A soil sample is partially saturated. Its natural moisture content was found to be 22%and bulk density 2 g/cm3. If the specific gravity of solid particles is 2.65 and the density of water be taken as 1 g/cm3, find out the degree of saturation and the void ratio.
Problem Statement (Civil Services Exam 1984)
A partially saturated soil sample has:
- Natural moisture content (\( w \)) = 22%
- Bulk density (\( \rho \)) = 2 g/cm³
- Specific gravity of solids (\( G \)) = 2.65
- Density of water (\( \rho_w \)) = 1 g/cm³
Determine the degree of saturation (\( S \)) and void ratio (\( e \)).
Solution
1. Calculate Void Ratio (\( e \))
Using the bulk density formula:
\( \rho = \frac{G(1 + w)\rho_w}{1 + e} \)
Rearranged to solve for \( e \):
\( e = \frac{G(1 + w)\rho_w}{\rho} – 1 \)
Substitute values:
\( e = \frac{2.65 \times (1 + 0.22) \times 1}{2} – 1 = \frac{3.233}{2} – 1 = 0.6165 \)
2. Calculate Degree of Saturation (\( S \))
\( S = \frac{wG}{e} = \frac{0.22 \times 2.65}{0.6165} \approx 0.9457 \, \text{(or 94.57%)} \)
Results:
- Void ratio: \( e = 0.6165 \)
- Degree of saturation: \( S \approx 94.57\% \)
Explanation
Key Formulas:
- Bulk Density: Relates moisture content, specific gravity, and void ratio.
- Degree of Saturation: Derived from the relationship \( S = \frac{wG}{e} \).
Steps:
- The void ratio (\( e \)) is calculated by rearranging the bulk density equation to isolate \( e \).
- The degree of saturation (\( S \)) is computed using the void ratio and the given moisture content.
Physical Meaning
1. Void Ratio (0.6165):
- Indicates moderate packing density. Soils with \( e < 1 \) are considered medium-dense.
2. Degree of Saturation (94.57%):
- Nearly saturated soil (close to 100%), meaning most voids are filled with water.
- Affects permeability and shear strength, critical for slope stability and foundation design.
Exam Context: Tests understanding of phase relationships in partially saturated soils, essential for geotechnical analysis in civil engineering projects.
