EBK PRINCIPLES OF FOUNDATION ENGINEERIN
8th Edition
ISBN: 9780100547056
Author: Das
Publisher: YUZU
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Chapter 2, Problem 2.21P
To determine
Calculate the shear stress parameters
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A = 45
B = 40
C = 15
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.
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
Chapter 2 Solutions
EBK PRINCIPLES OF FOUNDATION ENGINEERIN
Ch. 2 - Prob. 2.1PCh. 2 - Prob. 2.2PCh. 2 - Prob. 2.3PCh. 2 - Prob. 2.4PCh. 2 - Prob. 2.5PCh. 2 - Prob. 2.6PCh. 2 - Prob. 2.7PCh. 2 - Prob. 2.8PCh. 2 - Prob. 2.9PCh. 2 - Prob. 2.10P
Ch. 2 - Prob. 2.11PCh. 2 - Prob. 2.12PCh. 2 - Prob. 2.13PCh. 2 - Prob. 2.14PCh. 2 - Prob. 2.15PCh. 2 - For a normally consolidated soil, the following is...Ch. 2 - Prob. 2.17PCh. 2 - Prob. 2.18PCh. 2 - Prob. 2.19PCh. 2 - Prob. 2.20PCh. 2 - Prob. 2.21PCh. 2 - Prob. 2.22PCh. 2 - Prob. 2.23PCh. 2 - Prob. 2.24PCh. 2 - Prob. 2.25PCh. 2 - Prob. 2.26PCh. 2 - Prob. 2.27P
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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
- 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_forwardA consolidated-drained triaxial test on a sand yields the results:All-around confining pressure = σ3 = 30 lb/in2Added axial stress at failure = Δσ = 96 lb/in2Determine the shear stress parameters (i.e., Φ' and c')arrow_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_forward
- please Show complete solution and avoid rounding off intermediate values. Please write legibly. Athanksarrow_forward10-14. The state plane stress in a mass of dense cohesionless sand is described by the following stresses: Normal stress on horizontal plane Normal stress on vertical plane = 200 kPa Shear stress on horizontal and vertical planes 370 kPa 80 kPa %3! Determine by means of the Mohr circle the magnitude and direc- tion of the principal stresses. Is this state of stress safe against failure? (After A. Casagrande.)arrow_forward2. 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.arrow_forward
- A sample of dry sand is subjected to a tri- axial test. The angle of internal friction is 36.6°. The minor principal stress is 192 KPa. Compute the effective normal stress in KPa at the point where shear occurs at the failure plane.arrow_forward2. The state plane stress in a mass of dense cohesionless sand (c=0) is described by the following stresses: Normal stress on horizontal plane = 300 kPa Normal stress on vertical plane = 100 kPa Shear stress on horizontal and vertical planes = ± 50 kPa Determine by means of the Mohr circle the magnitude and direction of the principal stresses. Is this state of stress safe against failure assuming φ= 20o?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
- 5. [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_forwardTriaxial tests performed on samples of aeolin sand. The failure conditions in terms of effective stress are (ov, 0h) = (515, 100), (1250, 200), (3500, 400), and (5325, 800) kPa. Using (s, t) space, determine the cohesion and friction angle. What is the orientation of the major principal stress with respect to the failure plane? Determine this graphically.arrow_forwardU1arrow_forward
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