Question

A given cohesionless soil has e_max = 0.85 and e_min = 0.50. In the field, the soil is compacted to a mass density of 1800 kg/m³ at a water content of 8%. Take the mass density of water as 1000 kg/m³ and Gs as 2.7. The relative density (in %) of the soil is
1. 56.43
2. 60.25
3. 62.87
4. 65.71

Answer 1

Correct Answer - Option 4 : 65.71

Concept:

Relative density of a soil sample having natural void ratio (eo) , maximum Void ratio(emax) and minimum void ratio (emin) is given by:

\({I_D} = \left( {\frac{{{e_{max}} - {e_o}}}{{{e_{max}} - {e_{min}}}}} \right) \times 100\)

Calculation:

Given,

emax = 0.85, emin = 0.50

eo = ?

Soil is compacted to density 1800 kg/m³ @ water content 8%

mass density of water = 1000 kg/m³, G_s = 2.7 

w = 8%, ρb = 1800 kg/m3, ρw = 1000 kg/m3 

we need to determine natural void ratio by the equation 

\({{\rm{\rho }}_{\rm{b}}} = \frac{{{\rm{G}} + {\rm{wG}}}}{{1 + {\rm{e}}}} \times {{{\rho }}_{\rm{w}}}\)

⇒ \(1800 = \frac{{2.7 + 0.08 \times 2.7}}{{1 + {\rm{e}}}} \times 1000\)

e = 0.62

Relative density, \(\left( {{{\rm{I}}_{\rm{D}}}} \right) = \frac{{{{\rm{e}}_{{\rm{max}}}} - {\rm{e}}}}{{{{\rm{e}}_{{\rm{max}}}} - {{\rm{e}}_{{\rm{min}}}}}} = {\rm{\;}}\frac{{0.85 - 0.62}}{{0.85 - 0.5}} =65.71\%\)

The relative density (in %) of the soil is 65.71%