A direct shear test was conducted on a cohesionless soil (c = 0) specimen under a normal stress of 200 kN/m2. The specimen failed at a shear stress of 100 kN/m2. The angle of internal friction of the soil (degrees) is-A-26.6B-29.5C-30.0D-32.6-A-26.6

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APTITUDE CRACK · Accepted Answer

The correct option is  A 26.6 Shear stress = C + normal stress*tan (theta). Here given C=0, shear stress = 100 and normal stress = 200. After putting all these values in above formula we will get. theta= 26.56 .

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Q1: A direct shear test was conducted on a cohesionless soil (c = 0) specimen under a normal stress of 200 kN/m². The specimen failed at a shear stress of 100 kN/m². The angle of internal friction of the soil (degrees) is

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Q1: A direct shear test was conducted on a cohesionless soil (c = 0) specimen under a normal stress of 200 kN/m2. The specimen failed at a shear stress of 100 kN/m2. The angle of internal friction of the soil (degrees) is

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Q1: A direct shear test was conducted on a cohesionless soil (c = 0) specimen under a normal stress of 200 kN/m². The specimen failed at a shear stress of 100 kN/m². The angle of internal friction of the soil (degrees) is

For help Students Orientation
Mcqs Questions