---
title: "Road earthwork, grades from 90 m to 210 m (s)"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:road-earthwork-s-grade-90-210"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/road-earthwork-s-grade-90-210
site: BCT Engineering
---

# Road earthwork, grades from 90 m to 210 m (s)

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2067 Asar (old course), 7 marks
- Video: [Calculation of Volume of Earthwork from Cross Sections | Road Estimate | Earthwork Calculation](https://www.youtube.com/watch?v=zMl6OVLeKWM) — Civil Engineering Edu
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/estimating_and_costing/road-earthwork-s-grade-90-210
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Find out the quantity of earthwork of a portion of road to be constructed with the following data: Formation width of the road throughout = 10 m. Side slope in banking (2:1) and side slope in cutting (1:1). Downward grade 1 in 120 from distance 90 m to 120 m. While it remains in level from distance 120 m to 180 m and has again upward grade 1 in 90 from distance 180 m to 210 m. The formation level at distance 180 m = 1197.50 m. The ground levels are as under:

| R.L. of ground | 1198.65 | 1196.40 | 1199.30 | 1200.40 | 1198.10 |
|---|---|---|---|---|---|
| Distance (m) | 90 | 120 | 150 | 180 | 210 |

## Short answer

Determine formation levels from given grades and reference, compute depths. Using level-section area formula A=(B+s h)h, compute volumes by mean area method: 1,745.94 m³ cutting and 187.07 m³ filling.

## Answer

#### Given data and assumptions
- Formation width $10$ m throughout; banking slope $2:1$, cutting slope $1:1$; ground level across the road.
- Formation level at 180 m is 1197.50 m; level from 120 m to 180 m; falling 1 in 120 from 90 to 120 m, so level at 90 m = 1197.50 + 30/120 = 1197.75 m; rising 1 in 90 from 180 to 210 m, so level at 210 m = 1197.833 m.

#### Method
Level (plain-ground) sections are used. With centre height $h$, formation width $B$ and side slope $s:1$:

$$
A = (B + s\,h)\,h
$$

Volume between two sections = mean area × distance; where a section changes from cutting to filling, the length of each part is found by proportion of the depths and the volume of each part = ½ × end area × length.

| Chainage | GL (m) | FL (m) | Depth of cutting / height of filling (m) |
|---|---:|---:|---:|
| 0+090 | 1,198.650 | 1,197.750 | cutting 0.900 |
| 0+120 | 1,196.400 | 1,197.500 | filling 1.100 |
| 0+150 | 1,199.300 | 1,197.500 | cutting 1.800 |
| 0+180 | 1,200.400 | 1,197.500 | cutting 2.900 |
| 0+210 | 1,198.100 | 1,197.833 | cutting 0.267 |

#### Sectional areas
| Chainage | Centre ht. $h$ (m) | Cutting area (m²) | Filling area (m²) |
|---|---:|---:|---:|
| 0+090 | cut 0.90 | 9.810 | 0.000 |
| 0+120 | fill 1.10 | 0.000 | 13.420 |
| 0+150 | cut 1.80 | 21.240 | 0.000 |
| 0+180 | cut 2.90 | 37.410 | 0.000 |
| 0+210 | cut 0.27 | 2.738 | 0.000 |

#### Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---:|---:|---:|
| 0+090 – 0+120 | 30.0 | 66.22 | 110.71 |
| 0+120 – 0+150 | 30.0 | 197.75 | 76.36 |
| 0+150 – 0+180 | 30.0 | 879.75 | 0.00 |
| 0+180 – 0+210 | 30.0 | 602.22 | 0.00 |
| **Total** | 120.0 | **1,745.94** | **187.07** |

**Answer: Earthwork in cutting = 1,745.94 m³; earthwork in filling = 187.07 m³.**

## Check yourself

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

1. Formation is 1197.50 m at 180 m, level 120-180 m, falling 1 in 120 from 90 to 120 m. The level at 90 m is:
   A. 1197.25 m
   B. 1197.75 m
   C. 1197.50 m
   D. 1197.83 m
   Answer: B. It falls toward 120 m, so 90 m is higher: 1197.50 + 30/120 = 1197.75 m. (Key point: Formation level at start)

2. At 120 m the filling height is 1.10 m, banking slope 2:1, B = 10 m. The filling area is:
   A. 11.00 m^2
   B. 12.10 m^2
   C. 13.42 m^2
   D. 22.00 m^2
   Answer: C. A = (B + s h) h = (10 + 2 x 1.1) x 1.1 = 13.42 m^2. (Key point: Area of level section)

3. Between 120 m (filling 1.1 m) and 150 m (cutting 1.8 m), the filling part length is about:
   A. 11.4 m
   B. 18.6 m
   C. 15.0 m
   D. 30.0 m
   Answer: A. Proportion of depths: 30 x 1.1/(1.1 + 1.8) = 11.38 m; the rest (18.62 m) is cutting. (Key point: Cut-fill changeover)

## Close variants set in other exams

- [Road earthwork, grades 1 in 120 and 1 in 90 (0-120)](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Aroad-earthwork-r-grade-120-90) — Estimating and Costing, chapter 6, asked 2×
- [Road earthwork, formation level at 150 m](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Aroad-earthwork-t-grade-150) — Estimating and Costing, chapter 6, asked 2×

## Related in Estimating and Costing

- [Hill road earthwork, cutting 4 m / banking 6 m](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Ahill-road-k-cut4-fill6) — Estimating and Costing, chapter 6
- [Hill road earthwork, 0-90 m (8 m formation)](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Ahill-road-p-0-90) — Estimating and Costing, chapter 6
- [Hill road earthwork, 0 m to 90 m, 10/8 m widths](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Ahill-road-e-0-90) — Estimating and Costing, chapter 6, asked 2×
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- [Road earthwork, RD 0-120, gradient 1 in 150](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Aroad-earthwork-o-rds) — Estimating and Costing, chapter 6
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## Related in other subjects

- [Cutting volume, 120 m, upward slope 1 in 40](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-cutting-120m-formation-380) — Surveying I, chapter 9, asked 8×
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