The results of a direct shear test on a specimen of dry sand are as follows: Normal stress = 96.6 kPa; shear stress at failure = 67.7 kPa. find the magnitude and directions of the principal stresses acting on a soil element within the zone of failure.
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- 20. A cohesionless sand sample was subjected to a triaxial shear test. Failure occurred when the normal stress is 400 KPa and the shear stress is 250 KPa. a. What is major principal stress? b. What is the minor principal stressAssume 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?2. 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.
- Triaxial tests performed on samples of aeolin sand. The failure conditions in terms of effective stress are (ov, 0h) = (515, 100), (1250, 200), (3500, 400), and (5325, 800) kPa. Using (s, t) space, determine the cohesion and friction angle. What is the orientation of the major principal stress with respect to the failure plane? Determine this graphically.A triaxial shear test was performed on a well-drained sand sample. The normal stress on the failure plane and the shear stress on the failure plane, at failure were determined to be 6100 psf and 4600 psf, respectively. a. Determine the angle of internal friction of the sand? b. Determine the angle of the failure plane? c. Determine the maximum principal stress? Please answer this asap. For upvote. Thank you very muchThe 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 2.70
- 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 NAssume 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.Question Table 1 gives data from a standard shearbox(direct shear) test on a sample of 125g of dry sand. The initial dimensions of the sample were 60mmx60mm on plan x20mm in height. The test was caried out at a constant normal effective stress of 50k:Pa. Take the specific gravity of the soil grains Gs=2.65. Plot graphs of (a) shear stress t against shear strain y; (b) volumetric strain evol against shear strain y; (c) specific vohume v against shear strain y. (d) Comment on these graphs and estimate the peak and critical state effective angles of friction of the soil. Table 1: Shearbox Data Relative horizontal Upward vertical movement Shear stress t (kPa) displacement x (mm) of shearbox lid y (mm) 0.00 0.000 0.02 0.002 19 0.04 0.008 34 0.06 0.016 43- 0.08 0.026 47 0.20 0.064 56 0.32 0.128 51 0.48 0.192 46 0.64 0.256 41 0.80 0.288 37 0.96 0.320 34 1.12 0.321 33
- Table 1 gives data from a standard shearbox(direct shear) test on a sample of 125g of dry sand. The initial dimensions of the sample were 60mmx60mm on plan x20mm in height. The test was camried out at a constant normal effective stress of 50kPa. Take the specific gravity of the soil grains Gs=2.65. Plot graphs of (a) shear stress r against shear strain y; (b) volumetric strain evol against shear strain y; (c) specific volume v against shear strain y. (d) Comment on these graphs and estimate the peak and critical state effective angles of friction of the soil. Table 1: Shearbox Data Relative horizontal displacement x (mm) of shearbox lid y (mm) Upward vertical movement Shear stress t (kPa) 0.00 0.000 0.02 0.002 19 0.04 0.008 34 0.06 0.016 43- 0.08 0.026 47 0.20 0.064 56 0.32 0.128 51 0.48 0.192 46 0.64 0.256 41 0.80 0.288 37 0.96 0.320 34 1.12 0.321 33A specimen is subjected to a tri-axial test. The soil specimen is cohesionless. If the shear stress that cause failure is 3oo kPa and the normal stress at failure is only 475 kPa. 1. Determine the angle of shearing resistance. 2.Determine the angle that the failure plane measured from the major principal plane 3.Determine the maximum principal stress at which failure is expected to occur2. A triaxial shear test was performed on a well-drained sand sample. The normal stress on the failure plane and the shear stress on the failure plane, at failure was determined to be 6,300 psf and 4,200 psf, respectively. a. Determine the angle of internal friction of the sand. b. Determine the angle of failure plane. c. Determine the maximum principal stress.