Question 6 of 10: Factor of safety of a two-layer slope by the method of slices

Question 6: Factor of safety of a two-layer slope by the method of slices (24 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two-layer slope drawn on a 1 m grid. Taking the origin at the bottom left corner of the section, the base of Soil 2 is at $y = 0$, the Soil 1 / Soil 2 interface and the ground in front of the toe are at $y = 3$ m, the toe is at (7, 3) and the crest at (16, 10), so the face rises 7 m in 9 m. The trial circle read from the grid has its centre O at (8.5, 11.0) m and radius $R = 10$ m. No water table is shown, so pore pressures are taken as zero.

Find. The factor of safety of the given trial circle, using three slices.

[Figure not reproduced: Figure 1 (redrawn to the exam's 1 m grid). Trial circle centred on O (8.5 m, 11.0 m) with R = 10 m, and the three slices required by the hint: slice 1 lies wholly in Soil 2, slice 2 spans both layers, slice 3 lies wholly in Soil 1. See the official exam paper.]

Approach. Locate the two ends of the slip surface and the point at which it crosses the layer interface; those points, together with the toe, are exactly the slice boundaries the hint asks for. Compute each slice's weight from its area in each layer, take the base inclination at the slice mid-point, and assemble the ordinary (Fellenius) method of slices, in which the interslice forces are ignored so that the effective normal force on each base is $N' = W\cos\alpha$.

1. Write the slip surface and find where it meets the ground and the layer interface. With the centre at $(x_O, y_O) = (8.5,\ 11.0)$ m and $R = 10$ m$,$ the arc is $$y_s(x) = y_O - \sqrt{R^2 - (x-x_O)^2} = 11 - \sqrt{100 - (x-8.5)^2}$$Setting $y_s = 3$ m (the ground in front of the toe, which is also the layer interface) gives $(x-8.5)^2 = 100 - 64 = 36$, so $x = 2.5$ m and $x = 14.5$ m. Setting $y_s = 10$ m (the crest) gives $x = 8.5 + \sqrt{99} = 18.45$ m. The lowest point of the arc is at $y = y_O - R = 1.0$ m, which is 2 m above the base of Soil 2, so the whole slip surface stays inside the section.
2. Set the three slice boundaries. The hint fixes them without any further choice. Between $x = 2.5$ and $x = 7.0$ m the ground surface is the flat area in front of the toe, so the sliding mass there is wholly Soil 2. Between the toe at $x = 7.0$ m and $x = 14.5$ m the arc is still below the interface while the surface above is the slope face, so that slice contains both layers. Beyond $x = 14.5$ m the arc has risen above the interface and the slice is wholly Soil 1. The cuts are therefore at $x = 2.5,\ 7.0,\ 14.5$ and $18.45$ m, giving widths $b = 4.50,\ 7.50$ and $3.95$ m.
3. Compute the area of each layer within each slice. The area between a horizontal line $y = y_t$ and the arc, over $x_1 \le x \le x_2$, follows from the circular-segment integral with $u = x - x_O$: $$A = (y_t - y_O)(x_2-x_1) + \left[\frac{u}{2}\sqrt{R^2-u^2} + \frac{R^2}{2}\arcsin\frac{u}{R}\right]_{u_1}^{u_2}$$For slice 1, with $y_t = 3$ m, $u_1 = -6$, $u_2 = -1.5$, this gives $A = -36 + 41.231 = 5.231$ m$^2$ of Soil 2. Slice 2 has $11.119$ m$^2$ of Soil 2 below the interface, plus a triangle of Soil 1 above it of area $\tfrac{1}{2}(0 + 5.833)(7.5) = 21.875$ m$^2$. Slice 3 is entirely Soil 1, with $17.507$ m$^2$ between the ground surface and the arc.
4. Convert areas to slice weights. Each slice is 1 m thick normal to the section, so $W = \gamma_1 A_1 + \gamma_2 A_2$. Slice 1 gives $W_1 = 18(5.231) = 94.2$ kN/m; slice 2 gives $W_2 = 20(21.875) + 18(11.119) = 437.5 + 200.1 = 637.6$ kN/m; slice 3 gives $W_3 = 20(17.507) = 350.1$ kN/m.
5. Take the base inclination and base length of each slice. The tangent to the circle at the slice mid-point makes an angle $\alpha$ with the horizontal, where $$\begin{aligned}\sin\alpha &= \frac{x_m - x_O}{R} \\ l &= \frac{b}{\cos\alpha}\end{aligned}$$Mid-points $x_m = 4.75,\ 10.75$ and $16.475$ m give $\sin\alpha = -0.375,\ +0.225$ and $+0.7975$, that is $\alpha = -22.02^{\circ},\ +13.00^{\circ}$ and $+52.89^{\circ}$, and base lengths $l = 4.854,\ 7.697$ and $6.547$ m. The negative angle on slice 1 is real: that slice sits on the up-slope side of the circle's low point and its weight resists sliding.

   Slice geometry, weights and base inclinations

   Slice	x range (m)	b (m)	Area in Soil 1 (m2)	Area in Soil 2 (m2)	W (kN/m)	alpha (deg)	l (m)
   1	2.50 to 7.00	4.500	0.000	5.231	94.2	-22.02	4.854
   2	7.00 to 14.50	7.500	21.875	11.119	637.6	13.00	7.697
   3	14.50 to 18.45	3.950	17.506	0.000	350.1	52.89	6.547

6. Assemble the ordinary method of slices. With no pore water pressure the effective normal force on each base is $N' = W\cos\alpha$, and the factor of safety is the ratio of available shear resistance to the shear needed for equilibrium along the arc: $$F = \frac{\sum \left(c'_i\, l_i + W_i\cos\alpha_i\,\tan\phi'_i\right)}{\sum W_i \sin\alpha_i}$$The base of each slice takes the strength of the layer it actually passes through: slices 1 and 2 sit in Soil 2 ($c' = 10$ kPa, $\phi' = 20^{\circ}$), slice 3 in Soil 1 ($c' = 0$, $\phi' = 30^{\circ}$).

   Ordinary (Fellenius) method of slices — resisting and driving terms

   Slice	c′ (kPa)	phi′ (deg)	c′ l (kN/m)	W cos(alpha) tan(phi′) (kN/m)	Resisting (kN/m)	W sin(alpha) (kN/m)
   1	10	20	48.54	31.77	80.32	-35.31
   2	10	20	76.97	226.13	303.10	143.47
   3	0	30	0.00	121.96	121.96	279.23
   Sums					505.38	387.38

7. Evaluate the factor of safety. Substituting the two column sums, $$F = \frac{505.4}{387.4} = \boxed{1.30}$$The circle is stable but only marginally so. Slice 3, the steep upper wedge in the cohesionless Soil 1, supplies about 72 per cent of the total driving moment while contributing no cohesion at all; slice 1 works in the opposite sense and removes 35 kN/m of driving force.