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Slope Stability

Slope Stability. Crozet Tunnel. 4281’. Central America??. Gros Ventre Slide, WY, 1925 (pronounced “grow vahnt”). 50 million cubic yards. Earthquake Lake, MT, 1959. 29 fatalities. Nelson County, VA. Madison County, VA. Slope Stability. Stresses and Strength

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Slope Stability

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  1. Slope Stability Crozet Tunnel 4281’

  2. Central America??

  3. Gros Ventre Slide, WY, 1925 (pronounced “grow vahnt”) 50 million cubic yards

  4. Earthquake Lake, MT, 1959 29 fatalities

  5. Nelson County, VA

  6. Madison County, VA

  7. Slope Stability • Stresses and Strength • Applies to all sloping surfaces • Balancing of driving and resisting forces • If Resisting forces > Driving Forces: stability

  8. Slope Stability • Stresses and Strength • Applies to all sloping surfaces • Balancing of driving and resisting forces • If Resisting forces > Driving Forces: stability • Engineering Approach • Delineate the surface that is most at risk • Calculate the stresses • Calculate the Shear Strength

  9. A Friendly Review From Last Month…… Stress on an inclined plane to Force σ = Force / Area Where is Normal Force and Shear Force = ?? cos Θ= a =Fn h = Fg sin Θ = o = Fs h=Fg Fn = Fg cos Θ Fs = Fg sin Θ

  10. Shear Stress Analysis Fn = Fg cos Θ Fs = Fg sin Θ Find the Shear stress What is behind this pretty little box???

  11. Shear Stress Analysis Fn = Fg cos Θ Fs = Fg sin Θ

  12. Consider a planar slide whose failure surface is ‘linear’…..

  13. II. Planar Slide—case 1 Volume of Slice = MO x PR x 0.5 x 1 ft

  14. II. Planar Slide—case 1 Volume of Slice = MO x PR x 0.5 x 1 ft

  15. Sa = Shear Stress Sa = W sin β W = Fg Fn = Fg cos β Fs = Fg sin β

  16. Sa = W sin β Sr = Shear Resistance = (Friction + Cohesion)

  17. Sa = W sin β Fn = Fg cos β Fs = Fg sin β Sr = Friction + Cohesion = W cos β tan ϕ + c * (segment MO) Sr = W cos β tan ϕ + cL

  18. Factor of Safety Fs = Sr = W cos β tan ϕ + cL Sa = W sin β

  19. Sa = shear stress Sa = W sin β

  20. An Example….. • Slope of 23 degrees • Angle of internal friction of 30 degrees • Cohesion of 90 lbs/ft2 • Soil is 100lbs/ft3 • MO has a distance of 100 ft • PR has a distance of 22 ft • Determine the factor of safety!! Factor of Safety Fs = Sr = W cos β tan ϕ + cL Sa = W sin β

  21. An Example….. • Slope of 23 degrees • Angle of internal friction of 30 degrees • Cohesion of 90 lbs/ft2 • Soil is 100lbs/ft3 • MO has a distance of 100 ft • PR has a distance of 22 ft • Determine the factor of safety!! Factor of Safety Fs = Sr = W cos β tan ϕ + cL Sa = W sin β

  22. An Example….. • Slope of 23 degrees • Angle of internal friction of 30 degrees • Cohesion of 90 lbs/ft2 • Soil is 100lbs/ft3 • MO has a distance of 100 ft • PR has a distance of 22 ft • Determine the factor of safety!! Factor of Safety Fs = Sr = W cos β tan ϕ + cL Sa = W sin β

  23. An Example….. • Slope of 23 degrees • Angle of internal friction of 30 degrees • Cohesion of 90 lbs/ft2 • Soil is 100lbs/ft3 • MO has a distance of 100 ft • PR has a distance of 22 ft • Determine the factor of safety!! Factor of Safety Fs = Sr = W cos β tan ϕ + cL Sa = W sin β

  24. An Example….. • Slope of 23 degrees • Angle of internal friction of 30 degrees • Cohesion of 90 lbs/ft2 • Soil is 100lbs/ft3 • MO has a distance of 100 ft • PR has a distance of 22 ft • Determine the factor of safety!! Factor of Safety Fs = Sr = W cos β tan ϕ + cL Sa = W sin β

  25. An Example….. • Slope of 23 degrees • Angle of internal friction of 30 degrees • Cohesion of 90 lbs/ft2 • Soil is 100lbs/ft3 • MO has a distance of 100 ft • PR has a distance of 22 ft • Determine the factor of safety!! Factor of Safety Fs = Sr = W cos β tan ϕ + cL Sa = W sin β

  26. An Example….. • Slope of 23 degrees • Angle of internal friction of 30 degrees • Cohesion of 90 lbs/ft2 • Soil is 100lbs/ft3 • MO has a distance of 100 ft • PR has a distance of 22 ft • Determine the factor of safety!! Factor of Safety Fs = Sr = W cos β tan ϕ + cL Sa = W sin β

  27. An Example….. • Slope of 23 degrees • Angle of internal friction of 30 degrees • Cohesion of 90 lbs/ft2 • Soil is 100lbs/ft3 • MO has a distance of 100 ft • PR has a distance of 22 ft • Determine the factor of safety!! Factor of Safety Fs = Sr = W cos β tan ϕ + cL Sa = W sin β

  28. An Example….. • Slope of 23 degrees • Angle of internal friction of 30 degrees • Cohesion of 90 lbs/ft2 • Soil is 100lbs/ft3 • MO has a distance of 100 ft • PR has a distance of 22 ft • Determine the factor of safety!! Factor of Safety Fs = Sr = W cos β tan ϕ + cL Sa = W sin β

  29. An Example….. • Slope of 23 degrees • Angle of internal friction of 30 degrees • Cohesion of 90 lbs/ft2 • Soil is 100lbs/ft3 • MO has a distance of 100 ft • PR has a distance of 22 ft • Determine the factor of safety!! Factor of Safety Fs = Sr = W cos β tan ϕ + cL Sa = W sin β

  30. An Example….. • Slope of 23 degrees • Angle of internal friction of 30 degrees • Cohesion of 90 lbs/ft2 • Soil is 100lbs/ft3 • MO has a distance of 100 ft • PR has a distance of 22 ft • Determine the factor of safety!!

  31. Slides: Rotational (slump)

  32. III. Rotational Slide—case 1

  33. III. Rotational Slide—case 1 • The process • Determine volume of each slice • Determine the weight of each slice

  34. III. Rotational Slide—case 1 • The process • Determine volume of each slice • Determine the weight of each slice • Calculate the driving and resisting forces • of each slice • Sum ‘em up and let it rip!

  35. III. Rotational Slide—case 1 “should use a minimum of 6 slices”

  36. For Slice 1: 11’ x 20’ x 1’ = 220 ft3 220 ft3 x 100 lbs/ft3 = 22,000 lbs. For Slice 2: 30’ x 20’ x 1’ = 600 ft3 600 ft3 x 100 lbs/ft3 = 60,000 lbs Calculate the weight of each slice… For Slice 3: 38’ x 20’ x 1’ = 760 ft3 760 ft3 x 100 lbs/ft3 = 76,000 lbs For Slice 4: 25’ x 20’ x 1’ = 500 ft3 500 ft3 x 100 lbs/ft3 = 50,000 lbs

  37. The Driving Force:

  38. The Driving Force: (+) (+)

  39. The Driving Force: (-)

  40. The Driving Force: (+) (+) (+) (+) (+) (-) (+) (-) (-) (-/+)

  41. The Driving Force: Your turn! 50,000 lbs 76,000 lbs 60,000 lbs 22,000 lbs

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