---
title: "Channel earthwork and land area, 50 cm fall"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:channel-land-acquisition-h"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/channel-land-acquisition-h
site: BCT Engineering
---

# Channel earthwork and land area, 50 cm fall

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2078 Bhadra, 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-land-acquisition-h
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Calculate the quantity of earthwork and area of permanent land required for the land acquisition purpose for a portion of a channel from following data: Bed width = 4 m; Free board = 45 cm; Side slope in cutting = 1:1; Side slope in banking = 1.5:1; Full supply depth = 1 m; Top width of bank = 3 m left and 1.5 m right. There is 50 cm fall at chainage 800 m.

| Chainage | 800 | 850 | 900 | 950 | 1000 | 1050 |
|---|---|---|---|---|---|---|
| RL ground | 109.8 | 109.7 | 109.55 | 109.30 | 109.25 | 109.15 |
| RL Bed | 109.52 | Bed slope 1:250 | | | | |

## Short answer

Part cutting, part banking throughout; cutting = 1,155.86 m³, banking = 785.21 m³. Permanent land area from outer bank toes = 3,365.00 m².

## Answer

#### Given data and assumptions
- Bed width $4$ m; FSD $1$ m; free board $0.45$ m, so $D = 1.45$ m; top width of bank $3$ m (left) and $1.5$ m (right).
- Side slope in cutting $1:1$, in banking $1.5:1$ (inner and outer faces of banks).
- The ground falls with chainage, so the canal is taken to flow in the direction of increasing chainage, with the bed falling at 1 in 250. The printed bed level 109.52 m at chainage 800 is the level just upstream of the 0.50 m fall. The reach considered lies downstream of the fall, so the bed at 800 is $109.52 - 0.50 = 109.02$ m, falling $50/250 = 0.20$ m every 50 m: 109.02, 108.82, 108.62, 108.42, 108.22, 108.02 m.

#### Method
Total depth of canal section $D = \text{FSD} + \text{free board} = 1.00 + 0.45 = 1.45$ 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.

Here every station is in part cutting, part banking, so $A_c = (4 + e)e$ and $A_b = 4.5\,h_b + 3\,h_b^2$.

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+800 | 109.800 | 109.020 | 0.780 | part cutting, part banking | 3.728 | 4.362 |
| 0+850 | 109.700 | 108.820 | 0.880 | part cutting, part banking | 4.294 | 3.540 |
| 0+900 | 109.550 | 108.620 | 0.930 | part cutting, part banking | 4.585 | 3.151 |
| 0+950 | 109.300 | 108.420 | 0.880 | part cutting, part banking | 4.294 | 3.540 |
| 1+000 | 109.250 | 108.220 | 1.030 | part cutting, part banking | 5.181 | 2.419 |
| 1+050 | 109.150 | 108.020 | 1.130 | part cutting, part banking | 5.797 | 1.747 |

#### Volumes
| Between | Distance (m) | Cutting (m³) | Banking (m³) |
|---|---:|---:|---:|
| 0+800 – 0+850 | 50.0 | 200.57 | 197.54 |
| 0+850 – 0+900 | 50.0 | 221.98 | 167.27 |
| 0+900 – 0+950 | 50.0 | 221.98 | 167.27 |
| 0+950 – 1+000 | 50.0 | 236.88 | 148.97 |
| 1+000 – 1+050 | 50.0 | 274.45 | 104.16 |
| **Total** | | **1,155.86** | **785.21** |

#### Permanent land
Width of land at a section = distance between the outer toes of the two banks = canal width at ground level $(b + 2e)$ + inner and outer slopes of both banks $(4 \times 1.5\,h_b)$ + bank tops $(3 + 1.5)$.

| Station | Land width (m) |
|---|---:|
| 0+800 | 14.080 |
| 0+850 | 13.680 |
| 0+900 | 13.480 |
| 0+950 | 13.680 |
| 1+000 | 13.080 |
| 1+050 | 12.680 |

Area of land $=$ mean width × length, summed over the reaches $= 694.0 + 679.0 + 679.0 + 669.0 + 644.0 = 3,365.00$ m² $= 0.3365$ ha.

**Answer: Earthwork in cutting = 1,155.86 m³; earthwork in banking = 785.21 m³; total = 1,941.07 m³.** **Permanent land = 3,365.00 m².**

## Check yourself

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

1. With a bed slope of 1 in 250, how much does the bed rise in 50 m?
   A. 0.20 m
   B. 0.50 m
   C. 0.05 m
   D. 0.25 m
   Answer: A. 50/250 = 0.20 m, giving 109.02, 109.22, 109.42 and so on. (Key point: Bed gradient)

2. At 0+900, GL = 109.55 and bed = 109.42, D = 1.45 m. Which case applies?
   A. Full cutting, since GL is above the bed
   B. Part cutting, part banking, since 0 < e = 0.13 < D
   C. Full banking, since e is smaller than D
   D. Full cutting, since e is below 1 m
   Answer: B. For 0 < e < D the canal is partly cut and partly banked. (Key point: Section at 0+900)

3. What is D used here?
   A. FSD only = 1.00 m
   B. FSD - free board = 0.55 m
   C. FSD + free board = 1.00 + 0.45 = 1.45 m
   D. Bed width / 4 = 1.0 m
   Answer: C. Total depth is the full supply depth plus free board. (Key point: Total depth D)

## Close variants set in other exams

- [Channel earthwork and land area, bed 4 m](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Achannel-land-acquisition-f) — Estimating and Costing, chapter 6, asked 2×

## Related in Estimating and Costing

- [Canal earthwork, bed 3 m, free board 0.44 m](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Achannel-q-portion) — Estimating and Costing, chapter 6
- [Irrigation channel earthwork, bed 5 m, 1 in 3000](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Airrigation-channel-j) — Estimating and Costing, chapter 6
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- [Channel fully in banking, 0.5-1.5 km](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Acanal-fully-banking-i) — Estimating and Costing, chapter 6
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