Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN: 9781305635180
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Chapter 10, Problem 10.3P
To determine
Find the mass of the sand specimen.
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The size of a sand specimen in a direct shear test was 50 mm x 50 mm x 30 mm (height). It is known that, for the sand, tan phi prime = 0.65/e (where e = void ratio) and the specific gravity of soil solids Gs = 2.65. During the test, a normal stress of 140 kN/m^2 was applied. Failure occured at a shear stress of 105 kN/m^2. What was the mass of the sand specimen?
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 much
The 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
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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- Foundation: Direct Shear Testarrow_forwardFriction angle of dry sand is 35°. In a direct shear test on this sand, a normal stress of 150kPa was applied. What shear stress will cause the failure? (a) 150kPa (b) 50kPa (c) 78kPa (d) 105kPaarrow_forwardA consolidated-undrained test was conducted on a normally consolidated clay sample. The results are as follows: Undrained angle of friction = 38.65°, Drained angle of friction = 62.59°, Deviator stress at failure = 509 lb/ft^2. Determine the pore water stress at failure in lb/ft^2.arrow_forward
- A consolidated-undrained triaxial compression test was performed on a normalconsolidated clay sample. During the experiment, the confining pressure was 140 kPa, thedeviator stress was 125 kPa, and the pore water pressure was 75 kPa at the time of failure.According to the information given:i- Find the consolidated-undrained internal friction angle of the clay. ii- Find the drained friction angle of the clayarrow_forwardA consolidated-undrained test was conducted on a normally consolidated clay sample. The results are as follows: Undrained angle of friction = 37.16 degrees, Drained angle of friction = 63.71 degrees, Deviator stress at failure = 530 lb/ft^2. Determine the pore water stress at failure in lb/ft^2arrow_forward20. 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 stressarrow_forward
- Problem 4.4 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.arrow_forwardA dilatometer test was conducted in a clay deposit. The groundwater table was located at a depth of 3 m below the surface. At a depth of 8 m below the surface, the contact pressure was and the expansion stress was 350 kN/m2Determine the following:a. Coefficient of at-rest earth pressure,b. Overconsolidation ratio, OCRc. Modulus of elasticity,Assume stress at a depth of 8 m to be and 95 kN/m2 , poisson ratio=0.35.arrow_forwardQuestion 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 33arrow_forward
- 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.arrow_forwardA triaxial shear test was performed on a well-drained sand sample. The normal stress on the failure plane and the shearing stress on the failure plane were determined to be 75kPa and 42kPa, respectively. Determine the angle of internal friction of the sand, in degrees.Determine the axial stress applied to the specimen, in kPa.arrow_forward6 A consolidated-undrained test was conducted on a normally consolidated clay sample. The results are as follows: Undrained angle of friction = 36.90°, Drained angle of friction = 67.62°, Deviator stress at failure = 573 lb/ft^2. Determine the pore water stress at failure in lb/ft^2. Round off to three decimal places.arrow_forward
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