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Borrow pit to embankment, water added

Short answer

Excavate 105.9 m³ from the borrow pit; add 90 kN (approx. 9.17 m³ or 9170 litres) of water.

A soil in the borrow pit is at a dry density of 17 kN/m³ with a moisture content of 10%. The soil is excavated from this pit and compacted in an embankment to a dry density of 18 kN/m³ with a moisture content of 15%. Compute the quantity of soil to be excavated from the borrow pit and the amount of water to be added for 100 m³ of compacted soil in the embankment.

Given

Borrow pit: γd=17\gamma_d = 17 kN/m³, w=10%w = 10\%. Embankment: γd=18\gamma_d = 18 kN/m³, w=15%w = 15\%, V=100V = 100 m³. γw=9.81\gamma_w = 9.81 kN/m³.

Dry weight of soil needed

Ws=γd,emb×Vemb=18×100=1800 kNW_s = \gamma_{d,emb} \times V_{emb} = 18 \times 100 = 1800\ \text{kN}

Volume to be excavated

The same solids weight must come from the pit:

Vpit=Wsγd,pit=180017=105.9 m3V_{pit} = \frac{W_s}{\gamma_{d,pit}} = \frac{1800}{17} = 105.9\ \text{m}^3

Water to be added

Ww, pit=0.10×1800=180 kNWw, required=0.15×1800=270 kNWw, added=270−180=90 kN\begin{aligned} W_{w,\ pit} &= 0.10 \times 1800 = 180\ \text{kN} \\ W_{w,\ required} &= 0.15 \times 1800 = 270\ \text{kN} \\ W_{w,\ added} &= 270 - 180 = 90\ \text{kN} \end{aligned}

In volume: 909.81=9.17\dfrac{90}{9.81} = 9.17 m³ ≈9170\approx 9170 litres.

Answer: excavate 105.9 m³ from the borrow pit; add 90 kN (about 9.17 m³ or 9170 litres) of water.

✓ Answer checked by Claude Opus 5.5 (high effort)

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