---
title: "Hill road earthwork, chainage 12-15"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:hill-road-q-chain-12-15"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/hill-road-q-chain-12-15
site: BCT Engineering
---

# Hill road earthwork, chainage 12-15

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2070 Chaitra, 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-q-chain-12-15
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Estimate the quantities of earthwork for a portion of a hilly road from following data: Formation width = 10 m. Side slopes in cutting = 1:1 and in banking = 2:1 (H:V), length of chain = 30 m.

| Chainage | 12 | 13 | 14 | 15 |
|---|---|---|---|---|
| Depth of cut | 0.4 | 0.2 | - | - |
| Ht. of banking | - | - | 0.3 | 0.5 |
| Transverse slope of ground | 1:10 | 1:12 | 1:10 | 1:8 |

## Short answer

Using side-hill section formulas with formation width 10 m, side slopes 1:1 in cutting and 2:1 in filling, and given cross slopes, the earthwork is 149.91 m³ in cutting and 232.33 m³ in filling.

## Answer

#### Given data and assumptions
- Formation width $B = 10$ m; side slope in cutting $1:1$, in banking $2:1$. The length of one chain is 30 m, so the sections are 30 m apart (chainage numbers 12, 13, 14, 15 are chain numbers).

#### 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) |
|---|---:|---:|---:|
| 12 | cut 0.40 | 4.500 | 0.062 |
| 13 | cut 0.20 | 2.489 | 0.338 |
| 14 | fill 0.30 | 0.222 | 4.000 |
| 15 | fill 0.50 | 0.071 | 6.750 |

Working for one section:

Take chainage 12: centre cutting depth $h = 0.40$ m, $n = 10$, $B = 10.0$ m.
- Ground height above formation at the uphill edge $= 0.900$ m, at the downhill edge $= -0.100$ m.
- Toe of the side slope on the uphill side at $x = 6.000$ m from the centre line (ground above formation by 1.000 m); downhill toe at $x = 5.250$ m (ground below formation by 0.125 m).
- Area from the co-ordinates of the polygon: cutting $= 4.500$ m², filling $= 0.062$ m².

#### Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---:|---:|---:|
| 12 – 13 | 30.0 | 104.84 | 6.01 |
| 13 – 14 | 30.0 | 40.67 | 65.07 |
| 14 – 15 | 30.0 | 4.40 | 161.25 |
| **Total** | 90.0 | **149.91** | **232.33** |

**Answer: Earthwork in cutting = 149.91 m³; earthwork in filling = 232.33 m³.**

## Check yourself

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

1. The chainages are 12, 13, 14, 15 with a chain length of 30 m. What is the distance between consecutive sections?
   A. 1 m
   B. 12 m
   C. 30 m
   D. 15 m
   Answer: C. The numbers are chain numbers; each chain is 30 m, so sections are 30 m apart. (Key point: Spacing of sections)

2. At chainage 12: h = 0.40 m cutting, n = 10, B = 10 m. What is the ground height relative to formation at the up-hill edge?
   A. +0.900 m
   B. +0.400 m
   C. +1.000 m
   D. +0.500 m
   Answer: A. Up-hill edge = h + (B/2)/n = 0.40 + 5/10 = 0.90 m above formation. (Key point: Edge heights)

3. In section 14 the centre is in filling (0.30 m) but a small cutting area (0.222 m^2) is listed. What causes this?
   A. The cutting slope is flatter
   B. Ground on the up-hill side rises above formation level
   C. An error in the banking height
   D. The formation is narrower on the high side
   Answer: B. Cross slope raises the ground by (B/2)/n = 0.5 m at the up-hill edge, more than the 0.30 m filling at the centre, so part is in cutting. (Key point: Cutting section with a filling part)

4. Cutting areas at 12 and 13 are 4.500 and 2.489 m^2. The cutting volume between them is about:
   A. 209.7 m^3
   B. 104.8 m^3
   C. 69.9 m^3
   D. 135.0 m^3
   Answer: B. (4.500 + 2.489)/2 = 3.4945 m^2 x 30 m = about 104.84 m^3. (Key point: Mean area volume)

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