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

Moist unit weight = 19.192 kN/m³

dry unit weight = 17.29 kN/m³

degree of saturation = 53.6 %

Moisture content = 20.52 %

Step-by-step 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 =
`(Gs*rw(1+ \omega ))/(1+e)` .............1

put here value

Moist unit weight =
`(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 =
`(Gs*rw)/(1+e)` ...........2

put here value

dry unit weight =
`(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 =
`(\omega Gs)/(e)` ..............3

put here value

degree of saturation =
`(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 =
`(S*e)/(Gs)` .............4

put here value

Moisture content =
`(1*0.55)/(2.68)`

Moisture content = 20.52 %