18-Geol-A4 Structural Geology, May 2018, Q3

Question C (30 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

C(1) Angular shear, shear strain and elongation of the shear zone

Given. Shear-zone width \(w = 1\ \text{m}\); dextral slip of the marker \(\Delta x = 10\ \text{m}\); the marker was originally perpendicular to the zone boundaries.

Find. The angular shear \(\psi\), the shear strain \(\gamma\), and the elongation \(e\) of the offset marker.

Fig. C1 — original (dashed) vs. dextrally strained (solid) marker across the 1 m shear zone.

Approach. Treat the marker and the zone width as the two legs of a right triangle; the angular shear is the angle the marker rotates through, the shear strain is its tangent, and the elongation follows from the ratio of strained to original marker length.

Shear strain from the offset geometry. \(\gamma = \tan\psi = \dfrac{\Delta x}{w} = \dfrac{10\ \text{m}}{1\ \text{m}} = \boxed{10}\).
Angular shear. Substituting, \(\psi = \arctan(\gamma) = \arctan(10) = \boxed{84.3^{\circ}}\) — the marker has been rotated almost onto the shear-zone boundary itself.
Elongation of the offset marker. The strained marker is the hypotenuse of the same right triangle: \(L_1=\sqrt{w^2+\Delta x^2}=\sqrt{1^2+10^2}=10.05\ \text{m}\). Elongation \(e=\dfrac{L_1-L_0}{L_0}=\dfrac{10.05-1}{1}=\boxed{9.05}\) (about +905%).

C(1) results
Quantity	Value
Shear strain γ	10
Angular shear ψ	84.3°
Elongation e of the offset marker	9.05 (+905%)

Topic: Simple-shear strain geometry

Key relations: \(\gamma=\tan\psi\); strained length \(L_1=\sqrt{w^2+\Delta x^2}\); \(e=(L_1-L_0)/L_0\).

Why this works: A marker that starts perpendicular to the shear-zone walls and is dragged by simple shear traces out exactly a right triangle whose legs are the zone width and the slip; the shear strain is by definition the tangent of the angle that triangle's hypotenuse makes with the marker's original orientation, so no separate strain-ellipse construction is needed for a single straight marker.

Common pitfall: Reporting \(\gamma\) as an angle (it is dimensionless, the tangent of the angular shear, not the angle itself) or elongation as a percentage of the WIDTH rather than of the marker's own original length are the two most common slips.

Source: Fossen Ch.2 §2.6–2.8, simple shear; Davis & Reynolds Ch.4.

C(2) Brachiopod hinge angles: heterogeneous or homogeneous strain?

The observation implies heterogeneous strain across the bedding plane. A homogeneous strain, by definition, applies the SAME strain ellipse everywhere on the surface: every brachiopod, regardless of its position on the bedding plane, would be distorted by an identical amount, so the 90° living hinge angle would be shifted to the same new (non-90°) angle for EVERY fossil. Because some brachiopods on this bedding plane still preserve the original 90° hinge angle while others show a distinctly different angle, the amount of strain must vary from point to point across the surface — which is precisely the definition of heterogeneous strain. Physically this is expected near a fold hinge or fault, where strain intensity is concentrated in some domains (strongly rotated/distorted brachiopods) and dies away in others (undeformed, still-90° brachiopods) over a single outcrop-scale surface.

Topic: Homogeneous vs. heterogeneous strain, tested with a fossil strain marker

Key relations: Homogeneous strain: straight lines stay straight, parallel lines stay parallel, ONE strain ellipse describes the whole domain. Heterogeneous strain: the strain ellipse varies spatially.

Why this works: A population of originally-identical right-angle markers (brachiopod hinge/bisector angles) scattered over one surface is exactly the classic test for strain homogeneity: if every marker records the same angular change, the strain field is homogeneous over that domain; if the recorded angular change varies from marker to marker, the field cannot be described by one strain ellipse.

Common pitfall: Assuming that ANY deformed fossil population implies homogeneous strain simply because the deformation mechanism (e.g. pressure solution) is uniform in style — the STYLE of strain can be uniform while its MAGNITUDE still varies spatially, which is exactly the heterogeneous case being tested here.

Source: Davis & Reynolds Ch.4 (strain, strain markers); Fossen Ch.3.

C(3) Naming the fault and its stratigraphic throw at point A

(a) Name of the fault. The road-cut sketch shows the hanging wall (east side, right of the fault trace) displaced DOWN relative to the footwall (west side, left of the trace) — a dip-slip sense with no evidence of a strike-slip component. This is a normal fault. The fault trace curves from a gentle dip at depth (lower-left) to a much steeper dip near the surface (upper-right), the classic concave-up (spoon-shaped) geometry of a listric normal fault, downthrown to the east.

(b) Stratigraphic throw at point A.

Given. Stratigraphic column thicknesses, youngest (Cretaceous, top) to oldest (Jurassic, base): U1 = 239 m, U2 = 187 m, U3 = 77 m, U4 = 110 m, U5 = 175 m. Point A sits on the fault trace exactly at the U1/U2 contact, on the hanging-wall (down-dropped) side.

Find. The vertical stratigraphic throw of the fault at point A.

Fig. C3 — stratigraphic column and faulted road cut. Point A sits on the U1/U2 contact in the hanging wall; at the same elevation the footwall exposes the U3/U4 contact.

Approach. Trace point A's elevation horizontally into the footwall block: whichever contact sits at that SAME elevation on the undisplaced side marks how much section has been cut out by the fault. The throw equals the combined thickness of the units missing between the two juxtaposed contacts.

Identify the juxtaposed markers. At point A's elevation, the hanging wall exposes the U1/U2 contact while the footwall (traced horizontally, at the same real-world elevation) exposes the U3/U4 contact.
Sum the missing section. In an unfaulted column, the U3/U4 contact lies a full thickness of U2 and U3 stratigraphically below the U1/U2 contact: \(T_{\text{throw}} = t_{U2}+t_{U3} = 187\ \text{m} + 77\ \text{m} = \boxed{264\ \text{m}}\).

C(3) results
Quantity	Value
Fault type	Listric normal fault, downthrown to the east
Stratigraphic throw at point A	≈ 264 m

Topic: Stratigraphic throw from a faulted column of known unit thicknesses

Key relations: Throw = the combined thickness of the stratigraphic units cut out (never exposed on either block at the same elevation) across the fault; apparent (map or cross-section) omission of section is the diagnostic signature of a normal fault, exactly as repetition of section is diagnostic of a reverse/thrust fault.

Why this works: Because the hanging wall of a normal fault drops relative to the footwall, erosion to a common surface (or exposure along one road cut) strips progressively OLDER rock off the uplifted footwall while the downthrown hanging wall still preserves the YOUNGER cover — juxtaposing a young marker against an older one at the same elevation, with the intervening thickness missing from the section.

Common pitfall: Confusing throw (a stratigraphic-thickness quantity, read off the column) with the fault's true dip-slip displacement (a distance ALONG the fault plane, which also depends on the fault's dip and is not asked for here).

Source: Davis & Reynolds Ch.6 (fault separation and displacement); Fossen Ch.10.

C(4) Identifying the dislocations and their Burgers vectors

(a) The first sketch shows an extra half-plane of atoms inserted into the upper part of the crystal, terminating at a dislocation line marked with the standard ⊥ symbol — this is an edge dislocation. The glide (slip) plane is the horizontal lattice plane that passes through the base of the extra half-plane and contains the dislocation line. The Burgers vector b lies IN the glide plane, perpendicular to the dislocation line, with magnitude equal to one lattice spacing; its direction matches the applied shear arrow shown on the sketch, i.e. b points due west (W), the direction the upper block has glided relative to the lower block.

Fig. C4(a) — edge dislocation: extra half-plane, dislocation line (⊥), glide plane and Burgers vector.

(b) The second sketch shows a vertical dislocation line with a helicoidal (spiral-ramp) offset of the lattice around it, and a Burgers circuit traced from a point labelled N on the top face around to a point labelled S where the circuit fails to close — the defining signature of a screw dislocation. By definition the Burgers vector of a screw dislocation is PARALLEL to the dislocation line itself (not perpendicular to it, as for an edge dislocation); applying the right-hand (finish-to-start) circuit convention to the drawn N→S path gives b vertical, directed from S to N, parallel to the dislocation line.

Fig. C4(b) — screw dislocation: vertical dislocation line and the N→S Burgers circuit.

Topic: Edge vs. screw dislocations and the Burgers vector

Key relations: Edge dislocation: line ⊥ b, defined by an extra half-plane; glide plane contains both the line and b. Screw dislocation: line ∥ b, defined by a helicoidal lattice ramp; Burgers circuit (finish-to-start, right-hand convention) gives b's sense.

Why this works: The Burgers vector is the single closure-failure vector of a circuit drawn around the dislocation, and its geometric relationship to the dislocation LINE is exactly what distinguishes the two dislocation types: perpendicular defines an edge dislocation (an extra sheet of atoms with a well-defined glide plane), parallel defines a screw dislocation (no unique glide plane; it can cross-slip onto any plane containing the line).

Common pitfall: Reversing the Burgers-vector sense by tracing the circuit anticlockwise instead of the standard right-hand (clockwise, looking along the dislocation line) convention, or reporting an edge dislocation's b as parallel to the dislocation line instead of perpendicular to it.

Source: Fossen Ch.12 (crystal defects and dislocation creep, rock rheology).

C(5) Sense of shear from the SW–NE shear zone

Three DIFFERENT shear-sense indicator types are visible and mutually consistent in the sketch:

C′ shear bands (extensional crenulation cleavage). The short, oblique, hachured surfaces cut the main sub-horizontal foliation (S) at a small angle and are consistently inclined in the transport direction — the standard rule is that C′ planes always show the SAME sense of shear as the bulk zone.
Rotated (σ-type) porphyroclasts. The dark, tailed grain in the centre of the panel has asymmetric recrystallized tails trailing off consistently with the C′ sense.
Sigmoidal foliation/vein trails. The S-shaped internal trails show the same consistent curvature, sweeping into parallelism with the shear-zone boundary in the transport sense.

Fig. C5 — SW–NE shear zone with C′ shear bands, a σ-porphyroclast, and a sigmoidal vein trail, all indicating top-to-the-NE (dextral) shear.

All three indicators agree: this is a top-to-the-NE (dextral) shear sense.

Topic: Shear-sense indicators in a ductile shear zone

Key relations: C/S and C′ (extensional crenulation cleavage) fabrics; σ- and δ-type porphyroclasts; sigmoidal foliation/vein trails; mica fish.

Why this works: No single indicator is fully reliable in isolation (a porphyroclast tail can be ambiguous, a single vein can be a pre-existing feature reoriented rather than grown during shear), so the standard field/exam protocol is to identify at least three geometrically INDEPENDENT indicator types and confirm they agree before committing to a shear sense.

Common pitfall: Using three variations of the SAME underlying indicator (e.g. three different porphyroclasts) rather than three genuinely different indicator TYPES, or reading the C′ obliquity in the wrong sense relative to the main foliation.

Source: Davis & Reynolds Ch.15 (ductile shear zones); Fossen Ch.13 (shear-sense criteria).

C(6) Ductile fault zone: offset marker, fabric trajectory and strain ellipses

In the middle-to-deep crust, faults accommodate displacement by DISTRIBUTED ductile flow across a finite-width shear zone rather than by slip on a single discrete plane. An originally straight, planar marker (e.g. a dike) crossing such a zone is progressively deflected into a smoothly sigmoidal S-shape wherever shearing is most intense (the zone centre), while remaining essentially undeflected and rectilinear outside the zone margins — unlike a brittle fault, there is no discrete break, only continuous curvature of the marker that asymptotically straightens away from the centre. Strain ellipses constructed from an initially circular marker follow the same pattern: near-circular (low strain) at the margins, becoming progressively more elongate and more strongly rotated toward parallelism with the shear-zone boundary as the centre (highest cumulative shear strain) is approached.

Fig. C6 — ductile shear zone: sigmoidal offset dike (fabric trajectory) and strain ellipses that sharpen toward the zone centre.

Topic: Strain partitioning across a ductile shear zone

Key relations: Simple-shear strain profile \(\gamma(y)\) peaks at the zone centre and decays toward the margins; fabric trajectory = the locus of progressively rotated marker orientations; strain ellipse aspect ratio and rotation both scale with local γ.

Why this works: Ductile shear zones are not a single slip surface but a continuum in which strain rate (and hence cumulative shear strain for a constant total displacement) is highest where the zone is narrowest/most focused — typically the centre — so any marker threading through the zone records that spatial strain gradient directly as its own curvature, exactly analogous to velocity profiles in viscous channel flow.

Common pitfall: Drawing the offset marker with a sharp kink (a brittle-style discrete break) instead of continuous curvature, which is the single most common way this sketch loses marks — ductile offset must be smooth and asymptotic to the far-field orientation at the margins.

Source: Davis & Reynolds Ch.15; Fossen Ch.13 (shear zones, strain partitioning).

Question 3 of 5