CH-4 Slope Stability.pdf
- 1. 09-Mar-21 1 Soil Mechanics-II GAMBELLA UNIVERSITY By Fentahun A. MSc. In Geotechnical Eng. Lecture notes Lateral Earth Pressure Introduction The term slope as used in here refers to any natural or manmade earth mass, whose surface forms an angle with the horizontal. Hills and mountains, river banks, etc. are common examples of natural slopes. Examples of manmade slopes include fills, such as embankments, earth dams, levees; or cuts, such as highway and railway cuts, canal banks, foundations excavations and trenches. Natural forces (wind, rain, earthquake, etc.) change the natural topography often creating unstable slopes. Failure of natural slopes (landslides) and manmade slopes have resulted in much death and destruction. 2 In assessing the stability of slopes, geotechnical engineers have to pay particular attention to: geology, drainage, groundwater, and the shear strength of the soils. The most common slope stability analysis methods are based on simplifying assumptions and the design of a stable slope relies heavily on experience and careful site investigation. 3 Cont.… Definitions of Key Terms Slip plane or failure plane or slip surface or failure surface is the surface of sliding. Sliding mass is the mass of soil within the slip plane and the ground surface. Slope angle (or simply slope) is the angle of inclination of a slope to the horizontal. The slope angle is usually referred to as a ratio, for example, 2:1 (horizontal: vertical) 4
- 2. 09-Mar-21 2 Some Types of Slope Failure Slope failures depend on the soil type, soil stratification, groundwater, seepage, and the slope geometry. A few types of slope failure are shown in Fig. 4.1. Failure of a slope along a weak zone of soil is called a translational slide (Fig. 4.1 a). are common in coarse-grained soils. 5 A common type of failure in homogeneous fine-grained soils is a rotational slide. Three types of rotational slides often occur. 1. base slide: occurs by an arc enclosing the whole slope. A soft soil layer resting on a stiff layer of soil is prone to base failure 2. toe slide: whereby the failure surface passes through the toe of the slope 3. slope slide: whereby the failure surface passes through the slope A flow slide occurs when internal and external conditions force a soil to behave like a viscous fluid and flow down even shallow slopes, spreading out in several directions 6 Cont.… Figure 4.1: some types of slope failure 7 Cont.… Some Causes of Slope Failure Slope failures are caused in general by natural forces, human mismanagement and activities. As shown in Fig. 4.2, some of the most common causes of slope failures are: oerosion, orainfall, oearthquake, ogeological features, oexternal loading, oconstruction activities (ex. excavation & fill), and oreservoir rapid drawdown. 8
- 3. 09-Mar-21 3 Figure 4.2: some causes of slope failure 9 Cont.… 10 Cont.… Two-dimensional Slope Stability Analysis Slope stability can be analysed using one or more of the following: limit equilibrium method, limit analysis, finite difference method, and finite element method Limit equilibrium is the most widely used method for stability analysis. In the following sections, we will learn some of the commonly used slope stability analysis methods that are based on the limit equilibrium 11 Stability Analysis of Infinite Slopes Infinite slopes have dimensions that extend over great distances. In practice, the infinite slope mechanism is applied to the case when a soft material of very long length with constant slope may slide on a hard material (e.g. rock) having the same slope. Let’s consider a clean, homogeneous soil of infinite slope αs as shown in Figure 4.3. To use limit equilibrium method, we must first speculate on a failure of slip mechanism. We will assume the slip would occur on a plane parallel to the slope. If we consider a slice of soil between the surface of the soil and the slip plane, we can draw a free-body diagram of the slice as shown in Figure 4.3. 12
- 4. 09-Mar-21 4 Figure 4.3: forces on a slice of soil in an infinite slope. 13 Cont.… 14 The forces acting on the slice per unit thickness are: the weight bz W , the shear forces j X and 1 j X on the sides, the normal forces j E and 1 j E on the sides, the normal force N on the slip plane and the mobilized shear resistance of the soil, T , on the slip plane. We will assume that forces that provoke failure are positive. If seepage is present, a seepage force bz i J w s develops, o where i-is the hydraulic gradient. For a uniform slope of infinite extent, 1 j j X X and 1 j j E E . Cont.… To continue with the limit equilibrium method, we must now use the equilibrium equations to solve the problem. The general objective of infinite slope stability analysis is to determine either the critical slope or critical height, or alternatively, the factor of safety of the slope. 15 Cont.… Factor of Safety The factor of safety of a slope is defined as the ratio of the available shear strength, τf, to the minimum shear strength required to maintain stability (which is equal to the mobilized shear stress on the failure surface) τm, that is: The shear strength of the soil is governed by the Mohr- Coulomb failure criterion (Chapter 1). 16 m f FS (4.1)
- 5. 09-Mar-21 5 Stability of Infinite Slopes In Φu=0, Cu Soil For the u =0, cu soil, the Mohr-Coulomb shear strength is given by: u f c (4.2) From statics and using Figure 4.3, s W N cos and s W T sin (4.3) The shear stress per unit length on the slip plane is given by: s s s s s s m z b bz b W l T cos sin cos sin cos sin (4.4) The factor of safety is then, ) 2 sin( 2 cos sin s u s s u z c z c FS (4.5) 17 At limit equilibrium, FS = 1. Therefore, the critical slope is ) 2 ( sin 1 2 1 z cu c (4.6) and the critical depth is: ) 2 sin( 2 s u c c z (4.7) 18 Cont.… Stability Of Infinite Slopes In C’-ϕ’ Soil – With No Seepage For a c’, ' soil, the Mohr-Coulomb shear strength is given by: ' tan ' ' n f c (4.8) The factor of safety FS is then: m n m m n c c FS ' tan ' ' tan ' ' ' (4.9) The normal and shear stresses per unit length at the failure plane in reference to figure 4.3 are given by: l N n ' and l T m (4.10) 19 For a slope without seepage, Js=0. From Eqns. (4.4, 4.9 and 4.10) we get: s s s s s s s z c W W z c FS tan ' tan cos sin ' sin ' tan cos ' cos sin ' (4.11) At limit equilibrium FS = 1. Therefore, the critical depth zc is given by ' tan tan sec ' 2 s s c c z (4.12) 20 For the case where ' s , the factor of safety is always greater than 1 and is computed from Eqn. (4.11). This means that there is no limiting value for the depth z, and at an infinite depth, the factor of safety approaches to s tan / ' tan . Cont.…
- 6. 09-Mar-21 6 For a coarse-grained soil with c’ = 0, Eqn. (4.11) becomes: s FS tan ' tan (4.13) At limit equilibrium FS = 1. Therefore, the critical slope angle is: ' c (4.14) 21 The implication of Eqn. (4.14) is that the maximum slope angle of a coarse-grained soil with c’ = 0, cannot exceed ' . In other words, the case c’ = 0 and ' s is always unstable and can not be applied to practical situations. Cont.… Stability Of Infinite Slopes In C’-ϕ’ Soil –Steady State Seepage We will now consider groundwater at the ground surface and assume that seepage is parallel to the slope. The seepage force is: bz i J w s Since seepage is parallel to the slope, sin i . From statics, s s bz W N cos ' cos ' ' (4.15) and s sat s w s w s s s bz bz bz bz J W T sin sin ) ' ( sin sin ' sin ' (4.16) 22 Therefore, the shear stress at the slip plane is: s s sat s s sat m z b bz l T cos sin cos sin From the definition of factor of safety (Eqn. 4.3), we get: s sat s s sat s s sat s s s sat z c zb bz z c FS tan ' tan ' cos sin ' tan cos ' tan cos ' cos sin ' (4.17) 23 Cont.… At limit equilibrium, FS=1. Therefore, the critical height is: ' tan ' tan csc ' 2 s s c c z At infinite depth the factor of safety in Eqn. (4.17) becomes: s sat FS tan ' tan ' (4.19) Eqn. (4.19) can also be used for calculating the factor of safety for a coarse-grained soil with c’ = 0. At limit equilibrium FS = 1, and hence, the critical slope for a coarse- grained soil with c’ = 0 is given by: ' tan ' tan sat s (4.18) 24 Cont.…
- 7. 09-Mar-21 7 For most soils, 2 1 ' sat . Thus, seepage parallel to the slope reduces the limiting slope of a clean, coarse-grained soil by about one-half. If the groundwater level is not at the ground surface, weighted average unit weights have to be used in Eqns. (4.17 and 4.18). 25 Cont.… Rotational Slope Failure The infinite slope failure mechanism is reasonable for infinitely long and homogeneous slopes made of coarse-grained soils, where the failure plane is assumed to be parallel to the ground surface. But in many practical problems slopes have been observed to fail through a rotational mechanism of finite extent. As shown in Fig. (4.1), rotational failure mechanism involves the failure of a soil mass on a circular or non-circular failure surface. In the following sections, we will continue to use the limit equilibrium method assuming a circular slip surface. Those methods, which are based on non-circular slip surface, are beyond the scope of this course. 26 Stability Of Slopes In Cu, Φu = 0 Soil–circular Failure Surfaces The simplest circular analysis is based on the assumption that a rigid, cylindrical block will fail by rotation about its center and that the shear strength along the failure surface is defined by the undrained strength cu. Figure 4.4 shows a slope of height H and angle αs The trial circular failure surface is defined by its center C, radius R and central angle θ. The weight of the sliding block acts at a distance d from the center. 27 Figure 4.4: slope failure in cu, ϕu=0 28 Cont.…
- 8. 09-Mar-21 8 Taking moments of the forces about the center of the circular arc, we have: Where: L is the length of the circular arc, W is the weight of the sliding mass and d is the horizontal distance between the circle center, C, and the centroid of the sliding mass. If cu varies along the failure surface then: 29 0 0 2 180 Wd R c Wd LR c FS u u (4.19) 0 0 0 2 2 0 1 1 2 180 ) ... ( Wd c c c R FS n un u u (4.20) Cont.… Effect Of Tension Cracks Tension cracks may develop from the upper ground surface to a depth z0 that can be estimated using Eqn. (3.13). The effect of the tension crack can be taken into account by assuming that the trial failure surface terminates at the depth z0, thereby reducing the weight W and central angle θ Any external water pressure in the crack creates a horizontal force that must be included in equilibrium considerations. 30 Stability Of Slopes In C’-ϕ’ Soil – Method Of Slices The stability of a slope in a c’- ϕ’ soil is usually analysed by discretizing the mass of the failure slope into smaller slices and treating each individual slice as a unique sliding block (Fig. 4.5). This technique is called the method of slices. In the method of slices, the soil mass above a trial failure circle is divided into a series of vertical slices of width b as shown in Fig. 4.6 (a). For each slice, its base is assumed to be a straight line defined by its angle of inclination θ with the horizontal whilst its height h is measured along the centerline of the slice. 31 Figure 4.5: slice discretization and slice forces in a sliding mass ASTU/Soil Mechanics-II 32 Cont.…
- 9. 09-Mar-21 9 Figure 4.6: a) method of slices in c’-ϕ soil, b) forces acting on a slice. ASTU/Soil Mechanics-II 33 Cont.… The forces acting on a slice shown in Fig. 4.6 (b) are: W = total weight of the slice = ×h×b N = total normal force at the base = N’ + U, where N’ is the effective total normal force and U = ul is the force due to the pore water pressure at the midpoint of the base length l. T = the mobilized shear force at the base = l m , where m is the minimum shear stress required to maintain equilibrium and is equal to the shear strength divided by the factor of safety: FS f m . X1, X2 = shear forces on sides of the slice and E1, E2 = normal forces on sides the slice. The sum of the moments of the interslice or side forces about the center C is zero. 34 Cont.… Thus, for moment equilibrium about the center C (note the normal forces pass through the center): n i i i n i i i f n i i m n i i i R W FS l R l R R T 1 1 1 1 ) sin ( ) ( ) ( (4.20) where n is the total number of slices. Replacing f by the Mohr-Coulomb shear strength, we obtain: n i i i n i i i n i i i n i i i n W N l c W l c FS 1 1 1 1 ' ) sin ( ) ' tan ' ' ( ) sin ( ) ' tan ' ( (4.21) 35 Cont.… 36 The term c’l may be replaced by cos / 'b c . For uniform c’, the algebraic summation of c’l is replaced by c’L, where L is the length of the circular arc. The values of N’ must be determined from the force equilibrium equations. However, this problem is statically indeterminate – because we have six unknown variables for each slice but only three equilibrium equations. Therefore, some simplifying assumptions have to be made. In this chapter two common methods that apply different simplifying methods will be discussed. These methods are called the Fellenius method and Bishop simplified method. Cont.…
- 10. 09-Mar-21 10 Fellenius Or Ordinary Or Swedish Method The ordinary or Swedish method of slices was introduced by Fellenius (1936). This method assumes that for each slice, the interslice forces X1=X2 and E1=E2. Based on this assumption and from statics, the forces normal to each slice are given by: 37 ul W N ul N W N cos ' ' cos (4.22) Substituting N’ into Eqn. 4.21, we obtain: n i i i n i i i W ul W l c FS 1 1 ) sin ( ) ' tan ) cos ( ' ( (4.23) For convenience, the force due to pore water is expressed as a function of W: i i i u W b u r (4.24) Where ru is called the pore water pressure ratio. Consequently, we have: n i i i n i i i u W r W l c FS 1 1 ) sin ( ) ' tan ) sec (cos ' ( (4.25) 38 Cont.… The term ru is dimensionless because the term 1 b h ub w w represents the weight of water with a volume of 1 b hw . Furthermore, ru can be simplified as follows: h h hb b h W ub r w w w w u (4.26) 39 Cont.… Bishop Simplified Method This method assumes that for each slice X1=X2 but E1≠ E2. These assumptions are considered to make this method more accurate than the Swedish method. An increase of 5% to 20% in the factor of safety over the Swedish method is usually obtained. Referring to Figure 4.6 b, and writing the force equilibrium in vertical direction (in order to eliminate E1 and E2), the following equation for N’ can be found: 40 FS FS l c ul W N ' tan sin cos sin ' cos ' (4.27)
- 11. 09-Mar-21 11 In addition to the force in the vertical direction, Bishop Simplified method also satisfies the overall moment equilibrium about the center of the circle as expressed in Eqn. (4.21). Putting cos / b l and W r ub u , and substituting Eqn. (4.27) into Eqn. (4.21), we obtain: 41 i n i i u n i i i m r W b c W FS 1 1 ' tan ) 1 ( ' ) sin ( 1 (4.28) where, FS m ' tan sin cos (4.29) Cont.… Equation (4.29) is non-linear in FS (that is FS appears on both sides of the equations) and is solved by iteration. An initial value of FS is guessed (slightly greater than FS obtained by Fellenius’ method) and substituted to Eqn. (4.29) to compute a new value for FS. This procedure is repeated until the difference between the assumed and computed values is negligible. Convergence is normally rapid and only a few iterations are required. The procedure is repeated for number of trial circles to locate the critical failure surface with the lowest factor of safety. 42 Cont.… Thank you!!! 43