Question

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.

Answer 1

Answer:

The value of relative density is 75 % while that of degree of saturation is 54.82%.

Explanation:

Given Data:

Bulk density of Sandy Soil=$\gamma_b=18.8\ kN/m^3$

Void Ratio in Densest state=$e_{max}=0.88$

Void Ratio in Loosest state=$e_{min}=0.48$

Water content=$w=12\%$

To Find:

Relative Density=$D_R=\dfrac{e_{max}-e}{e_{max}-e_{min}} \times 100 \%$

Degree of Saturation=$S=\dfrac{w\times G_s}{e}$

Now all the other values are given except e. e is calculated as follows

e is termed as In situ void ratio and is given as

$e=\dfrac{\gamma_w \times G_s-\gamma_d}{\gamma_d}$

Here

γ_w is the density of water whose value is 1

G_s is the constant whose value is 2.65

γ_d is the dry density of the sandy soil which is calculated as follows:

$\gamma_d=\dfrac{\gamma_b}{1+\dfrac{w}{100}}$

Putting values

$\gamma_d=\dfrac{18.8}{1+\dfrac{12}{100}}\\\gamma_d=16.78\ kN/m^3=1.678 g/cc$

Putting this value in the equation of e gives

$e=\dfrac{1 \times 2.65-1.678}{1.678}\\e=0.579=0.58$

So the value of Relative density is given as

$D_R=\dfrac{e_{max}-e}{e_{max}-e_{min}} \times 100 \%\\D_R=\dfrac{0.88-0.58}{0.88-0.48} \times 100 \%\\D_R=75 \%$

So the value of relative density is 75 %

Now the value of degree of saturation is given as

$S=\dfrac{w\times G_s}{e}\\S=\dfrac{12\times 2.65}{0.58}\\S=54.82 \%$

The value of degree of saturation is 54.82%.

Answer 2

Answer:

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  $e_{min}$  = 0.48

Void ratio of  $e_{max}$ = 0.88

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

$\gamma =\dfrac{W}{V} = \dfrac{W_{w}+W_{s}}{V}$

Where, V = 1 m³

W = 18.8 KN

Bulk unit weight, γ = $\gamma_d$ × (1 + W)

∴ 18.8 =  $\gamma_d$ × (1 + 0.12)

$\gamma_d$ = 18.8/ (1.12) = 16.79 kN/m³

$\gamma_d =\dfrac{W_s}{V} = \dfrac{W_{s}}{1} = 16.79 \, kN/m^3$

$W_s$ = 16.79 kN

∴ $W_w$ = 18.8 kN - 16.79 kN = 2.01 kN

$m_w = 2.01/9.81 = 0.205 \, kg$

Volume of water = 0.205 m³

$\gamma = \dfrac{GS \times \gamma _{w}\times \left (1+w \right )}{1 + e} = \dfrac{GS \times 9.81\times \left (1+0.12 \right )}{1 + e} =18.8$

e + 1 = 0.58×GS = 0.58×2.65 =

e = 1.54 - 1 = 0.55

The relative density is given by the relation;

$Relative \ density, Dr=\dfrac{e_{max} - e}{e_{max} - e_{min}}$

$Relative \ density, Dr=\dfrac{0.88 - e}{0.88 - 0.48} = \dfrac{0.88 - 0.55}{0.4} = 0.83$

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%.