---
title: "Hill road earthwork, cutting 4 m / banking 6 m"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:hill-road-k-cut4-fill6"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/hill-road-k-cut4-fill6
site: BCT Engineering
---

# Hill road earthwork, cutting 4 m / banking 6 m

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2075 Chaitra, 10 marks
- Video: [ROAD Estimating Earth work Part 2](https://www.youtube.com/watch?v=28fEnXK7vr4) — MOHIYUDDIN KOLHAR
- 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-k-cut4-fill6
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Estimate the quantity of earthwork of a hill road when the formation width in cutting is 4 m and side slope is 2:1. The formation width in banking is 6 m and side slope 3:1. The ground and formation level at the centre of road and also the transverse slopes of ground surface are as below:

| Chainage (m) | 0 | 50 | 100 | 150 | 200 | 250 |
|---|---|---|---|---|---|---|
| RL of GL (m) | 1150.00 | 1150.60 | 1151.50 | 1150.80 | 1151.50 | 1152.00 |
| RL of FL (m) | 1149.20 | 1150.00 | 1150.80 | 1151.60 | 1151.50 | 1153.20 |
| Cross slope (m) | 1:10 | 1:1 | 1:14 | 1:12 | 0 | 1:10 |

## Short answer

Side-hill section formulas for different formation widths (cut: 4m, slope 2:1; fill: 6m, slope 3:1) and given cross slopes. Volumes by mean area with zero-point proportioning. Earthwork: cutting 428.42 m³, filling 606.34 m³.

## Answer

#### Given data and assumptions
- Formation width: cutting $4$ m with side slope $2:1$; banking $6$ m with side slope $3:1$ (H:V). Sections are 50 m apart.
- Centre height = GL − FL: 0.80, 0.60, 0.70 (cutting), −0.80 (filling), 0.00, −1.20 (filling).
- The cross slope printed as 1:1 at 50 m is steeper than the 2:1 cut slope (the cut face would never meet the ground), so it is taken as 1:10. At 200 m the cross slope is 0 (level ground) and the centre height is zero.

#### 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 0.80 | 4.750 | 0.000 |
| 0+050 | cut 0.60 | 3.333 | 0.000 |
| 0+100 | cut 0.70 | 3.900 | 0.000 |
| 0+150 | fill 0.80 | 0.000 | 7.368 |
| 0+200 | 0.00 | 0.000 | 0.000 |
| 0+250 | fill 1.20 | 0.000 | 12.956 |

Working for one section:

Take chainage 0+050: centre cutting depth $h = 0.60$ m, $n = 10$, $B = 4.0$ m.
- Ground height above formation at the uphill edge $= 0.800$ m, at the downhill edge $= 0.400$ m.
- Toe of the side slope on the uphill side at $x = 4.000$ m from the centre line (ground above formation by 1.000 m); downhill toe at $x = 2.667$ m (ground above formation by 0.333 m).
- Area from the co-ordinates of the polygon: cutting $= 3.333$ m², filling $= 0.000$ m².

Change-over between 0+100 (wholly in cutting, 0.70 m) and 0+150 (wholly in filling, 0.80 m): the zero point is $50 \times 0.70/1.50 = 23.33$ m from 0+100, leaving $26.67$ m in filling.
- Cutting $= \tfrac12 \times 3.900 \times 23.33 = 45.50$ m³
- Filling $= \tfrac12 \times 7.368 \times 26.67 = 98.24$ m³

#### Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---:|---:|---:|
| 0+000 – 0+050 | 50.0 | 202.08 | 0.00 |
| 0+050 – 0+100 | 50.0 | 180.84 | 0.00 |
| 0+100 – 0+150 | 50.0 (23.33 cut + 26.67 fill) | 45.50 | 98.24 |
| 0+150 – 0+200 | 50.0 | 0.00 | 184.20 |
| 0+200 – 0+250 | 50.0 | 0.00 | 323.90 |
| **Total** | 250.0 | **428.42** | **606.34** |

**Answer: Earthwork in cutting = 428.42 m³; earthwork in filling = 606.34 m³.**

## Check yourself

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

1. The centre height at each section is found from:
   A. GL minus FL (positive = cutting, negative = filling)
   B. FL plus GL
   C. Cross slope x width
   D. Side slope x width
   Answer: A. At 0+000, 1150.00 - 1149.20 = 0.80 m cutting; negative values indicate filling. (Key point: Centre height)

2. Why is the printed cross slope 1:1 at 50 m taken as 1:10?
   A. A ground slope as steep as the 2:1 cut face means the cut would never meet the ground
   B. 1:1 is not a valid ratio
   C. Ground slopes are always 1:10
   D. To make the numbers round
   Answer: A. x_u uses (n - s); with n close to or below s the toe does not exist, so the figure is treated as a misprint. (Key point: Reinterpreting the 1:1 cross slope)

3. Which parameters apply in filling sections?
   A. B = 6 m and side slope 3:1
   B. B = 4 m and side slope 2:1
   C. B = 6 m and side slope 2:1
   D. B = 4 m and side slope 3:1
   Answer: A. Banking has formation width 6 m and slope 3:1; cutting has 4 m and 2:1. (Key point: Different widths and slopes)

4. At 200 m the centre height is zero and cross slope is 0. The area is:
   A. Zero, since GL equals FL on level ground
   B. Equal to B squared
   C. Equal to B/2
   D. Undetermined
   Answer: A. With ground at formation level and no cross slope no earthwork is needed. (Key point: Level ground section)

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