---
title: "Hill road earthwork, 0-90 m (8 m formation)"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:hill-road-p-0-90"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/hill-road-p-0-90
site: BCT Engineering
---

# Hill road earthwork, 0-90 m (8 m formation)

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2071 Chaitra, 12 marks
- Video: [How To Calculate EarthWork Quantity Of A Road By Mid Section Method ? @CivilConstruction](https://www.youtube.com/watch?v=tYGRP1C1QH8) — CivilConstruction
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/estimating_and_costing/hill-road-p-0-90
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Calculate the quantities of earthwork of a portion of hill road from the following data: Formation width = 8 m, side slope in cutting and filling = (1:1) and (2:1).

| Distance | Depth of cut | Depth of fill | Cross slope of ground |
|---|---|---|---|
| 0 m | 0.30 | - | 10:1 |
| 30 m | 0.20 | - | 15:1 |
| 60 m | - | 0.50 | 12:1 |
| 90 m | - | 0.70 | 8:1 |

## Short answer

Areas computed with side-hill formulas for given cross slopes; transition from cut to fill between 30 m and 60 m handled by zero point. Total earthwork: cutting 74.58 m³, filling 261.08 m³.

## Answer

#### Given data and assumptions
- Formation width $B = 8$ m; side slope in cutting $1:1$, in filling $2:1$.
- Sections 30 m apart. Some of the cutting sections (small centre depth, steep cross slope) are partly in filling on the low side, and are calculated that way.

#### Method
The ground has a transverse (cross) slope $n:1$ (H:V), so each section is a side-hill section. With centre height $h$ (cutting +, filling −) above/below formation level, formation width $B$ and side slope $s:1$:

$$
\begin{aligned} x_u &= \frac{n\,(B/2 + s h)}{n - s},\quad d_u = h + \frac{x_u}{n}\\ x_d &= \frac{n\,(B/2 + s h)}{n + s},\quad d_d = h - \frac{x_d}{n}\\ A &= \tfrac12 B h + \tfrac12\left[\tfrac{B}{2}(d_u + d_d) + h\,(x_u + x_d - B)\right] \end{aligned}
$$

Here $x_u, x_d$ are the horizontal distances of the up-hill and down-hill toes from the centre line, and $d_u, d_d$ are the heights of the ground above formation at the toes (formulas for cutting). For a section wholly in filling the roles swap: the down-hill toe is the far one, $x_d = n(B/2 + s h)/(n - s)$ and $x_u = n(B/2 + s h)/(n + s)$, with $h$ the filling height and $s$ the filling slope. Where the ground line crosses the formation level inside the road width (part cutting, part filling), the cutting and filling areas are found separately from the co-ordinates of the section (toe points from the same intersection equations). Volume between sections = mean area × distance, taken separately for cutting and filling. Where the centre changes from cutting to filling between two sections and one type of area is zero at one end, the change-over point is found by proportion of the centre heights, and that area is taken over its own part of the length only: $V = \tfrac12 A \times l$.

#### Sectional areas
| Chainage | Centre ht. $h$ (m) | Cutting area (m²) | Filling area (m²) |
|---|---:|---:|---:|
| 0+000 | cut 0.30 | 2.722 | 0.063 |
| 0+030 | cut 0.20 | 1.750 | 0.038 |
| 0+060 | fill 0.50 | 0.000 | 4.857 |
| 0+090 | fill 0.70 | 0.000 | 7.552 |

Working for one section:

Take chainage 0+000: centre cutting depth $h = 0.30$ m, $n = 10$, $B = 8.0$ m.
- Ground height above formation at the uphill edge $= 0.700$ m, at the downhill edge $= -0.100$ m.
- Toe of the side slope on the uphill side at $x = 4.778$ m from the centre line (ground above formation by 0.778 m); downhill toe at $x = 4.250$ m (ground below formation by 0.125 m).
- Area from the co-ordinates of the polygon: cutting $= 2.722$ m², filling $= 0.063$ m².

Change-over between 0+030 (cutting 0.20 m at the centre) and 0+060 (wholly in filling, 0.50 m): the zero point is $30 \times 0.20/0.70 = 8.57$ m from 0+030.
- Cutting is zero at 0+060, so it is taken over the 8.57 m only: $\tfrac12 \times 1.750 \times 8.57 = 7.50$ m³
- Filling is present at both ends (0.038 and 4.857 m²), so the mean area is used: $\tfrac{0.038 + 4.857}{2} \times 30 = 73.43$ m³

#### Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---:|---:|---:|
| 0+000 – 0+030 | 30.0 | 67.08 | 1.51 |
| 0+030 – 0+060 | 30.0 (cutting over 8.57) | 7.50 | 73.43 |
| 0+060 – 0+090 | 30.0 | 0.00 | 186.14 |
| **Total** | 90.0 | **74.58** | **261.08** |

**Answer: Earthwork in cutting = 74.58 m³; earthwork in filling = 261.08 m³.**

## Check yourself

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

1. At 0+000, centre cutting h = 0.30 m, cross slope 10:1, B = 8 m. What is the ground height relative to formation at the down-hill edge?
   A. +0.100 m (above)
   B. -0.100 m (below)
   C. -0.700 m (below)
   D. +0.700 m (above)
   Answer: B. Down-hill edge = h - (B/2)/n = 0.30 - 4/10 = -0.10 m, i.e. ground is below formation. (Key point: Edge heights of ground)

2. Why does the section at 0+000, with centre cutting 0.30 m, also show a small filling area (0.063 m^2)?
   A. The filling slope is 2:1
   B. The ground on the low side falls below formation level within the road width
   C. The formation width is larger in filling
   D. Chainage 0 is always treated as filling
   Answer: B. The steep cross slope drops the ground below formation at the down-hill edge, creating a small filling portion. (Key point: Filling part within a cutting section)

3. Using cutting areas 2.722 m^2 at 0+000 and 1.750 m^2 at 0+030 (30 m apart), the cutting volume is about:
   A. 134.2 m^3
   B. 67.1 m^3
   C. 81.7 m^3
   D. 52.5 m^3
   Answer: B. Mean area = (2.722 + 1.750)/2 = 2.236 m^2; times 30 m gives about 67.08 m^3. (Key point: Mean-area volume)

4. Which side slopes are used for cutting and for filling in this problem?
   A. Cutting 2:1, filling 1:1
   B. Both 1:1
   C. Both 2:1
   D. Cutting 1:1, filling 2:1
   Answer: D. The data give 1:1 in cutting and 2:1 in filling (H:V). (Key point: Different side slopes)

5. Between 0+030 and 0+060 the cutting area goes from 1.750 to 0.000 m^2. What cutting volume results?
   A. 52.50 m^3
   B. 26.25 m^3
   C. 1.75 m^3
   D. 58.33 m^3
   Answer: B. Mean area = 0.875 m^2 x 30 m = 26.25 m^3. (Key point: Totals)

## Related in Estimating and Costing

- [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×
- [Hill road earthwork, cross slope 1 in 5 (0-60)](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Ahill-road-r-cross-slope-1-5) — Estimating and Costing, chapter 6
- [Side-long ground hill road, 0-400 m](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Aside-long-road-n) — Estimating and Costing, chapter 6
- [Hill road earthwork, chainage 12-15](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Ahill-road-q-chain-12-15) — Estimating and Costing, chapter 6
- [Hill road earthwork, 0+060 to 0+120](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Ahill-road-m-formation-10-8) — Estimating and Costing, chapter 6
- [Hill road earthwork with cross sections](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Ahill-road-u-cross-sections) — Estimating and Costing, chapter 6

## Related in other subjects

- [Cut and fill volume, 20 m formation width](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-road-20m-width-cut-fill-1-1-1-2) — Surveying I, chapter 9
- [Cutting volume, 60 m, GL 540.70, formation 538.20](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-cutting-60m-gl-540-70) — Surveying I, chapter 9, asked 7×
- [Earthwork volume, formation 1201.85, 1.65 m depth](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-earthwork-formation-1201-85) — Surveying I, chapter 9
- [Balancing depth: bed width 20 m](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Abalancing-depth-bed20-cut15-fill2) — Irrigation and Drainage Engineering, chapter 3, asked 3×