Problem Statement
An undisturbed sample of soil has a volume of 100 cm3 and mass of 190 g. On oven drying for 24 hours, the mass is reduced to 160 g. If the specific gravity of grains is 2.68, determine the water content, voids ratio, and degree of saturation of the soil.
Solution
1. Water Content
Mass of water:
\( M_w = M_{\text{wet}} – M_{\text{dry}} = 190 – 160 = 30 \, \text{g} \)
Water content:
\( w = \frac{M_w}{M_d} = \frac{30}{160} = 0.188 = 18.8\% \)
2. Bulk and Dry Density
Bulk density:
\( \rho = \frac{M}{V} = \frac{190}{100} = 1.9 \, \text{g/cm}^3 \)
Dry density:
\( \rho_d = \frac{\rho}{1 + w} = \frac{1.9}{1 + 0.188} = 1.6 \, \text{g/cm}^3 \)
3. Voids Ratio
\( e = \frac{G \cdot \gamma_w}{\gamma_d} – 1 = \frac{2.68 \cdot 9.81}{15.69} – 1 = 0.67 \)
4. Degree of Saturation
\( S = \frac{w \cdot G}{e} = \frac{0.188 \cdot 2.68}{0.67} = 0.744 = 74.4\% \)
Results:
- Water Content: 18.8%
- Voids Ratio: 0.67
- Degree of Saturation: 74.4%
Explanation
- Water Content: The water content represents the amount of water present in the soil compared to the dry weight of the soil.
- Voids Ratio: The voids ratio measures the volume of voids (air and water) compared to the volume of solid particles in the soil.
- Degree of Saturation: The degree of saturation indicates the proportion of voids filled with water relative to the total void space.
Physical Meaning
- Engineering Applications: These parameters are critical for understanding soil behavior under load, ensuring stability in foundations, and designing drainage systems.
- Practical Implications: High water content and saturation can lead to reduced shear strength, while voids ratio affects compaction and permeability.
