Maximum permissible slope in 60 mm gravel

Given: trapezoidal channel $y = 1.5$ m, $b = 10$ m, $z = 2$; gravel $d_{50} = 60$ mm; $\phi = 37^\circ$; $K_2 = 0.9$.

Assumptions: $\tau_c = 0.06(\gamma_s - \gamma)d$ with $\gamma_s = 2.65\gamma$; the design shear is $K_2\tau_c$; maximum shear on bed $\gamma y S_0$, on sides $0.75\gamma y S_0$; $n = 0.047\,d^{1/6} = 0.0294$ (Strickler).

Step 1: Critical shear stress

$$\tau_c = 0.06(1.65)(9810)(0.06) = 58.27\ \text{N/m}^2,\qquad K_2\tau_c = 52.44\ \text{N/m}^2$$

Step 2: Side slope factor

$$\theta = \tan^{-1}\frac{1}{2} = 26.57^\circ,\qquad K = \sqrt{1 - \frac{0.2}{\sin^2 37^\circ}} = \sqrt{1 - \frac{0.2}{0.3622}} = 0.669$$

Step 3: Maximum slope from bed and side conditions

• Bed: $\gamma y S_0 \le 52.44 \Rightarrow S_0 \le \dfrac{52.44}{9810 \times 1.5} = 0.00356$
• Sides: $0.75\gamma y S_0 \le K(52.44) \Rightarrow S_0 \le \dfrac{0.669 \times 52.44}{0.75 \times 9810 \times 1.5} = 0.00318$

The side governs: maximum permissible slope $S_0 \approx 0.0032$ (about 1 in 314).

Step 4: Discharge at this slope

$$A = (10 + 2 \times 1.5)(1.5) = 19.5\ \text{m}^2,\quad P = 10 + 2(1.5)\sqrt5 = 16.71\ \text{m},\quad R = 1.167\ \text{m}$$ $$Q = \frac{1}{n}AR^{2/3}S_0^{1/2} = \frac{1}{0.0294}(19.5)(1.167)^{2/3}(0.0564) = 41.4\ \text{m}^3/\text{s}$$

Answer: Maximum permissible slope $\approx 0.0032$; discharge $\approx 41\ \text{m}^3/\text{s}$ without disturbing the stability of the channel.