Stable channel: Q 20, particle 4 cm

Given: $Q = 20\ \text{m}^3/\text{s}$, $S = 0.001000$ (1 in 1000), particle size $d = 40$ mm, $\phi = 38^\circ$. Side slope $1.73H:1V$ (30.0° with horizontal); $\gamma = 9.81$ kN/m³, specific gravity of particles 2.65. $n = \dfrac{d^{1/6}}{21.1} = \dfrac{0.04^{1/6}}{21.1} = 0.0277$ (Strickler).

Critical shear stress of the bed (Shields)

$$\tau_c = 0.047(\gamma_s - \gamma)d = 0.047\times(2.65-1)\times 9.81\times 0.04 = 30.43\ \text{N/m}^2$$

Depth limited by the bed

$$\gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.03043}{9.81\times 0.001000} = 3.10\ \text{m}$$

Depth limited by the side slope

Reduction factor for the side:

$$K = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}} = \sqrt{1 - \frac{\sin^2 30.0^\circ}{\sin^2 38^\circ}} = 0.583$$

Maximum stress on the side is $0.75\gamma DS$, so:

$$D_{side} = \frac{K\tau_c}{0.75\gamma S} = \frac{0.583\times 0.03043}{0.75\times 9.81\times 0.001000} = 2.41\ \text{m}$$

The smaller value governs: stable depth $D = 2.41$ m (side controlled).

Bed width from Manning's equation

Using $D = 2.41$ m and $Q = \frac{1}{n}AR^{2/3}S^{1/2}$, trial gives $B = 2.00$ m, adopt $B = 2.0$ m. Check: $A = 14.88$ m², $R = 1.278$ m, $V = 1.344$ m/s, $Q = 20.00$ m³/s.

Answer: Stable depth $D \approx 2.41$ m, bed width $B \approx 2.0$ m, side slope 1.7H:1V.

This section is deep and narrow. Because every depth up to 2.41 m is stable, a wider and shallower section (larger $B$, smaller $D$) carrying the same discharge is equally stable and is usually preferred in practice.