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
I.
A.
Stresses and Strength
Applies to all sloping surfaces
•
•
Balancing of driving and resisting forces
If Resisting forces > Driving Forces:
stability
Slope Stability
I.
A.
Stresses and Strength
Applies to all sloping surfaces
•
•
B.
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
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
Shear Stress Analysis
Find the
Shear stress
Fn = Fg cos Θ
Fs = Fg sin Θ
What is behind this pretty little box???
Shear Stress Analysis
Fn = Fg cos Θ
Fs = 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 Stress
Sa = W sin β
W = Fg
Fn = Fg cos β
Fs = Fg sin β
Sa = W sin β
Sr = Shear Resistance = (Friction + Cohesion)
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
Factor of Safety
Fs = Sr = W cos β tan ϕ + cL
Sa = W sin β
Sa = shear stress
Sa = W sin β
An Example…..
Factor of Safety
•Slope of 23 degrees
•Angle of internal friction of 30 degrees
•Cohesion of 90 lbs/ft2
Fs = Sr = W cos β
•Soil is 100lbs/ft3
Sa = W sin β
•MO has a distance of 100 ft
•PR has a distance of 22 ft
Determine the factor of safety!!
tan ϕ + cL
An Example…..
Factor of Safety
•Slope of 23 degrees
•Angle of internal friction of 30 degrees
•Cohesion of 90 lbs/ft2
Fs = Sr = W cos β
•Soil is 100lbs/ft3
Sa = W sin β
•MO has a distance of 100 ft
•PR has a distance of 22 ft
Determine the factor of safety!!
tan ϕ + cL
An Example…..
Factor of Safety
•Slope of 23 degrees
•Angle of internal friction of 30 degrees
•Cohesion of 90 lbs/ft2
Fs = Sr = W cos β
•Soil is 100lbs/ft3
Sa = W sin β
•MO has a distance of 100 ft
•PR has a distance of 22 ft
Determine the factor of safety!!
tan ϕ + cL
An Example…..
Factor of Safety
•Slope of 23 degrees
•Angle of internal friction of 30 degrees
•Cohesion of 90 lbs/ft2
Fs = Sr = W cos β
•Soil is 100lbs/ft3
Sa = W sin β
•MO has a distance of 100 ft
•PR has a distance of 22 ft
Determine the factor of safety!!
tan ϕ + cL
An Example…..
Factor of Safety
•Slope of 23 degrees
•Angle of internal friction of 30 degrees
•Cohesion of 90 lbs/ft2
Fs = Sr = W cos β
•Soil is 100lbs/ft3
Sa = W sin β
•MO has a distance of 100 ft
•PR has a distance of 22 ft
Determine the factor of safety!!
tan ϕ + cL
An Example…..
Factor of Safety
•Slope of 23 degrees
•Angle of internal friction of 30 degrees
•Cohesion of 90 lbs/ft2
Fs = Sr = W cos β
•Soil is 100lbs/ft3
Sa = W sin β
•MO has a distance of 100 ft
•PR has a distance of 22 ft
Determine the factor of safety!!
tan ϕ + cL
An Example…..
Factor of Safety
•Slope of 23 degrees
•Angle of internal friction of 30 degrees
•Cohesion of 90 lbs/ft2
Fs = Sr = W cos β
•Soil is 100lbs/ft3
Sa = W sin β
•MO has a distance of 100 ft
•PR has a distance of 22 ft
Determine the factor of safety!!
tan ϕ + cL
An Example…..
Factor of Safety
•Slope of 23 degrees
•Angle of internal friction of 30 degrees
•Cohesion of 90 lbs/ft2
Fs = Sr = W cos β
•Soil is 100lbs/ft3
Sa = W sin β
•MO has a distance of 100 ft
•PR has a distance of 22 ft
Determine the factor of safety!!
tan ϕ + cL
An Example…..
Factor of Safety
•Slope of 23 degrees
•Angle of internal friction of 30 degrees
•Cohesion of 90 lbs/ft2
Fs = Sr = W cos β
•Soil is 100lbs/ft3
Sa = W sin β
•MO has a distance of 100 ft
•PR has a distance of 22 ft
Determine the factor of safety!!
tan ϕ + cL
An Example…..
Factor of Safety
•Slope of 23 degrees
•Angle of internal friction of 30 degrees
•Cohesion of 90 lbs/ft2
Fs = Sr = W cos β
•Soil is 100lbs/ft3
Sa = W sin β
•MO has a distance of 100 ft
•PR has a distance of 22 ft
Determine the factor of safety!!
tan ϕ + cL
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
weight of slice
100
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 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
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
Calculate the weight of each slice…
The Driving Force:
The Driving Force:
(+)
(+)
The Driving Force:
(-)
The Driving Force:
(+)
(+)
(+)
(-)
(-)
(-)
(-/+)
(+)
(+)
(+)
The Driving Force:
Your turn!
50,000 lbs
76,000 lbs
60,000 lbs
22,000 lbs
????
The Resisting Force:
cohesion
slice
weight
slope angle
angle of
Internal friction
The Resisting Force:
cohesion
slope angle
slice
weight
Angle of internal friction: 30 degrees
Cohesion: 50 lbs/ft2
Length of failure plane: 122 ft
angle of
Internal friction
(+)
(+)
(+)
(+)
The Resisting Force:
cohesion
Angle of internal friction: 30 degrees
Cohesion: 50 lbs/ft2
Length of failure plane: 122 ft
Your turn!!
slice
weight
slope angle
angle of
Internal friction
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/ft3
Angle of internal friction (ϕ): 30 degrees
Cohesion (c): 400 lbs/ft2
Another Example:
Unit Weight of Soil (γ): 130 lbs/ft3
Angle of internal friction (ϕ): 30 degrees
Cohesion (c): 400 lbs/ft2
Hc = 2 * c * tan(45 + ϕ/2)
γ
Another Example:
Unit Weight of Soil (γ): 130 lbs/ft3
Angle of internal friction (ϕ): 30 degrees
Cohesion (c): 400 lbs/ft2
Hc = 2 * c * tan(45 + ϕ/2)
γ
Determine the maximum depth of the trench that
will stand with the walls unsupported….
Another Example:
Unit Weight of Soil (γ): 130 lbs/ft3
Angle of internal friction (ϕ): 30 degrees
Cohesion (c): 400 lbs/ft2
Hc = 2 * c * tan(45 + ϕ/2)
γ
Hc = 2 * 400 lbs/ft2 * tan(45 + 30/2)
130 lbs/ft3
Another Example:
Unit Weight of Soil (γ): 130 lbs/ft3
Angle of internal friction (ϕ): 30 degrees
Cohesion (c): 400 lbs/ft2
Hc = 2 * c * tan(45 + ϕ/2)
γ
Hc = 2 * 400 lbs/ft2 * tan(45 + 30/2)
130 lbs/ft3
Hc = 2 * 400 lbs/ft2 * 1.732
130 lbs/ft3
Another Example:
Unit Weight of Soil (γ): 130 lbs/ft3
Angle of internal friction (ϕ): 30 degrees
Cohesion (c): 400 lbs/ft2
Hc = 2 * c * tan(45 + ϕ/2)
γ
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
Yet Another Example:
Unit Weight of Soil (γ): 110 lbs/ft3
Angle of internal friction (ϕ): 24 degrees
Cohesion (c): 600 lbs/ft2
Yet Another Example:
Unit Weight of Soil (γ): 110 lbs/ft3
Angle of internal friction (ϕ): 24 degrees
Cohesion (c): 600 lbs/ft2
Determine the safe depth of a vertical cut for a
Factor of Safety of 2
Yet Another Example:
Unit Weight of Soil (γ): 110 lbs/ft3
Angle of internal friction (ϕ): 24 degrees
Cohesion (c): 600 lbs/ft2
Determine the safe depth of a vertical cut for a
Factor of Safety of 2
Hc = 4 * cd * sin i * cos ϕd
γ (1 – cos(i – ϕd))
Eq. 14.42, Das, 5th edition
….and
Fc = c
cd
Fϕ = ϕ
ϕd
Yet Another Example:
Unit Weight of Soil (γ): 110 lbs/ft3
Angle of internal friction (ϕ): 24 degrees
Cohesion (c): 600 lbs/ft2
Determine the safe depth of a vertical cut for a
Factor of Safety of 2
Hc = 4 * cd * sin i * cos ϕd
γ (1 – cos(i – ϕd))
….and
2 = 600
cd
2 = 24
ϕd
Yet Another Example:
Unit Weight of Soil (γ): 110 lbs/ft3
Angle of internal friction (ϕ): 24 degrees
Cohesion (c): 600 lbs/ft2
Determine the safe depth of a vertical cut for a
Factor of Safety of 2
Hc = 4 * cd * sin i * cos ϕd
γ (1 – cos(i – ϕd))
….and
2 = 600
cd
cd = 300
2 = 24
ϕd
ϕd = 12
Yet Another Example:
Unit Weight of Soil (γ): 110 lbs/ft3
Angle of internal friction (ϕ): 24 degrees
Cohesion (c): 600 lbs/ft2
Determine the safe depth of a vertical cut for a
Factor of Safety of 2
Hc = 4 * cd * sin i * cos ϕd
γ (1 – cos(i – ϕd))
Hc = 4 * 300 lbs/ft2 * sin 90 * cos 12
110 lbs/ft3 (1 – cos(90 – 12))
Yet Another Example:
Unit Weight of Soil (γ): 110 lbs/ft3
Angle of internal friction (ϕ): 24 degrees
Cohesion (c): 600 lbs/ft2
Determine the safe depth of a vertical cut for a
Factor of Safety of 2
Hc = 4 * cd * sin i * cos ϕd
γ (1 – cos(i – ϕd))
Hc = 4 * 300 lbs/ft2 * (1) * (0.978)
110 lbs/ft3 (1 – 0.208)
Yet Another Example:
Unit Weight of Soil (γ): 110 lbs/ft3
Angle of internal friction (ϕ): 24 degrees
Cohesion (c): 600 lbs/ft2
Determine the safe depth of a vertical cut for a
Factor of Safety of 2
Hc = 4 * cd * sin i * cos ϕd
γ (1 – cos(i – ϕd))
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:
Unit Weight of Soil (γ): 110 lbs/ft3
Angle of internal friction (ϕ): 24 degrees
Cohesion (c): 600 lbs/ft2
Determine the safe depth of a vertical cut for a
Factor of Safety of 2
Hc = 4 * cd * sin i * cos ϕd
γ (1 – cos(i – ϕd))
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