Tractive force design: Q 45

Given: $Q = 45\ \text{m}^3/\text{s}$, $S = 1$ in 4500, $n = 0.0225$, $\tau_p = 0.0035\ \text{kN/m}^2$, side slope 0.5H:1V. The single permissible value is taken for both bed and sides (the sides carry only 0.75 of the bed stress, so they are safer).

Principle

A channel is stable when the tractive force exerted by the flowing water on the boundary does not exceed the permissible (critical) tractive force of the soil. For a wide channel the maximum tractive stress on the bed is $\tau_0 = \gamma D S$ and on the sides about $0.75\gamma D S$.

Depth from bed stability

$$D = \frac{\tau_p}{\gamma S} = \frac{0.0035}{9.81\times 0.000222} = 1.606\ \text{m}$$

Adopt $D = 1.60$ m (rounded down, so that the bed stress does not exceed the permissible value).

Bed width from Manning's equation

$Q = \frac{1}{n}AR^{2/3}S^{1/2}$ with $A = (B + 0.5D)D$, $P = B + 2D\sqrt{1+0.5^2}$. Solving by trial for $D = 1.60$ m gives $B = 31.96$ m, say 32.0 m.

Check with $B = 32.0$ m:

$$\begin{aligned} A &= 52.48\ \text{m}^2,\ P = 35.58\ \text{m},\ R = 1.475\ \text{m} \\ V &= \frac{1}{0.0225}\times 1.475^{2/3}\times 0.000222^{1/2} = 0.859\ \text{m/s} \\ Q &= AV = 45.06\ \text{m}^3/\text{s} \end{aligned}$$

Stresses: bed $\tau_0 = 9.81\times 1.60\times 0.000222 = 0.00349\ \text{kN/m}^2$ and sides $0.75\tau_0 = 0.00262\ \text{kN/m}^2$, both within the permissible value.

Answer: Depth $D = 1.60$ m, bed width $B \approx 32.0$ m, side slope 0.5H:1V, bed slope 1 in 4500, velocity 0.86 m/s (discharge 45.1 m³/s).