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Compacted soil 17 kN/m³, OWC 16%

Short answer

From γ_d = 17 kN/m³, w = 16%, G = 2.65: void ratio e = (2.65×9.81/17) - 1 = 0.529; degree of saturation S = 0.16×2.65/0.529 = 80.1%; zero air void dry unit weight = 2.65×9.81/(1+0.16×2.65) = 18.26 kN/m³.

The maximum dry unit weight of a compacted soil mass is found to be 17 kN/m³ with optimum water content being 16%. Determine the void ratio and degree of saturation of this soil after compaction. 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 solids as 2.65.

Given

γd=17\gamma_{d} = 17 kN/m³, w=16%w = 16\% (OWC), G=2.65G = 2.65, γw=9.81\gamma_w = 9.81 kN/m³.

Void ratio

e=Gγwγd−1=2.65×9.8117−1=0.529e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.65 \times 9.81}{17} - 1 = 0.529

Degree of saturation

S=wGe=0.16×2.650.529=0.801=80.1%S = \frac{wG}{e} = \frac{0.16 \times 2.65}{0.529} = 0.801 = 80.1\%

Zero air void unit weight at w=16%w = 16\%

γd,zav=Gγw1+wG=2.65×9.811+0.16×2.65=25.991.424=18.26 kN/m3\gamma_{d,zav} = \frac{G\gamma_w}{1 + wG} = \frac{2.65 \times 9.81}{1 + 0.16 \times 2.65} = \frac{25.99}{1.424} = 18.26\ \text{kN/m}^3

Answer: e=0.529e = 0.529, S=80.1%S = 80.1\%, γd,zav=18.26\gamma_{d,zav} = 18.26 kN/m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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