Question 6 of 10: Question 4, Part 2: Wedge Daylighting on a Rock Cut

Question 4, Part 2: Wedge Daylighting on a Rock Cut (10 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.

Reading the table. It is the unknown that part (a) asks for. The face strikes due N (000°) and dips EAST, so its dip direction is 090°, which is exactly the N90°E direction of part (b).

Given. Joint set 1: strike N25°E, dip 60° SE (dip direction 115°). Joint set 2: strike N70°W, dip 50° NE (dip direction 020°). Rock cut face: strike N (000°), dipping E (dip direction 090°), dip unknown.

Find. (a) the maximum dip the cut face can have without the joint-bounded wedge daylighting; (b) the apparent dip of each joint set on a N90°E-trending vertical section.

Approach. The wedge can only slide along the line where the two joints intersect. It daylights (becomes kinematically free) when (i) that line's trend points out of the face, within 90° of the face's dip direction, and (ii) the line plunges less steeply than the face does measured in that same trend direction (Markland's test). The limiting face dip therefore makes the face's apparent dip along the intersection trend equal to the intersection plunge: $\tan\psi_f\cos(\alpha_f-\alpha_i)=\tan\psi_i$.

1. Plane normals. For a plane with dip direction $\varphi$ and dip $\delta$, build the horizontal strike vector and the down-dip vector and cross them to get the normal: $\mathbf{n}_1$ (set 1, $\varphi=115^\circ,\delta=60^\circ$) and $\mathbf{n}_2$ (set 2, $\varphi=020^\circ,\delta=50^\circ$).
2. Line of intersection. $\mathbf{L} = \mathbf{n}_1\times\mathbf{n}_2$ (normalized, taken pointing downward) gives trend and plunge: $\mathbf{L}\Rightarrow$ trend $\approx \text{N}58^\circ\text{E}$, plunge $\approx \boxed{43.2^\circ}$.
3. (a) Daylighting condition. The intersection trends 058°, which is 32° from the face's dip direction (090°). It points out of the east-dipping face, so the wedge CAN daylight. The limiting face dip is $\psi_f=\arctan\left(\dfrac{\tan43.2^\circ}{\cos(90^\circ-57.9^\circ)}\right)=\arctan\left(\dfrac{0.939}{0.847}\right)$, so the maximum face dip is $\approx \boxed{48^\circ}$. Any steeper face exposes the intersection line, and the wedge can slide out along it. (The limit exceeds the 43.2° plunge because the wedge slides obliquely to the face; a face dipping directly along 058° would have to stay below 43.2°.)
4. (b) Apparent dips on the N90°E section. Using $\tan\delta' = \tan\delta\cos(\varphi-\alpha)$ with section azimuth $\alpha=090^\circ$: joint set 1 ($\varphi=115^\circ,\delta=60^\circ$) gives $\delta'_1=\arctan[\tan60^\circ\cos(25^\circ)] \approx \boxed{57.5^\circ}$; joint set 2 ($\varphi=020^\circ,\delta=50^\circ$) gives $\delta'_2=\arctan[\tan50^\circ\cos(-70^\circ)]\approx \boxed{22.2^\circ}$.