Question
Download Solution PDFA saturated soil sample has a dry unit weight of 18000 N/m3 and specific gravity 2.65. If unit weight of water is 9810 N/m3, determine the water content of the soil sample?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
\({{\rm{\gamma }}_{\rm{d}}} = \frac{{{\rm{G}}{{\rm{\gamma }}_{\rm{w}}}}}{{1 + {\rm{e}}}}{\rm{\;and\;e}} \times {{\rm{S}}_{\rm{r}}} = {\rm{w}} \times {\rm{G}}\)
Where,
𝝲d = Dry unit weight of soil, G = Specific gravity, 𝝲w = Unit weight of water, e = void ratio, and Sr = Degree of saturation
Calculation:
\({{\rm{\gamma }}_{\rm{d}}} = \frac{{{\rm{G}}{{\rm{\gamma }}_{\rm{w}}}}}{{1 + {\rm{e}}}}\)
∴ 18000 = (2.65 × 9810) / (1 + e)
e = 0.444
e × Sr = w × G
0.444 × 1 = w × 2.65
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