Geomechanics and Ground Control · Ground Control · 2 marks · Numerical answer
The 20 m high rock slope shown has a 45° face and a potential sliding joint at 30°. Rock unit weight is 20 kN/m³, joint cohesion 30 kPa and friction angle 22°. Find sliding safety factor, to two decimals. Assume a dry unsupported slope.
Work per metre perpendicular to the section. The upper wedge width is $20/\tan30^\circ-20=14.641$ m. Wedge area is half this width times the 20 m height, or 146.410 m², giving weight W = 2928.203 kN. Joint length is $20/\sin30^\circ=40$ m. Driving force is $W\sin30^\circ=1464.102$ kN. Resistance is $30(40)+W\cos30^\circ\tan22^\circ=2224.633$ kN. Divide resistance by driving force to obtain safety factor 1.51941, or 1.52.
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