Channel fully in banking, 0.5-1.5 km

Given data and assumptions

• Bed width $4.5$ m; FSD $1.5$ m; free board $0.5$ m, so $D = 2.0$ m; top width of each bank $2.5$ m; side slope of banks $1.5:1$ (inner and outer).
• Bed falls 1 in 5000: 0.10 m per 500 m, so bed = 1316.00, 1315.90, 1315.80 m at 500, 1000, 1500 m. The canal is wholly in banking.

Method

Total depth of canal section $D = \text{FSD} + \text{free board} = 1.50 + 0.50 = 2.00$ m. At each station the bed level is compared with the ground level; $e =$ GL − bed level.

• If $e \ge D$: full cutting, area $A_c = (b + s_c e)\,e$ (no banks needed).
• If $0 < e < D$: part cutting, part banking. Cutting area $A_c = (b + s_c e)e$; each bank stands $h_b = D - e$ above ground, so bank area (both banks) $A_b = (t_L + t_R)h_b + (s_i + s_o)h_b^2$ with $s_i, s_o$ the inner and outer bank slopes.
• If $e \le 0$: full banking, bed above ground. Bank top is $H = D - e$ above ground. Gross embankment $= (T + s_o H)H$ with $T = t_L + t_R + b + 2 s_i D$, and the canal space $(b + s_i D)D$ is deducted.

Volume between stations = mean area × distance (separately for cutting and banking).

Sectional areas

Volumes

Answer: Earthwork in cutting = 0.00 m³; earthwork in banking = 50,364.69 m³; total = 50,364.69 m³.