---
title: "Hill road earthwork, 0+060 to 0+120"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:hill-road-m-formation-10-8"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/hill-road-m-formation-10-8
site: BCT Engineering
---

# Hill road earthwork, 0+060 to 0+120

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2074 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-m-formation-10-8
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Calculate the quantity of earthwork for a portion of hill road from following data: Formation width = 10 m in banking and 8 m in cutting, side slope in cutting = 1:1, side slope in filling = 2:1.

| Chainage | Cut depth | Fill height | Transverse slope |
|---|---|---|---|
| 0+060 | 0.5 | - | 10:1 |
| 0+090 | 0.6 | - | 15:1 |
| 0+120 | - | 0.7 | 12:1 |

## Short answer

Side-hill section formulas using given cross slopes compute cutting and filling areas at each chainage. The transition between chainage 0+090 (cut) and 0+120 (fill) locates zero point 13.85 m from 0+090. Total earthwork: cutting 182.02 m³, filling 69.18 m³.

## Answer

#### Given data and assumptions
- Formation width: cutting $8$ m, banking $10$ m; side slope in cutting $1:1$, in filling $2:1$.
- Sections 30 m apart; cutting at 0+060 and 0+090, filling at 0+120.

#### 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+060 | cut 0.50 | 4.455 | 0.000 |
| 0+090 | cut 0.60 | 5.254 | 0.000 |
| 0+120 | fill 0.70 | 0.000 | 8.565 |

Working for one section:

Take chainage 0+060: centre cutting depth $h = 0.50$ m, $n = 10$, $B = 8.0$ 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 $x = 5.000$ m from the centre line (ground above formation by 1.000 m); downhill toe at $x = 4.091$ m (ground above formation by 0.091 m).
- Area from the co-ordinates of the polygon: cutting $= 4.455$ m², filling $= 0.000$ m².

Change-over between 0+090 (wholly in cutting, 0.60 m) and 0+120 (wholly in filling, 0.70 m): the zero point is $30 \times 0.60/1.30 = 13.85$ m from 0+090, leaving $16.15$ m in filling.
- Cutting $= \tfrac12 \times 5.254 \times 13.85 = 36.38$ m³
- Filling $= \tfrac12 \times 8.565 \times 16.15 = 69.18$ m³

#### Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---:|---:|---:|
| 0+060 – 0+090 | 30.0 | 145.64 | 0.00 |
| 0+090 – 0+120 | 30.0 (13.85 cut + 16.15 fill) | 36.38 | 69.18 |
| **Total** | 60.0 | **182.02** | **69.18** |

**Answer: Earthwork in cutting = 182.02 m³; earthwork in filling = 69.18 m³.**

## Check yourself

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

1. Which formation width applies at chainage 0+120?
   A. 10 m (banking)
   B. 8 m (cutting)
   C. 18 m
   D. 4 m
   Answer: A. 0+120 is in filling, and the formation width in banking is 10 m. (Key point: Widths)

2. For 0+060 (h = 0.5, n = 10, B = 8, s = 1), the uphill toe is x_u = 10 x (4 + 0.5)/(10 - 1) =
   A. 5.000 m
   B. 4.091 m
   C. 4.5 m
   D. 9 m
   Answer: A. 10 x 4.5 / 9 = 5.0 m; the downhill toe is 45/11 = 4.091 m. (Key point: Toe distances)

3. The cutting volume between 0+060 and 0+090 is 145.64 m3. It was obtained as:
   A. (4.455 + 5.254)/2 x 30
   B. 4.455 x 30
   C. (5.254 - 4.455) x 30
   D. 5.254 x 60
   Answer: A. Mean area 4.8545 x 30 = 145.64 m3. (Key point: Volume formula)

4. A transverse slope of 15:1 compared with 10:1 is:
   A. Flatter ground
   B. Steeper ground
   C. Same slope
   D. Vertical
   Answer: A. 15 horizontal per 1 vertical is gentler than 10 per 1. (Key point: Cross slope direction)

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