---
title: "Road earthwork with RL of GL and FL (0-600)"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:road-earthwork-b-gl-fl"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/road-earthwork-b-gl-fl
site: BCT Engineering
---

# Road earthwork with RL of GL and FL (0-600)

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2082 Baisakh, 6 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/road-earthwork-b-gl-fl
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Calculate earthwork for the following section of road from the given data.
i) Formation width = 8 m,
ii) Side slope in filling and cutting = 2:1 and 1½:1 respectively.

| Distance (m) | 0 | 100 | 200 | 300 | 400 | 500 | 600 |
|---|---|---|---|---|---|---|---|
| RL of GL | 99.5 | 100 | 99.8 | 101.6 | 102.3 | 101.1 | 103 |
| RL of FL | 100 | 1:200 (upward) | | | 1:400 (downward) | | |

## Short answer

Cutting = 2410.45 m³, filling = 1875.00 m³. Level-gound areas (B+sh)h; volumes averaged between sections; filling-to-cutting transition split by depth ratio.

## Answer

#### Given data and assumptions
- Formation width $B = 8$ m; side slope in filling $2:1$, in cutting $1.5:1$ (H:V). Ground is taken as level across the road (no cross slope given).
- Formation level at 0 m is 100.00 m. The gradient is read as 1 in 200 upward from 0 to 300 m and 1 in 400 downward from 300 m to 600 m (the printed table shows only the two gradients).

#### 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+000 | 99.500 | 100.000 | filling 0.500 |
| 0+100 | 100.000 | 100.500 | filling 0.500 |
| 0+200 | 99.800 | 101.000 | filling 1.200 |
| 0+300 | 101.600 | 101.500 | cutting 0.100 |
| 0+400 | 102.300 | 101.250 | cutting 1.050 |
| 0+500 | 101.100 | 101.000 | cutting 0.100 |
| 0+600 | 103.000 | 100.750 | cutting 2.250 |

#### Sectional areas
| Chainage | Centre ht. $h$ (m) | Cutting area (m²) | Filling area (m²) |
|---|---:|---:|---:|
| 0+000 | fill 0.50 | 0.000 | 4.500 |
| 0+100 | fill 0.50 | 0.000 | 4.500 |
| 0+200 | fill 1.20 | 0.000 | 12.480 |
| 0+300 | cut 0.10 | 0.815 | 0.000 |
| 0+400 | cut 1.05 | 10.054 | 0.000 |
| 0+500 | cut 0.10 | 0.815 | 0.000 |
| 0+600 | cut 2.25 | 25.594 | 0.000 |

#### Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---:|---:|---:|
| 0+000 – 0+100 | 100.0 | 0.00 | 450.00 |
| 0+100 – 0+200 | 100.0 | 0.00 | 849.00 |
| 0+200 – 0+300 | 100.0 | 3.13 | 576.00 |
| 0+300 – 0+400 | 100.0 | 543.44 | 0.00 |
| 0+400 – 0+500 | 100.0 | 543.44 | 0.00 |
| 0+500 – 0+600 | 100.0 | 1,320.44 | 0.00 |
| **Total** | 600.0 | **2,410.45** | **1,875.00** |

**Answer: Earthwork in cutting = 2,410.45 m³; earthwork in filling = 1,875.00 m³.**

## Check yourself

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

1. For ground level across the road, the sectional area is:
   A. A = B x h
   B. A = (B + s h) h
   C. A = (B/2 + s h) h
   D. A = B h + s h
   Answer: B. A level section is a trapezoid with area (B + s h) h, where s is the side slope and h the centre height. (Key point: Level section area)

2. With formation level 100.00 m at 0 m and a rising gradient of 1 in 200, the formation level at 200 m is:
   A. 100.20 m
   B. 100.50 m
   C. 101.00 m
   D. 101.50 m
   Answer: C. 200 m / 200 = 1.00 m rise, so FL = 101.00 m. (Key point: Formation level from gradient)

3. At 0+200, GL = 99.800 and FL = 101.000. The section is:
   A. Cutting 1.2 m
   B. Filling 1.2 m
   C. Cutting 0.2 m
   D. Filling 0.2 m
   Answer: B. The ground is below the formation, so earth must be filled; height = 101.0 - 99.8 = 1.2 m. (Key point: Cutting or filling)

4. Between 0+200 (filling 1.2 m) and 0+300 (cutting 0.1 m), how is the volume found?
   A. Mean area x 100 m, treating it all as filling
   B. The length of each part is found by proportion of depths, and each volume = 1/2 x end area x length
   C. Cutting and filling areas are subtracted
   D. The cutting is ignored as it is small
   Answer: B. Where the section changes type, a zero-depth point lies between; each part is a wedge with volume 1/2 x end area x its length found by proportion of the depths. (Key point: Section changing cut to fill)

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