Friction and Basic Slope Stability, Landslide Phenomena

Friction and basic slope stability, landslide phenomena

Robert C. Viesca and James R. Rice, Harvard University

11 June 2008 (based on notes for Mon 5 Nov 2007 lecture)

1. We introduced the notion of a static coeﬃcient of friction, µ:

We consider a block resting on a flat surface with force N normal to the block and surface and we ask what force T may we apply to the block tangential to the surface without causing motion? Introducing a static coefficient of friction mu that defines the maximum T : Tmax = µN. We must note that this statement is independent of area: if we keep N constant but change the proportions of the block so that the area facing the flat surface is now twice that before, then still Tmax = µN.

2. We describe the static coeﬃcient of friction as the tangent of a friction angle, φ :

Drawing a FBD of the unmoving block on the ﬂat surface, we have Tmax and N from (1) and a resisting force from the ﬂat surface with normal and tangential components equal in magnitude to and in the opposite directions of Tmax and N. The sum of these vectors creates a new vector inclined at an angle φ from vertical. This angle is the friction angle.

3. What is the maximum slope, θmax, upon which we can place a block whose friction angle with the surface, φ, is known?

The downslope component of the block’s weight is T and the slope-normal component is N. For a slope of inclination angle, theta, the downslope component of the force is T = N tan θ. If the maximum force allowable to keep the block static is Tmax = N tan φ, setting T = Tmax, we ﬁnd that θmax = φ.

4. Extension to inﬁnite slope:

Now instead of a block on an inclined surface, we consider a ﬂat slope, inclined at angle theta, with a potential failure plane parallel to the slope surface. We now have a “block” of soil of thickness b that inﬁnitely extends upslope and downslope. We consider the slope to be composed a cohesionless, granular soil.

1 Except, before where we would work in terms of force and write Tmax = N tan φ. Except, considering now we no longer have a discrete body, we can work in terms of stress (still maintaining that independence of area in our friction), writing τmax = σn tan φ, where σn is the stress normal to the potential failure plane and τmax is the maximum allowable shear stress for which the slope would be stationary. In the case where ﬂuid is present that statement would change to include the eﬀective normal stress σ¯n = σn − p where p is the pore pressure.

5. Maximum inclination angle for slope under “dry” conditions.

Consider a slope for which the water table is lower than the potential failure plane over the region considered. The soil has a speciﬁc weight γd (d indicating “dry”) and an friction angle φ.

Aside: If the failure plane occurs within the same soil material, this friction angle would be the friction angle “internal” to the soil, or “internal friction angle”; otherwise, if the potential plane is at an interface of two diﬀerent materials, the φ is merely the friction angle between material x and y)

Considering a column of soil with unit area extending from the surface to the failure plane (thickness b), the shear stress and the normal stress on the surface can be written as

τ = γdb sin θ

σn = γdb cos θ

or τ = σn tan θ so if (from (4)) τmax = σn tan φ then θmax = φ and the slope can be inclined as steeply as the friction angle along that failure surface.

6. Maximum inclination angle for slope saturated after heavy rain with downslope water ﬂux.

Now we consider a slope (again at some inclination θ) that has been saturated to the ground surface after much rain (any surface water has been able to runoﬀ, i.e., there is no

2 ponding). Additionally, an impermeable later exists parallel to the surface at some depth below a proposed failure plane, so that all water flux down slope is parallel to the slope. The soil has some specific weight γ (greater than γd to account for fluid saturation).

Consequently, lines of constant head run as a straight line between and perpendicular to the free surface and the impermeable surface. Taking the atmospheric pressure to be 0 as a reference value, the head on the free surface is merely zel.

We then choose an arbitrary datum at some point on the free surface inclined at an angle θ from the free surface (i.e., the datum is simply a horizontal line). At this datum point on the free surface, h = 0 and there is a line of constant head extending perpendicular to the free surface. To ﬁnd the pore pressure at the failure plane a distance b along this line of constant head we need consider h = 0 = zel + p/γw (or p = −γwzel), where zel is the vertical distance to the failure plane as measured from datum: zel = −b cos θ. Thus p|z=b = γwb cos θ.

Similar to the dry-slope case (but with diﬀerent speciﬁc weights), we can calculate the total normal stress at the surface as

σn = γb cos θ and the eﬀective normal stress at b is:

σ¯n = σn − p = (γ − γw) b cos θ

The shear stress is similar to the dry slope case (again, but with diﬀerent speciﬁc weights):

τ = γb sin θ

Aside: As mentioned in class, if you are concerned about an effect the downslope fluid flux may have on the shear stress, then you can show yourself (from the above work on calculating pore pressure at some depth b) that pressure with depth acting on upslope and downslope faces of some block has no net downslope component.

Taking the ratio of τ and σn: τ γ = tan θ σn γ − γw now as mentioned in (4) : τmax =σ ¯n tan φ

3 thus γ tan θmax = tan φ γ − γw γ tan θ = 1 − w tan φ max γ and since γw < γ, θmax is going to be less than φ.

◦ ◦ Example: φ = 30 , γ = 2γw ⇒ θmax = 16.1

7. Maximum inclination angle for submerged slope.

Now consider a slope submerged under water with hydrostatic conditions. Note that if we consider the pressure of the water acting on the faces of a slope element whose base lies on the potential failure plane, we see that the net force would effectively be the force required to keep in place a body of fluid of the same volume. So here the effective stress that holds the slope in place is in response to the buoyant weight.

So this problem is identical to the dry slope, except here the net speciﬁc weight is no longer γd, but γ − γw. The eﬀective normal stress is

σ¯n = (γ − γw) b cos θ and the shear stress is τ = (γ − γw) b sin θ

If the strength of a potential failure plane requires that τmax/σn = tan φ we see here that the maximum slope angle is θmax = φ. 8. Reading:

For a review of friction and its application to the study of basic slope stability, please see: Middleton and Wilcock, Mechanics in the Earth and Environmental Sciences, 1994, Sections 4.1 and 4.2 (friction), 4.16 (dry inﬁnite slope), and 6.14 (Eﬀect of pore pressure on slope instability)

For a general introduction to ”dry” friction, check out this nice website, down to and including Sect. 12.6: Example 1:

http://www.engin.brown.edu/courses/en3/Notes/Statics/friction/friction.htm

For a connection to ﬁeld cases of slope failure, I recommend taking a look at Rahn’s Engineering Geology, sections 6.1.3 (landslides) and 6.1.4 (debris ﬂows).