---
title: "Embankment excavation quantity (18 kN/m³)"
subject: "Soil Mechanics"
chapter: 2
key: "semester_4/soil_mechanics:embankment-excavated-quantity-2079"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/embankment-excavated-quantity-2079
site: BCT Engineering
---

# Embankment excavation quantity (18 kN/m³)

Soil Mechanics · Chapter 2: Solids- Water –Air Relations and Index properties of soils

- Asked in: 2079 Asoj, 3 marks
- Video: [Embankment fill is to be compacted at a density of 18 kN/m3. The soil of the borrow area is at a](https://www.youtube.com/watch?v=tAMd9ZE2Evg) — Unique Series - Civil Mantraa
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/embankment-excavated-quantity-2079
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-2

## Question

The in-situ field unit weight and water content of soil are 18 kN/m³ and 10% respectively. Soil excavated from this in-situ site is used for embankment construction. The dry unit weight and water content of the soil at the compaction site are 19 kN/m³ and 18% respectively. Determine the amount of soil to be excavated for 1 m³ of compaction. Assume necessary conditions.

## Short answer

In-situ dry unit weight is 18/1.10 equals 16.36 kN/m3. Solids needed for 1 m3 of fill are 19 kN. Excavated volume is 19 divided by 16.36, about 1.16 m3, assuming no soil loss.

## Answer

#### Given
In-situ: $\gamma = 18$ kN/m³, $w = 10\%$. Compacted fill: $\gamma_d = 19$ kN/m³, $w = 18\%$, volume = 1 m³.

#### Solution
Dry unit weight at the excavation site:
$$
\gamma_{d,field} = \frac{\gamma}{1 + w} = \frac{18}{1.10} = 16.36\ \text{kN/m}^3
$$
Weight of dry soil (solids) needed in 1 m³ of embankment:
$$
W_s = \gamma_{d,fill} \times V = 19 \times 1 = 19\ \text{kN}
$$
The weight of solids is the same at the borrow site and in the fill, so the volume to excavate is:
$$
V_{exc} = \frac{W_s}{\gamma_{d,field}} = \frac{19}{16.36} = 1.161\ \text{m}^3
$$

**Answer: about 1.16 m³ of soil must be excavated for each 1 m³ of compacted fill.**

Assumption: no soil is lost during hauling and spreading.

## Check yourself

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

1. The dry unit weight at the excavation site is:
   A. 14.4 kN/m3
   B. 19.8 kN/m3
   C. 18 kN/m3
   D. 16.36 kN/m3
   Answer: D. gamma_d = 18/1.10 = 16.36 kN/m3. (Key point: Dry unit weight of borrow)

2. The principle used to find the excavated volume is:
   A. Volume of soil is constant
   B. Weight of solids is the same at borrow site and in the fill
   C. Water content is constant
   D. Total weight is conserved after wetting
   Answer: B. Solids are only moved, so Ws is equal in both places. (Key point: Key principle)

3. Volume to be excavated per 1 m3 of fill is:
   A. 1.50 m3
   B. 0.86 m3
   C. 1.00 m3
   D. 1.16 m3
   Answer: D. V = 19/16.36 = 1.161 m3. (Key point: Final result)

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