Khosla: corrected pressures, slope correction 4%

Data and assumptions

Reading of the figure: pond level 157.00 m; with no flow the d/s water level is the d/s floor level 150.00 m, so $H = 7$ m. U/s pile at 0 m: floor 154.00 m (1 m thick), bottom 148.00 m ($d = 6$ m). Intermediate pile at 15.8 m: floor 152.00 m (1.3 m thick), bottom 149.00 m ($d = 3$ m). D/s pile at 40 m: floor 150.00 m (1.3 m thick), bottom 141.70 m as printed ($d = 8.3$ m). The 4% slope correction is multiplied by (slope length 3 m / $b'$), with $b' = 15.8$ m between the u/s and intermediate piles, and is negative at $C_1$ (up slope). Mutual interference is included. The permissible exit gradient is not given; 0.2 is assumed.

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), $b = 40$ 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 $H = 7.00$ 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, e.g. at $C_2$: $D = 152-141.7 = 10.3$ m, $d = 3$ m, $b' = 24.2$ m); slope = slope correction. The D points are not corrected. (The corrected $C_2$ comes out slightly above $E_2$ because the shallow intermediate pile receives large thickness and interference corrections.)

Exit gradient

$$G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{7.00}{8.30}\times\frac{1}{\pi\sqrt{2.961}} = 0.1560\ (\approx 1/6.4)$$

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   76.6% |###############################
 C1   70.0% |############################
 E2   57.7% |#######################
 D2   56.6% |#######################
 C2   58.0% |#######################
 E3   36.6% |###############
 D3   27.0% |###########
 C3    0.0% |

The hydraulic gradient line (HGL) drops 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).

Answer: Corrected pressures: u/s pile $D_1 = 76.6\%$, $C_1 = 70.0\%$; intermediate pile $E_2 = 57.7\%$, $D_2 = 56.6\%$, $C_2 = 58.0\%$; d/s pile $E_3 = 36.6\%$, $D_3 = 27.0\%$ (pressure heads in the table). Exit gradient $\approx 0.156$ (1 in 6.4), which is safe.