Principles of Foundation Engineering (MindTap Course List)
8th Edition
ISBN: 9781305081550
Author: Braja M. Das
Publisher: Cengage Learning
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Chapter 2, Problem 2.20P
To determine
Plot the shear stress at failure against normal stress.
Calculate the soil friction angle.
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Following data are given for a direct shear test conducted on dry sand:
Cylindrical specimen dimensions: diameter = 50 mm and height = 25 mm
Normal stress: 0.15 N/mm2
Shear force at failure: 276 N
Determine the angle of friction of this soil.
Normal Stress = 0.15 N/mm2
Shear Force = 276 N
Shear Force = 276 N
A direct shear test, when conducted on a remolded sample of sand, gave the following observations at the time of failure: Normal load = 288 N; shear load = 173 N. The cross-sectional area of the sample = 36 cm2. Determine the minor principal stress in kPa.
A direct shear test, when conducted on a remolded sample of sand, gave the following observations at the time of failure: Normal load = 288 N; shear load = 173 N. The cross-sectional area of the sample = 36 cm2. Determine the minor principal stress in kPa. Show diagram
Chapter 2 Solutions
Principles of Foundation Engineering (MindTap Course List)
Ch. 2 - Prob. 2.1PCh. 2 - Prob. 2.2PCh. 2 - Prob. 2.3PCh. 2 - Prob. 2.4PCh. 2 - Prob. 2.5PCh. 2 - Prob. 2.6PCh. 2 - Prob. 2.7PCh. 2 - Prob. 2.8PCh. 2 - Prob. 2.9PCh. 2 - Prob. 2.10P
Ch. 2 - Prob. 2.11PCh. 2 - Prob. 2.12PCh. 2 - Prob. 2.13PCh. 2 - Prob. 2.14PCh. 2 - Prob. 2.15PCh. 2 - For a normally consolidated soil, the following is...Ch. 2 - Prob. 2.17PCh. 2 - Prob. 2.18PCh. 2 - Prob. 2.19PCh. 2 - Prob. 2.20PCh. 2 - Prob. 2.21PCh. 2 - Prob. 2.22PCh. 2 - Prob. 2.23PCh. 2 - Prob. 2.24PCh. 2 - Prob. 2.25PCh. 2 - Prob. 2.26PCh. 2 - Prob. 2.27P
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- A triaxial shear test was performed on a well-drained sand sample. The normal stress on the failure plane and the shearing stress on the failure plane were determined to be 75kPa and 42kPa, respectively. Determine the angle of internal friction of the sand, in degrees.Determine the axial stress applied to the specimen, in kPa.arrow_forwardProblem # 5. The angle of friction of compacted dry sand is 37°. In a direct shear test on the sand, normal stress of 150 kN/m^2 was applied. The size of the specimen was 50mm x 50mm 30 mm (height): a. Compute the shearing stress.b. What shear force will cause will cause shear failure?c. Determine the shear stress at a depth 3m. If the void ratio of the soil is 0.60. Sp. Gr. of sand is 2.70.arrow_forwardA sample of dry sand in relatively loose condition is subject to a triaxial test. The sample fails when the confining stress (minor principal stress) is 50 kPa and the axial stress (major principal stress) is 170 kPa. What is the angle of internal friction for the soil?arrow_forward
- A tri-axial compression test on a cohesive sample cylindrical in shape yields the following effective stress. Major principal stress = 8 MN/m², Minor principal stress = 2 MN/m², Angle of failure plane θ=60°. Compute the cohesion of the soil sample.arrow_forwardA tri-axial compression test on a cohesive sample cylindrical in shape yields the following effective stress. Compute the cohesion of the soil sample. Major Principal Stress = 8 MN/m², Minor Principal Stress = 2 MN/m²arrow_forward: In a tri-axial test of a silty soil, the sample failed at normal stress of 475 kPa and a shear stress of 350 kPa. Which of the following gives the angle of internal friction? Compute the angle that the failure plane makes with the x-axis. Compute the maximum failure stress.arrow_forward
- 1 .For a direct shear test on a dry sand, the following are given:• Specimen size: 75 mm 75 mm 30 mm (height)• Normal stress: 200 kN/m2• Shear stress at failure: 175 kN/m2a. Determine the angle of friction, f b. For a normal stress of 150 kN/m2, what shear force is required to cause failure?in the specimearrow_forwardA direct shear test, when conducted on a remolded sample of sand, gave the following observations at the time of failure: Normal load = 288 N; shear load = 173 N. The cross-sectional area of the sample = 36 cm2. Determine the minor principal stress in kPa. Show the free body diagram. a.53.5 b.136.5 c.36.9 d.64.5arrow_forwardA dry sand sample in a triaxial test failed when the confining stress (minor principal stress) was 1,000 psf and the axial stress (major principal stress) was 4,000 psf. What is the angle of internal friction for this soil?arrow_forward
- Following data are given for a direct shear test conducted on dry silty sand:Specimen dimensions: diameter = 71 mm; height = 25 mmNormal stress: 150 kN/m2Shear force at failure: 276 Narrow_forwardThe 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/m2a) Determine the angle of friction,b) For a normal stress of 150 kN/m2, what shear force is required to cause failure?arrow_forwardA triaxial test was performed on a dry. cohesionless sand. The sample fails when the confining stress (minor principal) is 45 kPa and the axial stress (major principal) is 89 kPa. Determine the angle of internal friction Φ for the soil. Adjust your answer to TWO decimal places.arrow_forward
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