Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN: 9781305635180
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Chapter 10, Problem 10.2P
To determine
Find the angle of friction.
Find the required shear force to cause failure of the specimen.
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A direct shear test was conducted on a specimen of dry sand with a normal stress of 200 kN/m2. Failure occurred at a shear stress of 175 kN/m2. The size of the specimen testedwas 75 mm × 75 mm × 30 mm (height). Determine the angle of friction, . For a normal stress of 150 kN/m2, what shear force would be required to cause failure of the specimen?
Q # 3. Following data are given for a direct shear test conducted on dry sand:
Specimen dimensions: 63 mm x63 mm and 25 mm (height).
Normal stress: 105 kN/m?.
Shear force at failure: 300 N
a. Determine the angle of friction, for a normal stress of 180 kN/m?.
b. What shear force is required to cause failure?
Assume that both a triaxial shear test and a direct shear test are to be performed on a sample of dry
sand. When the triaxial shear test is performed, the specimen fails when the major and minor principal
stresses are 80 and 20 lb/in.2, respectively. When the direct shear test is performed, what shear
strength can be expected if the normal stress is 4000 lb/ft2?
Chapter 10 Solutions
Fundamentals of Geotechnical Engineering (MindTap Course List)
Ch. 10 - Prob. 10.1PCh. 10 - Prob. 10.2PCh. 10 - Prob. 10.3PCh. 10 - Prob. 10.4PCh. 10 - Prob. 10.5PCh. 10 - Prob. 10.6PCh. 10 - Prob. 10.7PCh. 10 - Prob. 10.8PCh. 10 - Prob. 10.9PCh. 10 - Prob. 10.10P
Ch. 10 - Prob. 10.11PCh. 10 - Prob. 10.12PCh. 10 - Prob. 10.13PCh. 10 - Prob. 10.14PCh. 10 - Prob. 10.15PCh. 10 - Prob. 10.16PCh. 10 - Prob. 10.17PCh. 10 - Prob. 10.18PCh. 10 - Prob. 10.19PCh. 10 - Prob. 10.20PCh. 10 - Prob. 10.21PCh. 10 - Prob. 10.22PCh. 10 - Prob. 10.23PCh. 10 - Prob. 10.24CTPCh. 10 - Prob. 10.25CTP
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- The angle of friction of a compacted dry sand is 37 degrees. In a direct shear test on the sand, anormal stress of 150 kN/m^2 was applied. The size of the specimen was 50mmx50mx30mm(height) SITUATION 1 a. Compute the shearing stress Your answer b. What shear force will cause shear failure? Your answer c. Determine the shear stress at a depth of 3m if the void ratio of the soil is 0.60. Gs Of sand is 2.70arrow_forwardQ # 3. Following data are given for a direct shear test conducted on dry sand: Specimen dimensions: 63 mm × 63 mm and 25 mm (height). Normal stress: 105 kN/m². Shear force at failure: 300 N Determine the angle of friction, for a normal stress of 180 kN/m². b. What shear force is required to cause failure? а.arrow_forward1 .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_forward
- 1- Following data are given for a direct shear test conducted on dry sand: Normal stress: 120 lb/ft² and shear stress at failure: 43 lb/ft² a) Determine the angle of friction, o' b) For a normal stress of 220 lb/ft², what shear force is required to cause failure? c) What are the principal stresses at failure? (for the loads from part b) d) What is the inclination of the major principal plane with the horizontal? (for the loads from part b)arrow_forwardAssume that both a triaxial shear test and a direct shear test are to be performed on a sample of dry sand. When the triaxial shear test is performed, the specimen fails when the major and minor principal stresses are 80 and 20 lbs/in^2, respectively. When the direct shear test is performed, what shear strength can be expected if the normal stress is 4000 lbs/ft^2? Show all work.arrow_forward20. A cohesionless sand sample was subjected to a triaxial shear test. Failure occurred when the normal stress is 400 KPa and the shear stress is 250 KPa. a. What is major principal stress? b. What is the minor principal stressarrow_forward
- A triaxial shear test was performed on a well-drained sand sample. The normal stress on the failure plane and the shear stress on the failure plane, at failure were determined to be 6100 psf and 4600 psf, respectively. a. Determine the angle of internal friction of the sand? b. Determine the angle of the failure plane? c. Determine the maximum principal stress? Please answer this asap. For upvote. Thank you very mucharrow_forwardStress, kPa 8. Following data are given for a direct shear test conducted on dry sand: Dimension of the cylindrical specimen: diameter = 71 mm; height = 25 mm; Normal stress 250 kN/m²; shear force at failure: 560 N. Complete the following a) What is the orientation of the failure plane in the specimen? b) Determine the effective stress angle of friction.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_forward
- 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 Narrow_forwardQuestion 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 (round off to 1st decimal place)arrow_forwardAfter conducting a tri-axial test on a sand sample, the normal and shearing stress on the failure plane at failure was found to be 475kPa and 350kPa, respectively. Which of the following most nearly gives the plunger stress?arrow_forward
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