Slope Stability of a Drainage Ditch — Taylor's Stability Chart: 18-Env-A3 Geotechnical and Hydrogeological Engineering, May 2013, Q4

Question 4: Slope Stability of a Drainage Ditch — Taylor's Stability Chart (20 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.

Given data
Quantity	Symbol	Value
Ditch depth	$H$	2 m
Side slope	—	2H : 1V ($\beta=26.57^\circ$)
Saturated unit weight	$\gamma$	18.5 kN/m³
Undrained cohesion	$c_u$	40 kN/m²
Undrained friction angle	$\phi_u$	≈ 0

Find. (a) factor of safety $F$ against shear failure of the current 2H:1V slope; (b) the side slope needed for a minimum $F=3$.

Approach. Taylor's chart gives $F=N_F\,c/(\gamma H)$, where the stability number $N_F$ is read off the curve for $\lambda_{c\phi}=\gamma H\tan\phi/c$. Because $\phi_u\approx0$, $\tan\phi_u=0$ regardless of $c$, so $\lambda_{c\phi}=0$ identically — the bottom-most curve on the chart applies directly, with no interpolation between curves needed.

[Figure not reproduced: Taylor's stability-number chart as printed with the exam (F = N_F·c/(γH); the λcφ=0 curve applies since φu≈0). See the official exam paper or the cited reference text.]

Taylor's stability-number chart as printed with the exam (F = N_F·c/(γH); the λcφ=0 curve applies since φu≈0).

Part (a) — read the stability number and compute $F$. With $c/(\gamma H)=40/(18.5\times2)=1.081$, the $\lambda_{c\phi}=0$ curve at $\beta=26.57^\circ$ reads $N_F\approx6.41$ (read against the chart's own log-scale axis; the curve reads 6.44 at $\beta=25^\circ$ and 6.31 at $\beta=30^\circ$, so 6.41 at 26.57° is a smooth, consistent interpolation). Then $$F=N_F\,\frac{c}{\gamma H}=6.41\times1.081=\boxed{6.93}.$$ The existing 2H:1V slope is very safe against undrained shear failure — nearly 2.3 times the usual minimum of 3.
Part (b) — check whether a steeper slope can be found that still just satisfies $F=3$. $F=3$ requires $N_F=3\gamma H/c=3/1.081=2.78$. Reading the $\lambda_{c\phi}=0$ curve across its full charted range, $N_F$ decreases only slowly and gently as $\beta$ increases — from about 7.1 at $\beta=5^\circ$ to about 5.8–6.0 even at the chart's steepest plotted angle, $\beta=45^\circ$ — and never drops anywhere near 2.78 within the chart. Extending the trend beyond the chart to the theoretical limit, a fully vertical unsupported cut ($\beta=90^\circ$) uses the well-known classical Taylor value $N_F=3.83$ for $\phi=0$ (Taylor, 1948), giving $$F_{vertical}=3.83\times1.081=\boxed{4.14},$$ which is still above 3.

Check: because $F_{vertical}\approx4.14>3$ even at the steepest geometrically possible slope angle, no side-slope angle exists that reduces the factor of safety to exactly 3 for this combination of a shallow ($H=2$ m) cut in a relatively strong clay ($c_u=40$ kPa). The ditch is short-term-stable at any practical slope, including vertical, purely on undrained-shear grounds; the governing consideration for a steep unsupported cut this size would instead be occupational excavation-entry safety (e.g. WorkSafeBC/CSA benching and shoring requirements for worker entry), not the calculated global factor of safety. If the intent of part (b) is instead "how much can the slope be steepened while keeping $F\ge3$," the answer is: any angle up to and including vertical satisfies that requirement, so the governing constraint becomes construction/workspace economy rather than shear stability.

Quantity	Value
Stability number, $N_F$ (2H:1V, $\lambda_{c\phi}=0$)	6.41
Factor of safety, current 2H:1V slope	F ≈ 6.93
Factor of safety at vertical cut ($\beta=90^\circ$, $N_F=3.83$)	F ≈ 4.14 (still > 3)
Slope satisfying $F=3$	none within 0–90° — not the binding case for this cut

Concept & Theory

Topic. Short-term (undrained, $\phi=0$) slope stability of a cut in saturated clay, via Taylor's stability-number chart.

Key relations. $F=N_F\,c/(\gamma H)$; $\lambda_{c\phi}=\gamma H\tan\phi/c$ selects which curve on the chart applies; for $\phi_u=0$, $\lambda_{c\phi}\equiv0$ regardless of $c$ or $H$.

Why this works. Taylor's chart is a pre-solved envelope of the critical (minimum-$F$) circular slip surface over every possible centre and radius, for a given slope angle and soil-strength ratio — reading $N_F$ replaces a manual trial-circle search. $F$ scales directly with $c/(\gamma H)$, so a shallow cut in a moderately strong clay is inherently very stable, and doubling the depth (not steepening the slope) is what would most efficiently erode that margin.

Common pitfall. Assuming a "required side slope for $F=3$" must always exist and force-fitting a shallower or steeper angle without first checking whether the CURRENT slope, or even a vertical cut, already exceeds the target — here it does, decisively.

Source. Das, Principles of Geotechnical Engineering, ch. 12 (slope stability, Taylor's method); Taylor, D.W. (1948), Fundamentals of Soil Mechanics.

Question 4 of 6: Slope Stability of a Drainage Ditch — Taylor's Stability Chart

Question 4: Slope Stability of a Drainage Ditch — Taylor's Stability Chart (20 marks)