07-Str-B1 · May 2013

Question 6 of 10: Factor of Safety of a Two-Layer Slope by the Method of Slices

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examinations — May 2013 — 07-Str-B1 Geotechnical Design. Three-hour, OPEN-BOOK exam; any non-communicating calculator permitted (the candidate must record its make and model). Format: Section A carries five short-answer questions of 7 marks each, of which any FOUR are to be answered; Section B carries the long design questions at 24 marks each, of which any THREE are to be answered. The paper instructs candidates to state any interpretive assumptions and to identify the source of every design chart or assumed value used. Every question in both sections is worked below, because the set is intended as a study resource.

Reference texts: Das, B.M., Principles of Foundation Engineering (9th ed., Cengage) — shallow foundations, consolidation settlement, sheet-pile walls, retaining walls and drilled shafts; Das, B.M., Principles of Geotechnical Engineering (9th ed., Cengage) — method of slices, lateral earth pressure, consolidation theory; Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM, 4th ed., 2006) — Canadian practice for SPT interpretation, pile design, limit-states design and tolerable settlement; Craig, R.F. / Knappett, J.A., Craig's Soil Mechanics (8th ed., CRC Press) — effective stress, shear strength and slope stability; Duncan, J.M., Wright, S.G. & Brandon, T.L., Soil Strength and Slope Stability (2nd ed., Wiley) — choice of strength parameters and factors of safety for short- and long-term analyses.

NOTE 1 — question numbering in the source. The printed paper labels the retaining-wall problem (Figure 4) and the drilled-pier problem (Figure 5) both as "Question 9", while the Section B heading reads "answer any THREE of the following FOUR questions". Section B therefore contains five printed problems under four numbers. They are set out below as Question 9 (retaining wall) and Question 10 (drilled pier) in printed order, so that each can be referred to unambiguously; the marks shown are those printed against each problem.

NOTE 2 — dimensions scaled from Figure 1. Figure 1 is a hand-drawn slope on a 1 m × 1 m grid with no written dimensions other than $R=10$ m. The geometry used in Question 6 was scaled from that grid: slope height 7 m over a 9 m horizontal run, a 3 m thick lower layer, and the centre of the trial circle 1.4 m horizontally beyond the toe and 8.0 m above it. Every one of these values reproduces the drawing to within about 0.2 m (one fifth of a grid square). Check against the original if the paper is used for marking rather than study.

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. A slope cut in two horizontal layers, with the trial slip circle of Figure 1 scaled from the 1 m grid.

Given data
Quantity	Value
Upper layer, Soil 1 — unit weight, friction angle, cohesion	$\gamma_1=20$ kN/m³, $\phi'_1=30^\circ$, $c'_1=0$
Lower layer, Soil 2 — unit weight, friction angle, cohesion	$\gamma_2=18$ kN/m³, $\phi'_2=20^\circ$, $c'_2=10$ kPa
Slope height and horizontal run (scaled from grid)	7.0 m rise over 9.0 m run, i.e. $\beta=37.9^\circ$
Thickness of Soil 2 below the toe level	3.0 m
Radius of the trial circle	$R=10.0$ m
Centre of circle relative to the toe	1.4 m horizontally beyond the toe, 8.0 m above it
Groundwater	Not shown on the figure; pore pressures taken as zero
Number of slices	3, divided as directed by the hint

Find. The factor of safety of the trial circle, using the ordinary (Fellenius, Swedish) method of slices with three slices divided so that slice 1 lies wholly in Soil 2, slice 2 spans both layers, and slice 3 lies wholly in Soil 1.

[Figure not reproduced: Figure 1 (redrawn to scale from the 1 m grid) — slope, layer boundary at the toe level and the trial circle of radius 10 m. Slice 1 lies wholly in Soil 2, slice 2 spans both layers, slice 3 lies wholly in Soil 1. Soil 1: γ=20 kN/m³, φ′=30°, c′=0. Soil 2: γ=. See the official exam paper.]

Approach. Set up coordinates on the toe, write the circle and the ground surface as equations, find the two points where the circle meets the ground surface and the point where it re-crosses the layer boundary; those three points and the toe give exactly the slice division the hint asks for. Then compute each slice's weight from its areas of Soil 1 and Soil 2, take the base inclination at the slice mid-point, and assemble the Fellenius ratio of resisting to driving moments about the centre.

Establish the geometry from the grid. Take the origin at the toe, $x$ positive into the slope and $z$ positive upwards. The ground surface is $z=0$ for $x\le0$, the slope face is $z=\tfrac{7}{9}x$ for $0\le x\le9$, and the crest is $z=7$ for $x\ge9$. The boundary between Soil 1 and Soil 2 is the horizontal plane $z=0$ through the toe, with Soil 2 occupying $-3\le z\le0$. The circle has centre $O=(1.4,\,8.0)$ and radius $R=10$ m, so its lower surface is $$z_{\text{arc}}(x)=8.0-\sqrt{100-(x-1.4)^2}$$ Its lowest point is $z=8.0-10.0=-2.0$ m, which lies 1.0 m above the base of Soil 2 — the circle stays inside the layer, as the hint requires.
Locate the ends of the failure arc. The arc meets the horizontal ground on the left where $z_{\text{arc}}=0$: $$(x-1.4)^2=100-8.0^2=36 \;\Rightarrow\; x=1.4-6.0=-4.60\ \text{m}$$ and it emerges through the crest where $z_{\text{arc}}=7$: $$(x-1.4)^2=100-1.0^2=99 \;\Rightarrow\; x=1.4+9.950=+11.35\ \text{m}$$ Since $11.35>9.0$, the arc does indeed daylight on the horizontal crest and not on the slope face.
Find the third division point. The arc rises back through the layer boundary $z=0$ at $$x=1.4+6.0=+7.40\ \text{m}$$ Beyond this the slip surface runs entirely within Soil 1. The four station points $x=-4.60,\ 0,\ +7.40,\ +11.35$ therefore give precisely the three slices the hint describes: slice 1 lies below the horizontal ground and wholly within Soil 2, slice 2 straddles the boundary and contains both soils, and slice 3 lies wholly above the boundary in Soil 1.
Compute the base inclination of each slice. For a slice whose mid-point is at $x_m$, the base of the slice makes an angle $\alpha$ with the horizontal given by $$\sin\alpha=\frac{x_m-x_O}{R}$$ taken positive when the mid-point lies on the driving side of the centre. For slice 1, $x_m=-2.30$ m gives $\sin\alpha_1=(-2.30-1.4)/10=-0.370$ and $\alpha_1=-21.72^\circ$: this slice sits behind the centre and its weight resists rotation. Slice 2 gives $\alpha_2=+13.30^\circ$ and slice 3 gives $\alpha_3=+52.89^\circ$. The base length of each slice is $\ell=b/\cos\alpha$.
Compute the area of each soil in each slice. The areas follow from integrating between the ground surface, the layer boundary and the arc. Slice 1 contains only Soil 2, of area $$A_{2,1}=\int_{-4.60}^{0}\big[0-z_{\text{arc}}(x)\big]\,dx=5.421\ \text{m}^2$$ Slice 2 contains a triangle of Soil 1 above the boundary, $$A_{1,2}=\int_{0}^{7.40}\tfrac{7}{9}x\,dx=\tfrac{7}{18}(7.40)^2=21.296\ \text{m}^2$$ together with $A_{2,2}=10.929$ m² of Soil 2 between the arc and the boundary. Slice 3 contains only Soil 1, $A_{1,3}=17.386$ m², bounded above by the slope face up to $x=9$ m and by the crest beyond it.
Assemble the slice weights. With $W=\gamma_1A_1+\gamma_2A_2$ per metre run, $$W_1=18(5.421)=97.6\ \text{kN/m}$$ $$W_2=20(21.296)+18(10.929)=425.9+196.7=622.6\ \text{kN/m}$$ $$W_3=20(17.386)=347.7\ \text{kN/m}$$
Identify the soil on each slice base. This is the step the hint exists to force. The base of slices 1 and 2 lies below $z=0$ and therefore in Soil 2, so those bases carry $c'=10$ kPa and $\phi'=20^\circ$. The base of slice 3 lies above $z=0$ and therefore in Soil 1, which is cohesionless with $\phi'=30^\circ$. The unit weights used for the slice weights are taken layer by layer; only the base parameters switch.
Apply the ordinary method of slices. With zero pore pressure the Fellenius expression for the factor of safety about the centre of rotation is $$FS=\frac{\sum\big(c'_i\ell_i+W_i\cos\alpha_i\tan\phi'_i\big)}{\sum W_i\sin\alpha_i}$$ Every term is a force on the base of a slice, multiplied by the common lever arm $R$, which therefore cancels. Slice by slice:

Slice computation — ordinary (Fellenius) method of slices, $R = 10$ m
Slice	$x$ range (m)	$b$ (m)	$\alpha$ (°)	$\ell$ (m)	$A_1$ (m²)	$A_2$ (m²)	$W$ (kN/m)	Base soil	$c'\ell+W\cos\alpha\tan\phi'$	$W\sin\alpha$
1	−4.60 to 0	4.60	−21.72	4.951	0	5.421	97.6	Soil 2	82.51	−36.10
2	0 to 7.40	7.40	+13.30	7.604	21.296	10.929	622.6	Soil 2	296.58	+143.21
3	7.40 to 11.35	3.95	+52.89	6.547	17.386	0	347.7	Soil 1	121.12	+277.30
Sums									500.21	384.41

Taking the sums into the Fellenius expression,

$$FS=\frac{500.21}{384.41}=\boxed{1.30}$$

The arithmetic of the two governing slices is worth setting out, since it shows where the resistance comes from. On slice 2, $c'\ell=10(7.604)=76.0$ kN/m and $W\cos\alpha\tan\phi'=622.6\cos13.30^\circ\tan20^\circ=220.5$ kN/m, giving 296.6 kN/m; on slice 3 the base is cohesionless, so the whole of its 121.1 kN/m comes from friction, $347.7\cos52.89^\circ\tan30^\circ$. Slice 1 contributes 82.5 kN/m of resistance and, because its base slopes the other way, a negative 36.1 kN/m of driving force — its weight helps to hold the mass back.

Question 6 — results
Quantity	Value
Entry and exit points of the trial circle (from the toe)	$x=-4.60$ m and $x=+11.35$ m
Slice boundaries	$x=-4.60,\ 0,\ +7.40,\ +11.35$ m
Slice weights $W_1,\ W_2,\ W_3$	97.6, 622.6, 347.7 kN/m
Base angles $\alpha_1,\ \alpha_2,\ \alpha_3$	−21.72°, +13.30°, +52.89°
Total resisting force $\sum(c'\ell+W\cos\alpha\tan\phi')$	500.2 kN/m
Total driving force $\sum W\sin\alpha$	384.4 kN/m
Factor of safety of the trial circle	$\boxed{FS=1.30}$

Check — what this number is and is not. $FS=1.30$ applies to the single trial circle drawn on Figure 1, not to the slope. A complete analysis searches over centres and radii for the minimum; the critical circle will generally give a lower value. Three slices is also a coarse discretisation — the question prescribes it to keep the arithmetic tractable, but ten to fifteen slices would be normal practice, and the ordinary method itself under-estimates $FS$ by roughly 5 to 15% relative to Bishop's simplified method on a circular surface of this kind because it ignores the interslice forces. The value obtained here should therefore be read as a conservative estimate for this one surface.

Topic. Slope stability of a layered slope by the ordinary (Fellenius, Swedish) method of slices.

Key relations. $FS=\dfrac{\sum\left[c'\ell+(W\cos\alpha-u\ell)\tan\phi'\right]}{\sum W\sin\alpha}$, with $\sin\alpha=(x_m-x_O)/R$ and $\ell=b/\cos\alpha$. Bishop's simplified method replaces the numerator term with $\dfrac{c'b+(W-ub)\tan\phi'}{m_\alpha}$, $m_\alpha=\cos\alpha\left(1+\tan\alpha\tan\phi'/FS\right)$, and must be iterated.

Why this works. Taking moments about the centre of the circle gives every base force the same lever arm $R$, so the moment equation reduces to a ratio of forces. The ordinary method obtains the normal force on each base by resolving the slice weight perpendicular to that base and neglecting the interslice forces entirely — which is why it needs no iteration, and why it is conservative.

Common pitfall. Applying one set of strength parameters to the whole surface. The unit weights are taken layer by layer for the weight of each slice, but $c'$ and $\phi'$ belong to the soil the slice base passes through — here Soil 2 for slices 1 and 2 and Soil 1 for slice 3. Using Soil 1 throughout would remove all the cohesion and drop $FS$ below 1.0; using Soil 2 throughout would raise it well above the true value. The second pitfall is forgetting the sign of $\alpha$ on slices behind the centre: their $W\sin\alpha$ is negative and must be subtracted from the driving sum, not added.

Source. Das, Principles of Geotechnical Engineering, Ch. 15 (slopes: method of slices, Bishop's simplified method, stability of slopes with layered soils); Craig's Soil Mechanics, Ch. 12 (slope stability, the Swedish and Bishop routines).

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