Principles of Foundation Engineering (MindTap Course List)
8th Edition
ISBN: 9781305081550
Author: Braja M. Das
Publisher: Cengage Learning
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Chapter 2, Problem 2.26P
To determine
Find the pore water pressure at failure
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8. For a normally consolidated clay specimen, the results of a drained triaxial test are as
follows:
Chamber-confining pressure = 125 kN/m?
Deviator stress at failure = 175 kN/m?
%3D
Determine the soil friction angle, o'
7.2 A specimen of normally consolidated clay (= 28°) was consolidated under a
chamber-confining pressure of 280 kN/m². For a drained test, by how much docs
the axial stress have to be reduced to cause failure by axial extension?
1
7.3 A normally consolidated clay specimen (= 31°) was consolidated under a
chamber-confining pressure of 132 kN/m². Failure of the specimen was caused by
an added axial stress of 158.1 kN/m in an undrained condition. Determine &u, Af
and the pore water pressure in the specimen at failure.
>
A consolidated-undrained triaxial test was conducted on a dense sand with a confining pressure of 138 kN/m2. Results showed that ϕ′=24 deg and ϕ=31 deg. Determine the deviator stress and the pore water pressure at failure.
Chapter 2 Solutions
Principles of Foundation Engineering (MindTap Course List)
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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- 2. 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_forwardQ: 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.arrow_forwardsample of normally consolidated clay was subjected to a consolidated undrained triaxial compression st that was carried out until the specimen failed at a deviator stress of 50 kN/m?. The pore water pressure at failure was recorded to be 20 kN/m? and confining pressure of 50 kN/m² was used in the test. Determine the consolidated undrained friction angle o and drained friction angle ¢cn: CD сиarrow_forward
- 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 failurearrow_forwardA consolidated-undrained test on a normally consolidate clay yield the following results: Confining Pressure = 160 KPa; Deviator Stress = 182 KPa; Pore Pressure = 73 KPa. Calculate the consolidated undrained friction angle.arrow_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
- A consolidated-undrained test on a normally consolidate clay yield the following results: Confining Pressure 167 KPa; Deviator Stress = 184 KPa; Pore Pressure = 74 KPa. Calculate the normal stress at the failure plane at drained condition in KPa.arrow_forwardA consolidated-drained triaxial test was conducted on a normally consolidated clay. The results kN o', = 276 kN (A0,), = 276 m² are as follows: Determine: (a) The angle of friction o; (b) The angle 0 that the failure plane makes with the major principal plane, and (c) The normal stress o’ and shear stress t, on the failure plane.arrow_forwardA 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.arrow_forward
- A 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_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_forwardA consolidated drained tri-axial test was conducted on normally consolidated clay. The results were: the confining pressure is 300 kPa and the deviators stress is 300 kPa. Compute the angle that the failure plane makes with the major principal axis in degrees. show free body diagram. a.54.74 b.16.3 c.24.72 d.19.47arrow_forward
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