LANDSLIDES: FORCE BALANCE AND SOIL WATER

Transcription

1 LANDSLIDES: FORCE BALANCE AND SOIL WATER STEPHEN T. LANCASTER. Introduction: It s Not That Hard Given the many potential variables involved in landsliding, making predictions may seem to be a hopeless task. However, consider this quote from the landslide master, Karl Terzaghi: Unfortunately the research activities in soil mechanics...diverted the attention of many investigators and teachers from the manifold limitations imposed by nature on the application of mathematics to problems in earthwork engineering. As a consequence, more and more emphasis has been placed on refinements in sampling and testing and on those very few problems that can be solved with accuracy. Yet, accurate solutions can be obtained only if the soil strata are practically homogeneous and continuous in horizontal directions. Furthermore, since the investigations leading to accurate solutions involve highly specialized methods of sampling and testing, they are justified only in exceptional cases. On the overwhelming majority of jobs no more than an approximate forecast is needed, and if such a forecast cannot be made by simple means it cannot be made at all. If it is not possible to make an approximate forecast, the behavior of the soil must be observed during construction, and the design may subsequently have to be modified in accordance with the findings. These facts cannot be ignored without defying the purpose of soil mechanics. Terzaghi, 996) [note: st edition was 948] In that spirit, we ll start with the simplest sort of formulation of the problem and then proceed to add a couple of important factors. 2. Force Balance for Dry Soil on an Infinite Slope The simplest case relevant to our discussion is a dry, shallow soil on an infinite slope Figure ). Under the infinite slope assumption, the earth pressure forces on the up- and downslope sides cancel e.g., in Figure, P P = ), and we neglect other lateral forces, such as friction, on the sides of the soil section. The inifinite slope assumption is generally valid if the distance to the top of the slope is much greater than the soil depth e.g., 2 ) and the slope is very wide. In Figure, the force of the soil s weight is W = dx dy h g, where is the soil bulk density, dx and dy are the down- and cross-slope dimensions of the slope element, h is the soil depth in the normal direction i.e., perpendicular to the surface), and g is the acceleration due to gravity. In the slope coordinate system, the weight, W, is decomposed into components, W sin θ and W cos θ, parallel to the x- and z-axes, respectively. Newton s Second Law and the fact that the soil is not being sucked into the bedrock dictate that Date: October 3, 23. Course Notes for GEO 322 Surface Processes, College of Earth, Ocean, and Atmospheric Sciences, Oregon State University.

2 2 STEPHEN T. LANCASTER there is a normal force equal and opposite to the slope-normal component of gravity, i.e., N = W cos θ. z y x θ P N tanφ W cosθ θ W N dx W sinθ Figure. Force balance for a section of dry soil on an infinite slope. Coordinate system is oriented to the slope. Finally, analogous to the normal force, a frictional force that is proportional to the normal force acts in the direction opposite to the downslope force, W sin θ, where the constant of proportionality, tan φ, is the coefficient of friction, and φ is also known as the angle of repose and the friction angle. This coefficient or angle or whatever term you prefer) is a material property that is generally measured with a tilt table or simply by making a pile of the stuff as steep as possible. Strictly speaking, the basal friction angle measured with a tilt table) and the internal friction angle measured by making a pile) may be different, e.g., if the material beneath the failure plane is different. Whereas we might wish to know about landslides along internal failure surfaces, such as the case of a collapsing river bank, we are often concerned with the failure, en masse, of the entire soil layer, i.e., all of the mobile regolith. In the latter case, we assume that the bedrock surface is the failure surface, and that bedrock surface might, say, be smooth enough that the basal friction angle would be smaller than the internal friction angle. For the sake of simplicity, I make no distinction between these angles in these notes hereafter. The force balance in the z-direction is trivial and stated above in defining the normal force, N. For the force balance in the x-direction, I write it as if the soil were on the verge of failure, i.e., the forces sum to zero, so that driving and resisting forces are equal and opposite: dx dy h g sin θ dx dy h g cos θ tan φ =. ) When considering a potential landslide on an infinite slope, rather than considering forces, it typically makes better sense to consider stresses, i.e., forces per unit area, on the failure plane. Inspection of ) reveals that we can divide everything by the basal area, dx dy, so that we express the balance as one of shear stresses i.e., force per unit area in a direction parallel to the failure surface): hg sin θ hg cos θ tan φ =. 2) In the typical notation of soil mechanics, the total shear stress driving slope failure i.e., landsliding) is represented by the Greek letter tau, where we have τ = hg sin θ; the total resisting stress, or strength, is represented by s, where we have s = hg cos θ tan φ. That dy P h

3 LANDSLIDES: FORCE BALANCE AND SOIL WATER 3 strength is often expressed in terms of the normal stress i.e., the force per unit area in the direction normal to the failure surface), represented by the Greek letter sigma, where we have σ = hg cos θ and s = σ tan φ. But what if these forces don t balance? Strictly speaking, if the soil mass is not accelerating, then Newton s Second Law says that the friction force cannot actually exceed the downslope force. Practically, though, we are more concerned about the risk of landsliding than with satisfying the letter of Newton s Laws, so we want to know how large the strength, s, could be. In soil mechanics, this risk, or the lack thereof, is measured by the factor of safety, which is just the ratio of resisting and driving stresses, i.e., F S = s/τ. When the magnitude of the potential friction is greater than the downslope driving force, the factor of safety is greater than one, and when friction is insufficient to halt landsliding, or slope failure, the factor of safety is less than one, or F S >, s > τ =, s = τ <, s < τ. 3) When geotechnical engineers say they want to have a large factor of safety, they re talking about F S. So what is the factor of safety for our simple case of dry soil on an infinite slope? It s the ratio of the magnitude of the second term in 2) to the first term: F S = hg cos θ tan φ hg sin θ cos θ tan φ = sin θ = tan φ tan θ, 4) where we have used, first, the fact that both numerator and denominator contain identical factors, hg, which cancel out, and second, the trigonometric identity, sin θ/ cos θ = tan θ so, cos θ/ sin θ = / tan θ). Note that tan θ is equivalent to what we have been calling the slope, i.e., this is just rise over run. Consider the implications of 4): it is dependent only on slope inclination, tan θ, and the material property, tan φ, and independent of soil depth and gravitational acceleration. Therefore, a pile of sand that is stable on Earth will also be stable on Mars. Our factor of safety is planet-independent! 3. Slope Stability in the Real World Yes, the real world includes sand dunes that are dry, at least on the surface, so equation 4) has utility in such cases, where the roles of water and cohesion are insignificant. In many, perhaps most, cases, however, these roles are not only significant but effectively determinative. That is, we are often concerned with the stability of slopes that we suspect might be unstable when the soil is very wet or root strength is absent or diminished.

4 4 STEPHEN T. LANCASTER 3.. Slope-Parallel Seepage. A common case is water flowing through a hillslope soil, so that the water table is approximately parallel to the surface and the soil-bedrock interface. In short, the effect of a water table within the potentially failing soil is to decrease the effective strength resisting failure. Whereas the total weight of a wet soil is greater than the weight of a dry soil, and the total normal and shear stresses are increased accordingly, the effective normal stress is less than the total because part of the weight of the soil is borne by the pore fluid, and only the soil particles produce significant friction. In other words, the effective weight of the soil beneath the water table is only its buoyant weight. equipotential line flow line z u = ρ w gh w cosθ θ θ dx s = σ tanφ τ = gh sinθ y x σ = gh cosθ σ = σ u ζ dy h h w /cosθ h w = mh h w cosθ The stresses are shown in Figure 2. Note that it s very similar to Figure, but the forces are already converted to stresses, and only the central section is shown, because we already know we can neglect the forces from adjacent sections. gh cosθ gh Figure 2. Stresses for a section of wet soil with parallel seepage on an infinite slope. Coordinate system is oriented to the slope. As before, gravity pulls the soil downward with a force per unit area of gh, and this gravitational stress is decomposed into components perpendicular and parallel to the slope, and these components give the normal stress, σ = gh cos θ, and the shear stress, τ =. Note that the bulk density,, is that of the wet soil and, for the same soil as in the previous example, will be greater than for dry soil. We ll assume, for the sake of simplicity, that the bulk density of the soil is the same above and below the water table. This is not a bad assumption if it is raining and all of the soil is wet. Due to the water table forming a layer of saturated soil of normal thickness, h w, pore pressure is positive at the soil-bedrock interface, but this part is a little tricky. Hydrostatic pressure i.e., the pressure of static, or still, water) at a point is determined by the vertical height of water above that point. However, this water is dynamic, i.e., flowing. The flow lines are parallel to the slope and the equipotential lines, or lines of equivalent total head, are perpendicular to the flow lines by definition. Total head has units of length and is defined as the head due to potential energy plus that due to the height of water above that point. In the hydrostatic case, total head is just elevation plus depth below the water surface. If the pressure were hydrostatic, then the total head at a point at the soil-bedrock interface at elevation, ζ, would be ζ + h w / cos θ. Remember, for the sort of angle shown, cos θ <, so h w / cos θ > h w. In this non-hydrostatic case, though, total head is still elevation plus pressure, but we have to contend with the fact that, unlike the hydrostatic case, the equipotential lines are not vertical. At the water table, the pressure is zero. As we move along the equipotential line toward the bedrock, the pressure increases due to the vertical component of the slope-normal thickness of the water above, so that, at the soil-bedrock interface, the pressure head is equal to the vertical component of the slope-normal thickness

5 LANDSLIDES: FORCE BALANCE AND SOIL WATER 5 of the saturated layer, h w cos θ, so the pore pressure is u = ρ w gh w cos θ Lambe and Whitman, 979). The effective normal stress, σ, is the part of the normal stress borne by the soil grains and therefore contributing to friction. This is simply the total normal stress minus the pore pressure, or σ = σ u = gh cos θ ρ w gh w cos θ, 5) and the shear strength is s = σ. So what s with the φ instead of just φ? Well, once we get the soil wet, we have to contend with the likelihood that there may be some surface tension holding things together. Or not. You might think of the notation, φ, as simply an acknowledgment that there s something other than simply the property of the material at play here. Now we can write the factor of safety, which is just F S = s/τ, as F S = gh cos θ ρ w gh w cos θ). 6) We can simplify this a little by pulling a couple of terms out of the parentheses and using tan θ = sin θ/ cos θ: F S = = ρ ) wh w gh cos θ h ρ ) wh w tan φ h tan θ Finally, if you ve looked carefully at Figure 2, you might have noticed that I snuck in another variable, so that we can represent the thickness of the saturated layer as h w = mh, where m is the fraction of that thickness that is saturated. Solving for m, we have m = h w /h. Substituting m for h w /h in 7), we can write the factor of safety as F S = 7) m ρ ) w tan φ tan θ. 8) Now, m is a fraction and is generally restricted so that m. The density of water is generally less than the bulk density of wet soil, so it is also true that < ρ w / <, and typically we might expect ρ w / /2. So, for example, for m =, so that the water table is at the soil surface, the factor of safety is F S /2) / tan θ). In effect, saturation has reduced the effective coefficient of friction by /2. You can see that water has a pretty strong effect on slope stability. In any given situation, we might use 8) to solve for critical conditions pertaining to stability by setting F S =, i.e., letting the shear strength and stress just balance one another. For a given hillslope, we might wish to know the fractional soil saturation that would trigger a

6 6 STEPHEN T. LANCASTER landslide. Setting 8) equal to and solving for m, m ρ ) w tan φ tan θ = m ρ w m ρ w m = ρ w = tan θ = tan θ tan θ ). 9) Note that, if we didn t like the answer, then we could always make m smaller by increasing the thickness, h, of the soil layer. Alternatively, for a given fractional saturation, we might wish to know the slope at which failure would occur: m ρ ) w tan φ tan θ = tan θ = m ρ ) w. ) Now, you might be saying to yourself that this still isn t all that realistic. What s missing? 3.2. Cohesion. Ah, yes, cohesion! Traditionally, cohesion is strength under conditions of zero normal stress. In the case of soils, cohesion arises from electrostatic bonds between clay and silt particles, and with rocks, cementation is typically the source De Blasio, 2). Cohesion is simply an extra strength term and is added to the shear strength so that s = +σ, where is the effective cohesion, which again accounts for the effect of water. We might also try to account for the effect of root strength by adding an apparent cohesion due to roots, c R, but for now let s consider only the effective cohesion,. Starting with 6), we simply add to the numerator to get F S = c + gh cos θ ρ w gh w cos θ), ) which is simple enough, but it does make things more complicated by adding a bit of a messy term to our simplified form, 8), where the equivalent with cohesion is F S = + m ρ ) w tan φ tan θ. 2)

7 LANDSLIDES: FORCE BALANCE AND SOIL WATER 7 Now, if we want to solve for the fractional saturation at failure, the algebra is a bit messier: + m ρ ) w tan φ tan θ = m ρ ) w tan φ tan θ = m ρ w m ρ w m = ρ w = tan θ = tan θ [ tan θ ) ) )]. 3) That cleaned up pretty well. This says that, in terms of the amount of water needed to make the slope fail, the effective slope is reduced by the ratio of effective cohesion to shear stress. We might, say, wish to solve for the effective cohesion at failure, given other factors: + gh cos θ ρ w gmh cos θ) = + ρ w m)gh cos θ = + ρ w m)gh cos θ = = ρ w m) gh cos θ. 4) That is, at failure, the effective cohesion is the difference between the shear stress and the frictional shear strength. Which is kind of a duh result. 4. Water on Hillslopes We ve covered the treatment of soil water in slope stability calculations, but how does the water get there? Aside from some special cases, rainfall is the source of the water. Although the special cases, such as seeps and springs, may be important, they are also unpredictable.) In short, at a given point in the landscape, the rainfall falls on the upslope area, which if convergent, funnels the flow to that point. In the particularly simple case in which that upslope area is paved, then all of the rainfall runs off over the surface, and all the flow at a point is overland flow, specifically, infiltration excess or Hortonian overland flow. But more on that later. In many, perhaps most, natural landscapes, especially vegetated ones, all rainfall from all but the most intense storms infiltrates into the soil and percolates downward until it reaches a water table, at which point the water begins to move laterally Figure 3). The relatively high hydraulic conductivity of the soil layer may be great enough, and the soil thick enough, to accommodate all of the rainfall, and all of the rainfall from upslope

8 8 STEPHEN T. LANCASTER areas, for most rain events. However, for more intense storms and further downslope, the transmissivity hydraulic conductivity times depth) may be insufficient, and the water table will break the surface, as illustrated in Figure 3. Now, in that figure, the landscape is simply a cross section, implying that the hillslope is planar. In some cases, this might be a reasonable approximation, and it is at least a reasonable place to start. So, for a planar slope the upslope area per unit contour width is simply the horizontal distance to the divide, which we will denote as x. P I P I Figure 3. Stresses for a section of wet soil with parallel seepage on an infinite slope. Coordinate system is oriented to the slope. I know, we earlier defined the x-direction as parallel to the slope, but now it s convenient to do otherwise. So there.) If we neglect water moving through the bedrock, again a good approximation in many cases, then the discharge per unit contour width at a point is simply the distance, x [L], times the rainfall rate, or q = P x, where q is discharge per unit width [L 2 T ], and P is the rainfall rate or intensity [L T ]. To find the depth of subsurface flow in the soil, we need to introduce Darcy s Law, which relates flow to the head gradient: h w q = K sat cos θ dh dx, 5) where K sat is the saturated hydraulic conductivity [L T ], which is a combination of material and fluid properties, H is the hydraulic head i.e., pressure head plus elevation head) [L], and dh/dx is the head gradient, which is simply the slope of the water table in our case. The cos θ term is because the thickness in Darcy s Law is vertical depth, and we previously defined h w as thickness measured normal to the land surface. To a reasonable approximation, we may assume that the water table is parallel to the ground surface e.g., Figure 2), and we ll assume that all subsurface flow is parallel to that surface this is not, strictly speaking, true, but it s a good place to start). In short, we can substitute the gradient of the land surface for the head gradient. Then, setting Darcy s Law 5) equal to the discharge from precipitation, we have P x = K sat sin θ, 6) cos θ and we can solve this for the saturated depth, h w : h w = h w P x K sat tan θ, 7) where we ve used again the fact that tan θ = sin θ/ cos θ. Given the soil depth, h, we can find the fractional saturation, m = h w /h: m = P x K sat h tan θ. 8) If we find that m > or h w > h, then we know that there will be overland flow, because when the water table reaches the surface, well, there s nowhere else for the water to go, and any additional rainfall just hits that water surface and runs off. We call this saturation overland flow, sometimes known as Dunne flow. P R

9 LANDSLIDES: FORCE BALANCE AND SOIL WATER 9 More generally, volume discharge, Q [L3 T ], is CB hw dh, 9) cos θ dx where w is the width through with the water flows. The steady-state discharge as a function of precipitation is just Q = P A, where A is the upslope area contributing to flow at a point, i.e., the contributing area, and the discharge per unit width is q = P A/w. Setting this equal to 5), we get hw dh PA = Ksat. 2) w cos θ dx Substituting with hw = mh, PA mh dh = Ksat, 2) w cos θ dx and solving for m, we get PA. 22) m= Ksat wh tan θ Q = qw = Ksat w 5m Slope stability and hydrology You might have noticed that, in addition to the solution for m in 8) and 22) based on precipitation rate and Darcy s Law, we solved the factor of safety equation for m, without cohesion in 9) and with cohesion in 3). As we do for the in-class exercise, we could compare values of m for saturation overland flow i.e., m = ) and marginal stability i.e., FS = ). We can also do things that are potentially a little more interesting. For example, in 255 class, precipitation rate is specified, but it need not be. In Figure 4. Topographic map -meter contour intervals) of CB Coos Bay fact, we may wish to link these equations to find the pre ) catchment at Mettman Ridge, near cipitation rate that might make the slope unstable Coos Bay in the Oregon Coast Range. Thick gray line marks catchment of a weir located at the downslope point. Thick black line marks edges of a landslide and debris flow that occurred in November, 996. Numbers in large circles are values of m, the fractional soil depth that was saturated during the storm that triggered the landslide. Red and blue contour segments are land and water table surface contours, respectively, within area marked by cyan dashed line. Magenta shading marks area predicted to have had safety factor of one FS = ). Figure adapted from Montgomery et al. 29). Let s start by setting 9) and 22) equal to one another: ρb tan θ PA =. 23) ρw tan φ Ksat wh tan θ Given topography, soil depth, and hydraulic conductivity, we can solve this for the precipitation: ρb Ksat h tan θ tan θ P =. 24) ρw A/w tan φ Note that I ve grouped contributing area and width as A/w, which is the contributing area per contour width: If the water table is parallel to the surface, then contours are also equipotential lines, and flow lines intersect them

10 STEPHEN T. LANCASTER at right angles. Flow in a real landscape, such as that in Figure 4, has a width, and the appropriate width is typically measured along a contour. And we can t really define a contributing area without also defining the width of the outlet, e.g., between the lines marking the edges of the landslide in Figure 4. As for the form of the expression, slopes can endure greater rainfall if bulk density, hydraulic conductivity, soil depth, slope, and friction angle are greater and contributing area per contour width and slope are smaller. Got that? Greater slopes make the slope stable and unstable. What have we forgotten? Well, we haven t included a limit on the value of m. It turns out that solutions linking increasing slope to increasing critical precipitation i.e., that which would lead to a failure) also have m > Figure 5). If slopes are stable when fully saturated, i.e., m cyan, magenta, and gold in Figure 5), we can consider them unconditionally stable. If the slope m x P m/s) 2.2 Figure 5. Three-dimensional graph of fractional soil saturation, m, and precipitation, P, vs. slope, tan θ, at the threshold for failure, i.e., F S =, assuming no cohesion as in equation 24). Three lines represent three different values of the friction slope, :.5 for blue/cyan,.75 for red/magenta, and. for green/gold. Blue, red, and green show the parts of the solution for which m, and cyan, magenta, and gold show where m >. is greater than the friction slope, i.e., tan θ >, then it is unstable no matter the level of saturation, so we can consider such slopes unconditionally unstable. Although the math is certainly messier, we can find the equivalent solution for cases with non-zero cohesion by setting the right sides of 3) and 22) equal to one another, [ tan θ )] P A = K sat wh tan θ, 25) ρ w and solving for precipitation rate, P, to get P = K sat h tan θ ρ w A/w [ tan θ.4 tanθ.6.8 )]. 26) hg sin θ We see, then, that the cohesion term is identical to that in 3). From Figure 6, we see that increasing cohesion increases both the amount of rainfall needed for failure and the slopes tan θ) at which failure can occur. In fact, for the largest cohesion = kpa), the soil is unconditionally stable at slopes exceeding those for unconditional instability for the other cases. For reference, the apparent cohesion due to roots in a mature conifer forest is generally

11 LANDSLIDES: FORCE BALANCE AND SOIL WATER thought to exceed kpa, and one study appears to indicate that apparent cohesion due to roots can approach or exceed 5 kpa in old-growth forests in the Oregon Coast Range. m Implications for Landslide Risk Should we believe in the accuracy of these solutions for the rainfall rates that would make slopes unstable? strictly speaking, probably not. But which assumptions lead to the greatest error and, therefore, uncertainty regarding the right answer? And what useful information can we gain from our simplified approach?.2 x 7.8 P m/s) tanθ Figure 6. Three-dimensional graph of fractional soil saturation, m, and precipitation, P, vs. slope, tan θ, at the threshold for failure, i.e., F S =, and including cohesion as in equation 24). Three lines represent three different values of the effective cohesion, :. for blue/cyan, kpa kilo-pascal) for red/magenta, and kpa for green/gold. Blue, red, and green show the parts of the solution for which m, and cyan, magenta, and gold show where m >. There are certainly some simplifications in our use of the factor of safety equation that lead to some uncertainty there. For example, although I left out the apparent cohesion due to root strength, c R, from the factor of safety equation, I asserted that it could simply be added to the effective cohesion,. Well, it may not be as simple as that, because root strength between the soil and the failure plane is typically small, often negligible. I know, we talked in class about how tree roots can effectively grip the bedrock so that they actually rip it apart as the wind blows and the tree sways. But we also noted that roots typically only penetrate to a depth of about m. In a place like the Oregon Coast Range, soils are, on average, typically thinner than m, but the exceptions are also typically the places where landslides occur: convergent topography i.e., large A/w) acts to accumulate flowing water as well as mobile regolith, so these locations of relatively deep soil are also more likely to become wet enough to fail as landslides. Lateral root strength, then, is typically much greater than the aforementioned vertical, anchoring root strength. The lateral root network reaches far from the trunk and can become intertwined with roots from neighboring trees and understory vegetation to create a nearly continuous mat of roots literally tying the various parts of the slope together. However, the assumption of an infinite slope implies that lateral stresses are negligible, and we can t necessarily just add them in. Specifically, we really need to find not only failure planes but perimeters. And that s complicated. There s an example of landslide post-diction as opposed to prediction) with a relatively sophisticated model in Figure 4, where the magenta.5

12 2 STEPHEN T. LANCASTER shading delineates an area for which F S =. It turns out, for example, that the size and shape of the potential failure area can be important. Some of our greatest simplifications are in the hydrology. Recall that we assumed that the subsurface flow is parallel to the slope and in a steady state relative to the rainfall. However, if rain is falling, then there must be vertical flow due to infiltration. And, whereas it does sometimes seem to rain constantly in western Oregon, the time to reach a true steady state between rainfall and subsurface flow might be on the order of weeks. Moreover, landslides do not typically occur during steady i.e., temporally constant) rainfall. At the site mapped in Figure 4, it had been raining for about 36 hours at about. mm/min., but the landslide occurred about an hour after a peak exceeding.7 mm/min. and nearly simultaneously with a second peak exceeding.4 mm/min. Piezometers near the magenta-shaded area Figure 4) showed that the total head was increasing when the landslide occurred, but Montgomery et al. 29) do not specify the time interval of the recorded measurements. It is possible that none of these piezometers actually recorded any response to the second rainfall peak. Weirs near the bottom of the area shown in Figure 4 did record a discharge peak apparently associated with the first peak, but measured discharge appeared to be falling at the time of the failure. It is likely, then, that the pore pressures that triggered this landslide were generated by vertically infiltrating rainfall rather than increased flow from upslope. It is also likely that this spike in pore pressure preceded the arrival of the actual water that fell during that second peak in the rainfall. As shown by Iverson 2), where the soil is already wet e.g., pore pressure equal to zero throughout the soil column) peaks in rainfall generate pressure waves that outpace the infiltrating water, and these pressure waves can generate pore pressure spikes that can trigger instability and failure. It seems likely, then, that our predicted critical precipitation rates are too large, at least when precipitation rates are averaged over times implied by the assumption of steady-state flow. In reality, then, lower average rainfall rates may set the stage for transient rainfall peaks that actually trigger failure. But if the timing of failure is dependent on such transient peaks, then failure must also depend on their location. The sorts of peaks in rainfall that trigger landslides are therefore nearly impossible to predict in either time or space without an effectively impossible fore-knowledge of internal storm dynamics. And that brings us back to the initial quote from Terzaghi 996), who advises to take simple approaches, because the complicated cases are impossible. So what exactly is the use of the simple approach we ve taken here? Simplified approaches alert us to places and conditions of heightened risk. For example, the predictions of Figure 6 suggest heightened risk in cases of low cohesion, especially where slopes tan θ) may approach or exceed the friction slope of unreinforced soil. The predictions of Figure 5 suggest heightened risk where variations in soil properties lead to variations in frictional strength. The appearance of contributing area per contour width, A/w, in the denominator of the equations predicting critical precipitation, 24) and 26), imply that landslide risk is greater on convergent slopes, an implication that is borne out by observations such as those of Montgomery et al. 29) Figure 4), among many others.

13 LANDSLIDES: FORCE BALANCE AND SOIL WATER 3 References De Blasio, F. V. 2), Introduction to the Physics of Landslides: Lecture Notes on the Dynamics of Mass Wasting, Springer, doi:.7/ Iverson, R. M. 2), Landslide triggering by rain infiltration, Water Resources Research, 36 7), 897-9, doi:.29/2wr99. Lambe, T. W., and R. V. Whitman 979), Soil Mechanics, SI Version, John Wiley & Sons, New York. Montgomery, D. R., K. M. Schmidt, W. E. Dietrich, and J. McKean 29), Instrumental record of debris flow initiation during natural rainfall: Implications for modeling slope stability, Journal of Geophysical Research, 4, doi:.29/28jf78. Terzaghi, K. 996), Soil Mechanics in Engineering Practice, 3rd ed., John Wiley & Sons, New York.