Question

(a) A sample of dry sand is tested in direet shear. Under an applied normal stress of 4,800 paf, the sample falls when the shear stress reaches 3,100 pof. What is the angle of internal friction for this soil? (b) A second sample of the same sand material is also to be tested in direct shear, but the applied compressive loading will be 3,500 pot instead of 4,800 pot. What shear stress is expected to fail the sample?

   (a) A sample of dry sand is tested in direet shear. Under an applied normal stress of 4,800 paf, the sample falls when the shear stress reaches 3,100 pof. What is the angle of internal friction for this soil?
(b) A second sample of the same sand material is also to be tested in direct shear, but the applied compressive loading will be 3,500 pot instead of 4,800 pot. What shear stress is expected to fail the sample?

Essentials of soil mechanics and foundations : basic geotechnics

David F. McCarthy 7th Edition

Chapter 11, Problem 3

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Step 1

Given: Normal stress (σn) = 4,800 paf Shear stress (τ) = 3,100 pof The formula for calculating the angle of internal friction (φ) is: tan(φ) = τ/σn Substitute the given values: tan(φ) = 3,100/4,800 tan(φ) ≈ 0.6458 Show more…

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(a) A sample of dry sand is tested in direet shear. Under an applied normal stress of 4,800 paf, the sample falls when the shear stress reaches 3,100 pof. What is the angle of internal friction for this soil? (b) A second sample of the same sand material is also to be tested in direct shear, but the applied compressive loading will be 3,500 pot instead of 4,800 pot. What shear stress is expected to fail the sample?

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