---
title: "Irrigation canal earthwork with vertical drop"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:canal-a-drop-0-400"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/canal-a-drop-0-400
site: BCT Engineering
---

# Irrigation canal earthwork with vertical drop

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2082 Bhadra, 7 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/canal-a-drop-0-400
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Estimate earthwork for an irrigation canal with following data: Bed width of canal = 5 m; F.S.L = 1 m (full supply depth) and free board = 0.5 m. Top width of both banks = 2 m. Outer bank side slope = 1:1 (H:V) and canal section slope = 2:1 (H:V). Canal bed is proposed at uniform gradient of 1 in 5,000. A vertical drop of 0.5 m is to be provided at 0+400 m chainage.

| Chainage (m) | R.L of Ground (m) | R.L of Canal Bed (m) |
|---|---|---|
| 0+000 | 129.24 | |
| 0+200 | 129.00 | 130.96 |
| 0+400 | 129.20 | |
| 0+600 | 129.12 | |

## Short answer

Canal is wholly in embankment. Banking areas are computed from bank dimensions and mean-area method yields 0.00 m³ cutting and 27,115.52 m³ banking.

## Answer

#### 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
| Station | GL (m) | Bed RL (m) | $e$ (m) | Condition | Cutting area (m²) | Bank area (m²) |
|---|---:|---:|---:|---|---:|---:|
| 0+000 | 129.240 | 131.000 | -1.760 | full banking | 0.000 | 47.528 |
| 0+200 | 129.000 | 130.960 | -1.960 | full banking | 0.000 | 51.872 |
| 0+400 (before drop) | 129.200 | 130.920 | -1.720 | full banking | 0.000 | 46.668 |
| 0+400 (after drop) | 129.200 | 130.420 | -1.220 | full banking | 0.000 | 36.198 |
| 0+600 | 129.120 | 130.380 | -1.260 | full banking | 0.000 | 37.018 |

#### Volumes
| Between | Distance (m) | Cutting (m³) | Banking (m³) |
|---|---:|---:|---:|
| 0+000 – 0+200 | 200.0 | 0.00 | 9,939.92 |
| 0+200 – 0+400 (before drop) | 200.0 | 0.00 | 9,854.00 |
| 0+400 (after drop) – 0+600 | 200.0 | 0.00 | 7,321.60 |
| **Total** | | **0.00** | **27,115.52** |

**Answer: Earthwork in cutting = 0.00 m³; earthwork in banking = 27,115.52 m³; total = 27,115.52 m³.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the total depth D of the canal section used for the banks?
   A. 1.0 m
   B. 1.5 m
   C. 0.5 m
   D. 2.0 m
   Answer: B. D = full supply depth + free board = 1.0 + 0.5 = 1.5 m. (Key point: Total depth)

2. Bed at 0+200 is 130.96 m, falling 1 in 5000. What is the bed just after the 0.5 m drop at 0+400?
   A. 130.42 m
   B. 130.92 m
   C. 130.46 m
   D. 131.42 m
   Answer: A. Bed at 0+400 before drop is 130.96 - 0.04 = 130.92 m; the 0.5 m drop gives 130.42 m. (Key point: Bed level with gradient and drop)

3. Bed is above the ground (e = GL - bed is negative) at every station. What condition applies?
   A. Full cutting
   B. Part cutting, part banking
   C. Full banking
   D. Cannot be determined
   Answer: C. When e is 0 or negative, the bed is above ground, so the canal is wholly in embankment. (Key point: Classification)

4. For full banking, how is the net bank area found?
   A. Gross embankment (T + s_o H)H minus the canal space (b + s_i D)D
   B. (b + s D) D only
   C. Sum of both bank top widths x D
   D. Cutting area x 2
   Answer: A. The gross embankment includes the canal space, which holds no earth and is deducted. (Key point: Embankment area)

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