FoS by φ = 0 analysis

The $\phi = 0$ analysis is a total stress method for a saturated clay under undrained (short-term) conditions, where $\tau_f = c_u$ and $\phi_u = 0$. It uses the Swedish (slip) circle.

Assumptions

• Homogeneous saturated clay with undrained strength $c_u$ and unit weight $\gamma$.
• Failure along a circular arc of radius $R$ with centre $O$.
• Moments are taken about $O$ per metre length of slope.

Method

1. Draw the slope to scale and choose a trial centre $O$ and radius $R$.
2. Find the area of the sliding mass and compute its weight $W = \gamma A$; locate the centroid at horizontal distance $x$ from $O$.
3. Measure the arc length $L_a = R\theta$ (with $\theta$ in radians).
4. Moments about $O$:

$$\text{Resisting moment} = c_u L_a R, \qquad \text{Driving moment} = Wx$$ $$F = \frac{c_u L_a R}{W x}$$

5. Repeat for several trial circles; the least value of $F$ is the factor of safety of the slope.

If a tension crack of depth $z_c = \dfrac{2c_u}{\gamma}$ is considered, $L_a$ is reduced by the length in the crack, and a hydrostatic force may be added to the driving moment.

Example (assumed data): $c_u = 30$ kN/m², $\gamma = 18$ kN/m³, $R = 12$ m, arc angle $80^\circ$ ($1.396$ rad), area of sliding mass $= 80$ m², $x = 2.5$ m.

$$\begin{aligned} L_a &= 12 \times 1.396 = 16.76\ \text{m} \\ W &= 18 \times 80 = 1440\ \text{kN/m} \\ F &= \frac{30 \times 16.76 \times 12}{1440 \times 2.5} = \frac{6034}{3600} = 1.68 \end{aligned}$$

Answer: $F \approx 1.68$ for this trial circle; the minimum $F$ over trial circles governs the design (should be $\geq 1.3$ to 1.5 for the short term).