--- title: "Khosla: uplift, exit gradient 0.15, floor at A-D" subject: "Irrigation and Drainage Engineering" chapter: 5 key: "semester_6/irrigation_and_drainage_engineering:khosla-uplift-exit-gradient-015-abcd" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/khosla-uplift-exit-gradient-015-abcd site: BCT Engineering --- # Khosla: uplift, exit gradient 0.15, floor at A-D Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks - Asked in: 2064 Jestha (old course), 16 marks - 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-exit-gradient-015-abcd - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5 ## Question Calculate the uplift pressure at key points in figure shown below using Khosla Theory. Check the exit gradient and thickness of the floor at A, B, C and D locations as shown in figure. The safe exit gradient is 0.15. [Figure: levels 105 m (pond), 101 m, 100 m, 99 m, 95 m, 94.5 m, 97 m, 95.5 m, 93 m; points A, B, C, D; horizontal distances 20 m, 5 m and 25 m] ## Short answer Pressures: C1 75.8%, E2 60.6%, C2 52.3%, E3 21.5%. Exit gradient 0.245 exceeds safe 0.15. Required thicknesses: A 2.19 m (unsafe), B 2.04 m, C 1.27 m, D 1.23 m (safe with provided 1.5 m). ## Answer #### Data and assumptions The figure is read as follows: pond level 105 m; with no flow the d/s water level is the d/s floor level 97 m, so $H = 8$ m. U/s pile at 0 m: floor top 100 m (bottom 99 m, 1 m thick), pile bottom 95 m ($d = 5$ m). Intermediate pile at 20 m: floor top 99 m, bottom 94.5 m ($d = 4.5$ m). D/s pile at 50 m: floor top 97 m, bottom 95.5 m (1.5 m thick), pile bottom 93 m ($d = 4$ m). Locations: A = u/s junction ($x = 0$), B = $x = 20$ m (at the intermediate pile), C = $x = 25$ m (foot of the step, floor top 99 m), D = d/s end ($x = 50$ m). Safe exit gradient 0.15, $G = 2.4$. #### 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), $b = 50$ 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 | 5.00 | 10.000 | 5.525 | 27.98 | 19.45 | | D/s pile | 4.00 | 12.500 | 6.770 | 25.11 | 17.52 | | Intermediate pile | d (m) | α1 | α2 | λ1 | λ | φE (%) | φD (%) | φC (%) | |---|---|---|---|---|---|---|---|---| | at 20 m | 4.50 | 4.444 | 6.667 | -1.093 | 5.648 | 62.08 | 56.20 | 50.52 | #### Pressures at the key points Pressure is expressed as a percentage of the head $H = 8.00$ m (no flow). | Point | Simple φ (%) | Corrections (%) | Corrected φ (%) | Pressure head (m) | |---|---|---|---|---| | E1 (top of u/s pile) | 100.00 | 0 | 100.00 | 8.00 | | D1 | 80.55 | - | 80.55 | 6.44 | | C1 | 72.02 | thk +1.70, int +2.09 | 75.82 | 6.07 | | E2 | 62.08 | int -1.44 | 60.64 | 4.85 | | D2 | 56.20 | - | 56.20 | 4.50 | | C2 | 50.52 | int +1.78 | 52.31 | 4.18 | | E3 | 25.11 | thk -2.85, int -0.71 | 21.55 | 1.72 | | D3 | 17.52 | - | 17.52 | 1.40 | | C3 (d/s end) | 0 | 0 | 0 | 0 | Corrections: thk = floor thickness, $+\dfrac{\phi_D-\phi_C}{d}t$ at C and $-\dfrac{\phi_E-\phi_D}{d}t$ at E (thickness at the intermediate pile not known, so neglected there); 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). 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; at B the higher value $E_2$ is used). | Location | x (m) | φ (%) | HG level (m) | Floor top (m) | Unbalanced head (m) | Required t (m) | Provided t (m) | Check | |---|---|---|---|---|---|---|---|---| | A (x = 0) | 0 | 75.82 | 103.07 | 100.00 | 3.07 | 2.19 | 1.00 | not safe | | B (x = 20) | 20 | 60.64 | 101.85 | 99.00 | 2.85 | 2.04 | - | provide at least 2.04 | | C (x = 25) | 25 | 47.18 | 100.77 | 99.00 | 1.77 | 1.27 | - | provide at least 1.27 | | D (x = 50) | 50 | 21.55 | 98.72 | 97.00 | 1.72 | 1.23 | 1.50 | safe | #### Exit gradient $$ G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{8.00}{4.00}\times\frac{1}{\pi\sqrt{6.770}} = 0.2447\ (\approx 1/4.1) $$ Permissible $= 0.1500$. Not safe; deepen the d/s pile or lengthen the floor. #### Hydraulic gradient line Bars show the corrected pressure (each # is 2.5%): ```text E1 100.0% |######################################## D1 80.5% |################################ C1 75.8% |############################## E2 60.6% |######################## D2 56.2% |###################### C2 52.3% |##################### E3 21.6% |######### D3 17.5% |####### 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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Pond level is 105 m and the d/s floor level is 97 m. What head is used? A. 8 m B. 10 m C. 5 m D. 4 m Answer: A. H = 105 - 97 = 8 m. (Key point: Head with no flow) 2. G_E = 0.2447 against a safe limit of 0.15. What is concluded? A. Safe, since 0.2447 is above 0.15 B. Not safe; deepen the d/s pile or lengthen the floor C. Unsafe only if G is below 2.4 D. Safe because the floor is 50 m long Answer: B. G_E must be below the permissible value; 0.2447 exceeds 0.15. (Key point: Exit gradient check) 3. Required thickness at a point is computed from which steps? A. t = h x (G-1) B. t = H/G C. t = phi x G D. HG level minus floor top gives h; then t = h/(G-1) Answer: D. The unbalanced head h = HG level - floor top level, and t = h/(G-1). (Key point: Unbalanced head and thickness) 4. At D the required thickness is 1.14 m and 1.5 m is provided. The check is A. Not safe because G is 2.4 B. Safe C. Cannot be decided D. Not safe Answer: B. Provided 1.5 m exceeds the required 1.14 m, so D is safe. 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