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 10.4P
To determine
Find the required shear force to cause failure of the specimen.
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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 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.
For a dry sand specimen in a direct shear test box, the following are given:
• Angle of friction: 38°
• Size of specimen: 2 in. X 2 in. X 1.2 in. (height)
• Normal stress: 20 lb/in.?
Determine the shear force required to cause failure.
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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- Friction angle of dry sand is 35°. In a direct shear test on this sand, a normal stress of 150kPa was applied. What shear stress will cause the failure? (a) 150kPa (b) 50kPa (c) 78kPa (d) 105kPaarrow_forwardCan TRIAXIAL TESTS be performed on sand samples? Briefly explain. The results of an undrained triaxial test on a clay soil sample are given in table below Cell Pressure (kN/m2) 200 400 600 Additional Axial Load at Failure (N) 342 388 465 Each sample, originally 76mm long and 38mm in diameter, experienced a vertical deformation of 5.1mm. Evaluate the stress at failure for each cell pressure given in table above. Draw Mohr Circles and determine the shear strength parameters of the soil.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 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?arrow_forwardA specimen of rock was subjected to a compressive force of (10 kN) which tends an axial deformation was equal to (0.01 mm) and lateral deformation was of (0.001 mm). If the sample dimensions (length 80 mm and diameter 40 mm), find: Axial strain, Lateral strain, Volumetric strain, Young's Modulus (kN/m²), Bulk Modulus (kN/m) and Shear Modulus (kN/m2).arrow_forwardA sand sample is subjected to direct shear testing at its normal (in situ) water content. Two tests are performed. For one of the tests, the sample shears at a stress of 3,000 psf when the normal stress is 4,000 psf. In the second test, the sample shears at a stress of 4,000 psf when the normal stress is 6,000 psf. From these data, determine the value of apparent cohesion and the corresponding angle of internal friction.arrow_forward
- A sand sample is subjected to direct shear testing at it's (in - situ) water content. Two tests are performed. For one of the tests, the sample shears at a stress of 400 kPa when the normal stress is 600 kPa. From these data,Determine the value of the apparent cohesion. Determine the corresponding angle of internal friction.arrow_forwardIn a direct shear test on a specimen of cohesionless soil (sand), the vertical normal stress on the specimen is 210 kPa and the horizontal shear stress at failure is 135 kPa. You may assume the uniform stress distribution within the failure zone. a) determine the magnitude and direction of the principal stresses at failure using a Mohr's circle b) briefly explain whether it's possible to determine the magnitude and direction of the principal stresses when the material is not failed.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
- Problem 4.4 The results of a direct shear test on a specimen of dry sand are as follows: Normal stress = 96.6 kPa; shear stress at failure = 67.7 kPa. find the magnitude and directions of the principal stresses acting on a soil element within the zone of failure.arrow_forwardAn undisturbed soil sample, 110 mm in diameter and 220 mm in height, was tested in a triaxial machine. The sample sheared under an additional axial load of 3.35 kN with a vertical deformation of 21 mm. The failure plane was inclined at 50˚ to the horizontal and the cell pressure was 300 kN/m2. i. Draw the Mohr circle diagram representing the above stress conditions, and from it determine: − Coulomb’s equation for the shear strength of the soil, in terms of total stress; − the magnitude and obliquity of the resultant stress on the failure planearrow_forwardThe 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_forward
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