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Important Relationships

Proctor's com...

Question

Proctor's compaction test for the maximum dry density of a certain soil gave the results as: 1.77 gm/cc and OMC 14.44%. The specific gravity of the clay soil grain was 2.66. What was the saturation degree for this soil?

A

44%

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B

55%

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C

66%

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D

77%

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Solution

The correct option is D 77%
(d)
${(γ_{d})}_{max} = 1.77 g / c c$

OMC = 14.44%
G = 2.66

$γ_{d} = \frac{G . γ_{w}}{1 + e}$

$\Rightarrow 1.77 = \frac{2.66 \times 1}{1 + e}$

e = 0.5028

$\Rightarrow$ Se = wG

$S = \frac{w G}{e} = \frac{0.1444 \times 2.66}{0.5028}$

= 0.7639 = 76.39%

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