A fully saturated clay sample has a mass of 101.5 g and a volume of 50 cm³. After oven drying, the clay has a mass of 84.5 g. Assuming that the volume does not change during drying, determine Specific gravity, Void ratio, Porosity, Dry unit weight.

Fully Saturated Clay Sample Analysis

Problem Statement

A fully saturated clay sample has a mass of 101.5 g and a volume of 50 cm³. After oven drying, the clay has a mass of 84.5 g. Assuming that the volume does not change during drying, determine:

Specific gravity (\( G \))
Void ratio (\( e \))
Porosity (\( n \))
Dry unit weight (\( \gamma_d \))

Solution

1. Calculate Water Content (\( w \))

\( w = \frac{M_w}{M_s} \times 100 = \frac{101.5 – 84.5}{84.5} \times 100 = \frac{17}{84.5} \times 100 = 20.12\% \)

2. Calculate Specific Gravity (\( G \))

For saturated soil, \( S = 1 \):
\( e = wG \)
From the volume relationship:
\( V_s = \frac{M_s}{G \cdot \rho_w} \)
\( V_v = V_{\text{total}} – V_s = 50 – \frac{84.5}{G \cdot 1} \)
Since \( e = \frac{V_v}{V_s} \), substitute \( e = wG \):
\( wG = \frac{50 – \frac{84.5}{G}}{\frac{84.5}{G}} \)
Solving for \( G \):
\( G = 2.55 \)

3. Calculate Void Ratio (\( e \))

\( e = wG = 0.2012 \times 2.55 = 0.51 \)

4. Calculate Porosity (\( n \))

\( n = \frac{e}{1 + e} \times 100 = \frac{0.51}{1.51} \times 100 = 33.8\% \)

5. Calculate Dry Unit Weight (\( \gamma_d \))

\( \gamma_d = \frac{M_s}{V_{\text{total}}} \cdot g = \frac{84.5}{50} \times 9.81 = 16.58 \, \text{kN/m}^3 \)

Final Results:

Specific gravity (\( G \)) = 2.55
Void ratio (\( e \)) = 0.51
Porosity (\( n \)) = 33.8%
Dry unit weight (\( \gamma_d \)) = 16.58 kN/m³

Explanation

1. Water Content (\( w \)):
The water content of 20.12% indicates the proportion of water in the soil relative to the solid mass. This value is typical for saturated clays.

2. Specific Gravity (\( G \)):
The specific gravity of 2.55 is calculated using the relationship between water content, void ratio, and volume. This value is typical for clay soils.

3. Void Ratio (\( e \)):
A void ratio of 0.51 indicates a moderately dense soil structure. This value is calculated using the relationship \( e = wG \).

4. Porosity (\( n \)):
The porosity of 33.8% represents the proportion of the total volume occupied by voids. This value is derived from the void ratio.

5. Dry Unit Weight (\( \gamma_d \)):
The dry unit weight of 16.58 kN/m³ represents the weight of soil solids per unit volume, excluding water. This value is important for assessing the soil’s load-bearing capacity.

Physical Meaning

1. Water Content and Saturation:
A water content of 20.12% indicates that the soil is fully saturated, meaning all voids are filled with water. This affects the soil’s compressibility and shear strength.

2. Specific Gravity and Soil Composition:
A specific gravity of 2.55 suggests that the soil solids are denser than water, which is typical for clay soils. This value helps in understanding the soil’s mineral composition.

3. Void Ratio and Soil Structure:
A void ratio of 0.51 indicates a moderately dense soil structure. This value affects the soil’s permeability and compressibility.

4. Porosity and Water Retention:
A porosity of 33.8% indicates that 33.8% of the soil volume is occupied by voids. This value is important for understanding the soil’s water retention and drainage characteristics.

5. Dry Unit Weight and Load-Bearing Capacity:
A dry unit weight of 16.58 kN/m³ suggests a moderately dense soil, which is suitable for many construction applications. This value is crucial for foundation design and stability analysis.