Stable depth and bed width: Q 5, d50 3 cm

Given: $Q = 5\ \text{m}^3/\text{s}$, $S = 0.002000$ (1 in 500), particle size $d = 30$ mm. The sides are well protected, so the angle of repose is not needed; a side slope of 1.5H:1V (33.7° 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

$$\gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.02282}{9.81\times 0.002000} = 1.16\ \text{m}$$

The sides are well protected, so only bed stability governs: $D = D_{bed}$.

Bed width from Manning's equation

Using $D = 1.16$ m and $Q = \frac{1}{n}AR^{2/3}S^{1/2}$, trial gives $B = 1.59$ m; adopt $B = 1.6$ m. Check: $A = 3.87$ m², $R = 0.670$ m, $V = 1.296$ m/s, $Q = 5.02$ m³/s.

Answer: Stable depth $D \approx 1.16$ m, bed width $B \approx 1.6$ m, side slope 1.5H:1V.

This section is deep and narrow. Because every depth up to 1.16 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.