Question

1.The moist unit weights and degrees of saturation of a soil are given: moist unit weight (1) = 16.62 kN/m^3, degree of saturation (1) = 50%; moist unit weight (2) = 17.71 kN/m^3, degree of saturation (2) = 75%. Determine a) e (void ratio) and b) Gs (specific gravity)

Answer 1

Answer:

Gs = 2.647

e = 0.7986

Explanation:

We know that moist unit weight of soil is given as

$\gamma_m \ or\ bulk\ density = \frac{(Gs+Se)\times \gamma_w}{(1+e)}$

where,  $\gamma_m$ = moist unit weight of the soil

Gs = specific gravity of the soil

S = degree of saturation

e = void ratio

$\gamma_w$ = unit weight of water = 9.81 kN/m3

From data given we know that:

At 50% saturation,$\gamma_m = 16.62 kN/m3$

puttng all value to get Gs value;

$16.62= \frac{(Gs+0.5*e)\timees 9.81}{(1+e)}$

Gs - 1.194*e = 1.694 .........(1)

for saturaion 75%, unit weight = 17.71 KN/m3

$17.71 = \frac{(Gs+0.75*e)\times 9.81}{(1+e)}$

Gs - 1.055*e = 1.805 .........(2)

solving both  equations (1) and (2), we obtained;

Gs = 2.647

e = 0.7986