Question

A soil sample has a moist unit weight of 117 pcf, a moisture content of 17 percent, and soil particles with a specific gravity of 2.70. Answer the following questions about this soil.
(A) Dry unit weight.
(B) Porosity.
(C) Degree of saturation.
(D) Weight of water, in pounds per cubic foot, to be added to reach full saturation.

Answer 1

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 =$\frac{\gamma }{1+w}$ = $\frac{18.379}{1+.17} = 15.71$ KN/m³

(B) $\gamma _d =\frac{G_{s}* \gamma _w }{1+e}$ therefore  $1+e = \frac{G_{s}* \gamma _w }{\gamma _d }$ $\frac{2.7*9.81}{15.71} =$ 1.69

Therefore e = 0.69 and

Porososity n = $\frac{e}{1+e}$ =$\frac{0.69}{1+0.69}$ × 100% = 40.82 %

(C)  $S_{e} = w*G_{s}$ therefore S =$\frac{w*G_{s}}{e}$ = $\frac{0.17*9.81}{0.69}$×100 = 66.52 %

(D) $\gamma _{sat} =\frac{(G_{s}+e) \gamma_{w} }{1+e}$ = $\frac{(2.7+0.69)9.81}{1+0.69}$ = 19.68 kN/m³

The required amount of water is found from γ =ρ×g

ρ mass of water = $\frac{\gamma}{g} = \frac{(19.68-18.38)\frac{kN}{m^{3} } }{9.81 kg\frac{m}{s^{2} } } *\frac{1000 N}{kN} *\frac{9.81 kg\frac{m}{s^{2} } }{N} = 1301 \frac{kg}{m^{3} }$

1 kg/m³ = 0.062 pcf therefore 1301 kg/m³ = 1301 ×0.062 pcf or 81.22 pcf