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Compacted soil 18 kN/m³, porosity and ZAV

Short answer

Given γ_d = 18 kN/m³, w = 15%, G = 2.7: void ratio e = 0.4715, porosity n = 0.4715/1.4715 = 32.0%, degree of saturation S = 85.9%. Zero air void dry unit weight = 18.85 kN/m³.

The maximum dry unit weight of a compacted soil mass is found to be 18 kN/m³ with optimum water content being 15%. Find the values of porosity and degree of saturation of this compacted soil. Also, find the value of the maximum dry unit weight on the zero air void line at that optimum water content. Take specific gravity of soil solid as 2.7.

Given

γd=18\gamma_{d} = 18 kN/m³, w=15%w = 15\%, G=2.7G = 2.7, γw=9.81\gamma_w = 9.81 kN/m³.

Void ratio and porosity

e=Gγwγd−1=2.7×9.8118−1=0.4715e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.7 \times 9.81}{18} - 1 = 0.4715 n=e1+e=0.47151.4715=0.320=32.0%n = \frac{e}{1+e} = \frac{0.4715}{1.4715} = 0.320 = 32.0\%

Degree of saturation

S=wGe=0.15×2.70.4715=0.859=85.9%S = \frac{wG}{e} = \frac{0.15 \times 2.7}{0.4715} = 0.859 = 85.9\%

Zero air void unit weight

γd,zav=Gγw1+wG=2.7×9.811+0.15×2.7=26.491.405=18.85 kN/m3\gamma_{d,zav} = \frac{G\gamma_w}{1 + wG} = \frac{2.7 \times 9.81}{1 + 0.15 \times 2.7} = \frac{26.49}{1.405} = 18.85\ \text{kN/m}^3

Answer: n=32.0%n = 32.0\%, S=85.9%S = 85.9\%, γd,zav=18.85\gamma_{d,zav} = 18.85 kN/m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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