Khosla: residual pressures and HGL, pond 260 m

Data and assumptions

Reading of the figure (not to scale): pond level 260 m; no flow, so the d/s water level is the d/s floor level 255 m and $H = 5$ m. U/s pile at 0 m: floor 256 m (1 m thick), bottom 250 m ($d = 6$ m). Then 3 m of level floor, the u/s 2:1 slope from 256 m up to the crest 258.5 m (5 m long) and 2 m to the intermediate pile at 10 m: floor 256 m, bottom 248 m ($d = 8$ m). Downstream of it: 2 m, the d/s 3:1 glacis from 258.5 m down to 255 m (10.5 m long) and 15 m of d/s floor to the d/s pile at 37.5 m: floor 255 m (1.2 m thick), bottom 248 m ($d = 7$ m). Both slopes end or start next to the intermediate pile, so their corrections are applied there, using $b'$ = distance between the two piles enclosing the slope: u/s slope (up, 6.5%): $-6.5\times 5/10$ at $E_2$; d/s glacis (down, 4.5%): $+4.5\times 10.5/27.5$ at $C_2$. Floor thickness correction uses 1 m (u/s) and 1.2 m (d/s); the thickness at the intermediate pile is neglected. Safe exit gradient (not given) assumed 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), $b = 37.5$ 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: 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, $b'$ = distance between the two piles); slope = slope correction. The D points are not corrected.

Exit gradient

$$G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{5.00}{7.00}\times\frac{1}{\pi\sqrt{3.225}} = 0.1266\ (\approx 1/7.9)$$

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   75.9% |##############################
 C1   73.2% |#############################
 E2   69.1% |############################
 D2   62.5% |#########################
 C2   56.0% |######################
 E3   32.0% |#############
 D3   25.8% |##########
 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 255 m): $C_1$ 258.66 m, $E_2$ 258.46 m, $C_2$ 257.80 m, $E_3$ 256.60 m.

Answer: The residual seepage pressures at the key points are the corrected values (percent of $H$ and metres of water) in the table; the subsoil HGL is plotted above, and the exit gradient is about 1 in 7.9 (safe).