Respuesta :
The following are the values:-
(a)Water content (W%) = 13.99%
(b) Total unit weight (Υ[tex]_{t}[/tex]) = 1778 kg/m³
(c) Dry unit weight (Y[tex]_{d}[/tex]) = 1559.75 kg/m³
(d) Void ratio (e) = 0.718
(e) Porosity (n) = 0.417
Volume of mold (V[tex]_{t}[/tex]) = 0.4 m³
Dry mass of sand (W[tex]_{s}[/tex]) = 623.9 kg
Wet mass of sand (W[tex]_{t}[/tex]) = 711.2 kg
Specific Gravity (G[tex]_{s}[/tex]) = 2.68
1) Water content (W%)
Water content (W%) = [tex]\frac{W_{w} }{W_{s} }[/tex] × [tex]100[/tex]
Water content (W%) = [tex]\frac{711.2 - 623.9}{623.9}[/tex] × [tex]100[/tex]
Water content (W%) = 13.99%
2) Total unit weight (Υ[tex]_{t}[/tex])
Total unit weight (Υ[tex]_{t}[/tex]) = [tex]\frac{W_{t} }{V_{t} }=\frac{711.2}{0.4}[/tex]
Total unit weight (Υ[tex]_{t}[/tex]) = 1778 kg/m³
3) Dry unit weight (Y[tex]_{d}[/tex])
Dry unit weight (Y[tex]_{d}[/tex]) = [tex]\frac{W_{s} }{V_{t} }=\frac{623.9}{0.4}[/tex]
Dry unit weight (Y[tex]_{d}[/tex]) = 1559.75 kg/m³
4) Void ratio(e)
Solid unit weight (Y[tex]_{s}[/tex]) = [tex]\frac{W_{s} }{V_{s} }[/tex]
Or
G[tex]_{s}[/tex] = [tex]\frac{Y_{s} }{Y_{w} }[/tex] ⇒ Y[tex]_{s}[/tex] = G[tex]_{s}[/tex]Y[tex]_{w}[/tex]
Y[tex]_{s}[/tex] = 2.68 × [tex]\frac{10^{-3} }{(10^{-2})^{3} }\frac{kg}{m}[/tex] = 2680 kg/m³
Now,
Void Ratio (e) = [tex]\frac{Y_{s} }{Yd}-1[/tex]
Void Ratio (e) = [tex]\frac{2680}{1559.72}-1[/tex]
Void Ratio (e) = 0.718
5) Porosity (n)
Porosity (n) = [tex]\frac{e}{1+e}[/tex]
Porosity (n) = [tex]\frac{0.718}{1+0.718}[/tex]
Porosity (n) = 0.417
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A laboratory sample of sand is formed inside a mold of 0.4 cubic meters. It took 711.2 kg of wet sand (dry mass of sand = 623.9 kg) to fill the mold. Assuming the specific gravity of the solid is 2.68, compute the:
(a)Water content (W%)
(b) Total unit weight (Υ[tex]_{t}[/tex])
(c) Dry unit weight (Y[tex]_{d}[/tex])
(d) Void ratio (e)
(e) Porosity (n)
