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A specimen of fine dry sand when subjected to triaxial compression test, failed at a deviator stress of 200 kN/m2. It failed with a pronounced failure plane with an angle of 36 degrees to the axis of the sample. The value of lateral pressure to which the specimen would have been subjected. (in kN/m2 , round up to 1 decimal place))
- a)223
- b)224.1
Correct answer is between '223,224.1'. Can you explain this answer?
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Most Upvoted Answer
A specimen of fine dry sand when subjected to triaxial compression te...
Triaxial Compression Test:
The triaxial compression test is a laboratory test used to determine the strength and stress-strain behavior of soil or rock specimens under different confining pressures. In this test, a cylindrical specimen is subjected to an axial load while confining pressure is applied externally to the lateral surface of the specimen.
Deviator Stress and Failure Plane:
Deviator stress is the difference between the major and minor principal stresses in a triaxial compression test. When a specimen fails, it typically forms a failure plane that is inclined at an angle to the axis of the sample.
Given Information:
- Deviator stress at failure = 200 kN/m2
- Angle of failure plane to the axis of the sample = 36 degrees
Calculating Lateral Pressure:
The lateral pressure is the confining pressure acting on the specimen. To calculate it, we need to find the magnitude of the major and minor principal stresses.
Step 1: Finding the Magnitude of the Major Principal Stress:
The major principal stress (σ1) is equal to the sum of the deviator stress (σd) and the mean stress (σm).
σ1 = σm + σd
Given that σd = 200 kN/m2, we need to find σm.
Step 2: Finding the Mean Stress:
The mean stress (σm) can be calculated using the following formula:
σm = (σ1 + 2σ3) / 3
Here, σ3 is the minor principal stress. To find σ3, we can use the angle of the failure plane.
Step 3: Finding the Magnitude of the Minor Principal Stress:
The minor principal stress (σ3) can be calculated using the following formula:
σ3 = σ1 - σd
Given that the angle of the failure plane is 36 degrees, we can calculate σ3 as follows:
σ3 = σ1 - σd = σ1 - 200
Now, we can substitute the value of σ3 in the formula for the mean stress to find σm.
Step 4: Finding the Mean Stress:
σm = (σ1 + 2σ3) / 3
Substituting the value of σ3, we get:
σm = (σ1 + 2(σ1 - 200)) / 3
Step 5: Finding the Lateral Pressure:
The lateral pressure is equal to the mean stress (σm).
Therefore, the lateral pressure to which the specimen would have been subjected is:
Lateral Pressure = σm
Numerical Calculation:
Let's perform the numerical calculation to find the value of the lateral pressure.
Given that σd = 200 kN/m2 and the angle of the failure plane is 36 degrees.
Step 1: Finding the Magnitude of the Major Principal Stress:
σ1 = σm + σd
σ1 = σm + 200
Step 2: Finding the Mean Stress:
σ3 = σ1 - σd
σ3 = σ1 - 200
Step 3: Finding the Magnitude of the Minor Principal Stress:
σ3 = σ1 - 200
Step 4: Finding the Mean Stress:
σm
The triaxial compression test is a laboratory test used to determine the strength and stress-strain behavior of soil or rock specimens under different confining pressures. In this test, a cylindrical specimen is subjected to an axial load while confining pressure is applied externally to the lateral surface of the specimen.
Deviator Stress and Failure Plane:
Deviator stress is the difference between the major and minor principal stresses in a triaxial compression test. When a specimen fails, it typically forms a failure plane that is inclined at an angle to the axis of the sample.
Given Information:
- Deviator stress at failure = 200 kN/m2
- Angle of failure plane to the axis of the sample = 36 degrees
Calculating Lateral Pressure:
The lateral pressure is the confining pressure acting on the specimen. To calculate it, we need to find the magnitude of the major and minor principal stresses.
Step 1: Finding the Magnitude of the Major Principal Stress:
The major principal stress (σ1) is equal to the sum of the deviator stress (σd) and the mean stress (σm).
σ1 = σm + σd
Given that σd = 200 kN/m2, we need to find σm.
Step 2: Finding the Mean Stress:
The mean stress (σm) can be calculated using the following formula:
σm = (σ1 + 2σ3) / 3
Here, σ3 is the minor principal stress. To find σ3, we can use the angle of the failure plane.
Step 3: Finding the Magnitude of the Minor Principal Stress:
The minor principal stress (σ3) can be calculated using the following formula:
σ3 = σ1 - σd
Given that the angle of the failure plane is 36 degrees, we can calculate σ3 as follows:
σ3 = σ1 - σd = σ1 - 200
Now, we can substitute the value of σ3 in the formula for the mean stress to find σm.
Step 4: Finding the Mean Stress:
σm = (σ1 + 2σ3) / 3
Substituting the value of σ3, we get:
σm = (σ1 + 2(σ1 - 200)) / 3
Step 5: Finding the Lateral Pressure:
The lateral pressure is equal to the mean stress (σm).
Therefore, the lateral pressure to which the specimen would have been subjected is:
Lateral Pressure = σm
Numerical Calculation:
Let's perform the numerical calculation to find the value of the lateral pressure.
Given that σd = 200 kN/m2 and the angle of the failure plane is 36 degrees.
Step 1: Finding the Magnitude of the Major Principal Stress:
σ1 = σm + σd
σ1 = σm + 200
Step 2: Finding the Mean Stress:
σ3 = σ1 - σd
σ3 = σ1 - 200
Step 3: Finding the Magnitude of the Minor Principal Stress:
σ3 = σ1 - 200
Step 4: Finding the Mean Stress:
σm
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Community Answer
A specimen of fine dry sand when subjected to triaxial compression te...
Σd = ∆σ = 200 kN/m2
We have
Angle of failure = 360 to the axis of the sample
θc = 90 - 36 = 54o
C = 0
σ1 = ∆σ + σ3 = 200 + σ3
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A specimen of fine dry sand when subjected to triaxial compression test, failed at a deviator stress of 200 kN/m2. It failed with a pronounced failure plane with an angle of 36 degrees to the axis of the sample. The value of lateral pressure to which the specimen would have been subjected. (in kN/m2 , round up to 1 decimal place))a)223b)224.1Correct answer is between '223,224.1'. Can you explain this answer?
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