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hjlf
3 years ago
6

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 lo

osest state of this soil are 0.51 and 0.87. Find the relative density and degree of saturation.
Engineering
1 answer:
Zinaida [17]3 years ago
7 0

Answer:

Relative density = 0.545

Degree of saturation = 24.77%

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=\frac{\gamma_t}{1+w}

=\frac{18}{1+0.05}

= 17.14 kN/m³

Also,

\gamma_d=\frac{G\gamma_w}{1+e}

here, G = Specific gravity = 2.7 for sand

17.14=\frac{2.7\times9.81}{1+e}

or

e = 0.545

Relative density = \frac{e_{max}-e}{e_{max}-e_{min}}

= \frac{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%

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