Khosla: uplift, exit gradient 0.15, floor at A-D

Data and assumptions

The figure is read as follows: pond level 105 m; with no flow the d/s water level is the d/s floor level 97 m, so $H = 8$ m. U/s pile at 0 m: floor top 100 m (bottom 99 m, 1 m thick), pile bottom 95 m ($d = 5$ m). Intermediate pile at 20 m: floor top 99 m, bottom 94.5 m ($d = 4.5$ m). D/s pile at 50 m: floor top 97 m, bottom 95.5 m (1.5 m thick), pile bottom 93 m ($d = 4$ m). Locations: A = u/s junction ($x = 0$), B = $x = 20$ m (at the intermediate pile), C = $x = 25$ m (foot of the step, floor top 99 m), D = d/s end ($x = 50$ m). Safe exit gradient 0.15, $G = 2.4$.

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), $b = 50$ 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 $H = 8.00$ 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 (thickness at the intermediate pile not known, so neglected there); 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.

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; at B the higher value $E_2$ is used).

Exit gradient

$$G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{8.00}{4.00}\times\frac{1}{\pi\sqrt{6.770}} = 0.2447\ (\approx 1/4.1)$$

Permissible $= 0.1500$. Not safe; deepen the d/s pile or lengthen the floor.

Hydraulic gradient line

Bars show the corrected pressure (each # is 2.5%):

 E1  100.0% |########################################
 D1   80.5% |################################
 C1   75.8% |##############################
 E2   60.6% |########################
 D2   56.2% |######################
 C2   52.3% |#####################
 E3   21.6% |#########
 D3   17.5% |#######
 C3    0.0% |

The hydraulic gradient line (HGL) drops vertically across each pile (from E to D to C) and runs as a straight line along the floor between the pile junctions (C of one pile to E of the next).