Khosla: average hydraulic gradient, uplift A-D

The figure is read as an impervious floor of length 30 m with a cut-off pile of depth 8 m at the upstream end and 10 m at the downstream end, under a head $H = 5$ m. Points A, B, C, D lie on the floor at 10, 15, 20 and 25 m from the upstream end. The creep (seepage) path is found first, then Bligh's linear loss of head is used.

Creep length and average hydraulic gradient

The water creeps down the upstream pile, along the floor and up the downstream pile.

$$\begin{aligned} L &= 2d_1 + b + 2d_2 = 2(8) + 30 + 2(10) = 66\ \text{m} \\ i_{avg} &= \frac{H}{L} = \frac{5}{66} = 0.0758 \approx \frac{1}{13.2} \end{aligned}$$

Uplift pressure at A, B, C, D

Head lost up to a point $= i_{avg}\times$ (creep length to the point). The upstream pile adds $2\times 8 = 16$ m before the floor starts. Residual (uplift) head $h = H - \text{head lost}$ and pressure $p = \gamma_w h$ with $\gamma_w = 9.81$ kN/m³.

Thickness of floor

Bligh's rule, with an extra 33 % for safety, for specific gravity $G = 2.24$:

$$t = \frac{4}{3}\,\frac{h}{G-1} = \frac{4}{3}\,\frac{h}{1.24}$$

For example at A: $t = \frac{4}{3}\times\frac{3.03}{1.24} = 3.26$ m.

Answer: average gradient $= 0.0758$ (1 in 13.2). Uplift heads at A, B, C, D $= 3.03$, 2.65, 2.27, 1.89 m of water (i.e. 29.7, 26.0, 22.3, 18.6 kN/m²). Floor thickness $= 3.26$, 2.85, 2.44, 2.04 m respectively.