Question

the maximum proctor dry density and optimum unit water content measured for a clayey sand in the lab were 1820 kg/m3 and 9%. assuming the soil in the field is compacted at a water content of 9%, calculate the density (not dry density) that the soil in the field should be compacted to if the contractors want a relative compaction of 95%. answer: 1884.61 kg/m3

Answer 1

The soil in the field should be compacted to a density of 1884.61 kg/m3 to achieve a relative compaction of 95% when compacted at a water content of 9%.

Step-by-step explanation:

To calculate the density that the soil in the field should be compacted to attain a relative compaction of 95%, we need to consider the maximum proctor dry density that was determined in the lab. Given that the maximum proctor dry density is 1820 kg/m3 and the optimum unit water content is 9%, we can proceed with the calculation.

First, we calculate the dry density at 95% relative compaction:

(0.95) × (1820 kg/m3) = 1729 kg/m3 (dry density at 95% compaction)

Then, to find the density when the water content is 9%, we use the formula:

Density = Dry Density ÷ (1 + Moisture Content)

By substituting the values, we get:

Density = 1729 kg/m3 ÷ (1 + 0.09) = 1729 kg/m3 ÷ 1.09 = 1884.61 kg/m3

The density that the soil should be compacted to in the field to achieve the desired relative compaction of 95% at a water content of 9% is 1884.61 kg/m3.