Lecture 17: The Role of Water in Slope Stability

Lecture 17: The Role of Water in Slope Stability Key Questions 1. How does the weight of water influence the factor of safety? 2. What does too much w...
Author: Tracy Barker
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Lecture 17: The Role of Water in Slope Stability Key Questions 1. How does the weight of water influence the factor of safety? 2. What does too much water do to the cohesion strength of the sediment? 3. How does water interact with clay minerals in the sediment? 4. How does water decrease the “friction force” and hence the factor of safety?

La Conchita, CA

Slope Model

. W

θ

FS  = 

resisting forces driving forces

W x cosθ x μ + c

friction + cohesion



= Fp

The slope fails if FS is less than “1”

.

W

Friction = Ff

FN

θ

Fp

cohesion = c

θ

W x sinθ

Water makes slopes unstable

1. Water adds weight

. W

θ

One gallon of water weighs 8 lbs

If a 100 by 200 foot slope soaks up 2 inches of rain, the slope weight increases by 200,000 lbs

Burien, WA near Shorewood Drive SW and 131st Street

FS  = 

resisting forces driving forces

W x cosθ x μ + c

friction + cohesion



= W x sinθ

Fp

water increases the weight of the slope  which increases the driving force Fp

.

W

Friction = Ff

FN

θ

Fp

cohesion = c

θ

The slope fails if FS is less than “1”

2. Too much water in the voids spaces reduces or eliminates the cohesion force holding the grains together. air water

grain Remember, the cohesion force requires air and water in the pores spaces.

FS  = 

resisting forces driving forces

W x cosθ x μ + c

friction + cohesion



= Fp

W x sinθ

too much water decreases cohesion

.

W

Friction = Ff

FN

θ

Fp

cohesion = c

θ

The slope fails if FS is less than “1”

3. Slopes with high clay contents are more unstable when wet.

. W

θ

Scanning electron microscope image of clay

Clay minerals have a “negative” surface charge

_________________________________

water is polar __

+

+

clay minerals attract water molecules to their surfaces

_________________________________

dry, more friction, hence more resistant to sliding

wet, less friction, hence lower strength

FS  = 

resisting forces driving forces

W x cosθ x μ + c

friction + cohesion



= Fp

wet clay reduces friction

.

W

Friction = Ff

FN

θ

Fp

cohesion = c

θ

The slope fails if FS is less than “1”

W x sinθ

4. Water buoys up the grains which reduces the friction force

. W

θ

Slope Model

. W

θ

shear plane

weight

friction force

FN = W x cosθ = normal force

.

W

θ

cosθ =

adjacent hypotenuse θ

weight

FN

friction force = normal force x coefficient of friction or friction force = FN x μ

friction force

What happens when the grains are immersed in water?

friction force

water surface

= P = water pressure ρw = water density g = acceleration of gravity

h

h = depth to the balloon

P = ρw·g·h = hydrostatic pressure

water surface

= P = water pressure ρw = water density g = acceleration of gravity

h1

h2

h = depth to the balloon

P1 = ρw·g·h1

P2 = ρw·g·h2

Cubic grain immersed in water (side length “L”)

Fluid pressure at the top = P = ρgh

L Fluid pressure at the bottom = P = ρg(h +L)

ρ = water density g = acceleration of gravity h = depth below the water surface

Cubic grain immersed in water (side length “L”)

Fluid pressure at the top = P = ρgh

ΔP = ρgL = change in pressure L Fluid pressure at the bottom = P = ρg(h +L)

ρ = water density g = acceleration of gravity h = depth below the water surface

Cubic grain immersed in water (side length “L”)

Fluid pressure at the top = P = ρgh L ΔP = ρgL = change in pressure

Fluid pressure at the bottom = P = ρg(h +L) Area = L2 Force = pressure x area

ρ = water density g = acceleration of gravity h = depth below the water surface

Cubic grain immersed in water (side length “L”)

Fluid pressure at the top = P = ρgh L ΔP = ρgL = change in pressure

Fluid pressure at the bottom = P = ρg(h +L) Area = L2 Force = ΔP x L2

ρ = water density g = acceleration of gravity h = depth below the water surface

Cubic grain immersed in water (side length “L”)

Fluid pressure at the top = P = ρgh L ΔP = ρgL = change in pressure

Fluid pressure at the bottom = P = ρg(h +L) Area = L2 Force = ρgL3 = ρgV = buaoyancy force = FB

Cubic grain immersed in water

FB = ρgV = the magnitude of buoyancy force is equal to the weight of the volume of fluid the grain displaces (Archimedes’ Principle)

Cubic grain immersed in water

The grain weight is reduced by FB

FB

Wnew = Wdry – buoyancy force

buoyancy force

buoyancy force = the weight of the volume of water the grain displaces

The total weight pushing on the shear plane is reduced

the weight of each grain is reduced by the buoyancy force

friction force

weight

FN

friction force = normal force x coefficient of friction or friction force = FN x μ

friction force

FS  = 

resisting forces driving forces

W x cosθ x μ + c

friction + cohesion



= Fp

water decreases friction and cohesion and increases Fp all of which reduce FS

.

W

Friction = Ff

FN

θ

Fp

cohesion = c

θ

The slope fails if FS is less than “1”

W x sinθ

FS  = 

resisting forces driving forces

friction + cohesion

=  Fp

FS  = 

resisting forces driving forces

friction + cohesion

=  Fp

FS  = 

resisting forces driving forces

friction + cohesion

=  Fp

FS  = 

resisting forces driving forces

friction + cohesion

=  Fp

FS  = 

resisting forces driving forces

friction + cohesion

=  Fp

FS  = 

resisting forces driving forces

friction + cohesion

=  Fp

USGS Experimental Debris –Flow Flume

Seattle Area, Washington Cumulative Precipitation Threshold

Seattle Area, Washington Precipitation Intensity-Duration Threshold

The Pe Ell landslide occurred on December 3rd, 2007, blocking State Highway 6 and  destroying three structures http://www.dnr.wa.gov/ResearchScience/Topics/GeologicHazardsMapping/Pages/landslides_dec07storm.aspx

Dec 11, 2004 - Massive landslide on the Sultan River, CA

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