--- title: "Canal earthwork, bed 3 m, free board 0.44 m" subject: "Estimating and Costing" chapter: 6 key: "semester_7/estimating_and_costing:channel-q-portion" source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/channel-q-portion site: BCT Engineering --- # Canal earthwork, bed 3 m, free board 0.44 m Estimating and Costing · Chapter 6: Detailed Estimate - Asked in: 2070 Chaitra, 10 marks - Video: [Calculate the quantity of earthwork in irrigation canal | calculate the permanent width | Type #3](https://www.youtube.com/watch?v=WkcdXvl0XiQ) — SBK Concept - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/estimating_and_costing/channel-q-portion - Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6 ## Question Calculate the quantity of earth work for a portion of channel with the following data: Bed width = 3 m; Free Board = 0.44 m; Side slope for digging = 1:1; Side slope for banking = 1:1½ (V:H); Fully supply depth = 1 m; Top width of bank = 1.5 m. | Chainage | 0 | 30 | 60 | 90 | 120 | 150 | |---|---|---|---|---|---|---| | RL of GL | 225.24 | 224.8 | 224.43 | 224.12 | 224.5 | 224.98 | | Proposed level | 224.00 | 223.94 | 223.88 | 223.82 | 223.76 | 223.7 | Also draw a typical X-section. ## Short answer Total depth D = 1.44 m, cutting slope 1:1, banking 1.5:1. All stations are part cutting/part banking. Volumes: cutting = 431.93 m³, banking = 579.68 m³, total = 1,011.61 m³. ## Answer #### 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) ```text 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 | Station | GL (m) | Bed RL (m) | $e$ (m) | Condition | Cutting area (m²) | Bank area (m²) | |---|---:|---:|---:|---|---:|---:| | 0 | 225.240 | 224.000 | 1.240 | part cutting, part banking | 5.258 | 0.720 | | 30 | 224.800 | 223.940 | 0.860 | part cutting, part banking | 3.320 | 2.749 | | 60 | 224.430 | 223.880 | 0.550 | part cutting, part banking | 1.953 | 5.046 | | 90 | 224.120 | 223.820 | 0.300 | part cutting, part banking | 0.990 | 7.319 | | 120 | 224.500 | 223.760 | 0.740 | part cutting, part banking | 2.768 | 3.570 | | 150 | 224.980 | 223.700 | 1.280 | part cutting, part banking | 5.478 | 0.557 | #### Volumes | Between | Distance (m) | Cutting (m³) | Banking (m³) | |---|---:|---:|---:| | 0 – 30 | 30.0 | 128.66 | 52.04 | | 30 – 60 | 30.0 | 79.08 | 116.93 | | 60 – 90 | 30.0 | 44.14 | 185.48 | | 90 – 120 | 30.0 | 56.36 | 163.33 | | 120 – 150 | 30.0 | 123.69 | 61.90 | | **Total** | | **431.93** | **579.68** | **Answer: Earthwork in cutting = 431.93 m³; earthwork in banking = 579.68 m³; total = 1,011.61 m³.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Banking slope is given as 1 V : 1.5 H. How is it used in the formulas? A. As 1.5:1 (H:V), so s = 1.5 B. As s = 1/1.5 = 0.67 C. As s = 1 D. As s = 2.5 Answer: A. 1 vertical to 1.5 horizontal is the same as a horizontal slope of 1.5 per unit of height. (Key point: Slope notation) 2. For a part cutting, part banking section, the area of both banks is A. A_b = (b + s e)e B. A_b = (t_L + t_R)h_b + (s_i + s_o)h_b^2 with h_b = D - e C. A_b = (t_L + t_R)D D. A_b = b D + s D^2 Answer: B. Each bank is a trapezoid of top width t and slope s on both faces, standing h_b = D - e above ground. (Key point: Bank area formula) 3. At 0, e = 1.24 m and D = 1.44 m. Bank area (t = 1.5 each, s = 1.5)? A. 1.24 m2 B. 3.60 m2 C. 0.72 m2 D. 0.36 m2 Answer: C. h_b = 0.20, area = 3.0 x 0.2 + 3.0 x 0.04 = 0.72 m2. (Key point: Bank at station 0) 4. How is the earthwork between two stations found? A. Larger of the two end areas x distance B. Sum of all areas x 30 m C. Area at the start x distance D. Average of the two end areas x chainage distance, separately for cutting and banking Answer: D. The mean area method is used for each of cutting and banking. 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