Khosla: thickness and exit gradient, 105 m head

Data and assumptions

Reading of the figure: water level 105 m; with no flow the d/s water level is the d/s floor level 100 m, so $H = 5$ m. U/s pile at 0 m: floor 101 m, bottom 98 m ($d = 3$ m). Intermediate pile at 5 m: floor 100 m, bottom 97.5 m ($d = 2.5$ m). D/s pile at 30 m ($5 + 5 + 20$): floor 100 m, bottom 97 m ($d = 3$ m). Floor thickness is not known, so the required thickness is computed and the thickness correction is neglected. $G = 2.4$. Safe exit gradient 0.2.

Khosla's constants

End piles: $\alpha = b/d$, $\lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-2}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-1}{\lambda}$ (in percent), $b = 30$ m; for the u/s pile $\phi_{D1} = 100-\phi_D$, $\phi_{C1} = 100-\phi_E$.

Intermediate pile ($b_1$ from the u/s end, $b_2$ from the d/s end): $\lambda_1 = \dfrac{\sqrt{1+\alpha_1^2}-\sqrt{1+\alpha_2^2}}{2}$, $\lambda = \dfrac{\sqrt{1+\alpha_1^2}+\sqrt{1+\alpha_2^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1-1}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1}{\lambda}$, $\phi_C = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1+1}{\lambda}$.

Pressures at the key points

Pressure is expressed as a percentage of the head $H = 5.00$ m (no flow).

Corrections: int = mutual interference, $19\sqrt{D/b'}\,\dfrac{d+D}{b}$, positive at C of the u/s pile and negative at E of the d/s pile of each pair ($D$ = depth of the neighbouring pile below the floor level at the pile being corrected, $b'$ = distance between the two piles). The D points are not corrected. ($E_2$ comes out slightly above $C_1$ because the two piles are only 5 m apart.)

Floor thickness

Unbalanced head $h$ = hydraulic-gradient level at the floor $-$ floor top level; required thickness $t = \dfrac{h}{G-1}$ with $G = 2.40$ (the HGL is taken straight between the corrected key points along the floor).

Exit gradient

$$G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{5.00}{3.00}\times\frac{1}{\pi\sqrt{5.525}} = 0.2257\ (\approx 1/4.4)$$

Permissible $= 0.2000$. Not safe; deepen the d/s pile or lengthen the floor.