Active pressure, sloping backfill and water table

Assumption: the wall height is not stated, so $H = 6$ m is assumed (the method is the same for any $H$). Rankine's theory for a sloping backfill ($\beta = 20^\circ$, $\phi = 30^\circ$) gives a pressure parallel to the slope.

Active coefficient

$$K_a = \cos\beta\,\frac{\cos\beta - \sqrt{\cos^2\beta - \cos^2\phi}}{\cos\beta + \sqrt{\cos^2\beta - \cos^2\phi}} = 0.4142$$

(a) Dry backfill

$$\begin{aligned} P_a &= \tfrac12 K_a\gamma H^2 = 0.5\times0.4142\times18\times6^2 = 134.2\ \text{kN/m} \end{aligned}$$

It acts at $H/3 = 2.00$ m above the base, inclined at $20^\circ$ to the horizontal (horizontal component 126.1 kN/m, vertical component 45.9 kN/m).

(b) Water table 2 m below the top

The top 2 m is dry ($\gamma=18$) and the lower 4 m is submerged ($\gamma'=12$).

• At 2 m: $p = 0.4142\times18\times2 = 14.91$ kN/m$^2$.
• At 6 m: $p = 0.4142\times(36 + 12\times4) = 34.79$ kN/m$^2$.
• Water pressure at the base: $9.81\times4 = 39.24$ kN/m$^2$.

The horizontal total is $114.32\cos20^\circ + 78.48 = 185.9$ kN/m, acting at

$$\bar y = \frac{107.43\times2.12 + 78.48\times1.333}{185.9} = 1.79\ \text{m}$$

The vertical component of the soil thrust is 39.1 kN/m (resultant 190.0 kN/m).

Answer:

• Dry: $P_a = 134.2$ kN/m at 2.00 m above the base.
• With the water table: the effective soil thrust falls to 114.3 kN/m (acting at 2.12 m), but with the water pressure (78.5 kN/m) the total horizontal thrust rises from 126.1 to 185.9 kN/m (+60 kN/m, about 47% more), and its point of application drops to 1.79 m above the base.