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 failure 12.19 The shear strength of a normally consolidated clay can be given by theequation 7, = o' tan 21°. The results of a consolidated-undrained test on theclay areChamber confining pressure 225 kN/m?Deviator stress at failure 112 kN/m²Determine:a. The consolidated-undrained (total stress) friction angleb. Pore water pressure developed in the specimen at failure

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The shear strength of a normally consolidated clay can be given by the equation (tau)f=(sigma prime)tan(21). The results of a consolidated-undrained test on the clay are Chamber confining pressure 5 225 kN/m2 Deviator stress at failure 5 112 kN/m2

Principles of Geotechnical Engineering (MindTap Course List)

9th Edition

ISBN:9781305970939

Author:Braja M. Das, Khaled Sobhan

Publisher:Braja M. Das, Khaled Sobhan

Chapter12: Shear Strength Of Soil

Section: Chapter Questions

Problem 12.19P

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Concept explainers

Test for Determination of Shear Strength

The term shear strength is often associated with the soil. Shear strength is defined as the resistance to the deformation of soil particles due to the continuous action of shear stress. In short, shear strength is the maximum shear stress that a soil can sustain.

Question

The shear strength of a normally consolidated clay can be given by the
equation (tau)f=(sigma prime)tan(21). The results of a consolidated-undrained test on the
clay are
Chamber confining pressure 5 225 kN/m2
Deviator stress at failure 5 112 kN/m2
Determine:
a. The consolidated-undrained (total stress) friction angle
b. Pore water pressure developed in the specimen at failure

12.19 The shear strength of a normally consolidated clay can be given by the
equation 7, = o' tan 21°. The results of a consolidated-undrained test on the
clay are
Chamber confining pressure 225 kN/m?
Deviator stress at failure 112 kN/m²
Determine:
a. The consolidated-undrained (total stress) friction angle
b. 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
[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 friction
Please answer 12.17
2. A series of consolidated, undrained triaxial tests were carried out on specimens of a saturated clay under no backpressure. The test data at failure are summarized: Confining Pressure (kPa) Deviator Stress (kPa) Pore Water Pressure (kPa) 150 192 80 300 341 154 450 504 222 a. Draw the Mohr circles and find the cohesion and friction angles in terms of effective stresses. b. Compute Skempton’s A-parameter at failure for all three specimens. c. Is the soil normally consolidated or overconsolidated? Why? d. Another specimen of the same clay that was consolidated under a cell pressure of 250 kPa was subjected to a consolidated, drained triaxial test. What would be the deviator stress at failure?
A 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 2014
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.
A 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 stress
Solve
x in red is equal to 39
Q-3: A consolidated-drained tri-axial test was conducted on a normally consolidated clay. The results are as follows: 03 = 276 kN/m² (Aod)f=276 kN/m² Determine a. Angle of friction, o' b. Angle that the failure plane makes with the major principal plane c. Find the normal stress o' and the shear stress ty on the failure plane d. Determine the effective normal stress on the plane of maximum shear stress Shear stress Š σí σí σ3 Effective stress failure envelope B σ3 = 276 kN/m² A Normal stress 20 σ = 552 kN/m² Ⓒ Cengage Learning 2014
please Show complete solution and avoid rounding off intermediate values. Please write legibly. Athanks
Solve this problem graphically and then analytically. A CU triaxial test was performed on a dense sand specimen at a confining pressure 03=40 kPa. The consolidated undrained friction angle of the sand is =39°, and the effective friction angle is d'=34°. Calculate: (a) the major principal stress at failure, o1, (b) the minor and the major effective principal stresses at failure, o'3f and oʻır, and (c) the excess pore water pressure at failure, (Aua)f.