Question

The proposed grading at a project site will consist of 25,100 m3 of cut and 23,300 m3 of fill and will be a balanced earthwork job. The cut area has an average moisture content of 8.3%. The fill will be compacted to an average relative compaction of 93% based on a maximum dry unit weight of 18.3 kN/m3 and an optimum moisture content of 12.9% obtained from the modified Proctor test. Compute the volume of water in kiloliters that will be required to bring these soils to the optimum moisture content.

Answer 1

We can use the formula for the moisture content of soil to solve for the volume of water required:

M = ((Ww / Ws) x 100)%

where M is the moisture content, Ww is the weight of water, and Ws is the weight of solids.

First, let's find the weight of solids for the cut area:

Ws = V x γ

where V is the volume and γ is the unit weight.

Ws = 25,100 m3 x 1.8 t/m3 x 1000 kg/t

Ws = 45,180,000 kg

Next, let's find the weight of solids for the fill area:

Ws = V x γ

where V is the volume and γ is the unit weight.

Ws = 23,300 m3 x 18.3 kN/m3 x 1000 N/kN

Ws = 425,190,000 N

Now, let's find the weight of water required for the cut area:

M = ((Ww / Ws) x 100)%

0.129 = ((Ww / 45,180,000) x 100)%

Ww = 58,402 kg

Finally, let's find the weight of water required for the fill area:

M = ((Ww / Ws) x 100)%

0.129 = ((Ww / 425,190,000) x 100)%

Ww = 548,991 kg

To find the total volume of water required, we need to convert the weight of water to volume using the density of water:

ρ = 1000 kg/m3

For the cut area:

Vw = Ww / ρ

Vw = 58,402 kg / 1000 kg/m3

Vw = 58.4 m3

For the fill area:

Vw = Ww / ρ

Vw = 548,991 kg / 1000 kg/m3

Vw = 548.991 m3

Therefore, the total volume of water required to bring the soils to the optimum moisture content is approximately 607.4 m3 or 607.4 kiloliters.