Toe width for overturning, submerged backfill

Reading of the data and assumptions: the wall is 6 m high with a 0.75 m thick base slab (so the stem is 5.25 m high and 0.75 m wide), heel 1.5 m and toe length $a$ (unknown). The water table is taken 1 m below the top of the backfill; the soil above it is saturated by capillary rise. The backfill is therefore at $\gamma_{sat}$ throughout, with water pressure below the 1 m level. Rankine active pressure on a vertical plane at the heel; passive resistance in front and base uplift are neglected; $\gamma_w = 9.81$.

1. Soil unit weights

$$\gamma_{sat} = \frac{(G+e)\gamma_w}{1+e} = \frac{(2.65+0.594)\times9.81}{1.594} = 19.96\ \text{kN/m}^3,\quad \gamma' = 10.15\ \text{kN/m}^3$$

$K_a = \tan^2(45^\circ-15^\circ) = 1/3$.

2. Active thrust and overturning moment (about the toe, arms measured above the base)

• At 1 m depth: $\sigma_v' = 19.96$, $p = 6.65$ kN/m$^2$.
• At the base: $\sigma_v' = 19.96 + 10.15\times5 = 70.74$, $p = 23.58$ kN/m$^2$.
• Water: $0$ at 1 m to $9.81\times5 = 49.05$ kN/m$^2$ at the base.

3. Resisting moment as function of $a$ (base width $B = a + 2.25$)

• Stem: $0.75\times5.25\times23 = 90.6$ kN/m at $a+0.375$.
• Slab: $(a+2.25)\times0.75\times23$ at $(a+2.25)/2$.
• Soil on heel: $1.5\times5.25\times19.96 = 157.2$ kN/m at $a+1.5$ from the toe.

Requiring $FS = M_r/M_o = 2$, i.e. $M_r = 751.7$ kN m/m, and solving for $a$ gives

$$a = 1.46\ \text{m}\quad (B = 3.71\ \text{m})$$

Check at $a=1.46$ m: stem 90.6 x 1.84 + slab 64.1 x 1.86 + soil 157.2 x 2.96 gives $M_r = 751.7$, and $M_r/M_o = 2.00$.

Answer: toe width $a \approx 1.46$ m (provide about 1.5 m) for $FS = 2$ against overturning with the backfill saturated and submerged.