Question 4 of 5: Faults, Folds, Deformation Mechanisms & Kinematic Analyses

Question D — Faults, Folds, Deformation Mechanisms & Kinematic Analyses (25 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.

D1(a) – Name the fault

Given. A vertical road cut through flat-lying Cretaceous units cut by one fault whose trace rises from lower left to upper right (so it dips toward the left), inclined only about 23° in the sketch. The reference column lists the units top-down: dotted (239 m), chevron-pattern (187 m), a blank unit (77 m), a faintly stippled unit (110 m) and a tick-pattern unit (175 m) above the Jurassic.

Find. The full descriptive name of the fault.

The hanging wall is the block above the inclined fault (upper left). In it, the chevron unit sits right at the ground surface, with the 77 m, 110 m and tick units below it. In the footwall (lower right), the chevron unit lies much deeper, beneath a cap of the dotted unit. The same unit is therefore structurally higher in the hanging wall, so the hanging wall has moved up relative to the footwall. That also places older rocks (the tick unit) directly above younger rocks (the 77 m and 110 m units) along the fault. Hanging wall up means a reverse dip-slip fault, and at a dip of less than about 30° it is a thrust fault: a low-angle reverse fault that dips toward the left of the section and shortens the sequence.

$$\boxed{\text{Thrust fault (low-angle reverse dip-slip fault, dip}\approx23^{\circ}\text{)}}$$

D1(b) – Stratigraphic throw at point A

Given. The column labels 239, 187, 77, 110 and 175 m are unit thicknesses, not elevations: each label sits at the middle of its unit, and the boxes are drawn to scale. Point A is where two contacts meet on the fault. On the hanging-wall side it is the base of the 77 m unit (the top of the 110 m unit). On the footwall side it is the dotted/chevron contact (the top of the 187 m chevron unit).

Find. Approximate stratigraphic throw at point A.

Approach. Stratigraphic throw is the stratigraphic thickness between the two contacts that the fault has juxtaposed at the point. Here the hanging-wall contact is lower in the sequence than the footwall contact by the whole chevron unit plus the whole 77 m unit.

1. Units separating the juxtaposed contacts. Going down the column from the footwall contact (top of the chevron unit) to the hanging-wall contact (base of the 77 m unit) passes through the 187 m chevron unit and the 77 m unit. $$\text{throw}=187+77$$ $$\boxed{\text{Stratigraphic throw at A} \approx 264\ \text{m}}$$
2. Graphical check. Measured directly across the fault in the sketch, the base of the chevron unit is ≈255 units higher in the hanging wall than in the footwall, on the same scale as the column. That agrees with 264 m to within the precision of a hand-drawn sketch.

D2 – Shear zone: angular shear, shear strain, elongation

Given. Shear-zone width w = 1 m; dextral slip along the zone = 10 m; a marker originally perpendicular to the shear-zone boundaries is sheared into an inclined line (the strained marker), shown in the source figure making an angle of ≈6° with the shear-zone boundary.

Find. (a) Angular shear ψ and shear strain γ; (b) elongation e of the offset marker.

1. Shear strain (a). For a marker originally perpendicular to the shear-zone boundaries, the shear strain is simply the ratio of slip to zone width: $$\gamma=\frac{\text{slip}}{w}=\frac{10}{1}$$ $$\boxed{\gamma = 10.0}$$
2. Angular shear (a). $$\psi=\arctan(\gamma)=\arctan(10.0)$$ $$\boxed{\psi \approx 84.3^{\circ}}$$ As a check, the strained marker's angle from the shear-zone boundary is \(90^{\circ}-\psi=\arctan(w/\text{slip})=\arctan(0.1)\approx5.7^{\circ}\), matching the source figure's printed 6° angle to within rounding.
3. Elongation (b). The strained marker's new length is the hypotenuse of the (width, slip) right triangle: $$L'=\sqrt{w^2+\text{slip}^2}=\sqrt{1^2+10^2}=\sqrt{101}$$ $$e=\frac{L'-w}{w}=\sqrt{101}-1$$ $$\boxed{e \approx 9.0}$$

D3 – Shear-sense indicators

Given. A SW–NE-trending outcrop line drawing showing internal fabric with several small hatched rotated-rectangle fragments, filled ovoid grains with attached tails, and gently sinuous foliation traces, cut by a steeper NE-side face.

Find. The overall sense of shear, supported by three named indicator types.

The foliation runs parallel to the length of the face, so the shear plane is sub-horizontal in the drawing. Three independent, mutually consistent shear-sense indicators are present: (i) σ-type porphyroclasts with stair-stepping tails — the small dark grains whose right-hand (NE) tail leaves from a higher level than the left-hand (SW) tail; (ii) asymmetric (Z-shaped) drag folds — the small folded layers whose upper hinge sits to the NE of their lower hinge, i.e. the folds are overturned toward the NE; (iii) shear bands / C′ extensional crenulation cleavage — the paired inclined lines dipping down toward the NE that cut and drag the foliation with a NE-side-down, synthetic offset. The small lens-shaped sigmoidal grains (mica-fish-like, rising toward the NE) agree. All of them record the upper part of the face moving toward the NE relative to the lower part: dextral (right-lateral), top-to-the-NE shear as the face is drawn.

D4 – Dislocations

Given. Two crystal-lattice sketches: (a) a simple-cubic lattice block (labelled W/E) with an extra half-plane of atoms inserted from the top face, terminating within the crystal at a dislocation line marked with the standard ⊥ symbol; (b) a simple-cubic lattice block (labelled N/S) that has been sheared with a partial cut (V-notch) into its top face, producing a helical ramp of atomic planes around a dislocation line that runs N–S along the base of the cut.

Find. (a) Dislocation type, glide plane, and Burgers vector (with direction); (b) dislocation type and Burgers vector.

D4(a)

The extra half-plane of atoms, inserted from above and terminating partway through the crystal, is the defining signature of an edge dislocation (the ⊥ symbol at the terminating edge is the standard crystallographic symbol for one). The dislocation line runs into the block, perpendicular to the W–E front face (i.e. N–S). The glide plane is the horizontal lattice plane passing through the dislocation line itself — the plane at the depth where the extra half-plane terminates, separating the compressed lattice immediately above it from the stretched lattice immediately below. The Burgers vector for an edge dislocation is perpendicular to the dislocation line and lies within the glide plane, with magnitude equal to one lattice repeat distance; here that is horizontal, oriented E–W (parallel to the W/E edge labelled on the block), pointing in the slip direction defined by closing a Burgers circuit around the dislocation in the real (imperfect) lattice against the same circuit in a perfect reference lattice.

$$\boxed{\text{Edge dislocation; glide plane horizontal; }\mathbf{b}\ (\text{E--W}),\ |\mathbf{b}|=1\ \text{lattice repeat}}$$

D4(b)

The lattice has been offset by a partial shear cut so that the atomic planes form a continuous helical (spiral) ramp winding around a single line running through the crystal — this is the defining geometry of a screw dislocation. Unlike an edge dislocation, a screw dislocation's Burgers vector is parallel to the dislocation line rather than perpendicular to it; with the dislocation line trending N–S (per the N/S labels and the sense of offset shown by the small displacement arrows), the Burgers vector is horizontal, trending N–S, magnitude one lattice repeat distance.

$$\boxed{\text{Screw dislocation; }\mathbf{b}\ \parallel\ \text{dislocation line (N--S)},\ |\mathbf{b}|=1\ \text{lattice repeat}}$$