---
title: "Channel fully in banking, 0.5-1.5 km"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:canal-fully-banking-i"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/canal-fully-banking-i
site: BCT Engineering
---

# Channel fully in banking, 0.5-1.5 km

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2076 Chaitra, 10 marks
- 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-fully-banking-i
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Workout the quantity of a portion of channel fully in banking with the following data:

| Distance | R.L. of Ground level | Proposed bed level |
|---|---|---|
| 500 | 1314.75 | 1316.00 |
| 1000 | 1314.90 | |
| 1500 | 1314.20 | |

The bed width of channel is 4.50 m. The bed slope is 1 in 5000. The full supply depth is 1.50 m and free board is 0.50 m. The top width of both side banks is 2.50 m in each bank. The side slope of banks is (1.5:1).

## Short answer

Total section depth is 2.0 m. All stations lie fully in banking, so cutting area is zero. Using mean area method over 500 m intervals, earthwork in cutting = 0.00 m³, banking = 50,364.69 m³.

## Answer

#### Given data and assumptions
- Bed width $4.5$ m; FSD $1.5$ m; free board $0.5$ m, so $D = 2.0$ m; top width of each bank $2.5$ m; side slope of banks $1.5:1$ (inner and outer).
- Bed falls 1 in 5000: 0.10 m per 500 m, so bed = 1316.00, 1315.90, 1315.80 m at 500, 1000, 1500 m. The canal is wholly in banking.

#### Method
Total depth of canal section $D = \text{FSD} + \text{free board} = 1.50 + 0.50 = 2.00$ 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) |
|---|---:|---:|---:|---|---:|---:|
| 500 | 1,314.750 | 1,316.000 | -1.250 | full banking | 0.000 | 51.219 |
| 1000 | 1,314.900 | 1,315.900 | -1.000 | full banking | 0.000 | 45.000 |
| 1500 | 1,314.200 | 1,315.800 | -1.600 | full banking | 0.000 | 60.240 |

#### Volumes
| Between | Distance (m) | Cutting (m³) | Banking (m³) |
|---|---:|---:|---:|
| 500 – 1000 | 500.0 | 0.00 | 24,054.69 |
| 1000 – 1500 | 500.0 | 0.00 | 26,310.00 |
| **Total** | | **0.00** | **50,364.69** |

**Answer: Earthwork in cutting = 0.00 m³; earthwork in banking = 50,364.69 m³; total = 50,364.69 m³.**

## Check yourself

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

1. With a bed slope 1 in 5000, how much does the bed fall between stations 500 m apart?
   A. 0.50 m
   B. 1.00 m
   C. 0.01 m
   D. 0.10 m
   Answer: D. 500/5000 = 0.10 m, so the bed is 1316.00, 1315.90, 1315.80 m. (Key point: Bed level)

2. In full banking, the height H of the bank top above ground is
   A. H = D - e, where e is negative (bed above ground)
   B. H = D only
   C. H = FSD + e
   D. H = e only
   Answer: A. For e = -1.25 m and D = 2.0 m, H = 3.25 m above ground. (Key point: Bank height for full banking)

3. Why is the canal space (b + s_i D)D deducted from the gross embankment area?
   A. Because the bank slopes are flatter than the cutting slopes
   B. Because the water channel inside the banks is not filled with earth
   C. To allow for shrinkage of the soil
   D. Because the bed width is double counted in free board
   Answer: B. The gross embankment area covers the whole trapezoid; the canal prism through which water flows has no earth in it. (Key point: Canal space deduction)

4. In the formula, T = t_L + t_R + b + 2 s_i D. What does T represent?
   A. Width of the embankment at ground level
   B. Water surface width alone
   C. Width of the embankment at canal top level (through bank tops and canal)
   D. Total width of land to acquire
   Answer: C. At D above the bed, the top width is two bank tops plus the canal top width b + 2 s_i D. (Key point: Width T of the top)

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