Khosla: HGL and exit gradient, pond 102.50

Data and assumptions

The figure is read as follows. Pond level 102.50 m; with no flow the d/s water level is taken at the d/s floor level 97.00 m, so $H = 5.50$ m. Piles (positions measured from the u/s pile): u/s pile at 0 m, floor level 100.00 m, bottom 94.00 m ($d = 6$ m); intermediate pile at 15 m, floor level 99.00 m, bottom 94.00 m ($d = 5$ m); d/s pile at 45 m, floor level 97.00 m, bottom 92.00 m ($d = 5$ m). Total floor length $b = 45$ m. Floor thickness is not given, so the thickness correction is neglected. The 2:1 slope correction is 6.5%, multiplied by (slope length / $b'$), where $b'$ is the distance between the two piles enclosing the slope: u/s slope (2 m, up slope, $b' = 15$ m): $-6.5\times 2/15$ at $C_1$; d/s glacis (101.00 to 98.00 m, 6 m long, down slope, $b' = 30$ m): $+6.5\times 6/30$ at $E_3$.

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), $b = 45$ 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 $H = 5.50$ 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); slope = slope correction. The D points are not corrected.

Exit gradient

$$G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{5.50}{5.00}\times\frac{1}{\pi\sqrt{5.028}} = 0.1562\ (\approx 1/6.4)$$

Permissible (assumed) $= 0.2000$. Safe (less than the permissible value).

Hydraulic gradient line

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

 E1  100.0% |########################################
 D1   77.8% |###############################
 C1   70.2% |############################
 E2   65.4% |##########################
 D2   60.2% |########################
 C2   55.6% |######################
 E3   29.7% |############
 D3   20.4% |########
 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). In levels (datum 97.00 m): $C_1$ 100.86 m, $E_2$ 100.60 m, $C_2$ 100.06 m, $E_3$ 98.63 m.