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31 August, 23:16

A certain well-graded sand deposit has an in-situ relative density of about 50%. A laboratory strength test on a sample of this soil produced an effective friction angle of 31. Does this test result seem reasonable? Explain the basis for your response.

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  1. 31 August, 23:24
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    The result seem reasonable.

    Explanation:

    Relative density shows us how heavy a substance is in comparison to water. It aids us in determine the density of an unknown substance by knowing the density of a known substance. Relative density is defined as the ratio of the density (mass of a unit volume) of a substance to the density of a given reference material. It is very important in the accurate determination of density.

    Relative density = emax - e/emax - emin * 100

    Relative density ranges between 35 and 65 because the soil is in medium state.

    For well graveled sand, maximum friction angle is uo^r

    and for minimum friction angle, minimum friction angle is 33°.

    Therefore, for given friction angle of 31° (shear strength criteria). This result seem reasonable.

    Therefore, it is worthy to note that if friction angle is more, the shear strength is more which implies that it is a densified soil, that is to say, void ratio decreases.
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