A moist soil sample is having a weight 3 52 . N .After oven drying its weight is reduced to 2.9 N. The specific gravity of solids and mass specific gravity are 2.65 and 1.85. resp. Determine water content, void ratio, porosity and degree of saturation. Unit weight of water is 10 kN/cu m.

(I) The mass of chunk of moist soil is 20 kg and its volume is 0.011$m^3$ . After drying in an oven , the mass reduces to 16.5kg.

Determine the water content, the density of moist soil,dry density , void ratio, porosity and degrees of saturation.

Solution Take G = 2.70

Mass of water = 20 - 16.5 = 3.50

(1) Water content (Moisture content) = $\dfrac{M_w}{M_s} = \dfrac{3.50}{16.50} = 0.212(21.21\% )$

(2) Water mass density $\rho = \dfrac{M}{V} = \dfrac{20}{0.011} = 1818.18 kg/m^3$

(3) Dry density $\rho_d = \dfrac{M_s}{V} = \dfrac{16.5}{0.011} = 1500kg/m^3$

(4) Void ratio = $\rho_d = \dfrac{G\times \rho_w}{1+e} = 1500 = \dfrac{2.70\times 1000}{1+e}$

$e = 0.8$

(5) Porosity $\eta = \dfrac{e}{1+e} = \dfrac{0.8}{1+0.8} = 0.44(44.44\%)$

(6) $e = \dfrac{WG}{S}$

$0.8 = \dfrac{0.212\times 2.70}{S}$

$S = 0.7155 (71.58 \%)$

(II) A soil speciman hs a water content of 10%. and a wet unit weight of 20kN/$m^3$ . If the specific gravity of solids is 2.70. Determine dry unit weight, void ratio and degree of saturation. Take $\gamma_w = 10kg/m^3$

Solution $\gamma_d = \dfrac{\gamma}{1+w} = \dfrac{20}{1+0.1} = 18.18kN/m^3$

$1+e = \dfrac{G\gamma_w}{\gamma_d}$

$1+e = \frac{2.70\times 10}{18.18} = 1.49$

or $ e = 0.49$

$e = \dfrac{WG}{S}$

$0.49 = \dfrac{0.1\times 2.70}{S}$

$S = \dfrac{0.1\times 2.70}{0.49} = 0.551 (55.1\% )$

(III) A sample of dry soil weights 68gm. Find the volume of voids if the total volume of sample is 40ml and the specific gravity of solid is 2.65. Also determine void ratio.

$\rho_d = \dfrac{M_s}{V} = \dfrac{68}{40} = 1.70gm/ml$

Volume of solids = $\dfrac{M_s}{G\rho_w} = \dfrac{68}{2.65\times 1} = 25.66m$

$V_v = V-V_s$

$= 40 - 25.66 = 14.34ml$

$e = \dfrac{V_v}{V_s} = 0.56$