Tractive force design in coarse alluvium, 20 m³/s

Given: $Q = 20\ \text{m}^3/\text{s}$, side slope 4H : 2V ($z = 2$), $S_0 = 0.0015$, $D_{75} = 2$ cm $= 20$ mm (moderately rounded), $n = 0.025$, $\phi = 40^\circ$, slightly sinuous channel.

Assumptions: For coarse non-cohesive material, Lane's permissible tractive force on the bed is $\tau_c = 0.4$ lb/ft² per inch of $D_{75}$, i.e. $\tau_c \approx 0.754\,D_{75}$ N/m² with $D_{75}$ in mm. For a slightly sinuous channel, $\tau_c$ is reduced to 90% (Lane). Maximum shear on the bed $= \gamma y S_0$; on the sides $= 0.75\gamma y S_0$.

Step 1: Permissible shear on the bed

$$\tau_c = 0.754 \times 20 = 15.08\ \text{N/m}^2,\qquad 0.9\,\tau_c = 13.57\ \text{N/m}^2$$

Step 2: Side slope factor

$$\theta = \tan^{-1}\frac{1}{2} = 26.57^\circ,\qquad K = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}} = \sqrt{1 - \frac{0.2}{0.4132}} = 0.718$$

Permissible side shear $= K(0.9\tau_c) = 0.718 \times 13.57 = 9.75\ \text{N/m}^2$.

Step 3: Depth

• Bed: $y \le \dfrac{13.57}{9810 \times 0.0015} = 0.922$ m
• Sides: $y \le \dfrac{9.75}{0.75 \times 9810 \times 0.0015} = 0.883$ m

The sides govern: $y = 0.883$ m.

Step 4: Base width. Manning's equation with $n = 0.025$, $S_0 = 0.0015$, $z = 2$, $y = 0.883$ m, $Q = 20$:

$$A = (b + 1.766)(0.883),\quad P = b + 2(0.883)\sqrt5 = b + 3.949$$

Trial $b = 15.45$ m: $A = 15.20\ \text{m}^2$, $P = 19.40$ m, $R = 0.784$ m.

$$Q = \frac{1}{0.025}(15.20)(0.784)^{2/3}(0.0387) = 20.0\ \text{m}^3/\text{s} \checkmark$$

Step 5: Froude number check

$$V = \frac{20}{15.20} = 1.32\ \text{m/s},\quad T = 15.45 + 4(0.883) = 18.98\ \text{m},\quad D = \frac{A}{T} = 0.801\ \text{m}$$ $$Fr = \frac{1.32}{\sqrt{9.81 \times 0.801}} = 0.47 < 1$$

The flow is subcritical, as required. (With $b/y \approx 17.5$, the channel is wide, so taking the bed shear as $\gamma y S_0$ is justified.)

Answer: $b \approx 15.5$ m, $y \approx 0.88$ m, side slope 2H : 1V (4H : 2V), $V = 1.32$ m/s, $Fr = 0.47$. Provide freeboard in addition.