Question 9.3: Design a straight trapezoidal channel for a design discharge......

For fine gravel, n = 0.024 and s = 3H : 1V. Therefore, \theta=\tan^{-1}  \frac{1}{3} = 18.4°
\quad\quadFrom Fig. 9-3, \phi = 24.° Hence,
\quad\quad\quad\quad K=\sqrt{1-\frac{\sin^{2}\theta}{\sin^{2}\phi}}=\sqrt{1-\frac{0.1}{0.16}}=0.63
From Fig. 9-4, the critical shear stress = 0.15 lbs/ft² = 7.18 N/m². Since the channel is straight, we do not have to make a correction for the alignment. The permissible shear stress for the channel side is 7.18× 0.63 = 4.52 N/m².
\quad\quadNow, the unit tractive force on the side = 0.76\gamma yS_{o} = 0.76×999×9.81y × 0.00025 = 1.862y. By equating the unit tractive force to the permissible stress, we obtain
\quad\quad\quad\quad1.862y = 4.52
or
\quad\quad\quad\quad y = 2.43  m
The channel bottom width, B_{o}, needed to carry 10 m³ /s may be determined from the Manning equation
\quad\quad\quad\quad \frac{1}{n}(B_{o}+sy)y(\frac{(B_{o}+sy)y}{B_{o}+2\sqrt{1+s^{2}}y})^{\frac{2}{3}}\sqrt{S_{o}}=Q
\quad\quadBy substituting n = 0.024, s = 3, y = 2.43, S_{o} = 0.00025, and Q = 10 m³ /s, and solving for Bo, we obtain
\quad\quad\quad\quad B_{o}=8.24  m
\quad\quadFor a selected freeboard of 0.75 m, the depth of section = 2.43 + 0.75 = 3.2 m. For ease of construction, select a bottom width, B_{o} = 8.25 m.