A dry sand specimen is put through a tri-axial test. The cell pressure is 50 kPa and the deviator stress at failure is 100 kPa. The angle of internal friction for the sand specimen is
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- A dry sand specimen is put through a tri-axial test. The cell pressure is 50 kPa and the deviator stress at failure is 100 kPa. The angle of internal friction for the sand specimen is3. In a drained tri-axial test, a specimen is confined under pressure and axially loaded until failure. The failure conditions of the specimen are analyzed for shear strength. What is the value of the minor principle stress? Half the value of the major principle stress Twice the value of the major principle stress . The same as the confining pressure The same as the cohesion value ● ● ●الشعبة In direct shear test, The area under shear gradually increases as the test progresses. ● ● the stress distribution on the failure plane is uniform none of them In direct shear test continuo talin الاسم/
- Find the angle of internal friction for the sand specimen, if the deviator stress at failure is 100kPa, the cell pressure is 50kPa.if the sample is a dry sand specimen put through a triaxial test. A 30' в 15° C 45' D 37'What is the difference in temperature tolerance between two specimens in Softening Point test? A. +2 °C В. +3 °C C. +1 °C D. +4 °CGive an example of a creep plot for some material. From the creep plot, determine (a). the steady-state creep rate (b) the rupture lifetime.
- A core sample of quartzite, 13 cm long and 6.5 cm in diameter, was tested in unconfined compression. The quartzite was known to have a modulus of elasticity of 5.7 x 10^3 MPa. a. At a pressure of 69 MPa, what would be the reduction in length of the sample? b. At this pressure what would be the vertical load on the sample? C. At 69 MPa stress the core showed an increase in diameter of 0.013 mm. What is the Poisson's Ratio for the sample?Stress condition in the Unconfined compression test represents Drained test Un-drained test Quick testI. Problem Solving: Answer what is being required. Show you complete solution. 1. Compute the tensile stress from the value acquired from the experiment if the breaking load is 105KN. The diameter of the specimen is 150mm and the height is 300mm. 2. Compute the flexural stress of the sample specimen if its dimension is 150mm x 150mm x 500mm in a 3 point loading. The effective length is 400mm. The specimen failed at 44,300N in the middle third. 3. Compression test was applied to a certain type of wood. Compute the stress based on the following dimensions. The grains are parallel with the length. а. Compression Test Dimension (mm) Load (N) Compressive stress (МРa)
- A core sample of granite, 12 cm long and 6 cm in diameter, was tested in unconfined compression. The granite was known to have a modulus of elasticity of 16 x 10^3 MPa. a. At a pressure of 69 MPa, what would be the reduction in length of the sample b. At this pressure what would be the vertical load on the sample? c. At 69 MPa stress the core showed an increase in diameter of 0.011 mm. What is the Poisson's Ratio for the sample?A cylindrical pressure vessel is fabricated from steel plating that has a thickness of 42 mm. The diameter of the pressure vessel is 450 mm and its length is 2.0 m. Determine the maximum internal pressure that can be applied if the longitudinal stress is limited to 140 MPa, and the circumferential stress is limited to 60 MPa. *The compressive strength obtained from the unconfined test is called the unconfined compressive strength (4,). True False Stress condition in the Unconfined compression test represents Drained test O Un-drained test O Quick test The unconfined compression test can be performed on : O dry or crumbly soils any granular soils O any fine cohesive soil any silts, peat, or fissured materials | (Ctrl) -