---
title: "Hill road earthwork, chainage 0-150 (50 m)"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:hill-road-j-formation-10"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/hill-road-j-formation-10
site: BCT Engineering
---

# Hill road earthwork, chainage 0-150 (50 m)

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2076 Asoj, 9 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-j-formation-10
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

A road is to be constructed in hilly area with formation width of 10 m, side slopes in banking and cutting (2:1) and (1:1). The height of banking or depth of cutting at the centre line of the road are given below. The cross slopes of ground are also given at different sections. Calculate the quantities of earthwork.

| Distance | Cutting | Filling | Cross slope of ground |
|---|---|---|---|
| 0 | 0.50 | -- | 12:1 |
| 50 | 0.60 | -- | 10:1 |
| 100 | -- | 0.40 | 15:1 |
| 150 | -- | 0.60 | 12:1 |

## Short answer

Side-hill section areas computed using formulas for ground cross slope n:1, formation width 10m, cut slope 1:1, fill slope 2:1. Volumes by mean area with zero-point proportion. Earthwork: cutting 403.61 m³, filling 343.58 m³.

## Answer

#### Given data and assumptions
- Formation width $B = 10$ m; side slope in cutting $1:1$, in banking $2:1$ (H:V).
- Sections 50 m apart; cutting at 0 and 50 m, filling at 100 and 150 m.

#### 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 0.50 | 5.462 | 0.000 |
| 0+050 | cut 0.60 | 6.677 | 0.000 |
| 0+100 | fill 0.40 | 0.000 | 4.624 |
| 0+150 | fill 0.60 | 0.000 | 7.269 |

Working for one section:

Take chainage 0+000: centre cutting depth $h = 0.50$ m, $n = 12$, $B = 10.0$ m.
- Ground height above formation at the uphill edge $= 0.917$ m, at the downhill edge $= 0.083$ 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 $x = 5.077$ m (ground above formation by 0.077 m).
- Area from the co-ordinates of the polygon: cutting $= 5.462$ m², filling $= 0.000$ m².

Change-over between 0+050 (wholly in cutting, 0.60 m) and 0+100 (wholly in filling, 0.40 m): the zero point is $50 \times 0.60/1.00 = 30.00$ m from 0+050, leaving $20.00$ m in filling.
- Cutting $= \tfrac12 \times 6.677 \times 30.00 = 100.15$ m³
- Filling $= \tfrac12 \times 4.624 \times 20.00 = 46.24$ m³

#### Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---:|---:|---:|
| 0+000 – 0+050 | 50.0 | 303.46 | 0.00 |
| 0+050 – 0+100 | 50.0 (30.00 cut + 20.00 fill) | 100.15 | 46.24 |
| 0+100 – 0+150 | 50.0 | 0.00 | 297.34 |
| **Total** | 150.0 | **403.61** | **343.58** |

**Answer: Earthwork in cutting = 403.61 m³; earthwork in filling = 343.58 m³.**

## Check yourself

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

1. Which values are used for the hill road in this problem?
   A. B = 10 m, cutting slope 1:1, banking slope 2:1
   B. B = 10 m, cutting 2:1, banking 1:1
   C. B = 8 m, both 1.5:1
   D. B = 10 m, both 2:1
   Answer: A. The question gives a 10 m formation with banking 2:1 and cutting 1:1. (Key point: Slopes and width)

2. For the cutting section at 0+000 (h = 0.5 m, n = 12), the uphill toe is farther from the centre line than the downhill toe because:
   A. The ground rises on the uphill side so the cut face must reach higher ground
   B. The road is wider uphill
   C. Cutting slope is flatter uphill
   D. Gravity moves earth uphill
   Answer: A. Uphill the ground rises, so the cut slope has to extend farther (x_u = 6.000 m) than downhill (x_d = 5.077 m). (Key point: Cross-slope effect)

3. Sections are 50 m apart. The volume between 0+000 and 0+050 (both cutting) is:
   A. Mean of 5.462 and 6.677 m2 x 50 m = 303.46 m3
   B. 5.462 x 50 m3
   C. (5.462 + 6.677) x 50 m3
   D. 6.677 x 100 m3
   Answer: A. Mean area (6.0695) x 50 = 303.46 m3. (Key point: Spacing)

4. Total earthwork is 470.38 m3 cutting and 412.95 m3 filling. These are:
   A. Kept as two separate totals
   B. Netted to 57.43 m3 cutting
   C. Added to 883.33 m3
   D. Averaged
   Answer: A. Cutting and filling are quoted separately. (Key point: Totals)

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