Khosla: cistern floor thickness from corrected phi

Data and assumptions

Corrected pressures: $\phi_{C1} = 82\%$ (u/s pile junction), $\phi_{C2} = 35\%$ (d/s pile junction). Full reservoir level 505 m, cistern floor 499 m, no tail water, so the head is $H = 505 - 499 = 6$ m. Floor length between piles $b = 40$ m, cistern length 15 m at the d/s end, $G = 2.4$.

Pressure along the floor

The HGL is a straight line between the junctions. The cistern starts $40 - 15 = 25$ m from the u/s pile:

$$\phi_{start} = 82 - (82-35)\times\frac{25}{40} = 52.62\%,\qquad \phi_{end} = 35\%$$

Uplift heads above the cistern floor level:

$$h_{start} = 52.62\%\times 6 = 3.157\ \text{m},\qquad h_{end} = 35\%\times 6 = 2.100\ \text{m}$$

Thickness of the cistern floor

$$t = \frac{h}{G-1}:\qquad t_{start} = \frac{3.157}{1.4} = 2.26\ \text{m},\qquad t_{end} = \frac{2.100}{1.4} = 1.50\ \text{m}$$

Exit gradient

The depth of the d/s pile is found from $\phi_{C2} = \phi_E$ of the d/s pile (the junction pressure):

$$\cos(\pi\times 0.35) = \frac{\lambda-2}{\lambda} \Rightarrow \lambda = \frac{2}{1-0.4540} = 3.663,\quad \alpha = \sqrt{(2\lambda-1)^2-1} = 6.246,\quad d = \frac{b}{\alpha} = 6.40\ \text{m}$$ $$G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{6}{6.40}\times\frac{1}{\pi\sqrt{3.663}} = 0.1558\ (\approx 1/6.4)$$

Answer: Minimum cistern floor thickness $\approx 2.26$ m at the beginning (u/s end of the cistern) and $\approx 1.50$ m at the end; exit gradient $\approx 0.156$ (about 1 in 6.4), below the usual safe value of 1/6.