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Slope Stability
4281’
Crozet Tunnel
Central America??
Gros Ventre Slide, WY, 1925(pronounced “grow vahnt”)
50 million cubic yards
Nelson County, VA
Madison County, VA
Slope Stability
I. Stresses and Strength
A. Applies to all sloping surfaces• Balancing of driving and resisting forces• If Resisting forces > Driving Forces:
stability
Slope Stability
I. Stresses and StrengthA. 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
Stress on an inclined plane to Force
σ = Force / Area
Where is Normal Force and Shear Force = ??
Fn = Fg cos ΘFs = Fg sin Θ
cos Θ = a = Fn h = Fg
sin Θ = o = Fs h = Fg
A Friendly Review From Last Month……
Shear Stress Analysis
What is behind this pretty little box???
Fn = Fg cos ΘFs = Fg sin Θ
Find theShear stress
Shear Stress Analysis
Fn = Fg cos ΘFs = Fg sin Θ
Consider a planar slide whose failure surface is ‘linear’…..
II. Planar Slide—case 1
Volume of Slice =MO x PR x 0.5 x 1 ft
II. Planar Slide—case 1
Volume of Slice =MO x PR x 0.5 x 1 ft
Sa = Shear StressSa = W sin β
Fn = Fg cos βFs = Fg sin β
W = Fg
Sa = W sin β
Sr = Shear Resistance = (Friction + Cohesion)
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 βFs = Fg sin β
Factor of Safety
Fs = Sr = W cos β tan ϕ + cL Sa = W sin β
Sa = shear stressSa = W sin β
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
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
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
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
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
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
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
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
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
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
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 length 100
failure height 22
volume 1100
unit weight 100
slope angle 23
angle of internal friction 30
cohesion 90
unit 100
weight of slice 110000
Sa 42980.42
Sr 67459.91
Factor of Safety = 1.56955
Slides: Rotational (slump)
III. Rotational Slide—case 1
III. Rotational Slide—case 1
A. The process• Determine volume of each slice• Determine the weight of each slice
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!
III. Rotational Slide—case 1
“should use a minimum of 6 slices”
For Slice 3:38’ x 20’ x 1’ = 760 ft3760 ft3 x 100 lbs/ft3 = 76,000 lbs
For Slice 4:25’ x 20’ x 1’ = 500 ft3500 ft3 x 100 lbs/ft3 = 50,000 lbs
For Slice 1:11’ x 20’ x 1’ = 220 ft3220 ft3 x 100 lbs/ft3 = 22,000 lbs.
For Slice 2:
30’ x 20’ x 1’ = 600 ft3600 ft3 x 100 lbs/ft3 = 60,000 lbs
Calculate the weight of each slice…
The Driving Force:
(+)
(+)
The Driving Force:
(-)
The Driving Force:
(-) (-) (-) (-/+) (+)(+)
(+)(+)
(+)
(+)
The Driving Force:
The Driving Force:
22,000 lbs
60,000 lbs
76,000 lbs
50,000 lbs
Your turn!
????
The Resisting Force:
cohesion sliceweight
slope angle angle of Internal friction
cohesion sliceweight
slope angle angle of Internal friction
Angle of internal friction: 30 degreesCohesion: 50 lbs/ft2Length of failure plane: 122 ft
The Resisting Force:
(+)
(+)
(+)
(+)
cohesion sliceweight
slope angle angle of Internal friction
Angle of internal friction: 30 degreesCohesion: 50 lbs/ft2Length of failure plane: 122 ft
The Resisting Force:
Your turn!!
Factor of Safety
Fs = Sr = W cos β tan ϕ + cL Sa = W sin β
Factor of Safety
Fs = Sr = 100,930 lbs Sa = 47,560 lbs
Factor of Safety
Fs = Sr = 100,930 lbs Sa = 47,560 lbs
Fs = 2.12
Another Example:
Unit Weight of Soil (γ): 130 lbs/ft3Angle of internal friction (ϕ): 30 degreesCohesion (c): 400 lbs/ft2
Another Example:
Hc = 2 * c * tan(45 + ϕ/2) γ
Unit Weight of Soil (γ): 130 lbs/ft3Angle of internal friction (ϕ): 30 degreesCohesion (c): 400 lbs/ft2
Another Example:
Hc = 2 * c * tan(45 + ϕ/2) γ
Unit Weight of Soil (γ): 130 lbs/ft3Angle of internal friction (ϕ): 30 degreesCohesion (c): 400 lbs/ft2
Determine the maximum depth of the trench thatwill stand with the walls unsupported….
Another Example:
Hc = 2 * c * tan(45 + ϕ/2) γ
Unit Weight of Soil (γ): 130 lbs/ft3Angle of internal friction (ϕ): 30 degreesCohesion (c): 400 lbs/ft2
Hc = 2 * 400 lbs/ft2 * tan(45 + 30/2) 130 lbs/ft3
Another Example:
Hc = 2 * c * tan(45 + ϕ/2) γ
Unit Weight of Soil (γ): 130 lbs/ft3Angle of internal friction (ϕ): 30 degreesCohesion (c): 400 lbs/ft2
Hc = 2 * 400 lbs/ft2 * tan(45 + 30/2) 130 lbs/ft3Hc = 2 * 400 lbs/ft2 * 1.732 130 lbs/ft3
Another Example:
Hc = 2 * c * tan(45 + ϕ/2) γ
Unit Weight of Soil (γ): 130 lbs/ft3Angle of internal friction (ϕ): 30 degreesCohesion (c): 400 lbs/ft2
Hc = 2 * 400 lbs/ft2 * tan(45 + 30/2) 130 lbs/ft3Hc = 2 * 400 lbs/ft2 * 1.732 130 lbs/ft3
Hc = 10.65 ft
Yet Another Example:
Unit Weight of Soil (γ): 110 lbs/ft3Angle of internal friction (ϕ): 24 degreesCohesion (c): 600 lbs/ft2
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/ft3Angle of internal friction (ϕ): 24 degreesCohesion (c): 600 lbs/ft2
Yet Another Example:
Hc = 4 * cd * sin i * cos ϕd γ (1 – cos(i – ϕd))
Determine the safe depth of a vertical cut for a Factor of Safety of 2
Unit Weight of Soil (γ): 110 lbs/ft3Angle of internal friction (ϕ): 24 degreesCohesion (c): 600 lbs/ft2
….and Fc = c Fϕ = ϕ
cd ϕd
Eq. 14.42, Das, 5th edition
Yet Another Example:
Hc = 4 * cd * sin i * cos ϕd γ (1 – cos(i – ϕd))
Determine the safe depth of a vertical cut for a Factor of Safety of 2
Unit Weight of Soil (γ): 110 lbs/ft3Angle of internal friction (ϕ): 24 degreesCohesion (c): 600 lbs/ft2
….and 2 = 600 2 = 24
cd ϕd
Yet Another Example:
Hc = 4 * cd * sin i * cos ϕd γ (1 – cos(i – ϕd))
Determine the safe depth of a vertical cut for a Factor of Safety of 2
Unit Weight of Soil (γ): 110 lbs/ft3Angle of internal friction (ϕ): 24 degreesCohesion (c): 600 lbs/ft2
….and 2 = 600 2 = 24
cd ϕd
cd = 300 ϕd = 12
Yet Another Example:
Hc = 4 * cd * sin i * cos ϕd γ (1 – cos(i – ϕd))
Determine the safe depth of a vertical cut for a Factor of Safety of 2
Unit Weight of Soil (γ): 110 lbs/ft3Angle of internal friction (ϕ): 24 degreesCohesion (c): 600 lbs/ft2
Hc = 4 * 300 lbs/ft2 * sin 90 * cos 12 110 lbs/ft3 (1 – cos(90 – 12))
Yet Another Example:
Hc = 4 * cd * sin i * cos ϕd γ (1 – cos(i – ϕd))
Determine the safe depth of a vertical cut for a Factor of Safety of 2
Unit Weight of Soil (γ): 110 lbs/ft3Angle of internal friction (ϕ): 24 degreesCohesion (c): 600 lbs/ft2
Hc = 4 * 300 lbs/ft2 * (1) * (0.978) 110 lbs/ft3 (1 – 0.208)
Yet Another Example:
Hc = 4 * cd * sin i * cos ϕd γ (1 – cos(i – ϕd))
Determine the safe depth of a vertical cut for a Factor of Safety of 2
Unit Weight of Soil (γ): 110 lbs/ft3Angle of internal friction (ϕ): 24 degreesCohesion (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
Yet Another Example:
Hc = 4 * cd * sin i * cos ϕd γ (1 – cos(i – ϕd))
Determine the safe depth of a vertical cut for a Factor of Safety of 2
Unit Weight of Soil (γ): 110 lbs/ft3Angle of internal friction (ϕ): 24 degreesCohesion (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