A sandy soil has a natural water content of 12% and bulk unit weight of 18.8 kN/m3 . The void ratios corresponding to the densest state (emin) and loosest state (emax) of this soil are 0.48 and 0.88, respectively. Find the relative density and degree of saturation for this soil.

The relative density = 0.83 which is equivalent to 83%

The degree of saturation, S = 0.58 which gives 58% saturation

Explanation:

The parameters given are;

Water content W% = 12%

Bulk unit weight, γ = 18.8 kN/m³

Void ratio of    = 0.48

Void ratio of   = 0.88

= Constant (As learnt from an answer to the question on the current page) = 2.65 for Sandy soil

Where, V = 1 m³

W = 18.8 KN

Bulk unit weight, γ = × (1 + W)

∴ 18.8 =   × (1 + 0.12)

= 18.8/ (1.12) = 16.79 kN/m³

= 16.79 kN

∴ = 18.8 kN - 16.79 kN = 2.01 kN

Volume of water = 0.205 m³

e + 1 = 0.58×GS = 0.58×2.65 =

e = 1.54 - 1 = 0.55

The relative density is given by the relation;

The relative density = 0.83

The relative density in percentage = 0.83×100 = 83%

S·e = GS×w = 0.12·2.65

S×0.55 = 0.318

The degree of saturation, S = 0.58

The degree of saturation, S in percentage = 58%.

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