---
title: "Tractive force design: Q 30"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:tractive-q30-slope-13700"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/tractive-q30-slope-13700
site: BCT Engineering
---

# Tractive force design: Q 30

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2078 Baisakh, 4 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-q30-slope-13700
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

Using tractive force approach, design a channel in alluvial soil for the following data: Discharge Q = 30 cumecs; Bed slope = 1/3700; Manning's n = 0.0225; Permissible tractive stress = 0.0035 kN/m²; Side slope = 0.5:1.

## Short answer

Depth 1.32 m from τ_p/(γS), bed width 26.6 m from Manning's trial with side slope 0.5:1. Bed slope 1 in 3700, velocity 0.83 m/s (discharge 30 m³/s).

## Answer

Given: $Q = 30\ \text{m}^3/\text{s}$, $S = 1$ in 3700, $n = 0.0225$, $\tau_p = 0.0035\ \text{kN/m}^2$, 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

$$
D = \frac{\tau_p}{\gamma S} = \frac{0.0035}{9.81\times 0.000270} = 1.320\ \text{m}
$$

Adopt $D = 1.32$ 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 $D = 1.32$ m gives $B = 26.62$ m, say **26.6 m**.

Check with $B = 26.6$ m:

$$
\begin{aligned}
A &= 35.98\ \text{m}^2,\ P = 29.55\ \text{m},\ R = 1.218\ \text{m} \\
V &= \frac{1}{0.0225}\times 1.218^{2/3}\times 0.000270^{1/2} = 0.833\ \text{m/s} \\
Q &= AV = 29.98\ \text{m}^3/\text{s}
\end{aligned}
$$

Stresses: bed $\tau_0 = 9.81\times 1.32\times 0.000270 = 0.00350\ \text{kN/m}^2$ and sides $0.75\tau_0 = 0.00262\ \text{kN/m}^2$, both not more than the permissible value (sides are safe).

**Answer:** Depth $D = 1.32$ m, bed width $B \approx 26.6$ m, side slope 0.5H:1V, bed slope 1 in 3700, velocity 0.83 m/s (discharge 30.0 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 = 1.32 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 30 m^3/s. (Key point: Bed width)

## Close variants set in other exams

- [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×
- [Tractive force design: Q 50](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Atractive-q50-slope-15000) — Irrigation and Drainage Engineering, chapter 4, asked 3×

## Related in Irrigation and Drainage Engineering

- [Sediment transport and tractive force approach](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Asediment-transport-tractive-force) — Irrigation and Drainage Engineering, chapter 4
- [Siphon aqueduct: Q 60, Manning 0.025](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Asiphon-aqueduct-q60-bed30-fsd14) — Irrigation and Drainage Engineering, chapter 8

## Related in other subjects

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- [Tractive force method of channel design](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Atractive-force-method-explain) — Hydraulics, chapter 10, asked 3×