Problem Statement
A Jodhpur-Mini-Compactor test provides:
- Wet soil mass = 612 g
- Volume = 300 cm³
- Moisture content (\( w \)) = 12.6%
- Specific gravity (\( G \)) = 2.68
Determine:
- Bulk unit weight (\( \gamma \))
- Dry unit weight (\( \gamma_d \))
- Void ratio (\( e \))
- Porosity (\( n \))
- Degree of saturation (\( S \))
Solution
1. Bulk Unit Weight (\( \gamma \))
\( \gamma = \frac{\text{Mass}}{\text{Volume}} \times g = \frac{612}{300} \times 9.81 = 20.01 \, \text{kN/m}^3 \)
2. Dry Unit Weight (\( \gamma_d \))
\( \gamma_d = \frac{\gamma}{1 + w} = \frac{20.01}{1.126} = 17.77 \, \text{kN/m}^3 \)
3. Void Ratio (\( e \))
\( e = \frac{G \cdot \gamma_w}{\gamma_d} – 1 = \frac{2.68 \times 9.81}{17.77} – 1 = 0.48 \)
4. Porosity (\( n \))
\( n = \frac{e}{1 + e} \times 100 = \frac{0.48}{1.48} \times 100 = 32.4\% \)
5. Degree of Saturation (\( S \))
\( S = \frac{wG}{e} \times 100 = \frac{0.126 \times 2.68}{0.48} \times 100 = 70.4\% \)
Final Results:
- Bulk unit weight = 20.01 kN/m³
- Dry unit weight = 17.77 kN/m³
- Void ratio = 0.48
- Porosity = 32.4%
- Degree of saturation = 70.4%
Explanation
- Bulk unit weight reflects total soil mass including water
- Void ratio < 0.5 indicates moderately dense compaction
- 70% saturation shows significant air voids despite compaction
Physical Meaning
1. Moderate Compaction:
Void ratio of 0.48 suggests balanced density for structural support.
2. Partial Saturation:
70% saturation indicates potential for further densification with water addition.
3. Engineering Significance:
These properties help predict settlement behavior and optimal compaction effort.
