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MindTap Engineering for Das/Sobhan's Principles of Geotechnical Engineering, 9th Edition, [Instant Access], 2 terms (12 months)
9th Edition
ISBN: 9781305971257
Author: Braja M. Das; Khaled Sobhan
Publisher: Cengage Learning US
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Question
Chapter 12, Problem 12.15P
(a)
To determine
Find the angle of friction
(b)
To determine
Find the angle such that failure plane makes with the major principal plane.
(c)
To determine
Find the normal stress
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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
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
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
Chapter 12 Solutions
MindTap Engineering for Das/Sobhan's Principles of Geotechnical Engineering, 9th Edition, [Instant Access], 2 terms (12 months)
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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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.Similar questions
- 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_forward5- 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 failurearrow_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
- 1.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_forward5. [Shear Strength] ( e). In a series of unconsolidated-undrained triaxial tests on speciments of a fully saturdated clay, the following results were obtained at failure. Minor Principal Stress (КРа) 200 400 600 Major Principal Stress 422 618 820 Calculate the following: a. Cohesion b. Angle of internal frictionarrow_forward3. [Mohr's Circle] (: ). For the stressed soil element shown below, determine a. Major Principal Stress in psi. b. Minor Principal Stress in psi. c. Normal Stress at Failure Plane AE in psi. d. Shear Stress at Failure Plane AE in psi. 150 psi 60 psi D. 90 psi 60 psi 45° A Barrow_forward
- 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.arrow_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_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_forward
- 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_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_forwardSolve 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.arrow_forward
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