---
title: "Khosla: inside corners and exit gradient (5 m head)"
subject: "Irrigation and Drainage Engineering"
chapter: 5
key: "semester_6/irrigation_and_drainage_engineering:khosla-two-corners-exit-gradient-79a"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/khosla-two-corners-exit-gradient-79a
site: BCT Engineering
---

# Khosla: inside corners and exit gradient (5 m head)

Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks

- Asked in: 2079 Asoj, 6 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-two-corners-exit-gradient-79a
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5

## Question

Calculate the uplift pressure at the two inside corner parts and also the exit gradient using Khosla theory for the following headworks.

[Figure: weir section with a 5 m head over the weir; labelled dimensions 3 m, 1 m, 2 m, 6 m and a total floor length of 35 m; sheet piles at both ends (figure details only partly legible in the scan)]

## Short answer

With H=5m, floor length 35m, u/s pile depth 3m, d/s pile 6m, Khosla theory with thickness and interference corrections yields uplift at u/s inside corner C1: 78.7% (3.93m), at d/s inside corner E2: 30.9% (1.55m). Exit gradient = 0.143 (1 in 7).

## Answer

#### Data and assumptions

The figure details are only partly legible, so the following reading is adopted: head $H = 5$ m; total floor length $b = 35$ m between the end piles; u/s pile 3 m deep with floor thickness 1 m; d/s pile 6 m deep with floor thickness 2 m (floor top flush with the bed). The inside corners are $C_1$ (floor and u/s pile) and $E_2$ (floor and d/s pile). The safe exit gradient is not given, so 1/6 is assumed.

#### Khosla's constants

$\alpha = b/d$, $\lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-2}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-1}{\lambda}$ (in percent), total floor length $b = 35$ m. For the u/s pile, $\phi_{D1} = 100-\phi_D$ and $\phi_{C1} = 100-\phi_E$.

| Pile | d (m) | α | λ | φE (%) | φD (%) |
|---|---|---|---|---|---|
| U/s pile | 3.00 | 11.667 | 6.355 | 25.97 | 18.10 |
| D/s pile | 6.00 | 5.833 | 3.459 | 36.14 | 24.83 |

#### Pressures at the key points

Pressure is expressed as a percentage of the head $H = 5.00$ m.

| Point | Simple φ (%) | Corrections (%) | Corrected φ (%) | Pressure head (m) |
|---|---|---|---|---|
| E1 (top of u/s pile) | 100.00 | 0 | 100.00 | 5.00 |
| D1 | 81.90 | - | 81.90 | 4.10 |
| C1 | 74.03 | thk +2.62, int +2.02 | 78.68 | 3.93 |
| E2 | 36.14 | thk -3.77, int -1.43 | 30.94 | 1.55 |
| D2 | 24.83 | - | 24.83 | 1.24 |
| C2 (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, $19\sqrt{D/b'}\,\dfrac{d+D}{b}$ with $b' = 35$ m (at $C_1$: $D = 6$, $d = 3$; at $E_2$: $D = 3$, $d = 6$), positive at $C_1$ and negative at $E_2$. The D points are not corrected.

#### Exit gradient

$$
G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{5.00}{6.00}\times\frac{1}{\pi\sqrt{3.459}} = 0.1426\ (\approx 1/7.0)
$$

Permissible $= 1/6 = 0.1667$. Safe (less than the permissible value).

**Answer:** Uplift at the u/s inside corner $C_1 \approx 78.7\%$ of $H$ = 3.93 m; at the d/s inside corner $E_2 \approx 30.9\%$ of $H$ = 1.55 m; exit gradient $G_E = 0.143$ (about 1 in 7).

## Check yourself

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

1. The two inside corners where uplift is required are:
   A. C1 (floor/u/s pile) and E2 (floor/d/s pile)
   B. E1 and C2
   C. D1 and D2
   D. E1 and E2
   Answer: A. The inside corners are where the floor meets the piles: C1 and E2. (Key point: Inside corners)

2. At C1 the simple phi 74.03% becomes 78.68% by adding:
   A. Interference (+2.02) and thickness (+2.62)
   B. Slope correction
   C. Only interference
   D. Subtracting thickness
   Answer: A. C1 lies upstream of the other pile (positive interference) and thickness correction is positive. (Key point: Corrections at C1)

3. At E2 the thickness correction is negative because:
   A. Pressure at the floor underside is lower than at the pile tip junction level
   B. E2 is upstream
   C. Floor is thick
   D. Slope is up
   Answer: A. Interpolating from E2 toward the lower pressure at the underside reduces the value, giving 32.37%. (Key point: Corrections at E2)

4. Exit gradient G_E = (H/d)(1/(pi sqrt(lambda))) uses which pile?
   A. The d/s pile (6 m)
   B. The u/s pile (3 m)
   C. Both averaged
   D. The floor thickness
   Answer: A. The exit gradient occurs at the d/s end, controlled by d/s pile depth: 0.1426 < 1/6. (Key point: Exit gradient)

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