Tractive force design: Q 50

Given: $Q = 50\ \text{m}^3/\text{s}$, $S = 1$ in 5000, $n = 0.022$, $\tau_p = 0.0025\ \text{kN/m}^2$, side slope 0.5H:1V. The single permissible value is taken for both bed and sides (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

$$D = \frac{\tau_p}{\gamma S} = \frac{0.0025}{9.81\times 0.000200} = 1.274\ \text{m}$$

Adopt $D = 1.27$ m.

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 $D = 1.27$ m gives $B = 53.01$ m, say 53.0 m.

Check with $B = 53.0$ m:

$$\begin{aligned} A &= 68.12\ \text{m}^2,\ P = 55.84\ \text{m},\ R = 1.220\ \text{m} \\ V &= \frac{1}{0.022}\times 1.220^{2/3}\times 0.000200^{1/2} = 0.734\ \text{m/s} \\ Q &= AV = 49.99\ \text{m}^3/\text{s} \end{aligned}$$

Stresses: bed $\tau_0 = 9.81\times 1.27\times 0.000200 = 0.00249\ \text{kN/m}^2$ and sides $0.75\tau_0 = 0.00187\ \text{kN/m}^2$, both not more than the permissible value (sides are safe).

Answer: Depth $D = 1.27$ m, bed width $B \approx 53.0$ m, side slope 0.5H:1V, bed slope 1 in 5000, velocity 0.73 m/s (discharge 50.0 m³/s).