Khosla: end pile pressure (L1 20 m)

Data and assumptions

Reading of the table: A = u/s water level 160 m; B = u/s floor level 143 m; C = crest 146 m; D = d/s floor level (at the intermediate pile) 142 m; E = bottom of u/s pile 138 m ($d_1 = 143-138 = 5$ m); F = bottom of the u/s floor 142 m (thickness 1.0 m); G = bottom of intermediate pile 139 m ($d_2 = 142-139 = 3$ m); I = d/s floor level 141 m and H = bottom of the d/s floor 140 m (thickness 1.0 m); J = bottom of d/s pile 135 m ($d_3 = 141-135 = 6$ m). With no flow the d/s water level is the d/s floor level 141 m, so $H = 160 - 141 = 19$ m. U/s pile at 0 m, intermediate pile at $L_1 = 20$ m, d/s pile at $L_1 + L_2 = 80$ m. The slope correction is not given and is neglected (the slopes are far from the end pile, so it would not change $E_3$ or $D_3$).

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), $b = 80$ 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 $H = 19.00$ m (no flow).

Corrections: thk = floor thickness, $+\dfrac{\phi_D-\phi_C}{d}t$ at C and $-\dfrac{\phi_E-\phi_D}{d}t$ at E; 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; at $E_3$: $D = 141-139 = 2$ m, $d = 6$ m, $b' = 60$ m). The D points are not corrected.

Floor thickness

Unbalanced head $h$ = hydraulic-gradient level at the floor $-$ floor top level; required thickness $t = \dfrac{h}{G-1}$ with $G = 2.40$.

Exit gradient

$$G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{19.00}{6.00}\times\frac{1}{\pi\sqrt{7.185}} = 0.3760\ (\approx 1/2.7)$$

Permissible $= 1/6 = 0.1667$. Not safe; deepen the d/s pile or lengthen the floor.

Answer: At the d/s (end) pile: floor-pile interface $E_3 = 22.8\%$ of $H$ (4.33 m); bottom end of the pile $D_3 = 17.0\%$ (3.23 m). Required floor thickness at the end $= 3.09$ m against 1.0 m provided, so the provided thickness is not sufficient. Exit gradient $G_E = 0.376$ against 1/6 = 0.167, so the exit gradient is not safe either (a deeper d/s pile or a longer floor is needed).