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.

Question

Engineering

Guest

25 March, 18:22

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.

+1

Answers (1)

Know the Answer?

Leroy Jefferson · Answer 1

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.