Question

The void ratio of an undisturbed soil sample is 0.55 and the moisture content is 11%. If Gs 5 2.68, determine: a. Moist unit weight b. Dry unit weight c. Degree of saturation d. Moisture content when the sample is fully saturated

Answer 1

Answer:

Moist unit weight = 19.192 kN/m³

dry unit weight = 17.29 kN/m³

degree of saturation = 53.6 %

Moisture content = 20.52 %

Explanation:

given data

void ratio e = 0.55

moisture content ω = 11%

Gs =  2.68

unit weight of water rw = 10 kN/m³

solution

we get here first Moist unit weigh that is

Moist unit weight = $\frac{Gs*rw(1+ \omega )}{1+e}$      .............1

put here value

Moist unit weight = $\frac{2.68*10(1+0.11)}{1+0.55}$  

Moist unit weight = 19.192 kN/m³

and

now we get dry unit weight that is

dry unit weight = $\frac{Gs*rw}{1+e}$   ...........2

put here value

dry unit weight = $\frac{2.68*10}{1+0.55}$  

dry unit weight = 17.29 kN/m³

and

now we get degree of saturation that is

degree of saturation = $\frac{\omega Gs}{e}$   ..............3

put here value

degree of saturation = $\frac{0.11*2.68}{0.55}$  

degree of saturation = 0.536 = 53.6 %

and

now we get Moisture content when the sample is fully saturated it mean S = 1

Moisture content  = $\frac{S*e}{Gs}$    .............4

put here value

Moisture content = $\frac{1*0.55}{2.68}$  

Moisture content = 20.52 %