--- title: "Khosla: uplift pressures, slope correction 6.5" subject: "Irrigation and Drainage Engineering" chapter: 5 key: "semester_6/irrigation_and_drainage_engineering:khosla-uplift-slope-correction-65-fig" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/khosla-uplift-slope-correction-65-fig site: BCT Engineering --- # Khosla: uplift pressures, slope correction 6.5 Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks - Asked in: 2077 Chaitra, 12 marks - Video: [Khosla's Theory (Part-1)। khosla method of independent variable। Correction in khosla theory।](https://www.youtube.com/watch?v=sUiW5y358Fs) — Ajay Verma - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/khosla-uplift-slope-correction-65-fig - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5 ## Question Using Khosla's method, calculate the uplift pressures at various key points in the figure below. Also, determine the exit gradient. Take the slope correction factor as 6.5. [Figure: not reproduced in the scan] ## Short answer Khosla method with slope correction 6.5% for head 5.50m and three piles yields corrected uplifts: C1 70.2% (3.86m), E2 65.4% (3.60m), C2 55.6% (3.06m), E3 29.7% (1.63m). Exit gradient = 0.156 (1 in 6.4). ## Answer 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): $\lambda_1 = \dfrac{\sqrt{1+\alpha_1^2}-\sqrt{1+\alpha_2^2}}{2}$, $\lambda = \dfrac{\sqrt{1+\alpha_1^2}+\sqrt{1+\alpha_2^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1-1}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1}{\lambda}$, $\phi_C = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1+1}{\lambda}$. | Pile | d (m) | α | λ | φE (%) | φD (%) | |---|---|---|---|---|---| | U/s pile | 6.00 | 7.500 | 4.283 | 32.10 | 22.20 | | D/s pile | 5.00 | 9.000 | 5.028 | 29.43 | 20.42 | | Intermediate pile | d (m) | α1 | α2 | λ1 | λ | φE (%) | φD (%) | φC (%) | |---|---|---|---|---|---|---|---|---| | at 15 m | 5.00 | 3.000 | 6.000 | -1.460 | 4.623 | 67.86 | 60.23 | 53.17 | #### Pressures at the key points Pressure is expressed as a percentage of the head $H = 5.50$ m (no flow). | Point | Simple φ (%) | Corrections (%) | Corrected φ (%) | Pressure head (m) | |---|---|---|---|---| | E1 (top of u/s pile) | 100.00 | 0 | 100.00 | 5.50 | | D1 | 77.80 | - | 77.80 | 4.28 | | C1 | 67.90 | int +3.20, slope -0.87 | 70.23 | 3.86 | | E2 | 67.86 | int -2.44 | 65.43 | 3.60 | | D2 | 60.23 | - | 60.23 | 3.31 | | C2 | 53.17 | int +2.45 | 55.62 | 3.06 | | E3 | 29.43 | int -1.07, slope +1.30 | 29.66 | 1.63 | | D3 | 20.42 | - | 20.42 | 1.12 | | C3 (d/s end) | 0 | 0 | 0 | 0 | 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). | Location | x (m) | φ (%) | HG level (m) | Floor top (m) | Unbalanced head (m) | Required t (m) | Provided t (m) | Check | |---|---|---|---|---|---|---|---|---| | C1 (u/s) | 0 | 70.23 | 100.86 | 100.00 | 0.86 | 0.62 | - | provide at least 0.62 | | E2 (int. pile) | 15 | 65.43 | 100.60 | 99.00 | 1.60 | 1.14 | - | provide at least 1.14 | | C2 (int. pile) | 15 | 55.62 | 100.06 | 99.00 | 1.06 | 0.76 | - | provide at least 0.76 | | E3 (d/s) | 45 | 29.66 | 98.63 | 97.00 | 1.63 | 1.17 | - | provide at least 1.17 | #### 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For no flow at pond 102.50 m with d/s floor 97.00 m, H is: A. 5.50 m B. 2.5 m C. 10.5 m D. 6 m Answer: A. H = 102.50 - 97.00 = 5.50 m. (Key point: Head) 2. How is the 6.5% slope correction applied? A. Multiplied by slope length / b; plus for the down slope, minus for the up slope B. Applied fully at all points C. Only at the pile tips D. Added at C3 Answer: A. E3 gets +6.5 x 6/45 = 0.87 and C1 gets -6.5 x 2/45 = 0.29. (Key point: Slope correction) 3. Interference corrections for the intermediate pile are negative at E2, D2, C2 because: A. Those piles are downstream of the interfering pile B. They are upstream C. Floor is thicker D. Slopes Answer: A. Piles downstream of an interfering pile get a negative correction. (Key point: Interference) 4. Exit gradient uses: A. H/d x 1/(pi sqrt(lambda)) with d/s pile d = 5 m, lambda = 5.028 B. u/s pile only C. Floor thickness D. Slope Answer: A. 5.5/5 x 1/(pi sqrt 5.028) = 0.156, below 0.2. (Key point: Exit gradient) 5. Why is the thickness correction neglected? A. Floor thickness was not given B. It is always zero C. Piles are deep D. Slopes cancel it Answer: A. Without a thickness there is nothing to interpolate. 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