---
title: "Khosla: uplift, HGL, thickness, exit gradient 1/5"
subject: "Irrigation and Drainage Engineering"
chapter: 5
key: "semester_6/irrigation_and_drainage_engineering:khosla-hgl-thickness-gradient-1-in"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/khosla-hgl-thickness-gradient-1-in
site: BCT Engineering
---

# Khosla: uplift, HGL, thickness, exit gradient 1/5

Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks

- Asked in: 2073 Bhadra, 8 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-hgl-thickness-gradient-1-in
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5

## Question

Calculate the uplift pressure at key points of the pile of the structure shown in figure below. Draw HGL and also check the thickness provided and safe exit gradient GE = 1/5.

[Figure: RL 103.0 m water level, 3 m head; floor RL 100.0 m; floor dimensions 1.0 m, 1.5 m, 1.25 m, 1.0 m; pile depths 2 m, 1.5 m and 3 m; 5.0 m; horizontal distances 10 m and 20 m]

## Short answer

Using Khosla's theory with thickness and interference corrections, corrected pressure heads at E1, D1, C1, E2, D2, C2, E3, D3, C3 are 3.00, 2.52, 2.44, 1.81, 1.82, 1.84, 0.73, 0.58, 0 m. Required thicknesses exceed provided at C1, E2 and C2 but not E3. Exit gradient is 0.1354 (about 1/7.4), less than safe 1/5.

## Answer

#### Data and assumptions

Reading of the figure: water level RL 103.0 m, floor RL 100.0 m, so the head is $H = 3$ m (d/s water at floor level). Pile depths 2 m (u/s, at 0 m), 1.5 m (intermediate, at 10 m) and 3 m (d/s, at 30 m); floor length $b = 10 + 20 = 30$ m. Floor thickness provided: 1.0 m (u/s), 1.25 m (intermediate), 1.0 m (d/s). $G = 2.4$. Safe exit gradient $1/5$.

#### 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), $b = 30$ 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 | 2.00 | 15.000 | 8.017 | 22.98 | 16.07 |
| D/s pile | 3.00 | 10.000 | 5.525 | 27.98 | 19.45 |

| Intermediate pile | d (m) | α1 | α2 | λ1 | λ | φE (%) | φD (%) | φC (%) |
|---|---|---|---|---|---|---|---|---|
| at 10 m | 1.50 | 6.667 | 13.333 | -3.315 | 10.056 | 64.12 | 60.69 | 57.39 |

#### Pressures at the key points

Pressure is expressed as a percentage of the head $H = 3.00$ m (no flow).

| Point | Simple φ (%) | Corrections (%) | Corrected φ (%) | Pressure head (m) |
|---|---|---|---|---|
| E1 (top of u/s pile) | 100.00 | 0 | 100.00 | 3.00 |
| D1 | 83.93 | - | 83.93 | 2.52 |
| C1 | 77.02 | thk +3.46, int +0.86 | 81.33 | 2.44 |
| E2 | 64.12 | thk -2.85, int -0.99 | 60.27 | 1.81 |
| D2 | 60.69 | - | 60.69 | 1.82 |
| C2 | 57.39 | thk +2.75, int +1.10 | 61.25 | 1.84 |
| E3 | 27.98 | thk -2.84, int -0.78 | 24.35 | 0.73 |
| D3 | 19.45 | - | 19.45 | 0.58 |
| 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; 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, $b'$ = distance between the two piles). The D points are not corrected. (Around the shallow intermediate pile the corrected values come out nearly equal, about 60 to 61%.)

#### Floor thickness

Unbalanced head $h$ = hydraulic-gradient level at the floor $-$ floor top level; required thickness $t = \dfrac{h}{G-1}$ with $G = 2.40$.

| Location | x (m) | φ (%) | HG level (m) | Floor top (m) | Unbalanced head (m) | Required t (m) | Provided t (m) | Check |
|---|---|---|---|---|---|---|---|---|
| C1 (x = 0) | 0 | 81.33 | 102.44 | 100.00 | 2.44 | 1.74 | 1.00 | not safe |
| E2 (x = 10) | 10 | 60.27 | 101.81 | 100.00 | 1.81 | 1.29 | 1.25 | not safe (marginal) |
| C2 (x = 10) | 10 | 61.25 | 101.84 | 100.00 | 1.84 | 1.31 | 1.25 | not safe (marginal) |
| E3 (x = 30) | 30 | 24.35 | 100.73 | 100.00 | 0.73 | 0.52 | 1.00 | safe |

#### Exit gradient

$$
G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{3.00}{3.00}\times\frac{1}{\pi\sqrt{5.525}} = 0.1354\ (\approx 1/7.4)
$$

Permissible $= 1/5 = 0.2000$. Safe (less than the permissible value).

#### Hydraulic gradient line

Bars show the corrected pressure (each # is 2.5%):
```text
 E1  100.0% |########################################
 D1   83.9% |##################################
 C1   81.3% |#################################
 E2   60.3% |########################
 D2   60.7% |########################
 C2   61.2% |########################
 E3   24.4% |##########
 D3   19.5% |########
 C3    0.0% |
```
The hydraulic gradient line (HGL) drops 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:** The u/s floor (1.74 m needed, 1.0 m provided) is not safe, the floor at the intermediate pile is marginally short (about 1.3 m needed, 1.25 m provided), and the d/s floor is safe. The exit gradient of 1/7.4 is within the safe value of 1/5.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Water level RL 103.0 m and floor RL 100.0 m. What is H?
   A. 3 m
   B. 1 m
   C. 5 m
   D. 10 m
   Answer: A. H = 103 - 100 = 3 m, with d/s water at floor level. (Key point: Head)

2. Permissible exit gradient 1/5 means
   A. 0.02
   B. 0.5
   C. 0.2, and G_E must be below it
   D. 5, and G_E must exceed it
   Answer: C. 1/5 = 0.2. Computed G_E = 0.1354 is below it, so it is safe. (Key point: Safe exit gradient 1/5)

3. At C1 required thickness 1.75 m and provided 1.0 m. Verdict?
   A. Cannot decide
   B. Safe since the head is only 3 m
   C. Not safe
   D. Safe
   Answer: C. Provided 1.0 m is less than 1.75 m required. (Key point: Thickness check at C1)

4. At E3 the required thickness is 0.54 m and 1.0 m is provided. Why is it safe?
   A. Because G_E is small
   B. Because the HGL is vertical
   C. Uplift there is small (phi about 25%) so little thickness is needed
   D. Because it is a pile
   Answer: C. Pressure is only 0.75 m of head near the d/s end, so required t = 0.54 m is below 1.0 m. (Key point: Thickness check at E3)

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