Irrigation canal earthwork, 0-600 m

Given data and assumptions

• Bed width $5$ m; FSD $1$ m; free board $0.6$ m, so $D = 1.6$ m; top width of each bank $2$ m; side slope in cutting $1:1$ and in banking $1.5:1$ (inner and outer faces of banks).
• Bed RL 324.00 m at 0, falling 1 in 3000: 324.00, 323.90, 323.80 m at 0, 300, 600 m. (No drop is mentioned in the data.)

Method

Total depth of canal section $D = \text{FSD} + \text{free board} = 1.00 + 0.60 = 1.60$ 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 = 3,285.68 m³; earthwork in banking = 2,560.72 m³; total = 5,846.40 m³.