---
title: "Hill road earthwork, chainage 130-280"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:hill-road-formation-8-table-a"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/hill-road-formation-8-table-a
site: BCT Engineering
---

# Hill road earthwork, chainage 130-280

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2082 Bhadra
- 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-formation-8-table-a
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

A road is to be constructed in hill areas with formation width of 8 m. Side slope in banking is 2:1 and side slope in cutting is 1.5:1. Calculate the quantities of earthwork from the following data:

| Chainage | Depth of cutting at centre of road | Height of filling at centre of road | Cross slope of ground |
|---|---|---|---|
| 130 | 0.7 m | - | 12:1 |
| 160 | 0.5 m | - | 15:1 |
| 190 | 0.4 m | - | 12:1 |
| 220 | - | 0.7 m | 10:1 |
| 250 | - | 0.6 m | 15:1 |
| 280 | - | 0.8 m | 12:1 |

## Short answer

Cutting = 309.82 m³, filing = 471.28 m³. Side-hill areas from given cross slopes and side slopes; volumes by average end-area, transition length prorated.

## Answer

#### Given data and assumptions
- Formation width $B = 8$ m in both cutting and filling; side slope in cutting $1.5:1$, in filling $2:1$ (H:V).
- Chainages are 30 m apart; cross slopes are taken as $n:1$ (H:V), e.g. 12:1 means 12 horizontal to 1 vertical.

#### 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+130 | cut 0.70 | 6.605 | 0.000 |
| 0+160 | cut 0.50 | 4.527 | 0.000 |
| 0+190 | cut 0.40 | 3.664 | 0.000 |
| 0+220 | fill 0.70 | 0.000 | 7.188 |
| 0+250 | fill 0.60 | 0.000 | 5.765 |
| 0+280 | fill 0.80 | 0.000 | 8.128 |

Working for one section:

Take chainage 0+220: centre filling height $h = 0.70$ m, $n = 10$, $B = 8.0$ m.
- Ground height above formation at the uphill edge $= -0.300$ m, at the downhill edge $= -1.100$ m.
- Toe of the side slope on the uphill side at $x = 4.500$ m from the centre line (ground below formation by 0.250 m); downhill toe at $x = 6.750$ m (ground below formation by 1.375 m).
- Area from the co-ordinates of the polygon: cutting $= 0.000$ m², filling $= 7.188$ m².

Change-over between 0+190 (wholly in cutting, 0.40 m) and 0+220 (wholly in filling, 0.70 m): the zero point is $30 \times 0.40/1.10 = 10.91$ m from 0+190, leaving $19.09$ m in filling.
- Cutting $= \tfrac12 \times 3.664 \times 10.91 = 19.98$ m³
- Filling $= \tfrac12 \times 7.188 \times 19.09 = 68.61$ m³

#### Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---:|---:|---:|
| 0+130 – 0+160 | 30.0 | 166.98 | 0.00 |
| 0+160 – 0+190 | 30.0 | 122.86 | 0.00 |
| 0+190 – 0+220 | 30.0 (10.91 cut + 19.09 fill) | 19.98 | 68.61 |
| 0+220 – 0+250 | 30.0 | 0.00 | 194.28 |
| 0+250 – 0+280 | 30.0 | 0.00 | 208.39 |
| **Total** | 150.0 | **309.82** | **471.28** |

**Answer: Earthwork in cutting = 309.82 m³; earthwork in filling = 471.28 m³.**

## Check yourself

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

1. Why are the sections in this problem treated as side-hill sections?
   A. The ground has a transverse (cross) slope, so the two sides of the road are at different levels
   B. The road has different side slopes in cutting and filling
   C. The chainages are 30 m apart
   D. The formation width is 8 m
   Answer: A. A cross slope n:1 makes the ground at the uphill and downhill edges differ in level, so the two toes are at different distances from the centre line. (Key point: Side-hill section)

2. A cross slope of 12:1 in this answer means:
   A. 12 vertical to 1 horizontal
   B. 12 horizontal to 1 vertical
   C. 12 percent grade
   D. 12 degrees to the horizontal
   Answer: B. Cross slopes are taken as n:1 (H:V), so 12:1 is 12 horizontal to 1 vertical. (Key point: Cross slope meaning)

3. For a cutting section the uphill toe distance is x_u = n(B/2 + s h)/(n - s). What happens to the uphill toe if the cross slope n approaches the cut side slope s?
   A. x_u becomes zero
   B. x_u becomes very large because the cut face never meets the ground
   C. x_u equals B/2
   D. x_u equals the downhill toe distance
   Answer: B. As n approaches s the denominator approaches zero, so the toe moves far away; the cut face is nearly parallel to the ground. (Key point: Toe distance formula)

4. How is the volume between two consecutive sections found, and how are cutting and filling handled?
   A. Mean area x distance, with cutting and filling kept separate
   B. Sum of areas x distance, cutting and filling netted off
   C. Last area x distance, cutting only
   D. Mean area x distance after adding cutting and filling areas together
   Answer: A. Volume = mean sectional area x distance, taken separately for cutting and filling; they are not netted. (Key point: Volume between sections)

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