A soil sample has water content 20%, specific gravity of solids 2.70 and void ratio 0.80. Find the degree of saturation, dry unit weight and bulk unit weight. Take γ_w = 9.81 kN/m³. What would the water content be at full saturation?
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Use Se = wG, then γ_d = Gγ_w/(1 + e).
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Worked answer.
Step 1: Degree of saturation. The basic relation between the four phase parameters is
S e = w G.
S = wG / e = (0.20 × 2.70) / 0.80 = 0.54 / 0.80 = 0.675, or 67.5%.
Step 2: Dry unit weight.
γ_d = G γ_w / (1 + e) = (2.70 × 9.81) / 1.80 = 26.487 / 1.80 ≈ 14.72 kN/m³.
Step 3: Bulk unit weight.
γ = γ_d (1 + w) = 14.715 × 1.20 ≈ 17.66 kN/m³.
Alternative formula: γ = (G + Se) γ_w / (1 + e) = (2.70 + 0.675 × 0.80) × 9.81 / 1.80 = (2.70 + 0.54) × 9.81 / 1.80 = 3.24 × 5.45 ≈ 17.66 kN/m³. Both methods agree.
Step 4: Water content at full saturation. With S = 1 and the same void ratio:
w_sat = e / G = 0.80 / 2.70 ≈ 29.6%.
The saturated unit weight would be γ_sat = (G + e) γ_w / (1 + e) = 3.50 × 9.81 / 1.80 ≈ 19.08 kN/m³.
Check. S must be between 0 and 1: 0.675 is valid. Also γ_d < γ < γ_sat (14.72 < 17.66 < 19.08), which is the expected order.
Other useful values
- Porosity n = e/(1 + e) = 0.80/1.80 ≈ 0.444 (44.4%).
- Submerged unit weight γ' = γ_sat − γ_w ≈ 19.08 − 9.81 = 9.27 kN/m³, used for effective stress below the water table.
Common mistake. Using w as 20 instead of 0.20 in Se = wG, or using porosity in place of void ratio in the unit weight formulas.
Likely follow-up. "Why does a lower dry unit weight matter in compaction?" Compaction quality is specified as a percentage of the maximum dry density from the Proctor test, so field dry density is compared against it.
Original UniCircle question. Last reviewed 28 Oct 2026.