12.001
LAB 11: Landslides
Due: Monday, May 10
A storm on January 4, 1982, triggered thousands of shallow landslides in the San Francisco Bay area. In
this lab, you will examine the conditions that caused one of these landslides in Marin County. You will
also explore general controls on slope stability.
Reconstruction of failure conditions for a specific landslide often involves a back-calculation procedure in
which several parameters are measured or estimated, and the remainder calculated. If the failure surface
is planar and parallel to the soil surface, if the width and length of the slide scar are much greater than the
depth of the failure, and if large strength differences exist in material above and below the failure plane,
then an "infinite-slope" stability analysis can be performed. In this case, effects at the ends and side of
the slides are justifiably ignored. We will use the infinite slope approximation throughout this exercise.
The general case is illustrated in Fig. 1. A landslide occurs when the shear stress () acting on the
potential failure plane exceeds the strength (S) of the material on the plane. The ratio S/ is defined as the
"Factor of Safety" (FS), and calculations normally assume that failure occurs when FS = 1. The shear
stress acting on the failure plane due to the weight of the overlying material is:
 sgz sincos


where

s = bulk density of the soil, including weight of water in pores (kg/m3)
g=
gravitational acceleration (9.81 m/s2)
z=

=
vertical thickness of the material above the failure plane (m)
failure plane angle (degrees)
The strength of the material is defined by the Coulomb criterion:
S=
C + (- u) tan 

where
C = effective cohesion (i.e., cohesion as influenced by the presence of water) (kPa)
 = normal stress on the slide plane due to the weight of the overlying material (kPa)
u=
pore water pressure due to a saturated zone above the failure plane (kPa)
 internal angle of friction of the material at the failure plane (degrees)
1
In this case,
 = sgz cos2


and
u = mzwg cos2

where

m = the fraction of the vertical thickness of the landslide that is saturated with water
w = density of water (1.0 g/cm3)
Other terms are defined as above. Combining (1), (2), (3) and (4) at FS = 1 yields:
sgz sincosC + (s - wm) gz cos2 tan


Note that if C = 0, then
  w m 
tan    s
tan 
 s 
(6)

Figure 1. Definition sketch for an infinite-slope stability analysis
Evaluation of Parameters:
(1)
Soil bulk density at time of failure, s, can be calculated from measured values of dry bulk density
and from observed or assumed water content. This requires calculating the volume of pore space,
which is normally done by assuming that the solid component of the sample has a density of 2.65
g/cm3, the density of quartz.
2
Here is an example calculation for a dry bulk density of 1.33 g/cm3. To calculate the volume of pore
space, it is helpful to draw a box dividing a unit of soil into pore space and solids, each with its own
volume and weight. Set the volume of the solid component at 1.0 cm3, and this gives you a weight
of 2.65 g for the unit of soil. Now you need the volume of pore space that will give you the
measured dry bulk density; in this case, this volume is identical to the void ratio, e (defined as
volume of voids/volume of solids).
Volume (cm3)
1.33 g/cm 3 
Weight (g)
e
Pore space
0
1.0
Solid
2.65
2.65  0 g
1 e cm3
Solving this equation gives e = 1.0, which implies that the soil is 50% pore spaces and 50% solids.
To calculate a saturated bulk density, fill the pore spaces with water and add this weight to that of
the solids.

(1.0 cm3) x (1.0 g/cm3) = 1.0 g of water
So the saturated bulk density is 3.65 g / 2.0 cm3 = 1.83 g/cm3
(2)
Vertical thickness of the material, z, and failure plane angle,  must be measured in the field.
(3)
Effective cohesion, C, and the internal angle of friction, , must be evaluated by laboratory or field
tests. The tests should be done with water contents similar to that at time of failure. For shallow
landslides in soil, it is normally assumed that the material was saturated, and the laboratory
experiments are performed on saturated samples. Often a direct shear box is used to evaluate C and
 in the laboratory. Undisturbed samples are placed in a split box, subjected to a confining
pressure, and then the upper box is pulled with sufficient force to cause failure in the material
between the two boxes. Values of effective normal stress and shear stress are then plotted as shown
in Fig. 2. According to Eq. (2), the intercept is the effective cohesion (C), and the slope of the line
is the tangent of the internal angle of friction (). In coarse-textured soils, the effective cohesion is
usually found to be zero in the laboratory, whereas stability analyses like the one in this lab indicate
that the soils must have had strength, presumably due to an apparent cohesion supplied by plant
roots. In forested landscapes like those in the Pacific Northwest, apparent cohesion due to plant
roots is a major component of soil strength.
3
kPa
25 20
50 40
100 80

tan 0.80

Figure 2.The y-intercept is the effective cohesion, and in this example the soil is cohesionless.
(4)
The fraction of the slide thickness that was saturated at the time of failure, m, has rarely been
observed. It is common to assume that in shallow, planar landslides, the entire mass that failed was
saturated (m = 1).
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Assignment
The following exercise is designed to provide familiarity with the above equations and to gain insight into
the significance of the various parameters contained in the equations. Show all work on separate sheets
of paper, and include these when you hand in your lab.
The landslide we will examine occurred on a 30° slope beneath a cover of California laurel and coast live
oak in a thick deposit of soil (Fig. 3). Landslides in the 1982 storm commonly occurred within the B
horizon, where there may have been differences in texture and permeability that made the upper portion
of the B horizon less stable. Engineering tests performed on the two horizons illustrated in Fig. 3 gave
the following results:
Dry Bulk Density:
Direct Shear Data:
(1)
A horizon:
B horizon:

1.29 g/cm3
1.68 g/cm3
(kPa)
A horizon
10.0
20.0
40.0
7.0
14.0
28.0
B horizon
15.0
30.0
50.0
15.5
31.1
51.8
Calculate the saturated bulk density as shown in the example.
A horizon s = 1.80 g/cm3
B horizon s = 2.04 g/cm3
(2)
Make a graph like Fig. 2 for each horizon, and compute the internal angles of friction.
A horizon = 35°
B horizon = 46°
Include these graphs when you hand in your lab. (See spreadsheet)
(3)
Assume that the soil was saturated to the surface. Compute the apparent cohesion necessary for FS
to equal 1 at saturation. First assume that the failure plane was at the boundary between the A and
B horizons (Case 1). Then repeat the calculation assuming that the failure plane was within the B
horizon, 2 m below the surface (Case 2, Fig. 3). For the second calculation you will need to
compute the normal and shear stress using both the bulk densities of the A and B horizons:
 = [s (A horizon) (1.0 m) + s (B horizon) (1.0 m)] g cos2
and
s (A horizon) (1.0 m) + s (B horizon) (1.0 m))] g sincos

5
Tabulate values of u, and C. Give your answers in kPa.


u
C
Case 1:
13.2
7.7
7.4
3.5
Case 2:
28.2
16.3
14.7
2.3
What percent of the total strength of the soil is due to cohesion in each case?
Case 1: 46%
Case 2: 14%
(4)
Assume the soil was cohesionless. Compute the level of saturation, m, necessary for failure for
each case.
Case 1: m = 0.32
Case 2: m = 0.85
(5)
Landslides play a major role in transporting debris from hillslopes in the California Coast Ranges.
Hillslopes may tend to be eroded to their maximum stable angle by landslides. Assume that these
landslides occur when the soil is saturated. What would be the maximum stable hillslope angle if:
a. The soil is cohesionless and has the internal angle of friction of the A horizon?
max = 17°
b. The soil is cohesionless and has the internal angle of friction of the B horizon?
max = 28°
Equation 5 can be rewritten for fully saturated soils (m = 1) as:
s
C
tan  
 tan 
s  w
s  w gz cos2 
Using this equation, construct a graph of maximum stable angle, max, vs. internal angle of friction,
. On the graph, show a line for each of the following:

1)
2)
3)
4)
C = 0, z = 0.5 m
C = 0, z = 2.0 m
C = 1500 Pa, z = 0.5 m
C = 1500 Pa, z = 2.0 m
The vertical axis of the graph should be max and should range from 0° to 70°, and the horizontal
axis of the graph should be . Use different markers to show your calculated values for the four
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cases, and label all lines. In all cases, let s = 1.8 g/cm3. Note that C = 1500 Pa may be typical of
shallow landslides in grasslands in Marin County. (See spreadsheet)
(6)
Discuss the general importance of cohesion, internal angle of friction, soil depth, and degree of soil
saturation in determining hillslope stability of potential shallow landslides. Contrast relative
stability in forested and grassland soils. Refer to calculations and graphs previously constructed.
Limit your discussion to two paragraphs or less.
The preceding results suggest a number of relationships between the parameters we have considered. In
Q3, we saw that cohesion accounts for 46% of material strength in the case of an A horizon failure, but
only 14% in a B horizon failure. This result, along with the regression performed in Q2, demonstrates that
cohesion is more important in materials with lower internal friction angles. This can easily be inferred
from the relationship S = C + (σ-u) tan φ. It also implies that cohesion makes a more important
contribution to the strength of shallow soil masses than it does to deeper ones. In Q4, we back-calculated
m values of 0.32 for a failure in A and 0.85 for a failure in B. As expected, the material with a lower angle
of internal friction requires less saturation to fail. This result also suggests that saturation has a greater
effect on shallower soils than on deeper soils.
For saturated soil with no cohesion, the first section of Q5 shows that the maximum stable angle is
proportional to the internal friction angle. The final plot of maximum stable angle vs. angle of internal
friction for several sets of C and z values reveals several other relationships. The two overlapping curves
in Figure 2 show that, under saturated conditions, the stability of cohesionless soil is independent of soil
depth. This is also obvious from the equation given, since the cohesion term is the only one that contains
z. The plot further tells us that, when cohesion is significant, thicker soils are less stable. This is intuitive,
since shear stress and normal stress both increase in proportion to soil depth, but cohesion does not.
In practice, there are limits to these relationships that are inherent to the factors influencing the different
parameters. For instance, consider the relative effects of cohesion in forested and grassland environments.
The main difference is the presence in forests of large, woody roots that penetrate deep below the surface.
The tensile strength of roots enhances the strength of the soil mass by increasing the effective cohesion
term. In shallow soil, this effect can be especially pronounced because the roots may penetrate the
underlying bedrock, anchoring the soil in place. Whereas root density in a grassland soil may be quite
high very close to the surface, the roots are not sufficiently deep to contribute significantly to the effective
cohesion of the entire soil mass. Note, however, that stability differences between forested and grassland
soils may not be as pronounced in deeper soil: since root depth is finite, fewer species will be able to form
“bedrock anchors” as soil depth increases, leaving a zone of weakness between the lower root boundary
and the soil-bedrock interface.
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Figure 3. Schematic cross section of upper portion of colluvial deposit on a 30° hillslope. The A horizon
is the uppermost, organic-rich zone in the soil. The B horizon contains less organic material, more rock
fragments, and more clay. The C horizon (not shown here) contains weathered bedrock that retains its
original structure and has not been transported downslope. In the scenario shown here, the boundary
between the A and B horizons is 1 m deep, and the landslide failure plane is 2 m deep.
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