equilibrum that have been used throughout this book. While this method of analysis is limited to a few simple cases of toppling failure, it provides a basic understanding of the factor that are important in toppling, and allows stabilization options to be evaluated. The stability of the slope. Thats is, as the base angle become flatter, the lengths of the blocks increase and there is a greater tendency of the taller blocks to topple resulting in decreased stability of the slope. If the base angle is coincident with the base of the block (..), then that geometry of toppling requires dilatancy of the block along the base plane and shearing on the faces of the block (Figure 9.8) However, if the base is stepped (i.e...) then each block can topple without dilatancy, provided there is displacement on the face-to-face contacts (figure 9.7). It is expected that more energy is required to dilate the rock mass than to develop shear along existing discontinuities, and so stepped base is more likely than a plannar base. Examination is base friction, centrifugal and numerical models (goodman and bray, 1976: Pritchard and Savigny, 1990. 1991: Adhikary et al 1997) show that base planes tend to be stepped and the approximate dip angle is in the range

It is consider that an appropriate stability analysis procedure for situation where the value of ( ) is unknown is to carry out a sensitivity analysis within the range given by and find the value that gives the least stable condition.

Based on the slope geometry shown in Figure 9.7, the number of blocks a making up the system is given by

(9.6)

The blocks are numbered from the roe of the slope upwards, with the lowest block being I and the upper block being a. In this idealized model, the height y of the ath block in a position below

the crest of the slope is v,, =n(a -h) (9.8) while above the crest = Yn—1 -a -b (9.9) The three constants at, a and b that are defined by the block and slope gcometry and are given by a = xtan(frf - (9.10) a2 = xtan(fr - *s) (9.11) b = xtan(t’b - (9.12)

9.4.2 Block stability Figure 9.7 shows the stability of a system of blocks subject to toppling, in which it is possible to distinguish three separate groups of blocks according to their mode of behavior: (a) A set of stable blocks in the upper part of the slope, where the friction angle of the base of the blocks is greater than the dip of this plane (i.e. p> and the height is limited so the center of gravity lies inside the base (v/tx <COtvp).:

(b) An intermediate set of toppling blocks where the center of gravity lies outside the base. (c) A set of blocks in the roe region, which are pushed by the toppling blocks above. Depending on the slope and block geometries, the toe blocks may be stable, topple or slide. Figure 9.9 demonstrates the terms used to define the dimensions of the blocks, and the position and direction of all the forces acting on the blocks during both toppling and sliding. Figure 9.9(a) shows a typical block (n) with the normal and shear forces developed on the base (R.,S;), and on the interfaces

with adjacent blocks (Pa, Q,,, P,-j, Q,!). When the block is one of the toppling set, the points of application of all forces are known, as shown in Figure 9.9(b). The points of application of the normal forces

P, are M and L on the upper and lower faces respectively of the block, and are given by the following. If the nth block is below the slope crest, then

For an irregular array of blocks, ye, L,, and M can be determined graphically. when sliding and toppling occurs, frictional forces are generated on the bases and sides of the blocks. In many geological environments, the friction angles on these two surfaces are likely to he differenr. For example, in a steeply dipping sedimentary sequence comprising sandstone beds separated by thin seams of shale, the shale will form the sides of the blocks, while joints in the sandstone will form the bases of the blocks. For these conditions, the friction angle of the sides of the blocks (dd) will be lower than friction angle on the bases (dn). These two friction angles can he incorporated into the I imnit equilibrium analysis as follows. For limiting friction on the sides of the block:

By resolving perpendicular and parallel to the base of a bloek with weipht , the normal and shear forces wting on the base of bloek n are,

Considering rotational equilibrium, it n found that the forie P, rhar as just suf&ienr to prevent toppling has the value

When the block under consideration is one of the sliding set (Figure 9.9(c)),

1Hlowever, the magnitudes of the forces PM-I and RA applied to the sides and base of the block, and their points of application L," and Kn. are unknown. Although the problem is indeterminate, the force P-1 required to prev vent sliding of block n can be determined if it is assumed that Q4-i = (tan4d . P.i). Then the shear force just sufficient to prevent sliding has the value.

9.43 Cakulation procedure for toppling stability of a system of blocks The calculation procedure for examining toppling stabiliry of a slope comprising a system of blocks dipping steeply inro the faces is as follows: (i) The dimensions of each block and the number of blocks are defined using equations (9.7)-(9.12). (ii) Values for the friction angles on the sides and base of the blocks (dj and ,) are assigned based on laboratory testing, or inspection. The friction angle on the base should be greater than the dip of the base to prevent sliding (i.e. > (iii) Starting with the top block, equation (9.2) is used to identify if toppling will occur, thar is, when y/Ax > cor For the upper toppling block, equations (9.23) and (9.25) are used to calculate the lateral forces required to prevent toppling and sliding, respectively. (iv) Let mnt be the uppermost block of the toppling set.

(v) Starting with block uni, determine the late eral forces P-j,, required to prevent topp llin,? and to prevent sliding. If P—ids the block is on the point of toppling and P4-1 is set equal to or if Pn-i,s s the block is on the point of sliding and P,1-1 is set equal to P,,-i,c.

In addition, a check is made that there is a normal force R on the base of the block, and that sliding does not occur on the base, that is R,, s-0 and (ISxI > R,1tan%), (vi) The next lower block (nj — 1) and all the lower blocks are treated in succession using

the same procedure. It may be found that a relatively short block that does not satisfy equation (9.2) for toppling, may still topple if the moment applied by the thrust force on the upper face is great enough to sarisfy the condition stated in (v) above. If the condit iion n-1I > P,,-1. is met for all blocks, then toppling extends down to block 1 and sliding does not occur. (vii) Eventually a block may be reached for which P-i,, s P,,l1,,. This establishes block 1n2, and for this and all lower blocks, the critical state is one of sliding. The stability of the sliding blocks is checked using equation (9.24), with the block being unstable if (S = R tanA). If block 1 is stable against both sliding and toppling (i.e. Po <0), then the overall slope is considered to be stable. If block I either topples or slides (i.e. P() > 0), then the overall slope is considered to be unstable.

9.4.4 Cable torce required to stabilize a slope If the calculation process described in Section 9.4.3 shows that block 1 is unstable, then a tensioned cable can be installed through this block and anchored in stable rock beneath the zone of toppling to prevent movemenr. The design parameters for anchoring are the bolt tension, the plunge of the anchor and its posirion on block 1 (Figure 9.9(c)). Suppose that an anchor is installed at a plunge angle T through block 1 at a distance L1 above its base. The anchor tension required to prevent toppling of block 1 is

while the required anchor tension to prevent sliding of block 1 is

When the force T is applied to block 1, the normal and shear force on the base of the block are, respectively,

The stability analysis for a slope with a tensioned anchor in block 1 is identical to that described in Section 9.4.3, apart from the calculations relating to block 1. The required tension is the greater of T and T defined by equations (9.26) and (9.27). 9.4.5 Iactor ot safrty ftr limiti,ig ejuilibriuiii analysis of toppling failures For both reinforced and unreinforced slopes, the factor of safety can be calculated by finding the friction angle for limiting equilibrium. The proc eedrre is first to carry out the limiting equilibrium stability analysis as described in Section 9.4.3

using the estimated values for the friction angles. If block 1 is unstable, then one or both of the frict iion angles are increased by increments until the value of P0 is very small. Conversely, if block 1 is stable, then the friction angles are reduced until P) is very small. These values of the friction angles are those required for limiting equilibriumn.

The limiting equilibrium friction angles are termed the required friction angles, while the actual friction angles of the block surfaces are termed the available friction angles. The factor of safery for toppling can be defined by dividing the tangent of the friction angle believed to apply to the rock layers (tan a’ajtahIe)’ by the tangent of the friction angle required for equilibrium (tan #rmlLjuircd)

The actual factor of safety of a toppling slope depends on the details of the geonietry of the topp llin? blocks. Figure 9.7 shows that once a column

overturns by a small amount, there are edge-to- face contacts between the blocks, and the friction required to prevenr further rotarion increases. Hence, a slope just at limiting equilibrium is metas stable. However, rotation equal to 2(vb — will convert the edge-to-face contacts along the sides of the columns into continuous face contacts and the friction angle required to prevent further rotation will drop sharply, possibly even below that required for initial equilibrium. The choice of factor of safety, therefore, depends on whether or not some deformation can be tolerated. The restoration of continuous face-to-face contact of toppled columns of rock is probably an important arrest mechanism in large-scale topp llin? failures. In many cases in the field, large surface displacements and tension crack forman oion can be observed and yet the volumes of rock that fall from the face are small.

9.4.6 Exam pie of liiir it equilibrium analysis of toppling The following is an example of the application of the Goodman and Bray limit equilibrium anal yysis to calculate the factor of safery and required bolting force of the toppling failure illustrated in Figure 9.10(a). A rock face 92.5 m high (H) is cur ar an angle of 56.60(*f) in a layered rock mass dipping ar 60° into the face (vd = 60°): the width of each block is 10 m (Ax). The angle of the slope above the crest of the cut is 4°(k), and the base of the blocks is stepped 1 mat every block (atn (1/10) = 5.7°, and V’h = (S.7+wc.) = 35.7°). Based

on this geometry, there are 16 blocks formed between the toe and crest of the slope (equation 9.7); block 10 is at the crest. Using equations (9.10)-(9.12), the constants are aj = 5.0 m, a = 5.2 m and b = 1.0m. These constants are used to calcul aaee the height v of each block, and the height to width ratio va/Ax as shown on the table imn Figure 9.10(b). The friction angles on the faces and bases of the blocks are equal and have a value of 38.1S°(daaiIahIc). The unit weighr of the rock is

25 kN/m3. It is assumed thut the slope is dry, and that there are no external forces acting. The stability analysis is started by exanining the oppling/sliding mode ot each block, starting at the crest. Since the

fnction angle on the base of the blocks is 38.150 and the dip of the base is 30. the upper blocks are stable against sliding. Equation 9.2) is then used to assess the toppling

mode. Since cots = 1.73, blocks 1r, 15 and 14 are stable, because for each the ratio y./Ax is