Stability of cantilever wall, 8.6 m

Assumptions (figure read from the scan): stem height 8.6 m, top width 0.4 m, front face vertical, back face inclined at $84^\circ$ to the horizontal, so the stem thickness at the base is $0.4 + 8.6/\tan84^\circ = 1.30$ m. Base slab 0.6 m thick and 4.7 m wide, toe 0.8 m, so heel $= 4.7-0.8-1.30 = 2.60$ m. Concrete 24 kN/m$^3$; backfill $\gamma=18$, $\phi=32^\circ$, $c=0$, sloping at $\beta=10^\circ$; foundation soil taken the same; base friction $\delta = \tfrac23\phi$. Rankine active thrust on a vertical plane through the heel, parallel to the slope.

1. Active thrust

Height of the plane: $H' = 9.2 + (4.7-1.2)\tan10^\circ = 9.82$ m.

$$K_a = \cos\beta\,\frac{\cos\beta-\sqrt{\cos^2\beta-\cos^2\phi}}{\cos\beta+\sqrt{\cos^2\beta-\cos^2\phi}} = 0.3210$$ $$P_a = \tfrac12K_a\gamma H'^2 = 0.5\times0.3210\times18\times9.82^2 = 278.4\ \text{kN/m}$$

at $H'/3 = 3.27$ m above the base: $P_h = 274.2$, $P_v = 48.3$ kN/m (acting at the heel end, $x=4.7$ m).

2. Vertical loads (about the toe)

$M_o = P_h\times H'/3 = 274.2\times3.27 = 897.2$ kN m/m.

3. Stability checks

• Overturning: $FS_{ot} = 2170.8/897.2 = 2.42$ (>2). Safe.
• Sliding: $\mu=\tan21.3^\circ = 0.391$,

$$FS_s = \frac{783.1\times0.391}{274.2} = 1.12\ (<1.5)$$

Not safe. The passive pressure on the 1.5 m embedment ($P_p = \tfrac12K_p\gamma D^2 = 65.9$ kN/m with $K_p=3.25$) gives $FS_s = 1.36$, still below 1.5, so a shear key under the stem (or a wider/longer heel) is required.

• Base pressure: $\bar x = (2170.8-897.2)/783.1 = 1.626$ m, $e = 2.35-1.626 = 0.724$ m $< B/6 = 0.783$ m (no tension).

$$q_{max} = 321\ \text{kN/m}^2\ (\text{toe}),\qquad q_{min} = 13\ \text{kN/m}^2\ (\text{heel})$$

• Bearing capacity (Vesic, $\phi=32^\circ$: $N_q=23.18$, $N_\gamma=30.21$, $B' = 4.7-2e = 3.25$ m, $D_f=1.5$ m): $q_u = \gamma D_fN_q + \tfrac12\gamma B'N_\gamma = 1510$ kN/m$^2$; $FS_b = 1510/321 = 4.7$ (>3). Safe.
• Overall stability: check a deep slip circle with Bishop's method ($FS\ge1.5$) if soft soil lies below.

Answer: overturning ($FS = 2.42$), eccentricity and bearing are safe, but sliding ($FS = 1.12$) is not; add a shear key.