# If Yd = dry density, Y = bulk density and w = water content, then the correct relationship of these will be

20 views
in General
closed
If Yd = dry density, Y = bulk density and w = water content, then the correct relationship of these will be
1. $Y = \dfrac{Y_d}{1-w}$
2. $Y_d = \dfrac{Y}{1+w}$
3. $Y_d = \dfrac{Y}{1-w}$
4. $Y = \dfrac{Yd}{1 + w}$

by (54.3k points)
selected

Correct Answer - Option 2 : $Y_d = \dfrac{Y}{1+w}$

Concept:

Water Content ( W ):

It is defined as the ratio of the weight of water to the weight of solid in a given soil mass. It is usually expressed as a percentage.

$w = \;\frac{{{W_W}}}{{{W_S}}} \times 100 = \;\frac{{W - {W_S}}}{{{W_S}}} \times 100$

Bulk density:

It is defined as the ratio of the total mass ( M ) of the soil to the total volume ( V ) of the soil. It is expressed in Kg / m3

$\rho = \;\frac{M}{V}$

Dry density:

It is defined as the ratio of the mass of Solids to the total Volume of the soil ( in moist condition ).

${\rho _d} = \;\frac{{{M_S}}}{V}$

Relation between dry unit weight, bulk unit weight, and water content ( w ).

We know that,

$w = \frac{{{W_W}}}{{{W_S}}}$

$w + 1 = \;\frac{{{W_W}}}{{{W_S} + 1}}$ =$\frac{{{W_{W + {W_S}}}}}{W}$

$w + 1 = \;\frac{W}{{{W_S}}}$

${W_S} = \;\frac{W}{{1 + w}}$

Dividing both sides by V, we get

$\frac{W_s}{V} = \frac{W}{V(1+w)}$

${\gamma _{d\;}} = \;\frac{\gamma }{{1 + w}}\;$