Stable depth: Q 20, d 40 mm

Given: $Q = 20\ \text{m}^3/\text{s}$, $S = 0.000667$ (1 in 1500), particle size $d = 40$ mm, $\phi = 36^\circ$, $n = 0.025$. The side slope is not given, so 2H:1V (26.6° with the horizontal) is assumed. Take $\gamma = 9.81$ kN/m³ and specific gravity of particles 2.65.

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

$$\gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.03043}{9.81\times 0.000667} = 4.65\ \text{m}$$

Depth limited by the side slope

Reduction factor for the side:

$$K = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}} = \sqrt{1 - \frac{\sin^2 26.6^\circ}{\sin^2 36^\circ}} = 0.649$$

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

$$D_{side} = \frac{K\tau_c}{0.75\gamma S} = \frac{0.649\times 0.03043}{0.75\times 9.81\times 0.000667} = 4.03\ \text{m}$$

The smaller value governs: stable depth $D = 4.03$ m (side controlled).

Section for the discharge

At $D = 4.03$ m even a very small bed width would carry more than 20 m³/s, so the stable depth is only the upper limit of the depth: any flow depth up to 4.03 m is stable. To pass the discharge, choose a practical bed width $B = 8$ m; Manning's equation then gives a flow depth

$$D = 1.57\ \text{m}\ (< 4.03\ \text{m}),\quad A = 17.49\ \text{m}^2,\quad R = 1.165\ \text{m},\quad V = 1.143\ \text{m/s}$$

and the bed stress $\gamma DS = 10.3$ N/m² is below $\tau_c = 30.4$ N/m².

Answer: Maximum stable depth $= 4.03$ m. Adopted section: $B = 8$ m, $D = 1.57$ m, side slope 2.0H:1V, $V = 1.14$ m/s.