Stability of an infinitely long slope

Consider the classical problem of determining the stability of an infinitely long slope, as illustrated in Figure 9.3. If the slope is underlain by a permeable stratum, as shown in the top half of Figure 9.3, any water in the slope flows vertically downwards into the underlying stratum and, as a result, pore water pressures do not build up in the soil. The stability of this slope is identical to that of a dry slope with the same gradient p.

Figure 9.3. Confined soil slopes overlying (top) permeable and (bottom) impermeable strata

Instead, if the slope is underlain by an impermeable stratum, as shown in the bottom half of Figure 9.3, then any water in the slope flows parallel to the underlying stratum, resulting in a hydraulic gradient parallel to the soil/rock interface.

9.4.1 Closed-form solution using global factor of safety

The global factor of safety against sliding F for both these situations is given by the 'infinite slope expression':3

r c' + ( 1 - ru )/H cos2 fitany r Ywhw F =-----with ru = -

where c' and 9 are the soil's effective cohesion and angle of shearing resistance; y and yw are the weight densities of the soil and water; ru is a pore pressure parameter; p is the slope angle; and the dimensions H and hw are defined in Figure 9.3. For a dry slope or a slope on a permeable stratum, ru = 0.

According to BS 6031,4 first-time slides with a good standard of investigation should be designed with a safety factor between 1.3 and 1.4; slides involving entirely pre-existing slip surfaces should be designed with F . 1.2.

The infinite slope expression can be re-arranged to give the stability number N necessary to provide a given factor of safety:

Figure 9.4 shows the values of N required to achieve various global factors of safety F, in the range 1.0 to 1.5, for a 1:3 slope (P = 18.4°) with ru = 0.5. Similar charts can be developed for other values of ru and p.

Figure 9.4. Stability number required to achieve various global factors of safety (F) against sliding for a 1:3 infinite slope with ru = 0.5

An example serves to illustrates how the chart is used. Imagine a 2.5m thick slope comprising soil with a weight density y = 20 kN/m3 and angle of shearing resistance 9 = 30°. From Figure 9.4, the stability number needed to achieve a factor of safety F = 1.3 is approximately N = 0.13. Hence the minimum effective cohesion c' that the soil must possess is:

9.4.2 Closed-form solution using partial factors from Eurocode 7

As discussed in Chapter 6, verification of strength according to Eurocode 7 involves demonstrating that design effects of actions Ed do not exceed the corresponding design resistance to those actions Rd:

Ed < Rd where the design effects are given by: Ed = sin p C°s p = rGrkH sin p c°s p and the design resistance by:

tan v

The subscripts d and k denote design and characteristic values, respectively.

Combining these expressions and re-arranging them produces the 'characteristic stability number' Nk:

' ( \ Nk =Y = (ygYcYk ) sin P cos p - Y (1 - if) cos2 p tan <pk

In deriving this expression, the partial factor Yg was applied to the characteristic vertical action (ovk = YkH), which appears in the equations for both Ed and Rd. Implicitly, we have assumed that the vertical action is unfavourable wherever it appears in this formulation. Alternatively, we could apply the partial factor to the characteristic action effects, taking account of whether those effects are favourable (using YG,fav) or unfavourable (using yg). This leads to a different expression for Nk:

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