Question

A 588 cm³ volume of moist sand weighs 1010 gm. Its dry weight is 918 gm and specific gravity of solids, G is 2.67. Assuming density of water as 1 gm/cm³, the void ratio is ___________.

Answer 1

Concept:

According to the three-phase system of soil mass,

V = V_V + V_S­

Where,

V_V = Total volume of voids = V_a + V_w

V_a = Volume of air, V_w = Volume of water

V = Total volume of soil, V_S = Volume of soil solids

Unit weight of soil

It is defined as the ratio of the total weight of soil to the total volume of soil mass

\({{\rm{γ }}_{\rm{s}}} = \frac{{\rm{W}}}{{\rm{V}}}\)

The specific gravity of soil

It is defined as the ratio of unit weight of a given volume of solids to the unit weight of an equivalent volume of water.

\({\rm{G}} = \frac{{{{\rm{γ }}_{\rm{s}}}}}{{{{\rm{γ }}_{\rm{w}}}}}\)

Voids ratio:

It is defined as the ratio of total volume of voids to the volume of solids in given soil mass.

\({\rm{e}} = \frac{{{{\rm{V}}_{\rm{v}}}}}{{{{\rm{V}}_{\rm{s}}}{\rm{\;}}}}\)

e > 0, voids ratio has no upper limit

Calculation:

Given,

W = 1010 gm, Ws = 918 gm

V = 588 cm³, G_s = 2.67

 γ_w = 1000 kg/m³ = 1 gm/cm³ 

\(\begin{array}{l} {G_s} = \frac{{{W_s}}}{{{V_s}{Y_w}}}\\ 2.67 = \frac{{918}}{{{V_s} \times 1}} \end{array}\)

V_s = 343.82 cm³

\(\begin{array}{l} {V_v} = V - {V_s} = 588 - 343.82 = 244.18{m^3}\\ e = \frac{{{V_v}}}{{{V_s}}} = \frac{{244.18}}{{343.82}} = 0.71 \end{array}\)