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Embankment excavation quantity (18 kN/m³)

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.

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.

Given

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

Solution

Dry unit weight at the excavation site:

γd,field=γ1+w=181.10=16.36 kN/m3\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:

Ws=γd,fill×V=19×1=19 kNW_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:

Vexc=Wsγd,field=1916.36=1.161 m3V_{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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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