Formula
e = Gs·γw·(1 + w) / γt − 1 n = e / (1 + e) S = w·Gs / e (S·e = w·Gs) γd = γt / (1 + w) = Gs·γw / (1 + e) γsat = (Gs + e)·γw / (1 + e) γ′ = γsat − γw
Water content w goes into the formulas as a decimal (20 % → 0.20). S·e = w·Gs is the relationship that comes up most often in exam problems: knowing any two of the three gives you the third.
Checking your inputs
- A degree of saturation above 100 % means the measured values do not agree. Recheck the water content or unit weight measurement.
- A void ratio of 0 or less means the bulk unit weight entered is unrealistically high.
- Void ratios are around 0.4 for dense sand and well above 1 for soft clay.
Worked example
Gs 2.65, water content 20 %, bulk unit weight 19 kN/m³, γw 9.81 kN/m³:
- e = 2.65 × 9.81 × 1.20 / 19 − 1 ≈ 0.642
- n = 0.642 / 1.642 ≈ 39.1 %
- S = 0.20 × 2.65 / 0.642 ≈ 82.6 %
- γd = 19 / 1.20 ≈ 15.83 kN/m³, γsat ≈ 19.67 kN/m³, γ′ ≈ 9.86 kN/m³
The calculator opens with these values.
References
- Das, B. M., Principles of Geotechnical Engineering, chapter on weight–volume relationships
- ASTM D854, Specific gravity of soil solids by water pycnometer; ASTM D2216, Laboratory determination of water content of soil (how the inputs are measured)