Canal earthwork, bed 3 m, free board 0.44 m

Given data and assumptions

• Bed width $3$ m; FSD $1$ m; free board $0.44$ m, so $D = 1.44$ m; top width of each bank $1.5$ m.
• Side slope for digging $1:1$; for banking 1 V : 1½ H, i.e. $1.5:1$ (H:V) on the inner and outer faces of the banks.

Method

Total depth of canal section $D = \text{FSD} + \text{free board} = 1.00 + 0.44 = 1.44$ 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).

Typical cross-section (part cutting, part banking)

      1.5 m                                  1.5 m
     _______                                _______   top of bank = bed + 1.44 m
    /       \ 1.5:1                  1.5:1 /       \     (FSD 1.0 m + FB 0.44 m)
   /  bank   \                            /  bank   \ 1.5:1
__/___________\________ G.L. ____________/___________\__
               \  1:1               1:1 /
                \      cutting         /    e = GL - bed level
                 \____________________/
                        b = 3 m

Sectional areas

Volumes

Answer: Earthwork in cutting = 431.93 m³; earthwork in banking = 579.68 m³; total = 1,011.61 m³.