---
title: "Meaning of shear stress reduction factor K"
subject: "Hydraulics"
chapter: 10
key: "semester_4/hydraulics:shear-reduction-factor-k-meaning"
source: https://www.bctengineer.com/q/semester_4/hydraulics/shear-reduction-factor-k-meaning
site: BCT Engineering
---

# Meaning of shear stress reduction factor K

Hydraulics · Chapter 10: Flow in mobile boundary channel

- Asked in: 2073 Magh, 3 marks; 2073 Bhadra, 2 marks; 2071 Bhadra, 2 marks (asked 3 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/shear-reduction-factor-k-meaning
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-10

## Question

Why is the shear stress reduction factor K necessary while designing the mobile boundary channel? Explain its physical meaning.

## Short answer

K is the ratio of critical side-slope shear stress to critical bed shear stress, always less than 1 because gravity assists flow on the slope. It is necessary to design channel banks: the permissible side shear is Kτc, preventing bank erosion even if the bed is stable; K links side slope steepness to stability.

## Answer

#### Physical meaning of $K$

$K$ is the ratio of the critical shear stress needed to start motion of a particle on the **side slope** to the critical shear stress needed on the **level bed**:

$$
K = \frac{\tau_{cs}}{\tau_{cb}} = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}} < 1
$$

where $\theta$ is the side slope angle and $\phi$ the angle of repose of the material. It tells us by what fraction the bed critical stress must be reduced to get the safe stress for the side slopes.

#### Why $K$ is needed in design

1. **Gravity helps the flow on the sides.** A particle on a side slope has a component of its weight down the slope in addition to the drag of the flow. It therefore moves at a lower shear stress than a particle on the flat bed.
2. **The sides are the critical part of the section.** Without $K$ a designer would size the channel using the bed critical stress only, and the banks would erode first, even though the shear stress on the banks (about $0.75\gamma yS_0$ for trapezoidal channels) is lower than on the bed (about $\gamma yS_0$).
3. **It links the side slope to the stability.** $K$ decreases as the side slope becomes steeper and falls to zero when $\theta = \phi$. By checking the side-slope stress against $K\tau_c$ the designer chooses a flatter side slope or a different cross-section so that both bed and banks are stable.
4. **It gives a uniform safety.** Stable-channel design needs the largest of $\tau_b/\tau_{cb}$ and $\tau_s/\tau_{cs}$ to be below 1; $K$ lets both be compared on the same basis.

## Check yourself

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

1. What does K physically express?
   A. The ratio of bed shear to side shear in the channel
   B. The ratio of water depth to side slope length
   C. The ratio of velocity on the side to velocity on the bed
   D. The ratio of critical shear stress on the side slope to that on the level bed, showing how much the bed value must be reduced for the sides
   Answer: D. K = tau_cs / tau_cb, the fraction by which bed critical stress is reduced for the side slope. (Key point: Physical meaning)

2. Why does a particle on a side slope move at a lower shear stress than one on the bed?
   A. Gravity adds a component of weight down the slope to the flow drag
   B. The water is faster on the side
   C. Side particles are smaller
   D. Side slopes have no friction
   Answer: A. The weight component down the slope helps the drag, so less flow shear is needed. (Key point: Why sides are weaker)

3. How does K change with the side slope?
   A. It increases as the slope gets steeper
   B. It decreases as the slope gets steeper and becomes zero at theta = phi
   C. It stays at 1 for all slopes
   D. It is zero for flat slopes
   Answer: B. K = sqrt(1 - sin^2(theta)/sin^2(phi)) falls with increasing theta and is zero at the angle of repose. (Key point: Variation of K)

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