Question 6 of 6: Cantilever Retaining Wall — Tipping and Sliding Stability

Question 6: Cantilever Retaining Wall — Tipping and Sliding Stability (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.

Find. (a) factor of safety against tipping about the toe; (b) factor of safety against sliding along the base.

Approach. Use the Rankine active-pressure ("virtual back") method: compute the active force on a vertical plane through the back of the heel over the full retained height $H$, resolve the wall+soil system's self-weight into the base slab, stem, and the soil columns riding on the heel and on the toe, then take moments about the toe (tipping) and compare base friction to the active thrust (sliding).

1. Resolve the figure's dimensions. The base width B was printed as 4.0 m alongside 0.5 m (toe) + 0.5 m (stem) + 3.0 m (heel) = 4.0 m — a self-consistent set once the earlier extraction's mis-read "toe width" is corrected to what the drawing actually shows: a 2.0 m VERTICAL dimension (front-grade depth above the footing), not a fourth horizontal width. All four numbers below are taken directly from the printed figure. Stem height above the base, $H-t_{base}=5.0-0.5=4.5$ m.
2. Active earth pressure (Rankine, virtual back at the heel). $$K_a=\frac{1-\sin\phi}{1+\sin\phi}=\frac{1-\sin38^\circ}{1+\sin38^\circ}=0.2379.$$ $$P_a=\tfrac{1}{2}K_a\gamma_{soil}H^2=\tfrac{1}{2}(0.2379)(18)(5.0)^2=\boxed{53.52\ \text{kN/m}},\quad\text{acting at }H/3=1.667\ \text{m above the base.}$$
3. Self-weight components (per metre of wall), moment arms measured from the toe's front (base) edge. $$W_1(\text{base slab})=B\,t_{base}\gamma_{conc}=(4.0)(0.5)(24)=48.0\ \text{kN/m},\ \ x_1=2.0\ \text{m};$$ $$W_2(\text{stem})=t_{stem}(H-t_{base})\gamma_{conc}=(0.5)(4.5)(24)=54.0\ \text{kN/m},\ \ x_2=0.75\ \text{m};$$ $$W_3(\text{soil on heel})=(3.0)(4.5)(18)=243.0\ \text{kN/m},\ \ x_3=2.5\ \text{m};$$ $$W_4(\text{soil on toe, depth }D_f)=(0.5)(2.0)(18)=18.0\ \text{kN/m},\ \ x_4=0.25\ \text{m}.$$ $$N=\sum W_i=\boxed{363.0\ \text{kN/m}}.$$
4. Part (a) — factor of safety against tipping (moments about the toe). $$M_{resist}=\sum W_i x_i=48.0(2.0)+54.0(0.75)+243.0(2.5)+18.0(0.25)=748.5\ \text{kN$\cdot$m/m},$$ $$M_{over}=P_a\left(\frac{H}{3}\right)=53.52(1.667)=89.21\ \text{kN$\cdot$m/m}.$$ $$FS_{tip}=\frac{M_{resist}}{M_{over}}=\frac{748.5}{89.21}=\boxed{8.39}.$$ (As a cross-check, the resultant lands only 0.18 m from the base centre — well inside the middle third, $B/6=0.67$ m — confirming no heel uplift, consistent with such a high $FS_{tip}$.)
5. Part (b) — factor of safety against sliding. With the water table 30 m down (no pore pressure at this shallow foundation) and using the base–soil friction angle $\delta=25^\circ$ (close to the common design guideline $\delta\approx\tfrac{2}{3}\phi=25.3^\circ$ for a cast-in-place base): $$FS_{slide}=\frac{N\tan\delta}{P_a}=\frac{363.0\tan25^\circ}{53.52}=\boxed{3.16}.$$ This conservatively neglects any passive resistance from the 2.0 m of soil embedded in front of the toe (standard practice, since that soil is often disturbed by utility trenching or erosion over the wall's service life); including it, $P_p=\tfrac12 K_p\gamma_{soil}D_f^2=151.3$ kN/m would raise the estimate to $FS_{slide}\approx5.99$.