Stable depth: Q 50, slope 1:1000

Given: $Q = 50\ \text{m}^3/\text{s}$, $S = 0.001000$ (1 in 1000), particle size $d = 50$ mm, $\phi = 36^\circ$. The side slope is not given, so 2H:1V (26.6° with the horizontal, flatter than $\phi$) 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

$$\gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.03804}{9.81\times 0.001000} = 3.88\ \text{m}$$

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:

$$D_{side} = \frac{K\tau_c}{0.75\gamma S} = \frac{0.649\times 0.03804}{0.75\times 9.81\times 0.001000} = 3.35\ \text{m}$$

The smaller value governs: stable depth $D = 3.35$ m (side controlled).

Bed width from Manning's equation

Using $D = 3.35$ m and $Q = \frac{1}{n}AR^{2/3}S^{1/2}$, trial gives $B = 2.58$ m; adopt $B = 2.6$ m. Check: $A = 31.16$ m², $R = 1.772$ m, $V = 1.610$ m/s, $Q = 50.15$ m³/s.

Answer: Stable depth $D \approx 3.35$ m, bed width $B \approx 2.6$ m, side slope 2.0H:1V.

This section is deep and narrow. Because every depth up to 3.35 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.