A soil sample with specific gravity of solids 3 has a mass specific gravity of 2. Assuming the soil to be perfectly dry, determine the void ratio.

Concept:

From a fundamental relationship, the density of soil is given by:

\({{\rm{\gamma }}_{{\rm{bulk}}}} = \frac{{\left( {{\rm{G}} + S × e} \right) × {\gamma _w}}}{{1 + e}} = \frac{{\left( {{\rm{G}} + w × G} \right) × {\gamma _w}}}{{1 + e}} = \frac{{\left( {1 + {\rm{w}}} \right) × G × {\gamma _w}}}{{1 + e}}\)

Where

\(\gamma\) = Density of soil, \(\gamma_w\) = Density of water, G = Specific gravity of soil, S = Degree of saturation, w = Water content, and e = void ratio

For the dry density of soil: Degree of saturation (S) = 0

The dry density of soil(\(\gamma_d\)) is given by

\({{\rm{\gamma }}_{\rm{d}}} = \frac{{\left( {\rm{G}} \right) × {\gamma _w}}}{{1 + e}}\)

Calculation:

Given data

The specific gravity of solids = 3

The mass-specific gravity = 2

The dry density of soil(\(\gamma_d\)) is given by

\({{\rm{\gamma }}_{\rm{d}}} = \frac{{\left( {\rm{G}} \right) × {\gamma _w}}}{{1 + e}}\)

\({\gamma_d \over \gamma_w} = {G \over 1+e}\)

\(G_m = {G \over 1+e}\)

Where G_m = Mass specific gravity

\(2 = \frac{{\left( {\rm{3}} \right)}}{{1 + e}}\)

\({{1 + e}} = 1.5\)

e = 0.5

The void ratio is 0.5.