Soil is to be excavated from a borrow pit which has a density of 1.75 g/cc and water content of 12%. The specific gravity of soil particles is 2.7. The soil is compacted so that water content is 18% and dry density is 1.65 g/cc. For 1000 cu.m of soil in fill, estimate (i) the quantity of soil to be excavated from the pit in cu.m; and (ii) the amount of water to be added. Also determine the void ratios of the soil in borrow pit and fill.

Soil Excavation and Compaction (Engg. Services Exam 1995)

Problem Statement (Engg. Services Exam 1995)

Soil is to be excavated from a borrow pit with the following properties:

Density (\( \gamma \)) = 1.75 g/cm³
Water content (\( w \)) = 12%
Specific gravity of soil particles (\( G \)) = 2.7

The soil is compacted to achieve:

Water content (\( w \)) = 18%
Dry density (\( \gamma_d \)) = 1.65 g/cm³

For 1000 m³ of soil in the fill, estimate:

The quantity of soil to be excavated from the borrow pit (in m³).
The amount of water to be added (in m³).

Also, determine the void ratios of the soil in the borrow pit and the fill.

Solution

1. Given Data

Borrow pit density (\( \gamma_1 \)) = 1.75 g/cm³
Borrow pit water content (\( w_1 \)) = 12% = 0.12
Specific gravity (\( G \)) = 2.7
Compacted dry density (\( \gamma_d \)) = 1.65 g/cm³
Compacted water content (\( w_2 \)) = 18% = 0.18
Volume of fill (\( V_2 \)) = 1000 m³

2. Calculate Void Ratio in Borrow Pit (\( e_1 \))

\( \gamma_1 = \frac{G \gamma_w (1 + w_1)}{1 + e_1} \)

Rearranging for \( e_1 \): \( e_1 = \frac{G \gamma_w (1 + w_1)}{\gamma_1} – 1 \)

Substituting values: \( e_1 = \frac{2.7 \times 1 \times (1 + 0.12)}{1.75} – 1 = 1.728 – 1 = 0.728 \)

3. Calculate Void Ratio in Fill (\( e_2 \))

\( \gamma_d = \frac{G \gamma_w}{1 + e_2} \)

Rearranging for \( e_2 \): \( e_2 = \frac{G \gamma_w}{\gamma_d} – 1 \)

Substituting values: \( e_2 = \frac{2.7 \times 1}{1.65} – 1 = 1.6364 – 1 = 0.6364 \)

4. Calculate Volume of Soil to be Excavated (\( V_1 \))

Since the volume of solids (\( V_s \)) remains constant: \( V_s = \frac{V_2}{1 + e_2} = \frac{1000}{1 + 0.6364} = 611.1 \, \text{m³} \)

Volume of soil in borrow pit: \( V_1 = V_s (1 + e_1) = 611.1 \times (1 + 0.728) = 1056 \, \text{m³} \)

5. Calculate Amount of Water to be Added (in m³)

Mass of dry soil (\( M_d \)): \( M_d = V_s \times G \times \gamma_w = 611.1 \times 2.7 \times 1 = 1649.97 \, \text{t} \)

Mass of water in borrow pit (\( M_{w1} \)): \( M_{w1} = M_d \times w_1 = 1649.97 \times 0.12 = 197.996 \, \text{t} \)

Mass of water in fill (\( M_{w2} \)): \( M_{w2} = M_d \times w_2 = 1649.97 \times 0.18 = 296.995 \, \text{t} \)

Water to be added (volume): \( \Delta V_w = \frac{M_{w2} – M_{w1}}{\rho_w} = \frac{296.995 – 197.996}{1} = 99 \, \text{m³} \)

Results:

Volume of soil to be excavated (\( V_1 \)) = 1056 m³
Water to be added = 99 m³
Void ratio in borrow pit (\( e_1 \)) = 0.728
Void ratio in fill (\( e_2 \)) = 0.6364

Explanation

Key Steps:

The void ratio in the borrow pit (\( e_1 \)) is calculated using the given density, water content, and specific gravity.
The void ratio in the fill (\( e_2 \)) is calculated using the dry density and specific gravity.
The volume of soil to be excavated is determined by maintaining the volume of solids constant.
The amount of water to be added is calculated by comparing the water content in the borrow pit and the fill, converting mass to volume (1 t = 1 m³).

Physical Meaning

1. Void Ratio:

The void ratio indicates the volume of voids relative to the volume of solids in the soil.
A higher void ratio means the soil is looser, while a lower void ratio indicates denser soil.

2. Water Addition (99 m³):

Water is added to achieve the desired compaction. The increase in moisture lubricates soil particles, allowing them to rearrange into a denser configuration.

3. Importance in Engineering:

Proper compaction reduces voids, improving soil strength and stability for foundations, roads, and embankments.
Water content adjustments ensure optimal compaction efficiency.

Exam Context: Tests understanding of soil phase relationships, compaction principles, and unit conversions.