1. Slope Stability 4281’ Crozet Tunnel

2. Central America??

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

5. Earthquake Lake, MT, 1959 29 fatalities

6. Nelson County, VA

7. Madison County, VA

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

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

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

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

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

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

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

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

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

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

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

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

24. Sa = shear stress 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!! Factor of Safety Fs = Sr = W cos β tan ϕ + cL Sa = W sin β

31. 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

32. 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

33. 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

34. 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

35. 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!! failure length100 failure height22 volume1100 unit weight100 slope angle23 angle of internal friction30 cohesion90 unit100 weight of slice110000 Sa42980.42 Sr67459.91 Factor of Safety =1.56955

38. Slides: Rotational (slump)

41. III. Rotational Slide—case 1

42. A.The process Determine volume of each slice Determine the weight of each slice

43. III. Rotational Slide—case 1 A.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!

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

45. 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 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…

46. The Driving Force:

47. (+) The Driving Force:

48. (-) The Driving Force:

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

50. 22,000 lbs 60,000 lbs 76,000 lbs 50,000 lbs Your turn!

51. ????

53. The Resisting Force: cohesionslice weight slope angleangle of Internal friction

54. cohesion slice weight slope angle angle of Internal friction Angle of internal friction: 30 degrees Cohesion: 50 lbs/ft2 Length of failure plane: 122 ft The Resisting Force: (+)

55. cohesion slice weight slope angle angle of Internal friction Angle of internal friction: 30 degrees Cohesion: 50 lbs/ft2 Length of failure plane: 122 ft The Resisting Force: Your turn!!

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

59. Factor of Safety Fs = Sr = 100,930 lbs Sa = 47,560 lbs

60. Factor of Safety Fs = Sr = 100,930 lbs Sa = 47,560 lbs Fs = 2.12

61. Another Example: Unit Weight of Soil (γ): 130 lbs/ft3 Angle of internal friction ( ϕ ): 30 degrees Cohesion (c): 400 lbs/ft2

62. Another Example: Hc = 2 * c * tan(45 + ϕ /2) γ Unit Weight of Soil (γ): 130 lbs/ft3 Angle of internal friction ( ϕ ): 30 degrees Cohesion (c): 400 lbs/ft2

63. Another Example: Hc = 2 * c * tan(45 + ϕ /2) γ Unit Weight of Soil (γ): 130 lbs/ft3 Angle of internal friction ( ϕ ): 30 degrees Cohesion (c): 400 lbs/ft2 Determine the maximum depth of the trench that will stand with the walls unsupported….

64. Another Example: Hc = 2 * c * tan(45 + ϕ /2) γ Unit Weight of Soil (γ): 130 lbs/ft3 Angle of internal friction ( ϕ ): 30 degrees Cohesion (c): 400 lbs/ft2 Hc = 2 * 400 lbs/ft2 * tan(45 + 30/2) 130 lbs/ft3

65. Another Example: Hc = 2 * c * tan(45 + ϕ /2) γ Unit Weight of Soil (γ): 130 lbs/ft3 Angle of internal friction ( ϕ ): 30 degrees Cohesion (c): 400 lbs/ft2 Hc = 2 * 400 lbs/ft2 * tan(45 + 30/2) 130 lbs/ft3 Hc = 2 * 400 lbs/ft2 * 1.732 130 lbs/ft3

66. Another Example: Hc = 2 * c * tan(45 + ϕ /2) γ Unit Weight of Soil (γ): 130 lbs/ft3 Angle of internal friction ( ϕ ): 30 degrees Cohesion (c): 400 lbs/ft2 Hc = 2 * 400 lbs/ft2 * tan(45 + 30/2) 130 lbs/ft3 Hc = 2 * 400 lbs/ft2 * 1.732 130 lbs/ft3 Hc = 10.65 ft

67. Yet Another Example: Unit Weight of Soil (γ): 110 lbs/ft3 Angle of internal friction ( ϕ ): 24 degrees Cohesion (c): 600 lbs/ft2

68. Yet Another Example: Determine the safe depth of a vertical cut for a Factor of Safety of 2 Unit Weight of Soil (γ): 110 lbs/ft3 Angle of internal friction ( ϕ ): 24 degrees Cohesion (c): 600 lbs/ft2

69. Yet Another Example: Hc = 4 * c d * sin i * cos ϕ d γ (1 – cos(i – ϕ d )) Determine the safe depth of a vertical cut for a Factor of Safety of 2 Unit Weight of Soil (γ): 110 lbs/ft3 Angle of internal friction ( ϕ ): 24 degrees Cohesion (c): 600 lbs/ft2 ….and Fc = c F ϕ = ϕ c d ϕ d Eq. 14.42, Das, 5 th edition

70. Yet Another Example: Hc = 4 * c d * sin i * cos ϕ d γ (1 – cos(i – ϕ d )) Determine the safe depth of a vertical cut for a Factor of Safety of 2 Unit Weight of Soil (γ): 110 lbs/ft3 Angle of internal friction ( ϕ ): 24 degrees Cohesion (c): 600 lbs/ft2 ….and 2 = 600 2 = 24 c d ϕ d

71. Yet Another Example: Hc = 4 * c d * sin i * cos ϕ d γ (1 – cos(i – ϕ d )) Determine the safe depth of a vertical cut for a Factor of Safety of 2 Unit Weight of Soil (γ): 110 lbs/ft3 Angle of internal friction ( ϕ ): 24 degrees Cohesion (c): 600 lbs/ft2 ….and 2 = 600 2 = 24 c d ϕ d c d = 300 ϕ d = 12

72. Yet Another Example: Hc = 4 * c d * sin i * cos ϕ d γ (1 – cos(i – ϕ d )) Determine the safe depth of a vertical cut for a Factor of Safety of 2 Unit Weight of Soil (γ): 110 lbs/ft3 Angle of internal friction ( ϕ ): 24 degrees Cohesion (c): 600 lbs/ft2 Hc = 4 * 300 lbs/ft2 * sin 90 * cos 12 110 lbs/ft3 (1 – cos(90 – 12))

73. Yet Another Example: Hc = 4 * c d * sin i * cos ϕ d γ (1 – cos(i – ϕ d )) Determine the safe depth of a vertical cut for a Factor of Safety of 2 Unit Weight of Soil (γ): 110 lbs/ft3 Angle of internal friction ( ϕ ): 24 degrees Cohesion (c): 600 lbs/ft2 Hc = 4 * 300 lbs/ft2 * (1) * (0.978) 110 lbs/ft3 (1 – 0.208)

74. Yet Another Example: Hc = 4 * c d * sin i * cos ϕ d γ (1 – cos(i – ϕ d )) Determine the safe depth of a vertical cut for a Factor of Safety of 2 Unit Weight of Soil (γ): 110 lbs/ft3 Angle of internal friction ( ϕ ): 24 degrees Cohesion (c): 600 lbs/ft2 Hc = 4 * 300 lbs/ft2 * (1) * (0.978) 110 lbs/ft3 (1 – 0.208) Hc = 1173.5 lbs/ft2 87.12 lbs/ft3

75. Yet Another Example: Hc = 4 * c d * sin i * cos ϕ d γ (1 – cos(i – ϕ d )) Determine the safe depth of a vertical cut for a Factor of Safety of 2 Unit Weight of Soil (γ): 110 lbs/ft3 Angle of internal friction ( ϕ ): 24 degrees Cohesion (c): 600 lbs/ft2 Hc = 4 * 300 lbs/ft2 * (1) * (0.978) 110 lbs/ft3 (1 – 0.208) Hc = 1173.5 lbs/ft2 87.12 lbs/ft3 Hc = 13.5 ft