Stable canal: Q 32, d 30 mm

Given: $Q = 32\ \text{m}^3/\text{s}$, $S = 0.000667$ (1 in 1500), particle size $d = 30$ mm, $\phi = 37^\circ$. The banks are unprotected, so side stability governs; a side slope of 2H:1V (26.6° with the horizontal) is assumed. Take $\gamma = 9.81$ kN/m³ and specific gravity of particles 2.65. $n = \dfrac{d^{1/6}}{21.1} = \dfrac{0.03^{1/6}}{21.1} = 0.0264$ (Strickler).

Critical shear stress of the bed (Shields)

$$\tau_c = 0.047(\gamma_s - \gamma)d = 0.047\times(2.65-1)\times 9.81\times 0.03 = 22.82\ \text{N/m}^2$$

Depth limited by the bed

$$\gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.02282}{9.81\times 0.000667} = 3.49\ \text{m}$$

Depth limited by the side slope

Reduction factor for the side:

$$K = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}} = \sqrt{1 - \frac{\sin^2 26.6^\circ}{\sin^2 37^\circ}} = 0.669$$

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

$$D_{side} = \frac{K\tau_c}{0.75\gamma S} = \frac{0.669\times 0.02282}{0.75\times 9.81\times 0.000667} = 3.11\ \text{m}$$

The smaller value governs: stable depth $D = 3.11$ m (side controlled).

Bed width from Manning's equation

Using $D = 3.11$ m and $Q = \frac{1}{n}AR^{2/3}S^{1/2}$, trial gives $B = 1.58$ m; adopt $B = 1.6$ m. Check: $A = 24.32$ m², $R = 1.568$ m, $V = 1.319$ m/s, $Q = 32.08$ m³/s.

Answer: Stable depth $D \approx 3.11$ m, bed width $B \approx 1.6$ m, side slope 2.0H:1V.

This section is deep and narrow. Because every depth up to 3.11 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.