# the following results were recorded during a shear box test on a cohesive soil. if the specimen size was 60mm x 60mm determine the apparent cohesion and angle of shearing resistance.1. the size of a sand specimen in a direct shear test was 50 mm × 50 mm × 30 mm (height). it is known that, for the sand, tan φ′ = 0.65/e (where e = void ratio) and the specific gravity of soil solids gs = 2.65. during the test, a normal stress of 140 kn/m2 was applied. Normal force, kN = 0.073          Shear Force, kN = 0.109 Normal force, kN = 0.191          Shear Force, kN = 0.139 Normal force, kN = 0.309          Shear Force, kN = 0.170 Normal force, kN = 0.427          Shear Force, kN = 0.197 Normal force, kN = 0.545          Shear Force, kN = 0.227

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the following results were recorded during a shear box test on a cohesive soil. if the specimen size was 60mm x 60mm determine the apparent cohesion and angle of shearing resistance.

1. the size of a sand specimen in a direct shear test was 50 mm × 50 mm × 30 mm (height). it is known that, for the sand, tan φ′ = 0.65/e (where e = void ratio) and the specific gravity of soil solids gs = 2.65. during the test, a normal stress of 140 kn/m2 was applied.

Normal force, kN = 0.073          Shear Force, kN = 0.109

Normal force, kN = 0.191          Shear Force, kN = 0.139

Normal force, kN = 0.309          Shear Force, kN = 0.170

Normal force, kN = 0.427          Shear Force, kN = 0.197

Normal force, kN = 0.545          Shear Force, kN = 0.227

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Step 1

Given data:

The side of the specimen is 60mm is equal to 0.06 m.

The result of the test are, help_outlineImage TranscriptioncloseNormal force, kN= 0.073 Shear Force, kN = 0.109 Shear Force, kN 0.139 Normal force, kN= 0.191 Normal force, kN= 0.309 Shear Force, kN 0.170 Shear Force, kN 0.197 Normal force, kN = 0.427 Normal force, kN = 0.545 Shear Force, kN = 0.227 fullscreen
Step 2

The area of the soil specimen is calculated as,

Step 3

The normal stress and shear ...

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### Geotechnical Engineering 