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An undrained triaxial test was carried on an undistribed sample of sand with cell pressure of 200 kN/m2. The additional axial load of failure is 342kN. The sample is originally 76mm long and 33mm in diameter is normal stress (σn) at 6m depth is 100 kN/m2 and vertical deformation is 5.1mm then the shear strength in (kN/m2) of sample is
- a)52
- b)45
- c)13
- d)56
Correct answer is option 'B'. Can you explain this answer?
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An undrained triaxial test was carried on an undistribed sample of san...
Σ1) at failure is 400 kN/m2?
We can use the formula for the stress ratio (K) in an undrained triaxial test:
K = (σ1 - σ3) / (σ1 - σ3) + 2Pc / 3
where σ3 is the minor principal stress (equal to zero in an undrained test) and Pc is the cell pressure.
We can rearrange this formula to solve for Pc:
Pc = (3K / 2) * (σ1 - σ3) - 1/2 * σ1
At failure, the axial stress (σ1) is equal to the applied load (342kN) divided by the cross-sectional area of the sample (π/4 * 0.033^2 = 8.56e-4 m2):
σ1 = 342 kN / 8.56e-4 m2 = 398,598 kN/m2
We are given that σ3 = 0, so we can plug in these values and solve for Pc:
K = (σ1 - σ3) / (σ1 - σ3) + 2Pc / 3
K = σ1 / (σ1 + 2Pc / 3)
0.5 = 398,598 kN/m2 / (398,598 kN/m2 + 2Pc / 3)
0.5 * (398,598 kN/m2 + 2Pc / 3) = 398,598 kN/m2
199,299 kN/m2 + 1/3 * Pc = 398,598 kN/m2
1/3 * Pc = 199,299 kN/m2
Pc = 597,897 kN/m2
Finally, we can use the stress ratio formula again to solve for the minor principal stress (σ3) at failure:
K = (σ1 - σ3) / (σ1 - σ3) + 2Pc / 3
0.5 = (398,598 kN/m2 - σ3) / (398,598 kN/m2 - σ3 + 2 * 597,897 kN/m2 / 3)
0.5 * (398,598 kN/m2 - σ3 + 2 * 597,897 kN/m2 / 3) = 398,598 kN/m2 - σ3
199,299 kN/m2 + 1/3 * 597,897 kN/m2 - 1/2 * σ3 = 398,598 kN/m2 - σ3
1/6 * σ3 = 199,299 kN/m2 - 1/3 * 597,897 kN/m2
σ3 = -199,299 kN/m2
This negative value for σ3 means that the sample experienced tensile failure in the radial direction, which is not uncommon in undrained tests on cohesive soils. However, the given problem statement specifies that the sample is sand, which is typically considered a non-cohesive soil. This discrepancy may be a mistake in the problem statement, or it may indicate that the sample was not truly representative of the sand deposit being tested.
We can use the formula for the stress ratio (K) in an undrained triaxial test:
K = (σ1 - σ3) / (σ1 - σ3) + 2Pc / 3
where σ3 is the minor principal stress (equal to zero in an undrained test) and Pc is the cell pressure.
We can rearrange this formula to solve for Pc:
Pc = (3K / 2) * (σ1 - σ3) - 1/2 * σ1
At failure, the axial stress (σ1) is equal to the applied load (342kN) divided by the cross-sectional area of the sample (π/4 * 0.033^2 = 8.56e-4 m2):
σ1 = 342 kN / 8.56e-4 m2 = 398,598 kN/m2
We are given that σ3 = 0, so we can plug in these values and solve for Pc:
K = (σ1 - σ3) / (σ1 - σ3) + 2Pc / 3
K = σ1 / (σ1 + 2Pc / 3)
0.5 = 398,598 kN/m2 / (398,598 kN/m2 + 2Pc / 3)
0.5 * (398,598 kN/m2 + 2Pc / 3) = 398,598 kN/m2
199,299 kN/m2 + 1/3 * Pc = 398,598 kN/m2
1/3 * Pc = 199,299 kN/m2
Pc = 597,897 kN/m2
Finally, we can use the stress ratio formula again to solve for the minor principal stress (σ3) at failure:
K = (σ1 - σ3) / (σ1 - σ3) + 2Pc / 3
0.5 = (398,598 kN/m2 - σ3) / (398,598 kN/m2 - σ3 + 2 * 597,897 kN/m2 / 3)
0.5 * (398,598 kN/m2 - σ3 + 2 * 597,897 kN/m2 / 3) = 398,598 kN/m2 - σ3
199,299 kN/m2 + 1/3 * 597,897 kN/m2 - 1/2 * σ3 = 398,598 kN/m2 - σ3
1/6 * σ3 = 199,299 kN/m2 - 1/3 * 597,897 kN/m2
σ3 = -199,299 kN/m2
This negative value for σ3 means that the sample experienced tensile failure in the radial direction, which is not uncommon in undrained tests on cohesive soils. However, the given problem statement specifies that the sample is sand, which is typically considered a non-cohesive soil. This discrepancy may be a mistake in the problem statement, or it may indicate that the sample was not truly representative of the sand deposit being tested.
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An undrained triaxial test was carried on an undistribed sample of sand with cell pressure of 200 kN/m2. The additional axial load of failure is 342kN. The sample is originally 76mm long and 33mm in diameter is normal stress (σn) at 6m depth is 100 kN/m2 and vertical deformation is 5.1mm then the shear strength in (kN/m2) of sample isa)52b)45c)13d)56Correct answer is option 'B'. Can you explain this answer?
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An undrained triaxial test was carried on an undistribed sample of sand with cell pressure of 200 kN/m2. The additional axial load of failure is 342kN. The sample is originally 76mm long and 33mm in diameter is normal stress (σn) at 6m depth is 100 kN/m2 and vertical deformation is 5.1mm then the shear strength in (kN/m2) of sample isa)52b)45c)13d)56Correct answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An undrained triaxial test was carried on an undistribed sample of sand with cell pressure of 200 kN/m2. The additional axial load of failure is 342kN. The sample is originally 76mm long and 33mm in diameter is normal stress (σn) at 6m depth is 100 kN/m2 and vertical deformation is 5.1mm then the shear strength in (kN/m2) of sample isa)52b)45c)13d)56Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An undrained triaxial test was carried on an undistribed sample of sand with cell pressure of 200 kN/m2. The additional axial load of failure is 342kN. The sample is originally 76mm long and 33mm in diameter is normal stress (σn) at 6m depth is 100 kN/m2 and vertical deformation is 5.1mm then the shear strength in (kN/m2) of sample isa)52b)45c)13d)56Correct answer is option 'B'. Can you explain this answer?.
An undrained triaxial test was carried on an undistribed sample of sand with cell pressure of 200 kN/m2. The additional axial load of failure is 342kN. The sample is originally 76mm long and 33mm in diameter is normal stress (σn) at 6m depth is 100 kN/m2 and vertical deformation is 5.1mm then the shear strength in (kN/m2) of sample isa)52b)45c)13d)56Correct answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An undrained triaxial test was carried on an undistribed sample of sand with cell pressure of 200 kN/m2. The additional axial load of failure is 342kN. The sample is originally 76mm long and 33mm in diameter is normal stress (σn) at 6m depth is 100 kN/m2 and vertical deformation is 5.1mm then the shear strength in (kN/m2) of sample isa)52b)45c)13d)56Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An undrained triaxial test was carried on an undistribed sample of sand with cell pressure of 200 kN/m2. The additional axial load of failure is 342kN. The sample is originally 76mm long and 33mm in diameter is normal stress (σn) at 6m depth is 100 kN/m2 and vertical deformation is 5.1mm then the shear strength in (kN/m2) of sample isa)52b)45c)13d)56Correct answer is option 'B'. Can you explain this answer?.
Solutions for An undrained triaxial test was carried on an undistribed sample of sand with cell pressure of 200 kN/m2. The additional axial load of failure is 342kN. The sample is originally 76mm long and 33mm in diameter is normal stress (σn) at 6m depth is 100 kN/m2 and vertical deformation is 5.1mm then the shear strength in (kN/m2) of sample isa)52b)45c)13d)56Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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