---
title: "Road earthwork, RD 0-120, gradient 1 in 150"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:road-earthwork-o-rds"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/road-earthwork-o-rds
site: BCT Engineering
---

# Road earthwork, RD 0-120, gradient 1 in 150

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2073 Shrawan, 16 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/road-earthwork-o-rds
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Prepare an estimate of earthwork for a road portion from the following data: Formation width = 8 m in cutting and 10 m in banking. Side slope in cutting = 1:1, side slope in banking = 2:1 (H:V).

| RD | 0 | 30 | 60 | 90 | 120 |
|---|---|---|---|---|---|
| RLS of ground | 507.0 | 507.95 | 507.30 | 506.90 | 506.50 |
| Formation level | 507.0 and upward gradient @ 1 in 150 | | | | |
| Cross slope of ground | 1:10 | 1:12 | 1:10 | 1:12 | 1:10 |

## Short answer

Using side-hill section formulas, areas at each RD are obtained. Volumes by mean area method and change-over point where cut transits to fill. Total cutting: 241.62 m³, filling: 573.42 m³.

## Answer

#### Given data and assumptions
- Formation width $8$ m in cutting and $10$ m in banking; side slope in cutting $1:1$, in banking $2:1$ (H:V).
- Formation level 507.00 m at RD 0, rising 1 in 150: 507.00, 507.20, 507.40, 507.60, 507.80 m. The section at RD 0 has zero centre height and is taken with the cutting width of 8 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: $V = \tfrac12 A \times l$.

| Chainage | GL (m) | FL (m) | Depth of cutting / height of filling (m) |
|---|---:|---:|---:|
| 0+000 | 507.000 | 507.000 | 0.000 |
| 0+030 | 507.950 | 507.200 | cutting 0.750 |
| 0+060 | 507.300 | 507.400 | filling 0.100 |
| 0+090 | 506.900 | 507.600 | filling 0.700 |
| 0+120 | 506.500 | 507.800 | filling 1.300 |

#### Sectional areas
| Chainage | Centre ht. $h$ (m) | Cutting area (m²) | Filling area (m²) |
|---|---:|---:|---:|
| 0+000 | 0.00 | 0.889 | 1.000 |
| 0+030 | cut 0.75 | 6.720 | 0.000 |
| 0+060 | fill 0.10 | 0.889 | 2.250 |
| 0+090 | fill 0.70 | 0.000 | 8.565 |
| 0+120 | fill 1.30 | 0.000 | 17.583 |

Working for one section:

Take chainage 0+000: centre cutting depth $h = 0.00$ m, $n = 10$, $B = 8.0$ m.
- Ground height above formation at the uphill edge $= 0.400$ m, at the downhill edge $= -0.400$ m.
- Toe of the side slope on the uphill side at $x = 4.444$ m from the centre line (ground above formation by 0.444 m); downhill toe at $x = 5.000$ m (ground below formation by 0.500 m).
- Area from the co-ordinates of the polygon: cutting $= 0.889$ m², filling $= 1.000$ m².

Change-over between RD 30 (wholly in cutting, 0.75 m) and RD 60 (filling 0.10 m at the centre): the zero point is $30 \times 0.75/0.85 = 26.47$ m from RD 30, leaving $3.53$ m in filling.
- Cutting is present at both ends (6.720 and 0.889 m²), so the mean area is used: $\tfrac{6.720 + 0.889}{2} \times 30 = 114.14$ m³
- Filling is zero at RD 30, so it is taken over the 3.53 m only: $\tfrac12 \times 2.250 \times 3.53 = 3.97$ m³

#### Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---:|---:|---:|
| 0+000 – 0+030 | 30.0 | 114.14 | 15.00 |
| 0+030 – 0+060 | 30.0 (filling over 3.53) | 114.14 | 3.97 |
| 0+060 – 0+090 | 30.0 | 13.33 | 162.23 |
| 0+090 – 0+120 | 30.0 | 0.00 | 392.23 |
| **Total** | 120.0 | **241.61** | **573.43** |

**Answer: Earthwork in cutting = 241.61 m³; earthwork in filling = 573.43 m³.**

## Check yourself

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

1. The formation level is 507.00 m at RD 0 and rises 1 in 150. What is the formation level at RD 30?
   A. 507.15 m
   B. 507.20 m
   C. 507.50 m
   D. 506.80 m
   Answer: B. A rise of 1 in 150 over 30 m gives 30/150 = 0.20 m, so the level is 507.00 + 0.20 = 507.20 m. (Key point: Formation level from gradient)

2. At RD 30 the ground is 507.95 m and the formation level is 507.20 m. What is the centre height?
   A. Filling 0.75 m
   B. Filling 1.15 m
   C. Cutting 0.75 m
   D. Cutting 1.15 m
   Answer: C. Ground is above formation by 507.95 - 507.20 = 0.75 m, so it is cutting. (Key point: Cutting or filling from levels)

3. For a side-hill section in cutting with ground cross slope n:1 and side slope s:1, which formula gives the up-hill toe distance from the centre line?
   A. n(B/2 + s h)/(n + s)
   B. n(B/2 + s h)/(n - s)
   C. (B/2 + s h)/n
   D. (B + s h) h
   Answer: B. On the up-hill side the ground rises and meets the side slope further out, giving the denominator (n - s); the down-hill side uses (n + s). (Key point: Side-hill toe distances)

4. At RD 0 the centre height is zero, yet the answer gives both a cutting area (0.889 m^2) and a filling area (1.000 m^2). Why?
   A. The formation width changes at RD 0
   B. The ground is above formation on the up-hill side and below it on the down-hill side
   C. The gradient is zero at RD 0
   D. Cutting and filling areas are always equal at zero height
   Answer: B. With a cross slope, the ground rises above formation on one side and falls below it on the other, so one section has both cutting and filling parts. (Key point: Why separate cutting and filling areas)

5. How is the volume between two consecutive sections found in this answer?
   A. Mean of the two areas x distance, separately for cutting and filling
   B. Sum of the two areas x distance
   C. Area of the first section x distance
   D. Combined net area (cut - fill) x distance only
   Answer: A. The mean-area method is used, and cutting and filling are calculated separately as they are different quantities. (Key point: Volume between sections)

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