Open-pit slope with a daylighting fault and water pressure
An iron ore open pit is 150 m deep with 12 m benches. In a 60 m high lower sector, the overall slope angle is 52°. A persistent fault dips 40° toward the pit and daylights in the lower bench. The fault has cohesion 25 kPa and friction angle 32°; rock unit weight is 26 kN/m3. A tension crack 25 m behind the crest and seepage from the fault indicate water pressure. Simplified planar analysis for a 45 m high wedge gives a dry factor of safety of about 1.0. The mine requires a design factor of safety of at least 1.3. Flattening the sector to 45° would increase waste by 0.8 Mm3; installing anchors would require about 2,200 kN/m of additional resisting force. The pit is scheduled to reach the sector in eight months. The geotechnical engineer must choose a design before the next pushback.
1. Calculate the dry factor of safety for the 45 m wedge and explain how water and the tension crack change it.
Read the worked answer
For a planar wedge, W = 0.5 x unit weight x H^2 x (cot plane dip - cot slope angle). Here W = 0.5 x 26 x 45^2 x (cot 40° - cot 52°) = 10.8 MN/m. Driving force = W sin 40° = 6.95 MN/m. Normal force = W cos 40° = 8.28 MN/m. Friction resistance = 8.28 x tan 32° = 5.17 MN/m. Cohesion area per metre = H / sin 40° = 70.0 m, so cohesion resistance = 25 x 70.0 = 1.75 MN/m. Dry factor of safety = (1.75 + 5.17) / 6.95 = 1.00.
Water in the tension crack and fault reduces the effective normal stress and adds a driving component, so the factor of safety falls below 1.0. A dry value of 1.0 is therefore unacceptable against the required 1.3.
2. Compare flattening to 45° with anchoring. Which do you recommend and why?
Read the worked answer
Flattening to 45° gives W = 0.5 x 26 x 45^2 x (cot 40° - cot 45°) = 5.05 MN/m. Driving force = 3.25 MN/m; normal force = 3.87 MN/m; friction = 3.87 x tan 32° = 2.42 MN/m; cohesion remains 1.75 MN/m. Dry factor of safety = (1.75 + 2.42) / 3.25 = 1.28, which exceeds 1.3 once drainage is installed.
For anchoring, required resistance is 1.3 x 6.95 = 9.04 MN/m. Existing dry resistance is 6.92 MN/m, so additional force is about 2.12 MN/m or 2,120 kN/m, close to the stated 2,200 kN/m. Anchors do not remove water, and the required force is large. Recommend flattening to 45° plus horizontal drains and seepage control; use anchors only locally if flattening is impossible. The extra 0.8 Mm3 waste is justified by the stability requirement.
3. What monitoring would you require during excavation?
Read the worked answer
Install slope prisms and an automated total station, inclinometers through the fault, piezometers and crack meters across the tension crack. Monitor rainfall and seepage flow. Set trigger levels for displacement rate, pore pressure and crack width, with a stop-work and evacuation plan. Survey the fault trace as each bench is exposed to confirm its persistence and dip. Review the factor of safety after each pushback and after heavy rain, because water pressure is the main short-term variable.