Foundation Design: Principles and Practices (3rd Edition)
3rd Edition
ISBN: 9780133411898
Author: Donald P. Coduto, William A. Kitch, Man-chu Ronald Yeung
Publisher: PEARSON
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Chapter 4, Problem 4.11QPP
A DMT was performed in a normally consolidated clay at a depth of 10 m, giving
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A consolidated drained tri-axial test was conducted on a normally consolidated
clay. The results are as follows:
Chamber confining presssure = 138 kPa
Deviator stress = 258 kPa
a. Compute the friction angle of the soil
b. Compute the normal stress at failure
c. Compute the shear stress at failure
A consolidated drained tri-axial test was conducted on normally consolidated clay. The results were: the confining pressure is 300 kPa and the deviators stress is 300 kPa. Compute the angle of shearing resistance.
A DMT was performed in a normally consolidated clay at a depth of 10 m, giving a KD of 2.1.The clay has a unit weight of 118 lb/ft3
, and the groundwater table is at the ground surface.
Estimate the undrained shear strength of the clay at a depth of 10 m.
Chapter 4 Solutions
Foundation Design: Principles and Practices (3rd Edition)
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- Three specimnens of clay having a small air-void content were tested in the shear box. Shear loading was started immediately after the application of the normal load and was completed in 8 minutes. The results are as follows: Normal Stress (kN/m2) Shear stress at failure (kN/m2) O Find the apparent cohesion of the clay. Find the angle of shearing resistance of the clay. What value of c would be obtained from an unconfined compression test on 145 241 337 103 117 131 the same soil?arrow_forwardA consolidated drained tri-axial test was conducted on a normally consolidated clay. The results were as follows: 03 = 300 kPa (confining pressure) Deviators stress = 300 kPa 0 Compute the angle of shearing resistance. Compute the angle that the failure plane makes with the major principal. stress. Compute the shear stress on the failure plane.arrow_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
- A sand sample is subjected to direct shear testing. Two tests areperformed. For test 1, The sample shears at a stress of 2500 psf whenthe normal stress is 4000 psf.Test 2, The sample shears at a stress of 3500 psf when the normalstress is 6000 psf. Determine the following:a) Angle of Internal frictionb) Value of cohesionc) Compute the shear stress at a depth of 12 ft. if the unit weight ofthe soil is 150 pcfarrow_forwardA 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_forwardTriaxial compression test (CU) is conducted on a normally consolidated clay at a confining pressure of 100 kPa. The deviator stress at failure is 80 kPa, and the pore-water pressure measured at failure is 50 kPa. Determine the angle of internal friction.arrow_forward
- 5. A consolidated, undrained triaxial test is being carried out on a normally consolidated clay where c - 0 and p'= 26'. The triaxial specimen was consolidated under a cell pressure of 300 kPa and backpressure of 80 kPa. Skempton's A parameter at failure is estimated to be 0.80. The drainage valve has since been dosed and the vertical deviator stress increased to failure. What would be the deviator stress and pore water pressure at failure?arrow_forwardA consolidated drained tri-axial test was conducted on a normally consolidated clay. The results as follows: Chamber confining pressure: 138 kPa Deviator Stress = 258 kPa Determine the normal stress at failure in kPa.arrow_forwardIn a drained triaxial test on consolidated clay the stress and angle are as follows: Deviator stress is 20 1b/in2 and friction angle is 210 . Calculate the effective confining pressure at failure.? A) 21 1b/in2 B) 22.2 1b/in2 C) 25.4 1b/in2 D) 17.9 1b/in2 make it fastarrow_forward
- A consolidated tri-axial test was conducted on a normally consolidated clay. The results were as follows: Chamber confining pressure = 304 kPa, Deviator stress = 419 kPa. Compute for the shear stress on the failure plane in kPa. Round off to two decimal places.arrow_forwardA specimen of normally consolidated clay was consolidated under a chamber confining pressure of 280kpa for a drained test by how much does the axial stress have to be reduced to cause failure by axial extension?arrow_forwardA CU triaxial test was carried out on a silty clay that was isotopically consolidated using a cell pressure of 125 kPa. The following data were obtained: Axial load (kPa) 0 5.5 11.0 24.5 28.5 35.0 50.5 85.0 105.0 120.8 Axial strain, &₁ (%) 0 0.05 0.12 0.29 0.38 0.56 1.08 2.43 4.02 9.15 Au (kPa) 0 4.0 8.6 19.1 29.3 34.8 41.0 49.7 55.8 59.0 (a) Plot the deviatoric stress vs. axial strain and excess porewater pressure vs. axial strain, respectively. (b) Determine the undrained shear strength (sµ). (Note: we assume that the sample reaches failure). (c) Determine the total principal stresses (0₁,03) and the effective principal stresses (0₁, 03') at the failure, respectively.arrow_forward
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