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?
Assume that both a triaxial shear test and a direct shear test are to be performed on a sample of dry
sand. When the triaxial shear test is performed, the specimen fails when the major and minor principal
stresses are 80 and 20 lb/in.2, respectively. When the direct shear test is performed, what shear
strength can be expected if the normal stress is 4000 lb/ft2?
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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- Q # 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_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_forwardA 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_forward
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- Following 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_forwardFor a dry sand specimen in a direct shear test box, the following are given: • Angle of friction: 38° • Size of specimen: 2 in. X 2 in. X 1.2 in. (height) • Normal stress: 20 lb/in.? Determine the shear force required to cause failure.arrow_forwardAssume that both a triaxial shear test and a direct shear test are to be performed on a sample of dry sand. When the triaxial shear test is performed, the specimen fails when the major and minor principal stresses are 80 and 20 lbs/in^2, respectively. When the direct shear test is performed, what shear strength can be expected if the normal stress is 4000 lbs/ft^2? Show all work.arrow_forward
- Problem # 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_forwardA sample of sand is sheared in the following manner. First an all around normal stresso 350 kPa is applied to the sample. Next a shear stress is applied to the horizontal pia while the normal stress remains constant. The shear stresses at failure are + 186 kPa. Ihe initial stress and stress condition at failure are shown below 350 kPa 350 kPa 186 kPa 350 kPa 350 kPa ー550 kPa 550 kPa 186 kPa 350 kPa 350 kPa Initial Conditions At Failure Determine the Following: fo A) Assuming c, = 0, draw a failure envelope for the soil. Define the friction angle ( the soil. B) Draw the Mohr's Circle at failure for the test. C) Show the location of the pole. D) Define the maximum and minimum principal stresses at failure. E) Define the normal stress corresponding to the maximum shear stressarrow_forward2. A sample of saturated clay of height 20 mm and water content 30% was tested in an oedometer. Loading and unloading of the sample were carried out. The thickness Hf of the sample at the end of each stress increment/decrement is shown in the table below. o: (kPa) H, (mm) 100 200 400 200 18.68 100 18.75 20 19.31 18.62 a. Plot the results as void ratio versus o'z (log scale). b. Determine Cc and Cr.arrow_forward
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