---
title: "Hill road earthwork, km 8.50 to km 9.00"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:hill-road-t-km-8-5"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/hill-road-t-km-8-5
site: BCT Engineering
---

# Hill road earthwork, km 8.50 to km 9.00

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2067 Poush (old course), 10 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/hill-road-t-km-8-5
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Calculate the quantity of earthwork in cutting and filling in a portion of a hill road from km 8.50 to km 9.00 having cross slope (transverse slope) of ground 1 in 5 with the following data. Formation width of road = 8 m. Side slope in cutting = (1:1). Side slope in filling = (2:1). Depth of cut at centre line at km 8.50 = 40 cm. Depth of cut at centre line at km 9.00 = 80 cm.

## Short answer

Earthwork in cutting = 3,125.00 m³; earthwork in filling = 166.67 m³.

## Answer

#### Given data and assumptions
- Formation width $B = 8$ m; side slope in cutting $1:1$, in filling $2:1$; cross slope of ground $1$ in $5$ ($n = 5$).
- Centre depth of cutting: 0.40 m at km 8.50 and 0.80 m at km 9.00; the length is 500 m. With only two sections, the mean-area method is used.

#### 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.

#### Sectional areas
| Chainage | Centre ht. $h$ (m) | Cutting area (m²) | Filling area (m²) |
|---|---:|---:|---:|
| km 8.50 | cut 0.40 | 4.500 | 0.667 |
| km 9.00 | cut 0.80 | 8.000 | 0.000 |

Working for one section:

Take chainage km 8.50: centre cutting depth $h = 0.40$ m, $n = 5$, $B = 8.0$ m.
- Ground height above formation at the uphill edge $= 1.200$ m, at the downhill edge $= -0.400$ m.
- Toe of the side slope on the uphill side at $x = 5.500$ m from the centre line (ground above formation by 1.500 m); downhill toe at $x = 5.333$ m (ground below formation by 0.667 m).
- Area from the co-ordinates of the polygon: cutting $= 4.500$ m², filling $= 0.667$ m².

#### Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---:|---:|---:|
| km 8.50 – km 9.00 | 500.0 | 3,125.00 | 166.67 |
| **Total** | 500.0 | **3,125.00** | **166.67** |

**Answer: Earthwork in cutting = 3,125.00 m³; earthwork in filling = 166.67 m³.**

## Check yourself

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

1. The portion runs from km 8.50 to km 9.00. What is the distance used in the volume calculation?
   A. 50 m
   B. 500 m
   C. 5 m
   D. 5000 m
   Answer: B. 0.50 km = 500 m. (Key point: Length of portion)

2. At km 9.00 (cut 0.80 m, n = 5, B = 8 m) the filling area is zero. Why?
   A. Filling slope is flat
   B. Down-hill edge height = 0.80 - 4/5 = 0, so ground does not fall below formation
   C. The formation width is zero
   D. Only km 8.50 can have filling
   Answer: B. The ground at the down-hill edge is exactly at formation level, so there is no filling part. (Key point: Why no filling at km 9.00)

3. With only two sections, how is cutting volume found?
   A. Area at km 8.50 x 500 m
   B. (4.500 + 8.000)/2 x 500 = 3125 m^3
   C. 8.000 x 500 = 4000 m^3
   D. (4.500 x 8.000) x 500
   Answer: B. The mean of the two cutting areas times the length gives 3125 m^3. (Key point: Mean area method)

## Related in Estimating and Costing

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- [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
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