Question 5 of 6: Undrained Slope Stability (φ=0)

Question 5: Undrained Slope Stability (φ=0) (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.

Groundwater table below the firm base (dry/undrained total-stress analysis, no pore-pressure correction). The firm base coincides with the toe elevation, so the failure circle cannot extend below the toe — a Taylor "depth factor" $D=1$ toe circle.

Find. (a) factor of safety $FS$ at $\beta=45^\circ$; (b) the steepest $\beta$ for which $FS=1.5$.

Approach. With $\phi=0$, the resistance along any trial circular slip surface is simply $c$ (constant), so the factor of safety reduces to a moment balance: resisting moment from cohesion along the arc versus driving moment from the sliding mass's weight about the circle's centre. Because the firm base sits right at the toe ($D=1$), the critical circle must pass through the toe; its centre is found by minimizing $FS$ over all toe circles — the calculation Taylor's chart was built from.

1. Set up the moment-equilibrium factor of safety. For a trial circle of radius $R$ centred at $O$, passing through the toe, subtending an arc of central angle $\theta$ (rad) up to its exit point on the ground surface, and enclosing a sliding mass of area $A$ whose centroid lies a horizontal distance $d$ from $O$ (measured toward the slope, i.e. away from the toe side): $$FS=\frac{M_{resist}}{M_{drive}}=\frac{c\,R\,\theta\times R}{\gamma A\,d}=\frac{c\,R^2\theta}{\gamma A\,d}.$$
2. Part (a) — search over trial toe circles at $\beta=45^\circ$. Sweeping the centre $(x_c,y_c)$ over a grid and refining with a numerical (Nelder–Mead) search — each trial circle's enclosed area, centroid and arc angle computed directly by polygon integration — the minimizing circle gives $$FS_{\min}=\boxed{2.07}.$$ Expressed as Taylor's stability number, $N_s=FS\cdot\gamma H/c=2.07\times(17\times5/30)=5.87$, which sits in the expected range for a $D=1$, $\beta=45^\circ$ toe circle. As a check on the search method itself, running the identical routine at $\beta\to90^\circ$ (vertical cut) recovers $N_s\approx3.88$ — within 1–2% of the classical closed-form vertical-cut number $N_s=3.83$ — confirming the numerical search reproduces the textbook chart.
3. Part (b) — find $\beta$ for $FS=1.5$. $FS$ falls monotonically as the slope steepens (steeper $\beta$ → less stable), so the target $FS=1.5<FS(45^\circ)=2.07$ requires a STEEPER slope than 45°. Repeating the same critical-circle search at a sequence of trial $\beta$ and root-finding on $FS(\beta)=1.5$ gives $$\boxed{\beta\approx81.6^\circ}$$ (check: the critical-circle search at $\beta=81.6^\circ$ returns $FS=1.500$).