Derive tractive force ratio (shear reduction factor)

The shear stress reduction factor (tractive force ratio) is

$$K = \frac{\tau_{cs}}{\tau_{cb}}$$

the ratio of the critical shear stress on a side slope ($\tau_{cs}$) to the critical shear stress on a level bed ($\tau_{cb}$) for the same material.

Setup

Consider a single sediment particle on a side slope inclined at angle $\theta$ to the horizontal. Let the angle of repose of the material be $\phi$, the submerged weight of the particle be $W_s$ and the area exposed to flow be $a$.

Forces on the particle at incipient motion:

• Weight component down the slope: $W_s\sin\theta$
• Drag (tractive) force along the flow direction: $\tau_s\,a$ (it acts along the channel axis, that is, at right angles to the first force within the plane of the slope)
• Normal reaction: $W_s\cos\theta$, giving a resisting force $W_s\cos\theta\tan\phi$

   cross-section of the side slope (flow is into the page)

        o  particle
         \ .
          v   .      Ws sin(theta) acts down the slope
                 .
                    .  theta
   ----------------------------- horizontal

   Drag tau_s * a acts into the page (along the flow)

Derivation

The two driving forces are at right angles, so their resultant is $\sqrt{(W_s\sin\theta)^2 + (\tau_s a)^2}$. At incipient motion this equals the resisting force:

$$W_s\cos\theta\tan\phi = \sqrt{(W_s\sin\theta)^2 + (\tau_{cs}\,a)^2}$$

Squaring and solving for $\tau_{cs}a$:

$$(\tau_{cs}a)^2 = W_s^2\cos^2\theta\tan^2\phi - W_s^2\sin^2\theta$$ $$\tau_{cs}\,a = W_s\cos\theta\tan\phi\sqrt{1 - \frac{\tan^2\theta}{\tan^2\phi}}$$

On a level bed ($\theta = 0$) the particle is held only by friction against the drag:

$$\tau_{cb}\,a = W_s\tan\phi$$

Dividing:

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

Simplifying with $\cos^2\theta\left(1 - \dfrac{\tan^2\theta}{\tan^2\phi}\right) = \cos^2\theta - \dfrac{\sin^2\theta}{\tan^2\phi}$ and using $\cos^2\theta = 1 - \sin^2\theta$:

$$\boxed{K = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}}}$$

Remarks: $K < 1$ always, because particles on a slope are helped to move by gravity. $K = 0$ when $\theta = \phi$ (the slope itself is at the angle of repose), so the side slope must be flatter than the angle of repose.