Stable canal: Q 40, d 5 cm

Given: $Q = 40\ \text{m}^3/\text{s}$, $S = 0.000833$ (1 in 1200), particle size $d = 50$ 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. $n = \dfrac{d^{1/6}}{21.1} = \dfrac{0.05^{1/6}}{21.1} = 0.0288$ (Strickler).

Critical shear stress of the bed (Shields)

$$\tau_c = 0.047(\gamma_s - \gamma)d = 0.047\times(2.65-1)\times 9.81\times 0.05 = 38.04\ \text{N/m}^2$$

Depth limited by the bed

$$\gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.03804}{9.81\times 0.000833} = 4.65\ \text{m}$$

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:

$$D_{side} = \frac{K\tau_c}{0.75\gamma S} = \frac{0.669\times 0.03804}{0.75\times 9.81\times 0.000833} = 4.15\ \text{m}$$

The smaller value governs: stable depth $D = 4.15$ m (side controlled).

Section for the discharge

At $D = 4.15$ m even a very small bed width would carry more than 40 m³/s, so the stable depth is only the upper limit of the depth: any flow depth up to 4.15 m is stable. To pass the discharge, choose a practical bed width $B = 15$ m; Manning's equation then gives a flow depth

$$D = 1.73\ \text{m}\ (< 4.15\ \text{m}),\quad A = 31.83\ \text{m}^2,\quad R = 1.401\ \text{m},\quad V = 1.257\ \text{m/s}$$

and the bed stress $\gamma DS = 14.1$ N/m² is below $\tau_c = 38.0$ N/m².

Answer: Maximum stable depth $= 4.15$ m. Adopted section: $B = 15$ m, $D = 1.73$ m, side slope 2.0H:1V, $V = 1.26$ m/s.