For a direct shear test on a dry sand, the following are given: • Specimen size: 75 mm x 75 mm x 30 mm (height) • Normal stress: 200 kN/m2 • Shear stress at failure: 175 kN/m i. Determine the angle of friction. ii. For a normal stress of 150 kN/m2 , what shear force is required to cause failure?

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Q: 1- Following data are given for a direct shear test conducted on dry sand: Normal stress: 120 lb/ft²…

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A: To find out the shear stress on failure plane.  Confining stress = 326 kPa Deviator stress = 414 kPa

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A: Given that: Shear stress at failure, τf=70 kN/m2Normal stress at failure, σf=96 kN/m2

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A: Given:   Specimen dimension - 63mm×63mm×25mmσNormal stress - 105 kN/m2v Shear force at failur - 300…

Q: 1. Which failure criterion is used to interpret the results from a direct shear test? Why? 2. Which…

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A: Sample Dia = 50 mm           height = 25 mm.

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Q: In a series of unconsolidated-undrained triaxial tests speciments of a fully saturdated clay, the…

A: To find cohesion (c) and angle of internal friction

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Q: # 3. Following data are given for a direct shear test conducted on dry sand: Specimen dimensions: 63…

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A: Using Mohr coulomb equation

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A: Ф=32 degree c=27kPa σn= 150kPa Area= 50*50 = 2500mm2 Height = 75 mm 1kPa=0.001N/mm2

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Q: SITUATION 5: A sand sample is subjected to direct shear testing at it's (in - situ) water content.…

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A: Normal stress=150kN/m2

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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.