---
title: "Canal earthwork by prismoidal method"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:canal-prismoidal-l"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/canal-prismoidal-l
site: BCT Engineering
---

# Canal earthwork by prismoidal method

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2075 Asoj, 6 marks
- Video: [Calculation of volume of earthwork by prismoidal and trapezoidal method #TU #IOE #2077 CHAITRA](https://www.youtube.com/watch?v=UsRKZCpwhsk) — ArkeyVerse
- 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-prismoidal-l
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Find the quantity of earth work of irrigation canal using prismoidal method from the following data:

| Distance (m) | 0 | 50 | 100 | 150 | 200 |
|---|---|---|---|---|---|
| RL of Ground (m) | 100.00 | 101.00 | 101.00 | 99.00 | 100.00 |
| RL of Formation (m) | 99.50 | 99.00 | 89.50 | 89.00 | 88.50 |

Formation bottom width of canal is 6 meter and side slope 1:1. (RL of formation at 100 m and 150 m as printed.)

## Short answer

All sections are in cutting with bottom width 6 m, side slope 1:1. Prismoidal (Simpson's) rule is applied to five sections at 50 m intervals. Volume of earthwork = 2,650.00 m³.

## Answer

#### Given data and assumptions

- Bottom width $b = 6$ m; side slope $1:1$; the canal is in cutting at every station. Depth $d$ = GL − formation level.
- The printed formation levels at 100 m, 150 m and 200 m (89.50, 89.00, 88.50) are not consistent with the first two (99.50, 99.00), so the pattern of a 0.5 m fall per 50 m is continued: 99.50, 99.00, 98.50, 98.00, 97.50 m.

#### Sectional areas

$A = (b + s\,d)\,d = (6 + d)\,d$

| Distance (m) | GL (m) | Formation RL (m) | Depth $d$ (m) | Area (m²) |
|---|---:|---:|---:|---:|
| 0 | 100.00 | 99.50 | 0.50 | 3.250 |
| 50 | 101.00 | 99.00 | 2.00 | 16.000 |
| 100 | 101.00 | 98.50 | 2.50 | 21.250 |
| 150 | 99.00 | 98.00 | 1.00 | 7.000 |
| 200 | 100.00 | 97.50 | 2.50 | 21.250 |

#### Volume by the prismoidal rule

There are 5 sections (4 equal intervals of $h = 50$ m, an even number), so Simpson's rule applies:

$$
\begin{aligned} V &= \frac{h}{3}\left[A_0 + 4(A_1 + A_3) + 2A_2 + A_4\right]\\ &= \frac{50}{3}\left[3.250 + 4(16.000 + 7.000) + 2(21.250) + 21.250\right]\\ &= \frac{50}{3} \times 159.00 = 2650.00\ \text{m}^3 \end{aligned}
$$

**Answer: volume of earthwork (prismoidal rule) = 2,650.00 m³.** (For comparison, the mean-area method gives 2,825.00 m³.)

## Check yourself

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

1. What does the prismoidal (Simpson's) rule require of the sections?
   A. An even number of sections
   B. An odd number of sections, i.e. an even number of equal intervals
   C. Only two sections
   D. Sections at unequal spacing
   Answer: B. Five sections give four equal intervals, an even number, so Simpson's rule applies. (Key point: Condition for Simpson's rule)

2. In V = (h/3)[A0 + 4(A1 + A3) + 2A2 + A4], which areas get weight 4?
   A. The first and last sections
   B. The even-numbered intermediate sections
   C. The odd-numbered intermediate sections
   D. All middle sections
   Answer: C. Odd-position intermediate areas get 4, even-position intermediate areas get 2, and end areas get 1. (Key point: Weights in the formula)

3. For b = 6 m and slope 1:1, what is the area at depth d = 2.0 m?
   A. 12.0 m2
   B. 8.0 m2
   C. 24.0 m2
   D. 16.0 m2
   Answer: D. A = (b + s d)d = (6 + 2)(2) = 16 m2. (Key point: Sectional area)

## Related in Estimating and Costing

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## Related in other subjects

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