A consolidated drained tri-axial test was conducted on a normally consolidated clay. The results were as foHows: 03 = 300 kPa (confining pressure) Deviators stress = 300 kPa Compute the angle of shearing resistance. Compute the angle that the failure plane makes with the major principal. stress. Compute the shear stress on the failure plane.
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- 4. Flow net is used for the determination of (a) Quantity of seepage (b) Hydrostatic pressure (c) Seepage pressure (d) All of them. 5. Both the shear stress and the normal stress on the plane of failure are measured directly in (a) CD triaxial shear test (b) CU triaxial shear test (c) Direct shear test (d) Unconfined compression test 6. In a consolidated undrained test on an over-consolidated clay, the volume of the soil sample during shear: (a) decreases (b) first decreases and then increases (c) first increases and then decreases (d) remains unchanged.8. A cohesionless (c' = 0 kPa) soil specimen was subjected to a consolidated-undrained triaxial test. The total stress path, the deviator stress, and the pwp are shown in the figures below. No scale is intended in the figure of the stress path. a. Determine o'₁ and 0's at failure. [o'₁ = 345 kPa, o's= 140 kPa] b. Determine the effective friction angle (') of the soil. [$' = 25°] C. Determine the 'A' pore water pressure parameter at failure. [A+ = 0.29] 250 -Deviator Stress q (kPa), 200 150 100 50 200 p (kPa) Deviator stress or PW/P (kPa) 5 10 Vertical strain (%) ---PWP 15 20A CID TXC test was performed on a sample of sand. At the end of consolidation, the back pressure was 140 kPa, and the cell pressure was 340 kPa. At failure, the deviator stress was (ơ1 – 03)f = 200 kPa. a. What was the total pore water pressure, u, at failure? b. Calculate the principle effective stresses, ơi' and os', at failure. c. Calculate the friction angle, d', for this soil.
- A series of unconsolidated undrained (UU) triaxial compression tests have been performed on two "identical" clay specimens. The test results are as follows: Test Number Confining Pressure (kPa)| Deviator Stress at Failure (kPa) 102 96 (i) (ii) 1 2 50 100 Plot the total stress Mohr circles for both specimens at failure (indicate the major and minor principal stresses on the plot). Draw the total stress Mohr-Coulomb failure envelop (i.e., the "p=0" envelop). Estimate the undrained shear strength.3. 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 ● ● ●The following results were obtained from a series of consolidation undrained (CU) shear test. Determine the cohesion and the angle of shearing resistance.
- A sample of dry sand is tested in direct shear. Theshear box holding the sample has a circular crosssection with a diameter of 50 mm. The normal(compressive) load imposed is 200 N. The sam-ple shears when the shear force is 130 N.(a) Determine the test normal stress and shearstress at failure.(b) Determine the angle of internal frictionfor this soil.(c) What is the probable condition of the testedsample (dense, loose, etc.)?The following results were obtained from a liquid limit test on a clay using the Casagrande cup device. Number of blows Moisture content (%) 6 12 20 28 32 52.5 47.1 43.2 38.6 37.0 a) Determine the liquid limit b) If the natural water content is 38% and the plastic limit is 23%, calculate the liquidity index. c) Would you expect a brittle failure? WHY?(a) In an unconfined compression test on a saturated clay, the maximum proving ring dial reading recorded was 240x103 mm, when the axial shortening of the specimen having an initial height of 70 mm and an initial diameter of 36 mm, is 12 mm. If the calibration factor of the proving ring is 3.2 N/10³ mm, calculate the unconfined compressive strength and the undrained shear strength of the clay.
- specimen In a consolidated-drained triaxial test on aclay, the failed at the deviator Stress of 124 kN/m² effective stress friction angle is known tod 31° 31 What was the effective confining pressure at failure? If the be -: 4.34 kg Weight of cement Weight of Fine aggregate kg 8.68 : 17.36 kg M20carede Weight of coarse aggregate Weight of Water 2.36 L. Slump Test Results Group No Diagram of your slump N/mm? Mix Strength 20 mm Slump 25 Type of SlumpTension specimens (diameter d0 0.500 in., gage length L0 2.00 in.) made of structural materials A and B are tested to failure in tension. (a) At failure the distances between the gage marks are LAf 2.90 in. and LBf 2.22 in.; the corresponding diameters at the failure cross sections are dAf 0.263 in. and dBf 0.471 in., respectively. Determine the percent elongation in 2 in. and the percent reduction in area for these two materials, and classify each material as either brittle or ductile. (b) From these tensile tests the following data are also obtained: EA 10.0 103 ksi, (Y)A 5 ksi, (U)A 13 ksi; EB 10.4 103 ksi, (Y)B 73 ksi, (U)B 83 ksi. From the data given here, make rough sketches (to scale) of the stress-strain diagrams of materials A and B.