The following data are given for a direct shear test conducted on dry sand: - Specimen dimensions: 75 mm X 75 mm X 30 mm (height) - Normal stress: 200 kN/m2 - Shear stress at failure: 175 kN/m2 a) Determine the angle of friction, b) For a normal stress of 150 kN/m2, what shear force is required to cause failure?

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Q: Following data are given for a direct shear test conducted on dry sand: Specimen dimensions: 74 mm x…

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Q: 1 .For a direct shear test on a dry sand, the following are given: • Specimen size: 75 mm 75 mm 30…

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Q: Following data are given for a direct shear test conducted on dry sand: • Specimen dimensions: 75 mm…

A: Given:- The dimension of the specimen = 75 mm ×75 mm×30 mm The normal stress = 250 kN/m2  The shear…

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A: Solution Given Data Dia of shear Box d=50 mm Normal compressive load Q=200 N Shear Force at Failure…

Q: pressure of 450 kPa. If the drained friction angle is 34°, determine:

A: Given data Confining pressure, σ3=450kPa friction angle, ϕ=34o

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Q: Friction angle of dry sand is 35°. In a direct shear test on this sand, a normal stress of 150kPa…

A: Given data : Normal stress (σn)=150 KPaFriction angle (ϕ)=35∘The given soil is sandy soil .Hence…

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Q: A sample of dry sand is subjected to a tri-axial test. The angle of internal friction is 33.2°. The…

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A: Given data - Vertical stress = 150 KN/m² Horizontal stress = 250 KN/m² Shear stress = 86.6 KN/m²…

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Q: SITUATION: The angle of friction of a compacted dry sand is 37°. In a direct shear test on the sand,…

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A: Solution; Calculate the shear strain: r=xho*100r=0.0020*100r=0.00%

Test for Determination of Shear Strength

The term shear strength is often associated with the soil. Shear strength is defined as the resistance to the deformation of soil particles due to the continuous action of shear stress. In short, shear strength is the maximum shear stress that a soil can sustain.