Problem Statement (GATE Exam 1994)
A soil sample has the following properties:
- Total unit weight (\( \gamma_t \)) = 18.8 kN/m³
- Specific gravity (\( G \)) = 2.67
- Water content (\( w \)) = 12%
Calculate:
- Dry unit weight (\( \gamma_d \))
- Void ratio (\( e \))
- Degree of saturation (\( S \))
Solution
1. Dry Unit Weight Calculation
\( \gamma_d = \frac{\gamma_t}{1 + w} = \frac{18.8}{1 + 0.12} \approx 16.786 \, \text{kN/m}^3 \)
2. Void Ratio Determination
\( e = \frac{G \cdot \gamma_w}{\gamma_d} – 1 = \frac{2.67 \times 9.81}{16.786} – 1 \approx 0.56 \)
3. Degree of Saturation
\( S = \frac{wG}{e} = \frac{0.12 \times 2.67}{0.56} \approx 57.21\% \)
Results:
- Dry unit weight: \( \gamma_d \approx 16.79 \, \text{kN/m}^3 \)
- Void ratio: \( e \approx 0.56 \)
- Degree of saturation: \( S \approx 57.2\% \)
Explanation
Key Formulas:
- Dry unit weight: Removes water weight from total unit weight
- Void ratio: Relates pore space to solid particles
- Degree of saturation: Quantifies water-filled void fraction
Physical Meaning
1. Dry Unit Weight (16.79 kN/m³):
- Indicates dense soil structure with low porosity
- Typical for well-compacted cohesive soils
2. Void Ratio (0.56):
- Moderate porosity (36% of total volume is voids)
- Balances permeability and compressibility
3. 57.2% Saturation:
- 42.8% air voids allow for further compaction
- Partial saturation explains why \( \gamma_d < \gamma_{\text{sat}} \)
Engineering Implications:
- Suitable for foundations with moderate drainage
- Potential for settlement if water content increases
- Demonstrates phase relationships in soil mechanics
