Khosla: intermediate pile pressure (L1 30 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 $L_1 = 30$ m, d/s pile at $L_1 + L_2 = 90$ m. Slope correction factor 3.3% (4:1 slope), multiplied by (slope length / $b'$) with $b' = 30$ m (both slopes lie between the u/s and intermediate piles): u/s slope 12 m (up slope, $-$ at $C_1$) and d/s slope 16 m (down slope, $+$ at $E_2$). Floor thickness at the intermediate pile is not given and is neglected there.

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), $b = 90$ 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, e.g. at $C_2$: $D = 142-135 = 7$ m, $d = 3$ m, $b' = 60$ m); slope = slope correction. The D points are not corrected.

Exit gradient

$$G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{19.00}{6.00}\times\frac{1}{\pi\sqrt{8.017}} = 0.3560\ (\approx 1/2.8)$$

Permissible $= 1/6 = 0.1667$. 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   85.3% |##################################
 C1   79.6% |################################
 E2   64.3% |##########################
 D2   60.8% |########################
 C2   59.3% |########################
 E3   21.5% |#########
 D3   16.1% |######
 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).

Answer: At the intermediate pile: floor-pile interfaces $E_2 = 64.3\%$ and $C_2 = 59.3\%$ of $H$ (12.21 m and 11.26 m); bottom end of the pile $D_2 = 60.8\%$ (11.54 m). Exit gradient $G_E = 0.356$, more than 1/6 = 0.167, so the structure is not safe against piping.