---
title: "Embankment volume, 160 m, rising gradient 1 in 100"
subject: "Surveying I"
chapter: 9
key: "semester_3/surveying_i:volume-embankment-formation-10m-ground-levels"
source: https://www.bctengineer.com/q/semester_3/surveying_i/volume-embankment-formation-10m-ground-levels
site: BCT Engineering
---

# Embankment volume, 160 m, rising gradient 1 in 100

Surveying I · Chapter 9: Computation of Area and Volume

- Asked in: 2076 Asoj, 6 marks
- Video: [(V6-MU3-BCEEM) Prob-4:  An embankment 400 m long with a formation width of 15 m is to be ....](https://www.youtube.com/watch?v=5NDNxnCjRSE) — AddOnn
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/volume-embankment-formation-10m-ground-levels
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-9

## Question

A roadway embankment of formation width of 10 m and side slope 2:1 is to be constructed. The ground level along the centre line is as follows.

| Chainage | 0+000 | 0+040 | 0+080 | 0+120 | 0+160 |
|---|---|---|---|---|---|
| GL (m) | 1115.70 | 1114.30 | 1116.75 | 1115.15 | 1118.45 |

The embankment has a rising gradient of 1 in 100 and the formation level at zero chainage is 114.95 m [?]. Assuming the ground level across the centre line, compute the volume of earth work.

## Short answer

For level cross-sections, fill heights computed with formation at 1116.95 m rising 1 in 100. Areas using (b+sh)h. Trapezoidal volume 4697.10 m³, prismoidal volume 5720.87 m³, correction -1023.77 m³.

## Answer

**Reading of the given data:** the scanned formation level "114.95 m" cannot be right for ground levels near 1115 m (it would put the formation about 1000 m below the ground). Read as 1114.95 m, the formation would switch between cutting and filling at every section, which does not fit an embankment. It is therefore taken here as **1116.95 m** at 0+000, which keeps the formation above the ground at every chainage. "Ground level across the centre line" is taken to mean **level cross-sections** (no transverse slope). Formation width $b = 10$ m, side slope $s = 2$ (2:1), spacing $L = 40$ m.

#### Step 1: Formation levels and heights of fill

The formation rises at 1 in 100, i.e. $40/100 = 0.40$ m per 40 m: $FL = 1116.95 + 0.40 \times (\text{number of intervals})$. Height of fill $h = FL - GL$.

| Chainage (m) | GL (m) | FL (m) | h (m) |
|---|---|---|---|
| 0 | 1115.70 | 1116.95 | 1.25 |
| 40 | 1114.30 | 1117.35 | 3.05 |
| 80 | 1116.75 | 1117.75 | 1.00 |
| 120 | 1115.15 | 1118.15 | 3.00 |
| 160 | 1118.45 | 1118.55 | 0.10 |

#### Step 2: Cross-sectional areas

For a level section of an embankment, the area is a trapezoid:

$$
A = (b + s h)\,h
$$

| Chainage (m) | h (m) | b + s·h (m) | A (m²) |
|---|---|---|---|
| 0 | 1.25 | 12.50 | 15.625 |
| 40 | 3.05 | 16.10 | 49.105 |
| 80 | 1.00 | 12.00 | 12.000 |
| 120 | 3.00 | 16.00 | 48.000 |
| 160 | 0.10 | 10.20 | 1.020 |

#### Step 3: Trapezoidal formula

$$
V_T = L\left[\frac{A_1 + A_{5}}{2} + A_2 + \dots + A_{4}\right] = 40\left[\frac{15.625 + 1.020}{2} + 49.105 + 12.000 + 48.000\right] = 4697.10\ \text{m}^3
$$

#### Step 4: Prismoidal formula

There are four equal intervals (an even number), so Simpson's rule applies.

$$
V_P = \frac{L}{3}\left[A_1 + A_{5} + 4(A_2 + A_4) + 2A_3\right] = \frac{40}{3}\left[15.625 + 1.020 + 4(49.105 + 48.000) + 2(12.000)\right] = 5720.87\ \text{m}^3
$$

#### Step 5: Prismoidal correction

$$
C_p = V_T - V_P = 4697.10 - 5720.87 = -1023.77\ \text{m}^3
$$

**Answer: Volume of earthwork (filling) = 4697.10 m³ by the trapezoidal formula and 5720.87 m³ by the prismoidal formula (prismoidal correction = -1023.77 m³).**

## Check yourself

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

1. The formation rises 1 in 100 and the spacing is 40 m. What is the formation level at 0+040, and the height of fill there (GL = 1114.30 m)?
   A. 1117.35 m; 3.05 m
   B. 1116.55 m; 2.25 m
   C. 1117.35 m; 1.25 m
   D. 1116.95 m; 2.65 m
   Answer: A. FL rises 0.40 m per 40 m to 1117.35 m, so h = 1117.35 - 1114.30 = 3.05 m. (Key point: Formation level and fill height)

2. For a level section with formation width b, side slope s:1 and fill height h, the area is:
   A. b x h
   B. (b + s h) h
   C. (b + h/s) h
   D. 0.5 (b + s h) h
   Answer: B. The section is a trapezoid with top width b and bottom width b + 2sh, so the area is (b + s h) h. (Key point: Level-section area of embankment)

3. Why can the prismoidal (Simpson) formula be applied to the five sections?
   A. Because the areas are all positive
   B. Because there is an even number of equal intervals (four)
   C. Because there is an odd number of intervals
   D. Because the side slope is 2:1
   Answer: B. Simpson's rule needs an even number of equal intervals; five sections give four intervals. (Key point: Simpson's rule applicability)

## Related in Surveying I

- [Cutting volume, 120 m, upward slope 1 in 40](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-cutting-120m-formation-380) — Surveying I, chapter 9, asked 8×
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- [Cutting volume, 60 m, formation level 279.00](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-cutting-60m-gl-281-50-formation-279) — Surveying I, chapter 9, asked 7×
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- [Filling volume, 60 m, formation 1281.49](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-filling-60m-gl-1280-50) — Surveying I, chapter 9

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