Problem Statement
A saturated inorganic clay specimen has:
- Volume = 19.2 cm³
- Natural mass = 32.5 g
- Oven-dry mass = 20.9 g
Find in natural state:
- Water content (\( w \))
- Specific gravity (\( G \))
- Void ratio (\( e \))
- Saturated unit weight (\( \gamma_{\text{sat}} \))
- Dry unit weight (\( \gamma_d \))
Solution
1. Water Content (\( w \))
\( w = \frac{M_w}{M_s} \times 100 = \frac{32.5 – 20.9}{20.9} \times 100 = \frac{11.6}{20.9} \times 100 = 55.5\% \)
2. Specific Gravity (\( G \))
\( V_w = \frac{M_w}{\rho_w} = 11.6 \, \text{cm}^3 \)
\( V_s = V_{\text{total}} – V_w = 19.2 – 11.6 = 7.6 \, \text{cm}^3 \)
\( G = \frac{M_s}{V_s \cdot \rho_w} = \frac{20.9}{7.6 \times 1} = 2.75 \)
\( V_s = V_{\text{total}} – V_w = 19.2 – 11.6 = 7.6 \, \text{cm}^3 \)
\( G = \frac{M_s}{V_s \cdot \rho_w} = \frac{20.9}{7.6 \times 1} = 2.75 \)
3. Void Ratio (\( e \))
\( e = \frac{V_v}{V_s} = \frac{11.6}{7.6} = 1.53 \)
4. Saturated Unit Weight (\( \gamma_{\text{sat}} \))
\( \gamma_{\text{sat}} = \frac{M_{\text{total}}}{V_{\text{total}}} \cdot g = \frac{32.5}{19.2} \times 9.81 = 16.6 \, \text{kN/m}^3 \)
5. Dry Unit Weight (\( \gamma_d \))
\( \gamma_d = \frac{M_s}{V_{\text{total}}} \cdot g = \frac{20.9}{19.2} \times 9.81 = 10.7 \, \text{kN/m}^3 \)
Final Results:
- Water content = 55.5%
- Specific gravity = 2.75
- Void ratio = 1.53
- Saturated unit weight = 16.6 kN/m³
- Dry unit weight = 10.7 kN/m³
Explanation
- High water content indicates significant pore water in saturated clay
- Void ratio >1 reflects loose, compressible soil structure
- Unit weights show density relationship between saturated and dry states
Physical Meaning
1. High Void Ratio (1.53):
Indicates highly compressible soil requiring pre-construction consolidation.
2. Water Content (55.5%):
Typical for soft clays, affects permeability and shear strength.
3. Unit Weight Contrast:
\( \gamma_{\text{sat}} > \gamma_d \) shows significant water contribution to total weight.
