---
title: "Hill road earthwork, cross slope 1 in 5 (0-60)"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:hill-road-r-cross-slope-1-5"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/hill-road-r-cross-slope-1-5
site: BCT Engineering
---

# Hill road earthwork, cross slope 1 in 5 (0-60)

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2068 Baisakh (old course), 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/hill-road-r-cross-slope-1-5
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Work out the quantity of earth work in cutting and filling of a portion of a hill road as per data given below: Cross slope = 1 in 5, formation width = 8 m, side slope in cutting = 1:1, side slope in filling = 2:1.

| R.L. of formation | 699.20 | 702.20 | 704.20 |
|---|---|---|---|
| R.L. of ground | 698.80 | 700.00 | 706.20 |
| Distance (m) | 0 | 30 | 60 |

## Short answer

Using side-hill section formulas with cross slope 1:5, formation width 8 m, side slopes 1:1 in cutting and 2:1 in filling, the earthwork is 330.00 m³ in cutting and 1,110.00 m³ in filling.

## Answer

#### Given data and assumptions
- Formation width $B = 8$ m; side slope in cutting $1:1$, in filling $2:1$; ground cross slope $1$ in $5$.

#### Method
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 $d_u, d_d$ 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.

| Chainage | GL (m) | FL (m) | Depth of cutting / height of filling (m) |
|---|---:|---:|---:|
| 0+000 | 698.800 | 699.200 | filling 0.400 |
| 0+030 | 700.000 | 702.200 | filling 2.200 |
| 0+060 | 706.200 | 704.200 | cutting 2.000 |

#### Sectional areas
| Chainage | Centre ht. $h$ (m) | Cutting area (m²) | Filling area (m²) |
|---|---:|---:|---:|
| 0+000 | fill 0.40 | 0.500 | 6.000 |
| 0+030 | fill 2.20 | 0.000 | 34.000 |
| 0+060 | cut 2.00 | 21.500 | 0.000 |

Working for one section:

Take chainage 0+000: centre filling height $h = 0.40$ m, $n = 5$, $B = 8.0$ m.
- Ground height above formation at the uphill edge $= 0.400$ m, at the downhill edge $= -1.200$ m.
- Toe of the side slope on the uphill side at $x = 4.500$ m from the centre line (ground above formation by 0.500 m); downhill toe at $x = 8.000$ m (ground below formation by 2.000 m).
- Area from the co-ordinates of the polygon: cutting $= 0.500$ m², filling $= 6.000$ m².

#### Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---:|---:|---:|
| 0+000 – 0+030 | 30.0 | 7.50 | 600.00 |
| 0+030 – 0+060 | 30.0 | 322.50 | 510.00 |
| **Total** | 60.0 | **330.00** | **1,110.00** |

**Answer: Earthwork in cutting = 330.00 m³; earthwork in filling = 1,110.00 m³.**

## Check yourself

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

1. At 0+000, ground is 698.80 m and formation is 699.20 m. What is the centre condition?
   A. Cutting 0.40 m
   B. Filling 0.40 m
   C. Cutting 0.80 m
   D. Filling 2.00 m
   Answer: B. Ground is below formation by 0.40 m, so it is filling. (Key point: Cutting or filling at centre)

2. At 0+000 (filling 0.40 m, n = 5, B = 8 m) the ground at the up-hill edge is above formation by:
   A. 0.800 m
   B. 1.200 m
   C. 0.400 m
   D. 0.000 m
   Answer: C. Up-hill edge = -0.40 + 4/5 = +0.40 m, so some cutting appears even though the centre is in filling. (Key point: Edge heights with cross slope)

3. In a filling section on a cross slope, which toe is the far one and which formula applies?
   A. Up-hill toe, n(B/2 + s h)/(n + s)
   B. Down-hill toe, n(B/2 + s h)/(n - s)
   C. Up-hill toe, n(B/2 + s h)/(n - s)
   D. Down-hill toe, n(B/2 + s h)/(n + s)
   Answer: B. In filling the roles swap: the down-hill toe is the far one with (n - s), giving 5(4 + 0.8)/3 = 8 m here. (Key point: Far toe in filling)

4. Between 0+030 (filling area 34.0) and 0+060 (cutting area 21.5), how are volumes found?
   A. Net area (34.0 - 21.5) x 30
   B. Cutting: 21.5/2 x 30 = 322.5; filling: 34.0/2 x 30 = 510
   C. Filling: 34.0 x 30
   D. Cutting: 21.5 x 30 = 645
   Answer: B. Cutting and filling are handled separately, each with mean area (the other end being zero) times 30 m. (Key point: Volume between filling and cutting sections)

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