Question

1). A loose, uncompacted sand fill 6 ft in depth has a relative density of 40%.
Laboratory tests indicated that the minimum and maximum void ratios of the sand are 0.46 and 0.90, respectively. The specific gravity of solids of the sand is2.65.

a). What is the dry unit weight of the sand?.
b). If the sand is compacted to a relative density of 75%, what is the decrease in thickness of the 6-ft fill?.

Answer 1

relative density = (e_max-e)/(e_max-e_min)

0.4=(0.9-e)/(0.9-0.46)

e=0.9- 0.44 X 0.4 = 0.724

dry unit wt. =γ_w(G_s-eS_r)/(1+e)

=10(2.65-0)/(1+0.724) =15.37 KN/m³

b)relative density = (e_max-e)/(e_max-e_min)

0.75=(0.9-e')/(0.9-0.46)

e'=0.9 - 0.44 X 0.75 = 0.57

new dry unit wt =γ_w(G_s-eS_r)/(1+e')

= 10(2.65)/(1+0.57)

= 16.88KN/m³

γ₁/γ₂ = (1/V₁)/(1/V₂)(γ is unit wt)

15.37/16.88=V2/V1

0.91 =h2/h1

h2=6 X 0.91 =5.46 ft