Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781305970939
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
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Chapter 12, Problem 12.19P
(a)
To determine
Find the consolidated undrained friction angle.
(b)
To determine
Find the pore water pressure developed in the specimen at failure.
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5- The shear strength of a normally consolidated clay can be given by the equation tf o'tan 27°. Following are the
results of a
consolidated-undrained
test on the clay.
• Chamber-confining pressure = 3,130 lb/ft²
Deviator stress at failure = 2,510 lb/ft²
a) Determine the consolidated-undrained friction angle
b) Pore-water pressure developed in the specimen at failure
A consolidated-undrained test determines that the shear strength of a normally consolidated clay is given by τf = σ’tan 27o. The confining pressure is 150 kPa, and deviator stress at failure is 120 kPa.• Find the consolidated-undrained friction angle• Pore water pressure developed in the specimen at failureFinal answer should be in 3 decimal places.
2. 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 was determined to be
6,300 psf and 4,200 psf, respectively.
a. Determine the angle of internal friction of the sand.
b. Determine the angle of failure plane.
c. Determine the maximum principal stress.
Chapter 12 Solutions
Principles of Geotechnical Engineering (MindTap Course List)
Ch. 12 - Prob. 12.1PCh. 12 - Prob. 12.2PCh. 12 - Prob. 12.3PCh. 12 - Prob. 12.4PCh. 12 - Prob. 12.5PCh. 12 - Prob. 12.6PCh. 12 - Prob. 12.7PCh. 12 - Prob. 12.8PCh. 12 - Prob. 12.9PCh. 12 - Prob. 12.10P
Ch. 12 - Prob. 12.11PCh. 12 - Prob. 12.12PCh. 12 - Prob. 12.13PCh. 12 - Following are the results of...Ch. 12 - Prob. 12.15PCh. 12 - Prob. 12.16PCh. 12 - Prob. 12.17PCh. 12 - Prob. 12.18PCh. 12 - Prob. 12.19PCh. 12 - Prob. 12.20PCh. 12 - Prob. 12.21PCh. 12 - Prob. 12.22PCh. 12 - Prob. 12.23PCh. 12 - Prob. 12.24PCh. 12 - Prob. 12.1CTP
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- The shear strength of a normally consolidated clay can be given by theequation (tau)f=(sigma prime)tan(21). The results of a consolidated-undrained test on theclay areChamber confining pressure 5 225 kN/m2Deviator stress at failure 5 112 kN/m2Determine:a. The consolidated-undrained (total stress) friction angleb. Pore water pressure developed in the specimen at failurearrow_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_forwardA clay specimen with drained shear strength parameters of c'=5 kPa and Φ=320 is consolidated under all around pressure of 140 kPa.The specimen was then loaded axially to failure under undrained condition and failed when the principal stress diffrences reached 60 kPa.Calculate the pore pressure(u) and the pore pressure coefficient D(=A*B=Δu/Δσv) of the clay.arrow_forward
- A series of previous CD tests conducted on a NC clay has established that the ultimate effective friction angle of the clay is 300. Now a CU test is conducted on this soil with an effective cell pressure of 100 kPa. The specimen fails at a deviatoric stress of 100 kPa. Estimate the pore pressure at failure. Given: σ3f′=Kaσ1f′whereKa=1−sinϕ1+sinϕ Please do fast .... ASAP Fastarrow_forwardA consolidated-drained tri-axial test was conducted on a normally consolidated clay. The results are as follows: 03 = 276 kN/m? (Aoa)f = 276 kN/m² Determine a. Angle of friction, o' b. Angle 0 that the failure plane makes with the major principal plane c. Find the normal stress o' and the shear stress tf on the failure plane d. Determine the effective normal stress on the plane of maximum shear stress of Effective stress failure envelope ↑ B of 20 276 kN/m2 A oj = 552 kN/m² Normal stress Shear stress © Cengage Learning 2014arrow_forwardA consolidated-drained tri-axial test was conducted on a normally consolidated clay. The results are as follows: O3 = 150 kN/m? (Aoa)f= 130 kN/m? Determine a. Angle of friction, o' b. Angle 0 that the failure plane makes with the major principal plane c. Find the normal stress o' and the shear stress tf on the failure plane d. Determine the effective normal stress on the plane of maximum shear stressarrow_forward
- 2- A three-axis compression test on a clay sample that is comprehensive and normal up to a stress of 300 kPa and a specific volume of 2 Consolidated is done, The values of q and volumetric strain are measured at axial strain, 0, 2 and 10 percent (failure) . The value of N is equal to 3 and A is equal to 0.2. Graph of all-round stress on initial all-round stress Draw the initial all-round stress in terms of sectionarrow_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_forward7.12 A sand specimen was subjected to a drained shear test using hollow cylin- der test equipment. Failure was caused by increasing the inside pressure while keeping the outside pressure constant. At failure, o, = 193 kN/m² and o; = 264 kN/m². The inside and outside radii of the specimen were 40 and 60 mm, respectively. (a) Calculate the soil friction angle. (b) Calculate the axial stress on the specimen at failure.arrow_forward
- [Shear Strength,. speciments of a fully saturdated clay, the following results were obtained at failure. In a series of unconsolidated-undrained triaxial tests on Major Principal Stress Minor Principal Stress (kPa) 200 422 618 400 600 820 Calculate the following: a. Cohesion b. Angle of internal frictionarrow_forward2. The shear strength of a normally consolidated clay can be given by the equation t = ơ' tan 28°. A consolidated-undrained test was conducted on the clay. Following are the results of the test. Chamber confining pressure = 105 KN/m? Deviator stress at failure = 97 KN/m² a. Compute the pore water pressure developed in the clay specimen at failure. b. Compute the consolidated-undrained friction angle 4. What would have been the deviator stress at failure if a drained test had been conducted with the same chamber confining pressure that is o3= 105 KN/m? C.arrow_forwardFor a normally consolidated clay, these are the results of a drained triaxial test: confining pressure = 112 kPa deviator stress at failure = 175 kPaa) Find the angle of internal friction, ø’.b) Determine the angle θ that the failure plane makes with the major principal plane.c) Find the normal stress σ’ and the shear stress τf on the failure plane.d) Determine the effective normal stress on the plane of maximum shear stress.arrow_forward
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