Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781305970939
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
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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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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.
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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.
Refer 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.
Chapter 12 Solutions
Principles of Geotechnical Engineering (MindTap Course List)
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
- Refer to the soil profile shown. Given H1 = 9.89 m., and H2 = 5.80 m. If the ground water table rises by 3.78 meters, determine the change in effective stress (numerical value only, in kPa) at the bottom of the clay layer. Properties of dry sand: Gs = 2.53, e = 0.63. Properties of clay: Gs = 2.75, e = 0.81arrow_forwardA consolidated-drained triaxial test was conducted on a normally consolidated clay with a chamber pressure, σ3 = 276 kN/m2. The deviator stress at failure is = 276 kN/m2. Determine: a. The angle of friction b. The angle θ that the failure plane makes with the major principal plane c. The normal stress, σf, and the shear stress, τf, on the failure planearrow_forwardConsolidation 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_forward
- Given H1 = 8.3 m., and H2 = 4.43 m. If the ground water table rises by 2.29 meters, determine the change in effective stress (numerical value only, in kPa) at the bottom of the clay layer. Properties of dry sand: Gs = 2.59, e = 0.61. Properties of clay: Gs = 2.74, e = 0.86. Round off to two decimal places. Answer: 13.95arrow_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_forwardThe effective internal friction angle of a normal compacted clay is 33°, the triaxial test of this clay specimen is carried out, and the applied confining pressure is 115 kN/m2. (1) If the compaction undrained test is carried out, the axial differential stress at failure is 102 kN/m2, and the water pressure in the clay test body at failure is determined. (2) If the compaction drainage test is carried out, the applied confining pressure is still 115 kN/m2, and the axial difference stress at the time of failure is determined.arrow_forward
- For a given consolidated clay: Gs= 2.71; LL= 45; In situ average effective overburden pressure = 120 kPa; In situ void ratio= 0.80 ; Thickness of clay layer = 4m ; average increase of effective stress on clay layer= 40 kPa; Effective pressure at the mid-height of clay layer= 60 kPa; Swell Index (Cs)= 1/5 Cc. Compute the pre-consolidation pressure in kPa. Show free body diagram a.92.9 b.100.5 c.49 d.87.64arrow_forward6. From the given soil profile, the groundwater table is located 2m below the ground level. (See picture below) Compute the horizontal effective stress at point A. (Answer: 53.36 kPa) Compute the lateral effective stress at point A if the lateral earth pressure coefficient at rest is K0=0.50. (Answer: 26.68 kPa) Compute the lateral total stress at point A. (Answer: 56.11 m)arrow_forwardA cylindrical sample of soil having a cohesion of 80 kN/m2 and an angle of internal friction of 20° is subjected to a cell pressure of 100 kN/m2. Determine: c. Determine the normal stress kn Kpa at the failure planearrow_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_forwardRefer to the soil profile shown. Given H1 = 8.93 m., and H2 = 5.32 m. If the ground water table rises by 2.01 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 = 0.65. Properties of clay: Gs = 2.72, e = 0.87. Round off to two decimal places. Selected Answer: 11.95 Correct Answer: 11.95 ± 1arrow_forwardin a drained triangle test on consolidated clay the stress and angle area as follows deviator stress is 20 lb/ in ^2 and friction angle is 21 degree calculate the effective confining pressure at failurearrow_forward
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