Respuesta :
Answer:
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 is 2.65.
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?
The answers to the question are
a. 15.37 kN/m³
b. 5.457 ft
Step-by-step explanation:
Relative density = [tex]\frac{e_{max}-e }{e_{max}-e_{min} }[/tex]
Where [tex]e_{max}[/tex] = maximum void ratio
[tex]e_{min}[/tex] = minimum void ratio
e = void ratio property
0.4 = (0.9 - e)/(0.9 - 0.46)
e = 0.9 - 0.4×0.44
= 0.724
Dry unit weight is given by
[tex]\gamma_{d} = \gamma_{w}\frac{G_{s} }{1+e}[/tex]
= 10×(2.65)/(1+0.724)
15.37 kN/m³
b. Relative density = [tex]\frac{e_{max}-e }{e_{max}-e_{min} }[/tex]
0.75 = (0.9 - e')/(0.9- 0.46)
or e' = 0.57
Hence the dry unit weight is given by
New dry unit weight wt =[tex]\gamma_{d} = \gamma_{w}\frac{G_{s} }{1+e'}[/tex]
= 10×2.65÷(1+0.57) = 16.88 kN/³m
[tex]\frac{\gamma_{1} }{\gamma_{2} } = (\frac{\frac{1}{V_{1} } }{\frac{1}{V_{2} } } ) (\gamma = unit weight)[/tex]
15.36/16.9 = V₂/V₁
0.9095 = h₂/h₁
h₂ = 0.9095 ×6 = 5.457 ft
