Khosla: uplift pressures, slope correction 6.5

The figure is not available, so the weir profile of the standard case (pond 102.50 m, 2:1 slopes, three pile lines) is used to show the method.

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.

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).

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).

Answer: The uplift pressures are the corrected values in the table (as % of $H$ and in metres of water): $C_1 = 70.2\%$, $E_2 = 65.4\%$, $D_2 = 60.2\%$, $C_2 = 55.6\%$, $E_3 = 29.7\%$, $D_3 = 20.4\%$. The exit gradient is about 1 in 6.4.