Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN: 9781305635180
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Question
Chapter 10, Problem 10.20P
(a)
To determine
Calculate the consolidated-undrained friction angle.
(b)
To determine
Calculate the pore water pressure developed in the clay specimen at failure.
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The stresses on a failure plane in a drained triaxial test on sand are as Follows: Normal Stress nf = 100 kN/m2
Shear Stress τf = 40 kN/m2
Required:
a) Determine the angle of internal friction and angle of failure plane
b) Find the major and minor principle stresses (1 , 3 )
Solve it quick?
Triaxial test (CU) is performed on a clay for a confining pressure of 110 kPa. If the deviator stress and pore-water pressure were 70 kpa and 55 kpa respectively then determine the internal friction angle.
For a normally consolidated clay, these are the results of a drained triaxial test: confining pressure = 112 kPa deviator stress at failure = 175 kPaa) Find the angle of internal friction, ø’.b) Determine the angle θ that the failure plane makes with the major principal plane.c) Find the normal stress σ’ and the shear stress τf on the failure plane.d) Determine the effective normal stress on the plane of maximum shear stress.
Chapter 10 Solutions
Fundamentals of Geotechnical Engineering (MindTap Course List)
Ch. 10 - Prob. 10.1PCh. 10 - Prob. 10.2PCh. 10 - Prob. 10.3PCh. 10 - Prob. 10.4PCh. 10 - Prob. 10.5PCh. 10 - Prob. 10.6PCh. 10 - Prob. 10.7PCh. 10 - Prob. 10.8PCh. 10 - Prob. 10.9PCh. 10 - Prob. 10.10P
Ch. 10 - Prob. 10.11PCh. 10 - Prob. 10.12PCh. 10 - Prob. 10.13PCh. 10 - Prob. 10.14PCh. 10 - Prob. 10.15PCh. 10 - Prob. 10.16PCh. 10 - Prob. 10.17PCh. 10 - Prob. 10.18PCh. 10 - Prob. 10.19PCh. 10 - Prob. 10.20PCh. 10 - Prob. 10.21PCh. 10 - Prob. 10.22PCh. 10 - Prob. 10.23PCh. 10 - Prob. 10.24CTPCh. 10 - Prob. 10.25CTP
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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
- Consolidation of an undrained clay occurs at an effective pressure of 100 kPa and a shear stress of 40 kPa. The ratio of effective pressure to shear stress in drained clay is 2 to 1. If the shear stress at zero effective pressure, To, is 10 kPa, what is the pore water pressure at failure? A. 10 kPa в. 40 кра с. 53 кра D. 67 kPaarrow_forwardRefer to the soil profile shown. Given H1 = 9.54 m., and H2 = 4.15 m. If the ground water table rises by 2.26 meters, determine the change in effective stress (numerical value only, in kPa) at the bottom of the clay layer. Properties of dry sand: Gs = 2.56, e = 0.68. Properties of clay: Gs = 2.72, e = 0.83. Round off to two decimal places.arrow_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_forward
- A consolidated-drained triaxial test was conducted on a normally consolidated clay.The results were as follows:σ3 = 276 kN/m2(Δσd)f = 276 kN/m2a. Find the angle of friction. ϕ'.b. What is the angle that the failure plane makes with the major principal stress?c. C. Determine the normal stress σ' and the shear stress τf on the failure plane.arrow_forwardThe following data are given for a direct shear test conducted on dry sand:- Specimen dimensions: 75 mm X 75 mm X 30 mm (height)- Normal stress: 200 kN/m2- Shear stress at failure: 175 kN/m2a) Determine the angle of friction,b) For a normal stress of 150 kN/m2, what shear force is required to cause failure?arrow_forwardRefer to the flexible loaded rectangular area shown in Figure 10.47. Using Eq. (10.36), determine the vertical stress increase below the center of the loaded area at depths z = 3, 6, 9, 12, and 15 m. Figure 10.47arrow_forward
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