Concept explainers
(a)
Find the angle of friction,
(a)
Answer to Problem 12.1P
The angle of friction,
Explanation of Solution
Given information:
The diameter (d) of specimen is 71 mm.
The height (h) of specimen is 25 mm.
Shear force
The normal stress
Calculation:
Show the expression of Mohr’s coulomb failure as follows:
Here,
Find the shear strength
Here, S is the shear force and
Substitute 276 N for shear force and 71 mm for d.
Find the angle of friction
Substitute
The value of
Thus, the angle of friction,
(b)
Find the shear force (S) required to cause failure.
(b)
Answer to Problem 12.1P
The shear force required (S) to cause failure is
Explanation of Solution
Given information:
The normal stress
Calculation:
Refer part (a).
Substitute 0 for
Find the shear force is required to cause failure as follows:
Substitute
Thus, the shear force (S) required to cause failure is
Want to see more full solutions like this?
Chapter 12 Solutions
Principles of Geotechnical Engineering (MindTap Course List)
- 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 diagramarrow_forwardA sample of ocean sand was retrieved from 7 m below the surface. The sample had been under a vertical stress Ov of 150 kN/m³, a horizontal stress Oh of 250 kN/m³, a shear stress of 86.6 kN/m. If the angle theta between the vertical stress and the principal stress is 60°, calculate the angle of internal friction of the soil. [Hint: you must sketch the Mohr circle of stresses] Please explain clearly and step by steparrow_forwardA 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_forward
- A sample was obtained from point A in the submerged clay layer shown below. It was determined that it had a w = 54%, and a Gs = 2.78. What is the effective vertical stress at A? Express your answer in kPa rounded to the nearest whole number.arrow_forwardFollowing 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_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
- Use Eq. (6.14) to determine the stress increase () at z = 10 ft below the center of the area described in Problem 6.5. 6.5 Refer to Figure 6.6, which shows a flexible rectangular area. Given: B1 = 4 ft, B2 = 6 ft, L1, = 8 ft, and L2 = 10 ft. If the area is subjected to a uniform load of 3000 lb/ft2, determine the stress increase at a depth of 10 ft located immediately below point O. Figure 6.6 Stress below any point of a loaded flexible rectangular areaarrow_forward1.A dry sand is known to have an angle of internal friction of 29. A triaxial test is planned, where the confining pressure will be 41 kPa. What is the maximum axial stress, in kPa, (major principal stress) that can be applied? Calculate the value to 1 decimal place. Do not provide units in your answer. 2.A clay soil is subjected to a triaxial test under unconsolidated-undrained conditions. At failure, the major and minor principal stresses are 8401 psf and 4875 psf, respectively. What is the shear strength of this soil if the confining pressure is doubled? Provide your answer in psf with no decimals.arrow_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_forward
- 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.arrow_forwardin a drained triangle test on consolidated clay the stress and angle area as follows deviator stress is 20 lb/ in ^2 and friction angle is 21 degree calculate the effective confining pressure at failurearrow_forward
- Fundamentals of Geotechnical Engineering (MindTap...Civil EngineeringISBN:9781305635180Author:Braja M. Das, Nagaratnam SivakuganPublisher:Cengage LearningPrinciples of Foundation Engineering (MindTap Cou...Civil EngineeringISBN:9781305081550Author:Braja M. DasPublisher:Cengage LearningPrinciples of Geotechnical Engineering (MindTap C...Civil EngineeringISBN:9781305970939Author:Braja M. Das, Khaled SobhanPublisher:Cengage Learning