Water pressure and blast vibration on a fault-controlled pit wall
A copper-gold open pit is 120 m deep and planned to 180 m. The west wall is cut in 12 m benches with 70° bench faces and a 45° overall angle. A continuous fault dips 38° into the pit and daylights in the wall; its residual friction angle is 35° and cohesion is 25 kPa. A 60 m long, 8 m thick wedge of rock (2D section, unit weight 25 kN/m³) is identified as a potential planar failure. The fault carries water, and a piezometer indicates an uplift force of 1,000 kN per metre of strike along the failure plane. Production blasting 250 m away produces a peak particle velocity of 12 mm/s at the wall. The site acceptance criterion is a factor of safety of 1.3 for permanent slopes. The geotechnical engineer must decide whether the wall can be left as designed while a pushback is delayed by six months.
1. Calculate the factor of safety for the wedge in the dry condition and with the measured water pressure. State whether the wall meets the acceptance criterion.
Read the worked answer
For a 2D wedge, weight per metre of strike W = 0.5 × 60 × 8 × 25 = 6,000 kN/m. The driving component is W sin 38° = 6,000 × 0.6157 = 3,694 kN/m. The normal component is W cos 38° = 6,000 × 0.7880 = 4,728 kN/m. Cohesion resistance is cL = 25 × 60 = 1,500 kN/m. Dry factor of safety: FS = [cL + W cos 38° tan 35°] / W sin 38° = [1,500 + 4,728 × 0.700] / 3,694 = (1,500 + 3,310) / 3,694 = 1.30. With uplift U = 1,000 kN/m, the effective normal force is 4,728 − 1,000 = 3,728 kN/m. FS = [1,500 + 3,728 × 0.700] / 3,694 = (1,500 + 2,610) / 3,694 = 1.11. The dry wall just meets 1.3; the water pressure reduces it to 1.11, below the criterion. The 12 mm/s blast vibration is also high for a permanent wall and can further reduce effective cohesion or disturb the fault, so the wall should not be left as designed for six months without treatment.
2. Evaluate whether dewatering alone can restore a factor of safety of 1.3. Recommend practical measures and support the recommendation with a calculation.
Read the worked answer
Dewatering alone cannot provide the required margin. To reach FS = 1.3, the numerator must be 1.3 × 3,694 = 4,802 kN/m. With cohesion of 1,500 kN/m, the friction term must be 3,302 kN/m, so the effective normal force must be 3,302 / 0.700 = 4,717 kN/m. Since the saturated normal force is 4,728 kN/m, the allowable uplift is only about 11 kN/m, essentially complete drainage. That is not a practical operating target with a 60 m long fault and a 1,000 kN/m uplift. Practical measures are therefore a combination: install horizontal drains or depressurisation wells to reduce uplift as far as possible; flatten the wall or remove the upper wedge to reduce driving force; add a toe buttress if space allows; and control blasting near the wall. A pushback delay of six months should be conditional on achieving a calculated FS of at least 1.3 with the revised geometry and measured pore pressure, not on dewatering alone.
3. Outline a monitoring and trigger-action plan for the six-month delay.
Read the worked answer
Monitor movement and pressure, not just visual inspection. Install survey prisms on the wall and bench crests, wireline extensometers across the fault, and piezometers in the fault and wall rock. Record blast vibration at the wall for every production blast and correlate it with movement. Inspect the fault trace and tension cracks after each blast and after heavy rain. Set trigger levels in advance: for example, a defined rate of prism movement, a defined cumulative displacement, a rise in piezometric level, or vibration above a set value should trigger a stop, re-assessment and remedial work. Rainfall should be logged because it raises pore pressure quickly. The plan should identify who reviews the data, how often, and what action follows each trigger. If movement accelerates or the calculated FS falls below 1.3, stop the wall and implement the buttress or pushback rather than relying on monitoring alone.