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Calculations and Summary of Results <br />Proposed 2:1 Slope, Pit and Slope Water <br />Level are Equal Slope Height <br />(ft) Slope Run <br />(ft) <br />Tan-8 Calculated <br />Slope Angle (e) <br />Degrees Friction Angle <br />(0) Degrees Factor of Safety for Infinite Slope = <br />tanO/tang <br />Proposed Excavation Limit 2:1 15 30 0.50 26.6 34 1.35 <br />Sloe to Concrete Ditch 15 55 0.27 15.3 34 2.47 <br />Slope to Slurry Wall 15 120 0.13 7.1 34 5.40 <br /> <br />Full Saturated Slope <br />Pit Dry <br />Slope Height <br />Slope Run <br />Tan 9 Calculated <br />Slope Angle (e) <br />Friction Angle Factor of Safety for Saturated <br />Circular Slope Failure, (b = 34° <br />, (ft) (ft) Degrees (?) Degrees see curves and calculations) <br />Slope at Excavation Limit 15 43.6 0.34 19.0 34 1.00 <br />Sloe to Concrete Ditch 15 55 0.27 15.3 34 1.35 <br />Slope to Slurry Wall 15 120 0.13 7.1 34 2.25 <br />Calculations for Factor of Safety (F) = 1 Tan 34/1 = 0.674507788 From Chart 5, e = 19 degrees <br />Tane=Tan 19=0.34 <br />Therefore a 15 foot slope height has a 43.6 slope run <br />Calculate F for Concrete Ditch Slope Height = 15 feet, distance from toe to concrete ditch is 55 feet, therefore slope angle e = 15.3 degrees <br />From Chart 5, Tan O/F = 0.5 for 9 = 15 degrees, solving for F, F = 1.35 <br />Calculate F for Slurry Wall Slope Height = 15 feet, distance from toe to concrete ditch is 120 feet, therefore slope angle 9 = 7.125 degrees <br />From Chart 5, Tan O/F = 0.3 for 9 = 7 degrees, solving for F, F = 2.25 <br />