Principles of Foundation Engineering, SI Edition
9th Edition
ISBN: 9781337672085
Author: Das, Braja M., SIVAKUGAN, Nagaratnam
Publisher: Cengage Learning
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Chapter 15, Problem 15.5P
To determine
Find the total likely swell beneath the foundation.
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A consolidated-drained tri-axial test was conducted on a normally consolidated clay. The results are as following: confining stress, σ3 = 250 kN/m² and deviator
stress, (Aod)f = 100 kN/m². i) Determine angle of friction, q; ii) Determine angle 0 that the failure plane makes with the major principal plane; iii) Find the normal
stress o and the shear stress tf on the failure plane; and iv) Determine the effective normal stress on the plane of maximum shear stress. Express your results in
two decimal places.
Q: A consolidated-drained tri-axial test was conducted on a normally consolidated clay. The
results are as following: confining stress, G3 = 150 kN/m² and deviator stress, (Aod)t = 150
kN/m². i) Determine angle of friction, q; ii) Determine angle 0 that the failure plane makes
with the major principal plane; iii) Find the normal stress and the shear stress ty on the failure
plane; and iv) Determine the effective normal stress on the plane of maximum shear stress.
Express your results in two decimal places.
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Principles of Foundation Engineering, SI Edition
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- Q.23 A normally consolidated clay settled by 2cm when the effective stress was increased from 100kPa to 200kPa. when the effective stress is increased to 400kPa then the total settlement will bearrow_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_forwardA sample of normally consolidated clay was subjected to a consolidated undrained triaxial compression test that was carried out until the specimen failed at a deviator stress of 50 kN/m2. The pore water pressure at failure was recorded to be 20 kN/m2 and confining pressure of 50 kN/m2 was used in the test. Determine the consolidated undrained friction angle.arrow_forward
- snarrow_forwardQ-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 2014arrow_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
- Question attachedarrow_forwardThe equation of the effective stress failure envelope for normally consolidated clayey soil is. a drained triaxial test was conducted with the same soil at a chamber confining pressure of 80 KN/m². Deviator at stress failure is (take angle = 250) A 197 KN/m² B 117 KN/m² C 234 KN/m² D 160 KN/m²arrow_forward(c) The shear strength of soil collected 4m below ground surface and tested under consolidated undrained test can be given by the equation t = 30 + o' tan 27º by Effective Stress (ES) analysis. A proposed structure of 80 kN/m² is to be constructed on over consolidated clay material with unit weight of 19 kN/m³. Determine: i. Pore water pressure developed in the specimen at failure ii. Minor (63) and major (61) effective principal stress iii. The expected cohesion value in Total Stress (TS) analysisarrow_forward
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