---
title: "Uplift at inside corners, 50 m floor weir"
subject: "Irrigation and Drainage Engineering"
chapter: 5
key: "semester_6/irrigation_and_drainage_engineering:weir-uplift-two-inside-corners-50m"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/weir-uplift-two-inside-corners-50m
site: BCT Engineering
---

# Uplift at inside corners, 50 m floor weir

Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks

- Asked in: 2079 Jestha, 8 marks; 2076 Bhadra, 8 marks (asked 2 times)
- 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/weir-uplift-two-inside-corners-50m
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5

## Question

A weir has a solid horizontal floor length of 50 m with two lines of cutoff of 8 m depth below the river bed at its two ends. The floor thickness is 1 m at upstream end and 2 m at downstream end, with its upper level being in flush with river bed. For an effective head of 5 m over the weir, calculate the uplift pressure at the two inside corner points and also the exit gradient.

## Short answer

Khosla's theory: α=6.25, λ=3.665, basic pressures u/s inside corner 65.01%, d/s 34.99%. After interference (+2.43% for u/s,–2.43% d/s) and thickness corrections (+1.36%,–2.73%), final u/s corner uplift 68.8% (3.44m), d/s corner 29.8% (1.49m). Exit gradient 0.104.

## Answer

Given: floor length $b = 50$ m, cutoffs at both ends, depth $d = 8$ m below the river bed (the floor top is flush with the bed), head $H = 5$ m, floor thickness $t_1 = 1$ m (u/s end) and $t_2 = 2$ m (d/s end). Khosla's theory is used. The inside corners are the junctions of the floor with the u/s pile ($C_1$) and with the d/s pile ($E_2$).

#### Khosla's constants

$$
\alpha = \frac{b}{d} = \frac{50}{8} = 6.25,\qquad \lambda = \frac{1 + \sqrt{1+\alpha^2}}{2} = 3.665
$$

$$
\begin{aligned}
\phi_E &= \frac{1}{\pi}\cos^{-1}\left(\frac{\lambda-2}{\lambda}\right) = 34.99\% \\
\phi_D &= \frac{1}{\pi}\cos^{-1}\left(\frac{\lambda-1}{\lambda}\right) = 24.09\%
\end{aligned}
$$

#### Pressures from the simple profile (thin floor)

| Point | Formula | Pressure (%) |
|---|---|---|
| $E_1$ (top of u/s pile) | 100 | 100 |
| $D_1$ (tip of u/s pile) | $100 - \phi_D$ | 75.91 |
| $C_1$ (u/s inside corner) | $100 - \phi_E$ | 65.01 |
| $E_2$ (d/s inside corner) | $\phi_E$ | 34.99 |
| $D_2$ (tip of d/s pile) | $\phi_D$ | 24.09 |
| $C_2$ (d/s end) | 0 | 0 |

#### Correction for mutual interference of piles

Both piles are 8 m deep ($d = D = 8$ m), spaced $b' = 50$ m apart, total floor length $b = 50$ m:

$$
C = 19\sqrt{\frac{D}{b'}}\left(\frac{d+D}{b}\right) = 19\sqrt{\frac{8}{50}}\times\frac{16}{50} = 2.43\%
$$

It is positive at $C_1$ (upstream of the interfering pile) and negative at $E_2$ (downstream of it).

#### Correction for floor thickness

The uplift acts on the underside of the floor, a distance $t$ below the key point, so the pressure is interpolated along the pile:

$$
\begin{aligned}
\text{at } C_1: &\ (\phi_{D_1} - \phi_{C_1})\frac{t_1}{d} = (75.91 - 65.01)\times\frac{1}{8} = +1.36\% \\
\text{at } E_2: &\ (\phi_{E_2} - \phi_{D_2})\frac{t_2}{d} = (34.99 - 24.09)\times\frac{2}{8} = -2.73\%
\end{aligned}
$$

#### Corrected pressures and uplift heads

$$
\begin{aligned}
\phi_{C_1}' &= 65.01 + 2.43 + 1.36 = 68.80\%,\qquad h_{C_1} = 0.6880\times 5 = 3.44\ \text{m} \\
\phi_{E_2}' &= 34.99 - 2.43 - 2.73 = 29.83\%,\qquad h_{E_2} = 0.2983\times 5 = 1.49\ \text{m}
\end{aligned}
$$

#### Exit gradient

$$
G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{5}{8}\times\frac{1}{\pi\sqrt{3.665}} = 0.1039\ (\approx 1/9.6)
$$

**Answer:** Uplift at the u/s inside corner $\approx 68.8\%$ of $H$ (3.44 m); at the d/s inside corner $\approx 29.8\%$ (1.49 m); exit gradient $G_E \approx 0.104$, which is less than the safe gradient of about 1/6 (0.167) for fine sand, so the structure is safe against piping.

## Check yourself

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

1. How is lambda found?
   A. lambda = b/d
   B. lambda = (1 + sqrt(1+alpha^2))/2, with alpha = b/d
   C. lambda = sqrt(1+alpha)
   D. lambda = 1/alpha
   Answer: B. alpha = b/d = 6.25, lambda = (1 + sqrt(1+alpha^2))/2 = 3.665. (Key point: Khosla constant)

2. Why is the pressure at C1 corrected for floor thickness?
   A. Uplift acts on the underside of the floor, a distance t below the key point
   B. The floor is made of concrete
   C. Lambda changes with thickness
   D. The head changes
   Answer: A. Uplift acts on the floor bottom, so pressure is interpolated along the pile by t/d. (Key point: Floor thickness correction)

3. What is the corrected uplift pressure at the u/s inside corner?
   A. 65.01%
   B. 34.99%
   C. 75.91%
   D. 66.37% of H, about 3.32 m
   Answer: D. phi_C1' = 65.01 + (75.91 - 65.01)(1/8) = 66.37%, and 0.6637 x 5 = 3.32 m. (Key point: Uplift at C1)

4. What is the formula and result for exit gradient?
   A. G_E = (H/d)(1/(pi sqrt(lambda))) = 0.104, less than the safe 1/6
   B. G_E = H/b = 0.1
   C. G_E = (H/d) x lambda = 2.3
   D. G_E = H/(pi d) = 0.2
   Answer: A. G_E = (5/8)/(pi sqrt(3.665)) = 0.104, below the safe value of about 0.167, so no piping. (Key point: Exit gradient)

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