---
title: "Maximum permissible slope in 60 mm gravel"
subject: "Hydraulics"
chapter: 10
key: "semester_4/hydraulics:gravel-channel-max-permissible-slope"
source: https://www.bctengineer.com/q/semester_4/hydraulics/gravel-channel-max-permissible-slope
site: BCT Engineering
---

# Maximum permissible slope in 60 mm gravel

Hydraulics · Chapter 10: Flow in mobile boundary channel

- Asked in: 2069 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/gravel-channel-max-permissible-slope
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-10

## Question

A trapezoidal channel 1.5 m deep, 10 m bed width, with 2:1 side slopes is excavated in gravel of median size of 60 mm. What is the maximum permissible channel slope and what discharge can the channel carry without disturbing its stability? Take angle of repose ($\phi$) = 37° and $K_2$ = 0.9.

## Short answer

Maximum permissible slope is 0.0032 (about 1 in 314), governed by side shear stress. The channel carries 41 m³/s without disturbing stability, from Manning’s equation with n=0.0294.

## Answer

**Given:** trapezoidal channel $y = 1.5$ m, $b = 10$ m, $z = 2$; gravel $d_{50} = 60$ mm; $\phi = 37^\circ$; $K_2 = 0.9$.

**Assumptions:** $\tau_c = 0.06(\gamma_s - \gamma)d$ with $\gamma_s = 2.65\gamma$; the design shear is $K_2\tau_c$; maximum shear on bed $\gamma y S_0$, on sides $0.75\gamma y S_0$; $n = 0.047\,d^{1/6} = 0.0294$ (Strickler).

**Step 1: Critical shear stress**

$$
\tau_c = 0.06(1.65)(9810)(0.06) = 58.27\ \text{N/m}^2,\qquad K_2\tau_c = 52.44\ \text{N/m}^2
$$

**Step 2: Side slope factor**

$$
\theta = \tan^{-1}\frac{1}{2} = 26.57^\circ,\qquad K = \sqrt{1 - \frac{0.2}{\sin^2 37^\circ}} = \sqrt{1 - \frac{0.2}{0.3622}} = 0.669
$$

**Step 3: Maximum slope from bed and side conditions**

- Bed: $\gamma y S_0 \le 52.44 \Rightarrow S_0 \le \dfrac{52.44}{9810 \times 1.5} = 0.00356$
- Sides: $0.75\gamma y S_0 \le K(52.44) \Rightarrow S_0 \le \dfrac{0.669 \times 52.44}{0.75 \times 9810 \times 1.5} = 0.00318$

The side governs: **maximum permissible slope $S_0 \approx 0.0032$** (about 1 in 314).

**Step 4: Discharge at this slope**

$$
A = (10 + 2 \times 1.5)(1.5) = 19.5\ \text{m}^2,\quad P = 10 + 2(1.5)\sqrt5 = 16.71\ \text{m},\quad R = 1.167\ \text{m}
$$

$$
Q = \frac{1}{n}AR^{2/3}S_0^{1/2} = \frac{1}{0.0294}(19.5)(1.167)^{2/3}(0.0564) = 41.4\ \text{m}^3/\text{s}
$$

**Answer: Maximum permissible slope $\approx 0.0032$; discharge $\approx 41\ \text{m}^3/\text{s}$** without disturbing the stability of the channel.

## Check yourself

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

1. What is the design shear for 60 mm gravel with K2 = 0.9?
   A. 52.44 N/m^2
   B. 58.27 N/m^2
   C. 64.7 N/m^2
   D. 35.1 N/m^2
   Answer: A. tau_c = 58.27 and K2 tau_c = 52.44 N/m^2. (Key point: Design shear)

2. Which slope limits come from bed and sides?
   A. Bed 0.00318, sides 0.00356
   B. Bed 0.00356, sides 0.00318
   C. Both 0.0032
   D. Bed 0.0356, sides 0.0318
   Answer: B. Bed: 52.44/(9810 x 1.5); sides use 0.75 and K = 0.669. (Key point: Slope limits)

3. Which slope is adopted?
   A. 0.00356 from the bed
   B. 0.0034 average
   C. 0.0032 from the side condition
   D. 0.0016
   Answer: C. The smaller value (side) is the maximum permissible slope. (Key point: Governing slope)

4. How is the discharge found at the maximum slope?
   A. Q = V x b
   B. Q = gamma y S0
   C. Q = 10 x 1.5
   D. Manning's equation with n = 0.0294, A = 19.5 m^2, R = 1.167 m, giving about 41 m^3/s
   Answer: D. Q = (1/n) A R^(2/3) S^(1/2) = 41.4 m^3/s. (Key point: Discharge)

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