Slope stability and strength of materials
When we look around at the countryside in the British Isles we see slopes that are stable; they are not continuously moving down–slope and failures are triggered only occasionally. Even in high mountain areas with steep slopes it is only the very steepest mountain walls that are falling down. In other words, slopes have a strength which is normally sufficient to maintain their stability. Here, we consider the strength of materials and how this is overcome when slopes 'fail', as well as how strength can be measured and the role of pore water pressure.
We can consider slope materials to be of three main kinds which roughly correspond to their nature; sands and gravels, clays and rock. Here we shall consider mainly the first two, as individual components and as a mixture of both – which is what many slopes consist of: gravel, sands and clay.
First, consider walking up a sand dune face. It is quite difficult; feet slip and more sand comes cascading down, even though the maximum slope may be only about 30º. Climbing a fine scree slope is hardly any easier. Although a coarse scree slope might be steeper – 35º or more – it is often less tiring than the sand dune. In all cases the stability of the slope is provided by friction of particle against particle. This is characteristic of granular materials (dry salt and sugar behave in exactly the same way). In nature, scree slopes have a maximum angle of about 40º, although they may be hundreds of metres high.
A dry clay slope, however, may be near-vertical, as is the case for some till cliffs at many coastal cliff sites. Most clay slopes are not this steep of course; they have water in them, and if there is too much water they fail at low angles. We call the strength provided by clays cohesion. How can we measure these components: friction and cohesion?
Figure 1: A representation of a direct shear box. The soil sample (usually 5cm square) is held in the box which is split horizontally so that the top half moves smoothly across the bottom half by applying a shear stress (equal and opposite, shown by the half arrows). The normal load n is applied to the top of the soil sample via a plate. The shear stress is measured via the force applied acting over the area of the sample; the distance moved (strain) is also recorded.<end>
A simple test is the direct shear box. Figure 1 shows the arrangement of a metal box with a removable top and which is split in half in the middle and parallel to the base. This box is filled with the test soil material (with known mass and volume so we know its density) and the top plate added together with a load (weight) on top. This is the normal load (or stress, n) acting over the horizontal area of the box. This would be stable no matter what value of n we had (which is measured in Megapascals, MPa). We can make it unstable by applying a shear stress (also in MPa) by pushing the top half of the box against the lower. We are forcing the material to fail across a shear plane in the box. The movement of the top of the box relative to the lower produces a strain or deformation in the soil but it requires an increasing applied shear stress to do so. The curve (a) in Figure 2 shows this relationship. From the origin (no applied shear stress, no strain) this rises to a peak. This is the maximum shear strength of the material for a given value of normal stress. After the peak value (P1) the strain continues but the measured shear stress falls and levels off. This is the residual strength of the soil. If we then repeat the test with some new clay/sand mixture and an increased normal stress (n)then we get the same sort of curve shape but with a higher peak strength (P2 on curve b). A third test at higher n would give a higher strength. With these three values of shear strength at failure (f ) corresponding to normal load we can plot one against the other, as in Figure 3. Joining up the points gives the line as shown with an intercept, known as the cohesion, c (with stress units of MPa, see below for explanation). The gradient (slope, ) of the line is the angle of internal friction. The equation of the curve is thus s = c + n tan where s is the peak strengthof the soil. This the Mohr-Coulomb equation for total stresses. Returning to Figure 2 and replacing the clay sand mixture in the box with sand alone, then repeating the first test, we would obtain curve c. Doing the tests again with different n we would get the plot going through the origin, i.e. c=0 with just a value of . This is what we would expect for sand – a friction value ; remember the dune slope, and no cohesion as there is no clay.
Figure 2: Deformation curves obtained from a direct shear box as in Figure 1. Two curves are shown for a clay sample, a and b, each for a different normal load, n. The peak value P1 for the red curve a is recorded and used as in Figure 3 and similarly for curve b and P2. If deformation continues by applying the shear stress then a 'residual strength' value is reached. Curve c shows a typical curve for a sand, showing friction but no cohesion. The maximum value achieved in this case depends upon the density of the sand in the box.
Figure 3:The Mohr-Coulomb equation and its graphical representation from shear box tests. The points P1, P2, etc. are derived from individual tests as in Figure 2 and plotted. The line joining them gives the equation of a straight line with the intercept being the cohesion c (clay component) and the gradient the angle of internal cohesion.
This gives us a basic understanding of slope behaviour. However, it does not tell us anything about water in the soil, especially the pore water pressure. A simple modification of our expression gives us: s = c + (n – u)tan where u is the pore water pressure. A pressure is a force per unit area, like a stress, so the introduction of u reduces the value of n and thus the value of peak shear strength s. Given the rather complex nature of soils this is a surprisingly simple result. In general, therefore, the higher the pore water pressure the nearer to failure we come.
Now going back to our mud flow or mud slide, which will contain granular as well as cohesive material, we can see why achieving a critical pore water pressure and why antecedent (pore water pressure) conditions are so significant.
At Minnis North and along the Antrim coast road the road itself is usually at the bottom of the slope and when a slope failure (flowslide) occurs the mud flows over the road. In other places, however, a road crosses a flowslide area in the middle. One such road, not far from Minnis North but about 100m higher up slope, has probably been there for a thousand years and is not active in the way that Minnis North behaves, with mud flow-slides from time to time. However, there is creep of materials in the slope, and the road, which runs directly across the hillside, has deformed.
Going back to Figure 2 we can also consider the residual strength – for clayey soils. The clay supplies some strength but not as much as at the peak value. It is the residual value that is mobilised once failure has been initiated in the slope. Subsequently, and this may be hundred or thousands of years after initial failure, a rise in pore water pressure will allow a small amount of movement, perhaps only a few cm, to take place.
Figure 4: Rupture of the road surface on the A57 (Snake Pass) in January 2008. This was one tear, another, some 100m away, represented the other side of the failure zone.
Figure 5: Looking directly downhill from the road shown in Figure 4. The white area shows the 'gap' in the hillside made by the progressive movement of the slope failure, about 120m wide and 3-4m deep. The inset shows the right-hand shear plane on the surface of the field linking the shear on the road with the right and edge of the failure zone in the distance.
Figure 4 shows a portion of the Snake Pass road (A57 between Sheffield and Glossop). After heavy rain on 19-20 January 2008 some movement of the road was reported. On the morning of 23 January there was sufficient movement (with a hole between two portions of the tear) to require the road to be closed. The newly laid road surface has 'torn' as a form of brittle fracture. Figure 5 is a view down-slope from the road. You can see that the edge of the valley slope has a 'lowered' section, about 120m across and up to 4m deep. This is due to the long-term flow of material from this slide area. The two main buckled areas on the road correspond to the width of the flow area and the lowered section of Figure 5.
Note on units: a force, measured in newtons (N), can be considered as a weight. A smallish eating apple weighs about 100g which equals 1N, i.e. not a lot. With slopes we are dealing with mega newtons, abbreviated to MN. The effects of the force will depend on the area to which it is applied. This gives us a stress (or pressure), i.e. stress = force/area. We give the name pascal (Pa) to 1Nm-2. For slopes, this is rather small for convenience so we will be dealing with mega pascals (MPa).
Some recent works looking at slope behaviour under rainfall conditions are:
Floris, M. and Bozzano, F.(2008) ‘Evaluation of a landslide reactivation: a modified rainfall threshold model based on historical records of rainfall and landslides’, Geomorphology, 94, pp. 40-57.
This article uses field monitoring of rainfall intensity and amount and applies this to numerical modelling of a complex landslide area in the Apennines. The authors use this to estimate the probability of reactivation of landslides by rainfall events.
Moore, R., Clark, A.R. and Lee, M.E. (1998) ‘Coastal cliff behaviour and management: Blackgang, Isle of Wight’, in Maund, J.G. and Eddleston, M. (eds) Geohazards in Engineering Geology. Geological Society of London, Special Publication 15, pp. 49-59.
This article summarises previous work and shows the importance of coastal environments involved with removal of basal debris in slide areas.
Waltham, A.C. (2002) Foundations of Engineering Geology. London. Spon.
A guide to a variety of ground conditions, including several on slopes, with clear diagrams.
© Geography 2008