---
title: "Side-long ground hill road, 0-400 m"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:side-long-road-n"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/side-long-road-n
site: BCT Engineering
---

# Side-long ground hill road, 0-400 m

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2074 Asoj, 10 marks
- Video: [Quantity Of Cutting and Filling for Hill Road | Earthwork Calculation |Estimating and Valuation](https://www.youtube.com/watch?v=AwiFNxDCO5g) — OPengineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/estimating_and_costing/side-long-road-n
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Calculate the quantities of earthwork of a hill road in side long ground from 0 m to 400 m partly in cutting and partly in filling with the following data: width of road = 10 m, side slope in cutting and filling = (1:1) and (2:1). The road has a downward gradient of 1 in 200. The cross slope of ground = 1 in 5. Formation level at 0 m = 1203.50 m.

| Ground level | 1202.50 | 1201.97 | 1202.35 | 1199.66 | 1200.50 |
|---|---|---|---|---|---|
| Distance | 0 | 100 | 200 | 300 | 400 |

## Short answer

Formation levels are computed from downward gradient 1 in 200, centre heights determined. With cross-slope 1 in 5, side-hill areas calculated and mean area method yields volumes. Earthwork: cutting 225.78 m³, filling 8262.11 m³.

## Answer

#### Given data and assumptions
- Formation width $B = 10$ m; side slope in cutting $1:1$, in filling $2:1$; cross slope of ground $1$ in $5$ ($n = 5$) throughout.
- Formation level at 0 m is 1203.50 m and falls 1 in 200, so it is 1203.50, 1203.00, 1202.50, 1202.00, 1201.50 m at 0, 100, 200, 300, 400 m.
- Ground levels are at the centre line.

#### 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.

| Chainage | GL (m) | FL (m) | Depth of cutting / height of filling (m) |
|---|---:|---:|---:|
| 0+000 | 1,202.500 | 1,203.500 | filling 1.000 |
| 0+100 | 1,201.970 | 1,203.000 | filling 1.030 |
| 0+200 | 1,202.350 | 1,202.500 | filling 0.150 |
| 0+300 | 1,199.660 | 1,202.000 | filling 2.340 |
| 0+400 | 1,200.500 | 1,201.500 | filling 1.000 |

#### Sectional areas
| Chainage | Centre ht. $h$ (m) | Cutting area (m²) | Filling area (m²) |
|---|---:|---:|---:|
| 0+000 | fill 1.00 | 0.000 | 16.667 |
| 0+100 | fill 1.03 | 0.000 | 17.169 |
| 0+200 | fill 0.15 | 2.258 | 5.510 |
| 0+300 | fill 2.34 | 0.000 | 43.275 |
| 0+400 | fill 1.00 | 0.000 | 16.667 |

Working for one section:

Take chainage 0+000: centre filling height $h = 1.00$ m, $n = 5$, $B = 10.0$ m.
- Ground height above formation at the uphill edge $= 0.000$ m, at the downhill edge $= -2.000$ m.
- Toe of the side slope on the uphill side at $x = 5.000$ m from the centre line (ground above formation by 0.000 m); downhill toe at $x = 11.667$ m (ground below formation by 3.333 m).
- Area from the co-ordinates of the polygon: cutting $= 0.000$ m², filling $= 16.667$ m².

#### Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---:|---:|---:|
| 0+000 – 0+100 | 100.0 | 0.00 | 1,691.77 |
| 0+100 – 0+200 | 100.0 | 112.89 | 1,133.96 |
| 0+200 – 0+300 | 100.0 | 112.89 | 2,439.28 |
| 0+300 – 0+400 | 100.0 | 0.00 | 2,997.10 |
| **Total** | 400.0 | **225.78** | **8,262.11** |

**Answer: Earthwork in cutting = 225.78 m³; earthwork in filling = 8,262.11 m³.**

## Check yourself

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

1. Formation level at 0 m is 1203.50 m and falls 1 in 200. FL at 300 m is:
   A. 1202.00 m
   B. 1202.50 m
   C. 1201.50 m
   D. 1203.00 m
   Answer: A. 300/200 = 1.5 m fall, giving 1202.00 m. (Key point: Formation levels)

2. At 0+200 the centre is in filling 0.15 m but the section also has 2.258 m2 of cutting. Why?
   A. The uphill edge is at a ground level above formation (cross slope 1 in 5), so that side is in cutting
   B. The cut slope is flatter
   C. Error in the table
   D. Filling slope is 1:1
   Answer: A. The ground rises 1 m over 5 m uphill, so at the uphill edge it is 0.85 m above formation, giving a cutting wedge. (Key point: Why a cutting area appears)

3. At 0+000 (h = 1 m filling, n = 5, B = 10, s = 2), the far downhill toe is x_d = n(B/2 + s h)/(n - s) =
   A. 11.667 m
   B. 5.000 m
   C. 7.0 m
   D. 3.333 m
   Answer: A. 5 x (5 + 2)/(5 - 2) = 35/3 = 11.667 m; the uphill toe is 35/7 = 5.0 m. (Key point: Section at 0+000)

4. Why is the filling at 0+300 (h = 2.34 m) so large (43.275 m2)?
   A. Greater filling height and the cross slope extends the downhill toe far out
   B. The cut slope is used
   C. The area is doubled
   D. The width is 20 m
   Answer: A. A greater centre fill height plus steep cross slope (1 in 5) increases the fill triangle on the downhill side. (Key point: Large filling)

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