Irrigation canal earthwork with vertical drop

Given data and assumptions

• Bed width $b = 5$ m; FSD $= 1$ m; free board $= 0.5$ m, so $D = 1.5$ m; top width of each bank $2$ m.
• Canal side slope $2:1$ (inner slope of banks) and outer bank slope $1:1$ (H:V).
• Bed RL falls 1 in 5000 in the direction of chainage: bed at 0+200 is 130.96 m, so it is 131.00 m at 0+000 and 130.92 m at 0+400. The drop of 0.5 m at 0+400 lowers the bed to 130.42 m, then 130.38 m at 0+600.
• The bed is above the ground everywhere, so the canal is wholly in embankment.

Method

Total depth of canal section $D = \text{FSD} + \text{free board} = 1.00 + 0.50 = 1.50$ 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 = 27,115.52 m³; total = 27,115.52 m³.