---
title: "Tractive force design: Q 50"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:tractive-q50-slope-15000"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/tractive-q50-slope-15000
site: BCT Engineering
---

# Tractive force design: Q 50

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2071 Magh, 5 marks
- Video: [Unit:4 | Design of Canals | Tractive force approach of canal design | Irrigation Engineering | TU,PU](https://www.youtube.com/watch?v=YCJu1hWKmeI) — Prashant YT
- 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/tractive-q50-slope-15000
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

Design an unlined channel in alluvial soil by the tractive force approach for a discharge of 50 cumecs from the following data: i) Bed slope = 1/5000 ii) Side slopes = 0.5:1 iii) Manning's n = 0.022 iv) Permissible tractive stress = 0.0025 kN/m².

## Short answer

Depth 1.27 m determined by τ_p/(γS), bed width 53.0 m from Manning's equation with side slope 0.5:1. Bed slope 1 in 5000, velocity 0.73 m/s (discharge 50 m³/s).

## Answer

Given: $Q = 50\ \text{m}^3/\text{s}$, $S = 1$ in 5000, $n = 0.022$, $\tau_p = 0.0025\ \text{kN/m}^2$, side slope 0.5H:1V. The single permissible value is taken for both bed and sides (sides carry only 0.75 of the bed stress, so they are safer).

#### Principle

A channel is stable when the tractive force exerted by the flowing water on the boundary does not exceed the permissible (critical) tractive force of the soil. For a wide channel the maximum tractive stress on the bed is $\tau_0 = \gamma D S$ and on the sides about $0.75\gamma D S$.

#### Depth from bed stability

$$
D = \frac{\tau_p}{\gamma S} = \frac{0.0025}{9.81\times 0.000200} = 1.274\ \text{m}
$$

Adopt $D = 1.27$ m.

#### Bed width from Manning's equation

$Q = \frac{1}{n}AR^{2/3}S^{1/2}$ with $A = (B + 0.5D)D$, $P = B + 2D\sqrt{1+0.5^2}$. Solving by trial for $D = 1.27$ m gives $B = 53.01$ m, say **53.0 m**.

Check with $B = 53.0$ m:

$$
\begin{aligned}
A &= 68.12\ \text{m}^2,\ P = 55.84\ \text{m},\ R = 1.220\ \text{m} \\
V &= \frac{1}{0.022}\times 1.220^{2/3}\times 0.000200^{1/2} = 0.734\ \text{m/s} \\
Q &= AV = 49.99\ \text{m}^3/\text{s}
\end{aligned}
$$

Stresses: bed $\tau_0 = 9.81\times 1.27\times 0.000200 = 0.00249\ \text{kN/m}^2$ and sides $0.75\tau_0 = 0.00187\ \text{kN/m}^2$, both not more than the permissible value (sides are safe).

**Answer:** Depth $D = 1.27$ m, bed width $B \approx 53.0$ m, side slope 0.5H:1V, bed slope 1 in 5000, velocity 0.73 m/s (discharge 50.0 m³/s).

## Check yourself

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

1. How is the depth fixed here?
   A. D = tau_p x gamma x S
   B. D = Q/(B V)
   C. D = 0.75 tau_p/S
   D. D = tau_p/(gamma S), setting the bed stress gamma D S equal to the permissible value
   Answer: D. Maximum bed stress is gamma D S, so D = 1.27 m. (Key point: Depth from bed stress)

2. Why are the sides not the critical part of the check?
   A. Maximum stress on the sides is only about 0.75 of the bed stress
   B. The sides are always lined
   C. The sides carry 1.5 times the bed stress
   D. Side stress is zero for side slope 0.5:1
   Answer: A. Sides carry about 0.75 gamma D S, which is below the permissible value. (Key point: Sides vs bed)

3. How is the bed width found once D is fixed?
   A. From B = tau_p/gamma
   B. By Manning's equation Q = (1/n) A R^(2/3) S^(1/2), solving for B by trial
   C. From Lacey's P = 4.75 sqrt(Q)
   D. From B = 2D
   Answer: B. Manning with A = (B + 0.5D)D fixes B so that the discharge is 50 m^3/s. (Key point: Bed width)

## Close variants set in other exams

- [Tractive force design: Q 30](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Atractive-q30-slope-13700) — Irrigation and Drainage Engineering, chapter 4, asked 3×
- [Tractive force design: Q 45](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Atractive-q45-slope-14500) — Irrigation and Drainage Engineering, chapter 4, asked 3×

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