Find the solid units weight in pcf and KN/m3, dry unit weight in pcf and kN/m3, Void Ratio, Porosity, Degree of Saturation
Silty-Sand Specific Gravity: 2.65
Poorly Graded Sand Specific Gravity: 2.66
Sandy Lean Clay Specific Gravity: 2.67
Poorly Graded Sand with Silt: 2.65
The Correct Answer and Explanation is:1
To calculate the solid unit weight, dry unit weight, void ratio, porosity, and degree of saturation, more information is typically needed such as:
- Moisture content (w)
- Bulk unit weight (γ)
- Dry unit weight (γ_d) or wet density (ρ)
- Water content (Saturation or % of water)
- Volume or mass relationships (e.g., weight of solids, volume of voids)
However, since you’ve only provided specific gravity values for different soil types, we can explain and calculate typical values assuming standard conditions.
📌 Assumptions:
We will perform calculations for Poorly Graded Sand (G_s = 2.66) under standard lab conditions assuming:
- Standard water unit weight, γ_w = 62.4 pcf = 9.81 kN/m³
- Dry density (ρ_d) ≈ 100 pcf = 1602 kg/m³ (typical for sands)
- Assume 80% degree of saturation (S = 0.8)
📊 Calculations:
- Solid Unit Weight (γ_s):
γs=Gs×γw=2.66×62.4=166.0 pcf=2.66×9.81=26.1 kN/m3γ_s = G_s \times γ_w = 2.66 \times 62.4 = 166.0 \, \text{pcf} = 2.66 \times 9.81 = 26.1 \, \text{kN/m}^3
- Dry Unit Weight (γ_d) (assumed typical for sand):
γd=100 pcf=15.7 kN/m3γ_d = 100 \, \text{pcf} = 15.7 \, \text{kN/m}^3
- Void Ratio (e):
e=(Gs×γwγd)−1=(2.66×62.4100)−1=1.66−1=0.66e = \left( \frac{G_s \times γ_w}{γ_d} \right) – 1 = \left( \frac{2.66 \times 62.4}{100} \right) – 1 = 1.66 – 1 = 0.66
- Porosity (n):
n=e1+e=0.661.66≈0.3976 or 39.8%n = \frac{e}{1+e} = \frac{0.66}{1.66} ≈ 0.3976 \text{ or } 39.8\%
- Degree of Saturation (S) (assumed 80% or computed if w is known):
S=w×Gse→S=0.8 (given/assumed)S = \frac{w \times G_s}{e} \rightarrow S = 0.8 \text{ (given/assumed)}
✅ Final Answers (for Poorly Graded Sand):
| Parameter | Value (pcf) | Value (kN/m³) |
|---|---|---|
| Solid Unit Weight | 166.0 pcf | 26.1 kN/m³ |
| Dry Unit Weight | 100.0 pcf | 15.7 kN/m³ |
| Void Ratio (e) | – | 0.66 |
| Porosity (n) | – | 39.8% |
| Degree of Saturation (S) | – | 80% |
📘 Explanation
In geotechnical engineering, understanding the physical properties of soil is crucial for predicting its behavior under load. One of the key parameters is the solid unit weight (γ_s), which represents the weight per unit volume of soil solids alone. It is calculated using the specific gravity (G_s) of soil particles and the unit weight of water (γ_w).
The dry unit weight (γ_d) refers to the weight per unit volume of the soil mass when completely dry, excluding water content. It’s a critical indicator of compaction and strength. In the case of poorly graded sand (G_s = 2.66), a dry unit weight of 100 pcf is considered typical.
The void ratio (e) expresses the ratio of the volume of voids (air and water) to the volume of solids. It’s a dimensionless quantity used to describe how tightly the soil particles are packed. Lower void ratios imply denser soils.
The porosity (n) is the ratio of void volume to total volume, showing how much of the soil volume is occupied by voids. It is directly related to the void ratio and provides similar information but in a more intuitive percentage form.
The degree of saturation (S) tells us what portion of the void space is filled with water. A fully saturated soil has S = 100%, while dry soil has S = 0%. Understanding S helps engineers evaluate drainage, compressibility, and shear strength of soils.
Collectively, these parameters are used to assess soil behavior for foundations, slopes, and earthworks, making their estimation and interpretation a cornerstone of soil mechanics.
