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Channel earthwork with drop at 0+090

Short answer

Mean-area method applied across a drop; all sections have part cutting and part banking. Cutting = 1,139.85 m³, banking = 953.55 m³, total = 2,093.40 m³.

Calculate the quantity of the earthwork of a channel with the following data: Bed width = 6 m, free board = 50 cm, slope of cutting 1:1, side slope of banking 1.5:1, full supply depth is 2 m, width of the banks is 1.5 m. There is vertical drop of 0.5 m at ground surface at chainage 0+090. The longitudinal slope of canal is 1 in 600.

ChainageGround Level (m)Bed Level (m)
0+000225.3224.00
0+030225.2
0+060225.1
0+090225.2 / 224.7 (as printed, before and after drop)
0+120224.8

Given data and assumptions

  • Bed width 66 m; FSD 22 m; free board 0.50.5 m, so D=2.5D = 2.5 m; width (top) of each bank 1.51.5 m.
  • Side slope in cutting 1:11:1; banks 1.5:11.5:1 on both faces.
  • Bed falls 1 in 600: 30 m gives 0.05 m, so the bed is 224.00, 223.95, 223.90, 223.85 m up to the drop at 0+090. The 0.5 m drop gives 223.35 m after the drop, and 223.30 m at 0+120. Ground level at 0+090 is 225.2 m before and 224.7 m after the drop.

Method

Total depth of canal section D=FSD+free board=2.00+0.50=2.50D = \text{FSD} + \text{free board} = 2.00 + 0.50 = 2.50 m. At each station the bed level is compared with the ground level; e=e = GL − bed level.

  • If e≥De \ge D: full cutting, area Ac=(b+sce) eA_c = (b + s_c e)\,e (no banks needed).
  • If 0<e<D0 < e < D: part cutting, part banking. Cutting area Ac=(b+sce)eA_c = (b + s_c e)e; each bank stands hb=D−eh_b = D - e above ground, so bank area (both banks) Ab=(tL+tR)hb+(si+so)hb2A_b = (t_L + t_R)h_b + (s_i + s_o)h_b^2 with si,sos_i, s_o the inner and outer bank slopes.
  • If e≤0e \le 0: full banking, bed above ground. Bank top is H=D−eH = D - e above ground. Gross embankment =(T+soH)H= (T + s_o H)H with T=tL+tR+b+2siDT = t_L + t_R + b + 2 s_i D, and the canal space (b+siD)D(b + s_i D)D is deducted.

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

Sectional areas

StationGL (m)Bed RL (m)ee (m)ConditionCutting area (m²)Bank area (m²)
0+000225.300224.0001.300part cutting, part banking9.4907.920
0+030225.200223.9501.250part cutting, part banking9.0628.438
0+060225.100223.9001.200part cutting, part banking8.6408.970
0+090 (before drop)225.200223.8501.350part cutting, part banking9.9227.418
0+090 (after drop)224.700223.3501.350part cutting, part banking9.9227.418
0+120224.800223.3001.500part cutting, part banking11.2506.000

Volumes

BetweenDistance (m)Cutting (m³)Banking (m³)
0+000 – 0+03030.0278.29245.36
0+030 – 0+06030.0265.54261.11
0+060 – 0+090 (before drop)30.0278.44245.81
0+090 (after drop) – 0+12030.0317.59201.26
Total1,139.85953.55

Answer: Earthwork in cutting = 1,139.85 m³; earthwork in banking = 953.55 m³; total = 2,093.40 m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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