Problem Statement (Civil Services Exam 1986)
A natural soil sample has:
- Bulk density (\( \rho \)) = 2 g/cm³
- Initial water content (\( w \)) = 6%
- Specific gravity of solids (\( G \)) = 2.67
Determine:
- Amount of water to add per m³ to raise water content to 15% (void ratio constant).
- Resulting degree of saturation (\( S \)).
Solution
1. Calculate Dry Density (\( \rho_d \))
\( \rho_d = \frac{\rho}{1 + w} = \frac{2}{1 + 0.06} \approx 1.887 \, \text{g/cm}^3 \)
2. Mass of Solids (\( M_d \)) in 1 m³
\( M_d = \rho_d \times V = 1.887 \, \text{g/cm}^3 \times 10^6 \, \text{cm}^3 = 1.887 \times 10^6 \, \text{g} \)
3. Initial and Final Water Mass
Initial \( M_w = 0.06 \times 1.887 \times 10^6 = 113,220 \, \text{g} \)
Final \( M_w = 0.15 \times 1.887 \times 10^6 = 283,050 \, \text{g} \)
Additional water required = \( 283,050 – 113,220 = 169,830 \, \text{g} = 169.83 \, \text{liters} \)
4. Void Ratio (\( e \))
\( e = \frac{G \rho_w}{\rho_d} – 1 = \frac{2.67 \times 1}{1.887} – 1 \approx 0.415 \)
5. Degree of Saturation (\( S \)) at 15% Water Content
\( S = \frac{wG}{e} = \frac{0.15 \times 2.67}{0.415} \approx 0.9652 \, \text{(or 96.52%)} \)
Results:
- Water to add: \( \approx 169.83 \, \text{liters/m}^3 \)
- Degree of saturation: \( S \approx 96.52\% \)
Explanation
Key Steps:
- Dry Density: Removes water’s effect to isolate solids’ contribution.
- Water Mass: Calculated as a percentage of solids’ mass at 6% and 15%.
- Void Ratio: Derived from dry density and specific gravity.
- Degree of Saturation: Increased water fills voids, raising \( S \).
Physical Meaning
1. Water Addition (169.83 liters/m³):
- Significant water required to increase saturation while maintaining void structure.
2. Degree of Saturation (96.52%):
- Near-full saturation, reducing air voids and increasing soil weight and stability.
Exam Context: Tests understanding of soil-water relationships for compaction and stability in civil engineering projects.
