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Pycnometer data, water content, γsub

Short answer

Water content=20% (from γ/γd−1). Void ratio e=0.733 yields degree of saturation S=wG/e=72.3%. Submerged unit weight=γsat−γw=19.15−9.81=9.34 kN/m³.

From the pycnometer test, the specific gravity of soil solid of the soil specimen is found to be 2.65. Also, the dry unit weight of this soil specimen is found to be 15 kN/m³. If one cubic meter of this soil specimen weighs 18 kN/m³, determine (i) water content, (ii) degree of saturation, and (iii) submerged unit weight of this soil specimen.

Given

G=2.65G = 2.65, γd=15\gamma_d = 15 kN/m³, γ=18\gamma = 18 kN/m³, γw=9.81\gamma_w = 9.81 kN/m³.

(i) Water content

w=γγd−1=1815−1=0.20=20%w = \frac{\gamma}{\gamma_d} - 1 = \frac{18}{15} - 1 = 0.20 = 20\%

(ii) Degree of saturation

Void ratio:

e=Gγwγd−1=2.65×9.8115−1=0.733e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.65 \times 9.81}{15} - 1 = 0.733 S=wGe=0.20×2.650.733=0.723=72.3%S = \frac{wG}{e} = \frac{0.20 \times 2.65}{0.733} = 0.723 = 72.3\%

(iii) Submerged unit weight

Imagine the soil saturated at the same void ratio:

γsat=(G+e)γw1+e=(2.65+0.733)×9.811.733=19.15 kN/m3\gamma_{sat} = \frac{(G+e)\gamma_w}{1+e} = \frac{(2.65+0.733) \times 9.81}{1.733} = 19.15\ \text{kN/m}^3 γ′=γsat−γw=19.15−9.81=9.34 kN/m3\gamma' = \gamma_{sat} - \gamma_w = 19.15 - 9.81 = 9.34\ \text{kN/m}^3

Answer: w=20%w = 20\%, S=72.3%S = 72.3\%, γ′=9.34\gamma' = 9.34 kN/m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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