The other shows large new grains formed by migration of grain boundaries across the old grains. Relative to the first, the second records deformation at:
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Answer: A — a higher temperature, because grain-boundary migration takes over from subgrain rotation above roughly 500 °C in quartz
In quartz the recrystallisation mechanism changes with temperature from bulging (BLG, roughly 300 to 400 °C) to subgrain rotation (SGR, roughly 400 to 500 °C) to grain-boundary migration (GBM, above about 500 °C), so the second sample was deformed hotter. Bulging, the tempting cold-end alternative, produces small new grains along old boundaries, not large ones. Strain amount alone does not fix the mechanism.Which of the following are reliable way-up (younging) criteria in deformed sedimentary and volcanic rocks?
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Answer: A — Graded bedding, in which grains are coarse at the base of a bed and fine at its top; B — Mud cracks that taper downward from the top of a bed
Coarse-to-fine grading upward and mud cracks that narrow downward both show the way up. Flute casts are sole marks found on the base of a bed, not its upper surface, and vesicles concentrate at the top of a lava flow where gas rises, not at its base, so those two statements would invert the sequence.At an outcrop, bedding strikes north–south and dips 70° west, and the axial-planar slaty cleavage also strikes north–south and dips 45° west. The beds are:
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Answer: D — overturned, because bedding dips more steeply than the axial-planar cleavage in the same direction
With bedding and cleavage dipping the same way, bedding steeper than cleavage (70° against 45°) marks the overturned limb of an inclined fold, because that limb is steeper than the axial surface. Dipping the same way alone does not make a limb upright: upright beds would be less steep than the cleavage. The bedding–cleavage relation does settle the way-up without a graded bed.A fold has an interlimb angle of 25°, an axial surface dipping 5° and a hinge plunging 4°. In Fleuty's descriptive scheme it is:
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Answer: C — a tight, recumbent, horizontal fold
An interlimb angle of 25° falls in the tight class (30° to 0°), an axial surface dipping 5° is recumbent (0° to 10°) and a plunge of 4° is horizontal (0° to 10°). A close fold would have an interlimb angle between 70° and 30°, and isoclinal means 0°.In which type of fold does the layer thickness measured parallel to the axial surface stay constant, giving thick hinges and thin limbs?
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Answer: C — A similar fold, Ramsay class 2, whose dip isogons are parallel
A similar fold (class 2) has parallel dip isogons and constant thickness measured parallel to the axial surface, so the orthogonal thickness falls on the limbs: thick hinges, thin limbs. A parallel fold keeps constant orthogonal thickness instead, which is the tempting alternative, and class 1A has thin hinges, the opposite pattern.A competent layer 0.5 m thick lies in a much less viscous matrix, with viscosity ratio η1/η2 = 48. Using the Biot–Ramberg dominant wavelength λd = 2π h (η1/(6 η2))^(1/3), where h is the layer thickness, λd is ____ m (to two decimal places).
Numerical answer — type the value.
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Answer: 6.28
η1/(6 η2) = 48/6 = 8 and 8^(1/3) = 2, so λd = 2π × 0.5 × 2 = 2π = 6.28 m. Using the viscosity ratio without dividing by 6 gives 48^(1/3) = 3.63 and a wavelength of 11.4 m, which is the usual slip.Which of the following statements about fold mechanisms are correct?
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Answer: A — In flexural-slip folding the bed-parallel slip is zero at the hinge and greatest on the limbs; B — Slickenlines on flexural-slip surfaces lie perpendicular to the fold hinge; C — In tangential longitudinal strain buckling the outer arc of the competent layer is stretched and the inner arc is shortened
Flexural slip is nil at the hinge and greatest on the limbs, leaves slickenlines at right angles to the hinge, and tangential longitudinal strain stretches the outer arc while shortening the inner arc, so the statements on slip, slickenlines and the outer and inner arcs are correct. Constant axial-parallel thickness describes a similar fold, which forms by passive shear flow and not by flexural slip, whose product is a parallel fold.Slaty cleavage in a deformed mudrock lies parallel to which plane of the finite strain ellipsoid?
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Answer: C — The XY plane, which is perpendicular to the direction of greatest shortening
Cleavage and foliation form perpendicular to the shortening axis Z, so they lie in the XY plane of the strain ellipsoid, and the stretching lineation within it is parallel to X. A plane containing Z would be perpendicular to the cleavage, not parallel to it.The line of intersection between bedding and an axial-planar cleavage is a lineation that is parallel to which feature of the fold?
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Answer: A — The fold hinge, because both surfaces contain the hinge line of the fold
Axial-planar cleavage contains the hinge direction and bedding is folded about it, so their intersection is parallel to the hinge (fold axis). The intersection lineation is not the stretching lineation, which is the tempting confusion, and it lies in the bedding, so it cannot be along the bedding pole.