Problem Statement (GATE Exam 2002)
A compacted embankment soil has the following properties:
- Bulk density (\( \rho \)) = 2.15 Mg/m³ = 2.15 g/cm³
- Water content (\( w \)) = 12%
- Specific gravity (\( G \)) = 2.65
Estimate:
- Dry density (\( \rho_d \))
- Void ratio (\( e \))
- Degree of saturation (\( S \))
- Air content (\( a_c \))
Solution
Given Data
- Bulk density: \( \rho = 2.15 \, \text{g/cm}^3 \)
- Water content: \( w = 12\% = 0.12 \)
- Specific gravity: \( G = 2.65 \)
- Water density: \( \rho_w = 1 \, \text{g/cm}^3 \)
(a) Dry Density (\( \rho_d \))
\( \rho_d = \frac{\rho}{1 + w} = \frac{2.15}{1 + 0.12} = 1.92 \, \text{g/cm}^3 \)
(b) Void Ratio (\( e \))
\( e = \frac{G \cdot \rho_w}{\rho_d} – 1 = \frac{2.65 \times 1}{1.92} – 1 = 0.38 \)
(c) Degree of Saturation (\( S \))
\( S = \frac{wG}{e} \times 100 = \frac{0.12 \times 2.65}{0.38} \times 100 = 83.68\% \)
(d) Air Content (\( a_c \))
\( a_c = (1 – S) \times 100 = (1 – 0.8368) \times 100 = 16.32\% \)
Final Results:
- Dry density = 1.92 g/cm³
- Void ratio = 0.38
- Degree of saturation = 83.68%
- Air content = 16.32%
Explanation
Fundamental Relationships:
- Dry density represents soil mass excluding water
- Void ratio compares void volume to solid volume
- Degree of saturation shows water-filled pore percentage
- Air content complements saturation in void analysis
Physical Meaning
1. High Dry Density (1.92 g/cm³):
Indicates effective compaction – crucial for embankment stability.
2. Low Void Ratio (0.38):
Suggests dense packing of soil particles, reducing settlement potential.
3. Partial Air Content (16.32%):
Shows significant air voids despite high saturation, affecting compressibility.
Engineering Implications:
These parameters help predict embankment performance under loads and water infiltration potential. The high saturation despite low water table suggests capillary water retention.
