Respuesta :
Answer:
(A) Dry unit weight. = 15.71 kN/m³ =100.91 pcf
(B) Porosity = 40.82 %
(C) Degree of saturation = 66.52 %
(D) Weight of water, in pounds per cubic foot, to be added to reach full saturation = 81.22 pcf
Explanation:
(A) γd =[tex]\frac{\gamma }{1+w}[/tex] = [tex]\frac{18.379}{1+.17} = 15.71[/tex] KN/m³
(B) [tex]\gamma _d =\frac{G_{s}* \gamma _w }{1+e}[/tex] therefore [tex]1+e = \frac{G_{s}* \gamma _w }{\gamma _d }[/tex] [tex]\frac{2.7*9.81}{15.71} =[/tex] 1.69
Therefore e = 0.69 and
Porososity n = [tex]\frac{e}{1+e}[/tex] =[tex]\frac{0.69}{1+0.69}[/tex] × 100% = 40.82 %
(C) [tex]S_{e} = w*G_{s}[/tex] therefore S =[tex]\frac{w*G_{s}}{e}[/tex] = [tex]\frac{0.17*9.81}{0.69}[/tex]×100 = 66.52 %
(D) [tex]\gamma _{sat} =\frac{(G_{s}+e) \gamma_{w} }{1+e}[/tex] = [tex]\frac{(2.7+0.69)9.81}{1+0.69}[/tex] = 19.68 kN/m³
The required amount of water is found from γ =ρ×g
ρ mass of water = [tex]\frac{\gamma}{g} = \frac{(19.68-18.38)\frac{kN}{m^{3} } }{9.81 kg\frac{m}{s^{2} } } *\frac{1000 N}{kN} *\frac{9.81 kg\frac{m}{s^{2} } }{N} = 1301 \frac{kg}{m^{3} }[/tex]
1 kg/m³ = 0.062 pcf therefore 1301 kg/m³ = 1301 ×0.062 pcf or 81.22 pcf
