Problem Statement
A natural soil deposit has a bulk unit weight of 18.44 kN/m3 and water content of 5%. Calculate the amount of water required to be added to 1 cubic metre of soil to raise the water content to 15%. Assume the voids ratio to remain constant. What will then be the degree of saturation? Assume \( G = 2.67 \).
Solution
1. Dry Unit Weight
\( \gamma_d = \frac{\gamma}{1 + w} = \frac{18.44}{1 + 0.05} = 17.56 \, \text{kN/m}^3 \)
2. Mass of Water in the Soil
Mass of dry soil:
\( W_d = \gamma_d \cdot V = 17.56 \cdot 1 = 17.56 \, \text{kN} \)
Mass of water at \( w = 5\% \):
\( W_w = w \cdot W_d = 0.05 \cdot 17.56 = 0.878 \, \text{kN} \)
Volume of water:
\( V_w = \frac{W_w}{\gamma_w} = \frac{0.878}{9.81} = 0.0895 \, \text{m}^3 \)
3. Mass of Water at \( w = 15\% \)
Mass of water:
\( W_w = w \cdot W_d = 0.15 \cdot 17.56 = 2.634 \, \text{kN} \)
Volume of water:
\( V_w = \frac{W_w}{\gamma_w} = \frac{2.634}{9.81} = 0.2685 \, \text{m}^3 \)
Additional water required:
\( V_{\text{additional}} = 0.2685 – 0.0895 = 0.179 \, \text{m}^3 = 179 \, \text{litres} \)
4. Degree of Saturation
Voids ratio:
\( e = \frac{G \cdot \gamma_w}{\gamma_d} – 1 = \frac{2.67 \cdot 9.81}{17.56} – 1 = 0.49 \)
Degree of saturation:
\( S = \frac{w \cdot G}{e} = \frac{0.15 \cdot 2.67}{0.49} = 0.817 = 81.7\% \)
Results:
- Additional Water Required: 179 litres
- Degree of Saturation: 81.7%
Explanation
- Dry Unit Weight: Indicates the weight of soil solids per unit volume.
- Water Content: Increasing water content requires adding a specific volume of water to maintain the voids ratio constant.
- Degree of Saturation: Represents the fraction of void space filled with water.
Physical Meaning
- Field Application: Helps in determining the volume of water required for achieving desired soil moisture levels in construction projects.
- Engineering Relevance: Understanding saturation levels ensures proper soil behavior under load and prevents failures due to inadequate compaction.
