---
title: "Tractive force design for 10 cumecs, 3 cm particles"
subject: "Hydraulics"
chapter: 10
key: "semester_4/hydraulics:tractive-force-10-cumecs-3cm-particles"
source: https://www.bctengineer.com/q/semester_4/hydraulics/tractive-force-10-cumecs-3cm-particles
site: BCT Engineering
---

# Tractive force design for 10 cumecs, 3 cm particles

Hydraulics · Chapter 10: Flow in mobile boundary channel

- Asked in: 2068 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/tractive-force-10-cumecs-3cm-particles
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-10

## Question

A channel which is to carry 10 m³/s through moderately rolling topography on a slope of 0.0016 is to be excavated in coarse alluvium with 50% of particles being 3 cm or more in diameter. Assume that channel is to be unlined and of trapezoidal section. Find suitable value of base width and side slope. Take $\phi$ = 34° and $K_2$ (ratio between bed shear stress and critical shear stress) = 0.75. Use tractive force method.

## Short answer

Suitable trapezoidal section: base width 4.6 m, side slope 2H:1V, depth 1.11 m. Tractive force method; side shear governs permissible depth, Manning’s equation with n=0.0262 yields 10 m³/s.

## Answer

**Given:** $Q = 10\ \text{m}^3/\text{s}$, $S_0 = 0.0016$, 50% of particles $\ge 3$ cm ($d_{50} = 0.03$ m), trapezoidal, $\phi = 34^\circ$, $K_2 = 0.75$ (ratio of permissible bed shear to critical shear).

**Assumptions:** $\tau_c = 0.06(\gamma_s - \gamma)d$ with $\gamma_s = 2.65\gamma$; maximum shear on bed $\gamma y S_0$, on sides $0.75\gamma y S_0$; Strickler's $n = 0.047(0.03)^{1/6} = 0.0262$.

**Step 1: Critical and permissible shear on the bed**

$$
\tau_c = 0.06(1.65)(9810)(0.03) = 29.14\ \text{N/m}^2,\qquad K_2\tau_c = 0.75(29.14) = 21.85\ \text{N/m}^2
$$

**Step 2: Side slope.** Choose $z = 2$ ($\theta = 26.57^\circ < \phi = 34^\circ$):

$$
K = \sqrt{1 - \frac{\sin^2 26.57^\circ}{\sin^2 34^\circ}} = \sqrt{1 - \frac{0.2}{0.3127}} = 0.600
$$

**Step 3: Depth**

- Bed: $y \le \dfrac{21.85}{9810 \times 0.0016} = 1.39$ m
- Sides: $y \le \dfrac{0.600 \times 21.85}{0.75 \times 9810 \times 0.0016} = 1.11$ m

The sides govern: **$y = 1.11$ m**.

**Step 4: Base width** from Manning's equation ($n = 0.0262$, $S_0 = 0.0016$):

$$
A = (b + 2.23)(1.114),\quad P = b + 2(1.114)\sqrt5 = b + 4.98
$$

Trial $b = 4.62$ m: $A = 7.63\ \text{m}^2$, $P = 9.61$ m, $R = 0.795$ m.

$$
Q = \frac{1}{0.0262}(7.63)(0.795)^{2/3}(0.04) = 10.0\ \text{m}^3/\text{s} \checkmark
$$

**Step 5: Check**

$$
V = 1.31\ \text{m/s},\quad T = 4.62 + 4(1.114) = 9.08\ \text{m},\quad Fr = \frac{1.31}{\sqrt{9.81 \times 7.63/9.08}} = 0.46 < 1
$$

**Answer: Base width $b \approx 4.6$ m, side slope 2H : 1V, depth $y \approx 1.11$ m** (subcritical flow, $V = 1.31$ m/s).

## Check yourself

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

1. What is tau_c for 3 cm particles?
   A. 29.14 N/m^2
   B. 21.85 N/m^2
   C. 48.56 N/m^2
   D. 0.60 N/m^2
   Answer: A. 0.06 x 1.65 x 9810 x 0.03 = 29.14 N/m^2. (Key point: Critical shear)

2. What bed shear is permitted with K2 = 0.75?
   A. 29.14 N/m^2
   B. 21.85 N/m^2
   C. 38.9 N/m^2
   D. 16.4 N/m^2
   Answer: B. K2 x tau_c = 0.75 x 29.14 = 21.85 N/m^2. (Key point: Permissible bed shear)

3. What governs the depth?
   A. Bed, y about 1.39 m
   B. Sides, y about 1.39 m
   C. Sides, y about 1.11 m
   D. Bed, y about 1.11 m
   Answer: C. Bed limit 1.39 m; side limit 1.11 m is smaller. (Key point: Governing depth)

4. Which is the final design result?
   A. b about 1.1 m, y 4.6 m
   B. b about 9 m, supercritical
   C. b about 2 m, z = 4
   D. b about 4.6 m, z = 2, y about 1.11 m, subcritical
   Answer: D. Manning's equation gives b = 4.62 m and Fr = 0.46. (Key point: Result)

## Close variants set in other exams

- [Tractive force design for 15 cumecs, 5 cm particles](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Atractive-force-15-cumecs-5cm-particles) — Hydraulics, chapter 10, asked 2×

## Related in Hydraulics

- [Tractive force design in coarse alluvium, 20 m³/s](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Atractive-force-trapezoid-20m3s-d75-2cm) — Hydraulics, chapter 10

## Related in other subjects

- [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×