--- title: "Khosla: HGL and d/s floor thickness, three piles" subject: "Irrigation and Drainage Engineering" chapter: 5 key: "semester_6/irrigation_and_drainage_engineering:khosla-corrected-phi-three-piles-hgl" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/khosla-corrected-phi-three-piles-hgl site: BCT Engineering --- # Khosla: HGL and d/s floor thickness, three piles Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks - Asked in: 2064 Kartik (old course), 8 marks - Video: [Problem on Khosla Theory.](https://www.youtube.com/watch?v=2NEHgwXCa-U) — Civil Giri - 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-corrected-phi-three-piles-hgl - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5 ## Question Following corrected φ values were computed from Khosla's curves in a barrage placed on permeable foundation. U/S sheet pile: φE1 = 100%, φD1 = 90%, φC1 = 85% Intermediate pile: φE2 = 80%, φD2 = 70%, φC2 = 65% D/S sheet pile: φE3 = 55%, φD3 = 45%, φC3 = 0% Distance between the U/S and intermediate piles is 20 m and that between the intermediate and D/S piles is 40 m. Assuming that the floor is horizontal throughout, draw the HGL for the subsoil flow. If the net head is 10 m, determine the thickness of D/S floor at a distance of 20 m and 30 m away from the intermediate sheet pile. Assume G for the floor material equal to 2.2. The symbols has usual meanings. ## Short answer Between intermediate and d/s piles, pressure drops linearly from 65% to 55% of total head. With 10 m net head, required floor thickness at 20 m is 5.00 m and at 30 m is 4.79 m. ## Answer #### Given Corrected pressures (percent of head): u/s pile $E_1 = 100$, $D_1 = 90$, $C_1 = 85$; intermediate pile $E_2 = 80$, $D_2 = 70$, $C_2 = 65$; d/s pile $E_3 = 55$, $D_3 = 45$, $C_3 = 0$. Net head $H = 10$ m, $G = 2.2$. Piles at 0, 20 m and 60 m (20 m and 40 m apart). The floor is horizontal, with its level taken as the datum (d/s water level). #### Hydraulic gradient line The HGL is a vertical drop across each pile and a straight line along the floor between piles: | Point | Pressure (%) | Head above floor (m) | |---|---|---| | $E_1$ | 100 | 10.0 | | $D_1$ | 90 | 9.0 | | $C_1$ | 85 | 8.5 | | $E_2$ | 80 | 8.0 | | $D_2$ | 70 | 7.0 | | $C_2$ | 65 | 6.5 | | $E_3$ | 55 | 5.5 | | $D_3$ | 45 | 4.5 | | $C_3$ | 0 | 0 | ```text 100 E1 |############################# 90 D1 |########################## 85 C1 |######################## (floor, 20 m) 80 E2 |####################### 70 D2 |#################### 65 C2 |################## (floor, 40 m) 55 E3 |################ 45 D3 |############# 0 C3 | ``` #### Thickness of the d/s floor (between intermediate and d/s piles) On the 40 m floor the pressure falls linearly from 65% at $C_2$ to 55% at $E_3$: $$ \phi(x) = 65 - (65-55)\frac{x}{40} $$ Unbalanced head $h = \phi H/100$ and thickness $t = \dfrac{h}{G-1} = \dfrac{h}{1.2}$: | Distance from intermediate pile | $\phi$ (%) | $h$ (m) | $t$ (m) | |---|---|---|---| | 20 m | 60.0 | 6.00 | 5.00 | | 30 m | 57.5 | 5.75 | 4.79 | **Answer:** Required floor thickness $\approx 5.00$ m at 20 m and $\approx 4.79$ m at 30 m from the intermediate sheet pile. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Between the intermediate and d/s piles (40 m), the pressure falls from 65% at C2 to 55% at E3. How is phi at x from the intermediate pile found? A. Linear interpolation: phi = 65 - 10 x/40 B. phi = 55 + 10 x C. phi = 65 - 55 x/40 D. Constant 65% Answer: A. The HGL is straight along the floor, so pressure varies linearly between C2 and E3. (Key point: Linear HGL on the floor) 2. What is phi at 20 m from the intermediate pile? A. 62.5% B. 60% C. 55% D. 65% Answer: B. 65 - 10 x 20/40 = 60%. (Key point: Pressure at 20 m) 3. For h = 6.0 m at 20 m and G = 2.2, the required thickness is A. 3.00 m B. 5.00 m C. 7.20 m D. 2.73 m Answer: B. t = h/(G-1) = 6.0/1.2 = 5.00 m. (Key point: Thickness formula) 4. Why is the thickness at 30 m smaller than at 20 m? A. The net head changes with distance B. The floor is thicker there C. G increases with distance D. Pressure falls along the floor (57.5% against 60%), so the uplift head is smaller Answer: D. phi is 57.5% so h = 5.75 m and t = 5.75/1.2 = 4.79 m. 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