Irrigation channel earthwork, bed 5 m, 1 in 3000

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 $1.5:1$ for cutting and banking (inner and outer).
• Bed RL 1395.50 m at 0, falling 1 in 3000: 1395.50, 1395.40, 1395.30, 1395.20 m at 0, 300, 600, 900 m.

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 = 8,831.25 m³; earthwork in banking = 1,828.50 m³; total = 10,659.75 m³.