Question

A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m3. The void ratios corresponding to the densest and loosest state of this soil are 0.51 and 0.87. Find the relative density and degree of saturation.

Answer 1

Relative density = 0.545

Degree of saturation = 24.77%

Step-by-step explanation:

Data provided in the question:

Water content, w = 5%

Bulk unit weight = 18.0 kN/m³

Void ratio in the densest state,
`e_(min)` = 0.51

Void ratio in the loosest state,
`e_(max)` = 0.87

Now,

Dry density,
`\gamma_d=(\gamma_t)/(1+w)`

`=(18)/(1+0.05)`

= 17.14 kN/m³

Also,

`\gamma_d=(G\gamma_w)/(1+e)`

here, G = Specific gravity = 2.7 for sand

`17.14=(2.7*9.81)/(1+e)`

or

e = 0.545

Relative density =
`(e_(max)-e)/(e_(max)-e_(min))`

=
`(0.87-0.545)/(0.87-0.51)`

= 0.902

Also,

Se = wG

here,

S is the degree of saturation

therefore,

S(0.545) = (0.05)()2.7

or

S = 0.2477

or

S = 0.2477 × 100% = 24.77%