Question 2 of 5: Open-pit wedge factor-of-safety derivation

Question 2: Open-pit wedge factor-of-safety derivation (20 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.

Given. A wedge bounded above by the horizontal ground surface, at the back by the vertical pit face of height H, below/ahead by joint system A (dip ψ, cohesion c, friction angle φ) which is the basal sliding surface, and laterally by two parallel joint-B release planes spaced W apart, cohesionless (C = 0) and assumed to carry no net resisting force on the block (their sole role is to isolate a finite-width wedge).

Find. An equation for the factor of safety, FoS, against sliding on Joint A, in terms of the wedge/excavation geometry (H, ψ, W) and the shear-strength parameters (c, φ) of Joint A.

Approach. Build the wedge as a triangular prism (cross-section area from the geometry of H and ψ, length W into the pit face), resolve its self-weight into components normal and parallel to the basal Joint-A plane, sum the resisting force (Joint A's cohesion plus friction on the resolved normal force — Joint B contributes nothing, since it is cohesionless and treated as a pure release surface) against the driving (down-dip) component of weight, and simplify.

1. Wedge cross-sectional area. With no tension crack, the wedge cross-section is the right triangle bounded by the vertical pit face (height H), the horizontal ground surface, and Joint A running from the toe of the pit face up to the ground surface at dip ψ. The horizontal run of Joint A is H/tanψ, so $$A_{xsec}=\tfrac{1}{2}H\cdot\frac{H}{\tan\psi}=\frac{H^2}{2\tan\psi}$$
2. Wedge weight and Joint-A plane area. Extending the cross-section a distance W into the pit face (the Joint-B spacing) gives the full 3D wedge: $$W_{wedge}=\gamma\,A_{xsec}\,W=\frac{\gamma H^2 W}{2\tan\psi}$$ The length of Joint A within the cross-section is H/sinψ, so its area is $$A_{JointA}=\frac{HW}{\sin\psi}$$
3. Resolve the weight on Joint A. The component of weight normal to Joint A drives frictional resistance; the down-dip component is the driving force: $$N=W_{wedge}\cos\psi,\qquad T=W_{wedge}\sin\psi$$
4. Resisting force. Joint A supplies both cohesion (over its full area) and friction (on the resolved normal force); Joint B is cohesionless and, acting purely as a lateral release surface, is taken to contribute no net resistance to sliding: $$R=c\,A_{JointA}+N\tan\phi=\frac{cHW}{\sin\psi}+\frac{\gamma H^2W\cos\psi}{2\tan\psi}\tan\phi$$
5. Factor of safety and simplification. $$FoS=\frac{R}{T}=\frac{\dfrac{cHW}{\sin\psi}+\dfrac{\gamma H^2W\cos\psi}{2\tan\psi}\tan\phi}{\dfrac{\gamma H^2W\sin\psi}{2\tan\psi}}$$ Every term carries a factor of W, which cancels between numerator and denominator — the wedge width, and hence the Joint-B spacing, drops out of the factor of safety entirely. Simplifying the remaining H, γ and trig terms gives the closed form $$\boxed{FoS=\frac{2c+\gamma H\cos^2\psi\tan\phi}{\gamma H\sin\psi\cos\psi}=\frac{2c}{\gamma H\sin2\psi}+\frac{\tan\phi}{\tan\psi}}$$