Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781305970939
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
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Chapter 12, Problem 12.2P
(a)
To determine
Find the principle stress at failure of the specimen.
(b)
To determine
Find the inclination of the major principle plane with the horizontal.
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Q # 3. Following data are given for a direct shear test conducted on dry sand:
Specimen dimensions: 63 mm x63 mm and 25 mm (height).
Normal stress: 105 kN/m?.
Shear force at failure: 300 N
a. Determine the angle of friction, for a normal stress of 180 kN/m?.
b. What shear force is required to cause failure?
please answer my homework problems. I will leave good review. thank you!
1- Following data are given for a direct shear test conducted on dry sand: Normal stress: 120 lb/ft² and shear stress at
failure: 43 lb/ft²
a) Determine the angle of friction, o'
b) For a normal stress of 220 lb/ft², what shear force is required to cause failure?
c) What are the principal stresses at failure? (for the loads from part b)
d) What is the inclination of the major principal plane with the horizontal? (for the loads from part b)
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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- A direct shear test was conducted on a specimen of dry sand with a normal stress of 200 kN/m2. Failure occurred at a shear stress of 175 kN/m2. The size of the specimen testedwas 75 mm × 75 mm × 30 mm (height). Determine the angle of friction, . For a normal stress of 150 kN/m2, what shear force would be required to cause failure of the specimen?arrow_forwardplease answer my homework problem. I’m struggling show all work so I can learn I will leave thumbs up!!arrow_forwardQ # 3. Following data are given for a direct shear test conducted on dry sand: Specimen dimensions: 63 mm × 63 mm and 25 mm (height). Normal stress: 105 kN/m². Shear force at failure: 300 N Determine the angle of friction, for a normal stress of 180 kN/m². b. What shear force is required to cause failure? а.arrow_forward
- Soil Mechanismarrow_forward12.3arrow_forwardThe angle of friction of a compacted dry sand is 37 degrees. In a direct shear test on the sand, anormal stress of 150 kN/m^2 was applied. The size of the specimen was 50mmx50mx30mm(height) SITUATION 1 a. Compute the shearing stress Your answer b. What shear force will cause shear failure? Your answer c. Determine the shear stress at a depth of 3m if the void ratio of the soil is 0.60. Gs Of sand is 2.70arrow_forward
- 1 .For a direct shear test on a dry sand, the following are given:• Specimen size: 75 mm 75 mm 30 mm (height)• Normal stress: 200 kN/m2• Shear stress at failure: 175 kN/m2a. Determine the angle of friction, f b. For a normal stress of 150 kN/m2, what shear force is required to cause failure?in the specimearrow_forward5,The following result were obtained from a series of undrained triaxial test carried out on undisturbed samples of a compacted soil: Each sample is originally 76mm Long and 38mm in diameter, experienced a vertical deformation 2 mm Draw the strength envelop & Determine coulomb equation for the shear strength of the soil in terms of total stresses, Cell pressure (kPa) Additional axial load at Failure (N) 100 350 250 400 400 500arrow_forwardFollowing data are given for a direct shear test conducted on dry sand: Cylindrical specimen dimensions: diameter = 50 mm and height = 25 mm Normal stress: 0.15 N/mm2 Shear force at failure: 276 N Determine the angle of friction of this soil. Normal Stress = 0.15 N/mm2 Shear Force = 276 N Shear Force = 276 Narrow_forward
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