---
title: "Tractive force design for 15 cumecs, 5 cm particles"
subject: "Hydraulics"
chapter: 10
key: "semester_4/hydraulics:tractive-force-15-cumecs-5cm-particles"
source: https://www.bctengineer.com/q/semester_4/hydraulics/tractive-force-15-cumecs-5cm-particles
site: BCT Engineering
---

# Tractive force design for 15 cumecs, 5 cm particles

Hydraulics · Chapter 10: Flow in mobile boundary channel

- Asked in: 2076 Baisakh, 5 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-15-cumecs-5cm-particles
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-10

## Question

A channel which is to carry 15 cumecs through moderately rolling topography on a slope of 0.0015 is to be excavated in coarse alluvium with 50% of particles being 5 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 internal frictional angle as 34° and ratio between bed shear stress and critical shear stress as 0.7. Use tractive force method.

## Short answer

Critical shear 48.56 N/m^2, design bed shear 33.99 N/m^2, side factor K=0.600, depth 1.85 m. Manning's equation with n=0.0285 yields base width 2.1 m, side slope 2H:1V, subcritical flow Fr=0.42.

## Answer

**Given:** $Q = 15\ \text{m}^3/\text{s}$, $S_0 = 0.0015$, 50% of particles $\ge 5$ cm ($d_{50} = 0.05$ m), trapezoidal, $\phi = 34^\circ$, ratio of bed shear to critical shear $K_2 = 0.7$.

**Assumptions:** $\tau_c = 0.06(\gamma_s - \gamma)d$ with $\gamma_s = 2.65\gamma$; the design bed shear is limited to $K_2\tau_c$; maximum shear on bed $\gamma y S_0$ and on sides $0.75\gamma y S_0$; Strickler's $n = 0.047\,d^{1/6} = 0.0285$.

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

$$
\tau_c = 0.06(1.65)(9810)(0.05) = 48.56\ \text{N/m}^2,\qquad K_2\tau_c = 0.7(48.56) = 33.99\ \text{N/m}^2
$$

**Step 2: Side slope.** Choose $z = 2$ (the side slope must be flatter than $\phi = 34^\circ$, i.e. $z > 1.48$):

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

**Step 3: Depth**

- Bed: $y \le \dfrac{33.99}{9810 \times 0.0015} = 2.31$ m
- Sides: $y \le \dfrac{0.600 \times 33.99}{0.75 \times 9810 \times 0.0015} = 1.85$ m

The sides govern: **$y = 1.85$ m**.

**Step 4: Base width** from Manning's equation ($n = 0.0285$):

$$
A = (b + 3.70)(1.85),\quad P = b + 2(1.85)\sqrt5 = b + 8.27
$$

Trial $b = 2.14$ m: $A = 10.79\ \text{m}^2$, $P = 10.40$ m, $R = 1.037$ m.

$$
Q = \frac{1}{0.0285}(10.79)(1.037)^{2/3}(0.03873) = 15.0\ \text{m}^3/\text{s} \checkmark
$$

**Step 5: Check**

$$
V = 1.39\ \text{m/s},\quad T = 2.14 + 4(1.85) = 9.54\ \text{m},\quad Fr = \frac{1.39}{\sqrt{9.81 \times 10.79/9.54}} = 0.42 < 1
$$

**Answer: Base width $b \approx 2.1$ m, side slope 2H : 1V ($\theta = 26.6^\circ$), depth $y \approx 1.85$ m.** The narrow base is a consequence of the high permissible depth; the section may be widened (with lower depth) if a shallower channel is preferred.

## Check yourself

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

1. How is the design bed shear obtained?
   A. K2 x tau_c = 0.7 x 48.56 = 33.99 N/m^2
   B. tau_c / 0.7
   C. tau_c = 48.56 only
   D. 0.75 x 48.56
   Answer: A. The design shear is limited to the ratio K2 times the critical stress. (Key point: Permissible bed shear)

2. Why must z exceed 1.48 for phi = 34 degrees?
   A. To get Fr below 1
   B. The side slope angle must be flatter than the angle of repose
   C. To reduce Manning's n
   D. To increase depth
   Answer: B. tan(34) = 0.675, so z must be greater than 1/0.675 = 1.48 for theta < phi. (Key point: Side slope limit)

3. Which governs and what depth?
   A. Bed, y = 2.31 m
   B. Sides, y = 2.31 m
   C. Sides, y = 1.85 m
   D. Bed, y = 1.85 m
   Answer: C. Side limit 1.85 m is smaller than bed limit 2.31 m. (Key point: Governing depth)

4. How is the base width found?
   A. From b = 2y
   B. From the Shields diagram
   C. From the Froude number
   D. From Manning's equation with the chosen y and z, by trial, b about 2.1 m
   Answer: D. With y fixed, Manning's equation for Q = 15 m^3/s gives b by trial. (Key point: Base width)

## Close variants set in other exams

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