Slope Stability

4281’

Crozet Tunnel

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. Stresses and StrengthA. 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:

stabilityB. 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