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A dense dry sand sample under tri-axial test has the confining pressure of 20 t/m2. The angle of internal! friction is 350. The sample will fail at normal pressure of about:
- a)30.5 t/m2
- b)31.5 t/m2
- c)32.5 t/m2
- d)33.5 t/m2
Correct answer is option 'B'. Can you explain this answer?
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A dense dry sand sample under tri-axial test has the confining pressur...
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A dense dry sand sample under tri-axial test has the confining pressur...
Given:
Confining pressure = 20 t/m2
Angle of internal friction = 35°
To find:
Normal pressure at failure
Solution:
The normal pressure at failure can be determined using the equation:
σ_n = σ_c + (σ_c - σ_p) * tan^2(ϕ/2)
Where:
σ_n = Normal pressure at failure
σ_c = Confining pressure
σ_p = Pore pressure
ϕ = Angle of internal friction
In this case, the pore pressure is not given, so we can assume σ_p = 0.
Step 1: Calculate the effective confining pressure
The effective confining pressure (σ'_c) can be calculated using the formula:
σ'_c = σ_c - σ_p
Since σ_p = 0, σ'_c = σ_c = 20 t/m2
Step 2: Calculate the normal pressure at failure
Using the formula:
σ_n = σ_c + (σ_c - σ_p) * tan^2(ϕ/2)
Substituting the given values:
σ_n = 20 + (20 - 0) * tan^2(35/2)
Using a scientific calculator:
tan^2(35/2) ≈ 0.8409
σ_n ≈ 20 + (20 - 0) * 0.8409
≈ 20 + 20 * 0.8409
≈ 20 + 16.818
≈ 36.818 t/m2
Step 3: Convert the result to the nearest option
The normal pressure at failure is approximately 36.818 t/m2. The nearest option is 31.5 t/m2 (option B).
Therefore, the correct answer is option B) 31.5 t/m2.
Confining pressure = 20 t/m2
Angle of internal friction = 35°
To find:
Normal pressure at failure
Solution:
The normal pressure at failure can be determined using the equation:
σ_n = σ_c + (σ_c - σ_p) * tan^2(ϕ/2)
Where:
σ_n = Normal pressure at failure
σ_c = Confining pressure
σ_p = Pore pressure
ϕ = Angle of internal friction
In this case, the pore pressure is not given, so we can assume σ_p = 0.
Step 1: Calculate the effective confining pressure
The effective confining pressure (σ'_c) can be calculated using the formula:
σ'_c = σ_c - σ_p
Since σ_p = 0, σ'_c = σ_c = 20 t/m2
Step 2: Calculate the normal pressure at failure
Using the formula:
σ_n = σ_c + (σ_c - σ_p) * tan^2(ϕ/2)
Substituting the given values:
σ_n = 20 + (20 - 0) * tan^2(35/2)
Using a scientific calculator:
tan^2(35/2) ≈ 0.8409
σ_n ≈ 20 + (20 - 0) * 0.8409
≈ 20 + 20 * 0.8409
≈ 20 + 16.818
≈ 36.818 t/m2
Step 3: Convert the result to the nearest option
The normal pressure at failure is approximately 36.818 t/m2. The nearest option is 31.5 t/m2 (option B).
Therefore, the correct answer is option B) 31.5 t/m2.
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Question Description
A dense dry sand sample under tri-axial test has the confining pressure of 20 t/m2. The angle of internal! friction is 350. The sample will fail at normal pressure of about:a)30.5 t/m2b)31.5 t/m2c)32.5 t/m2d)33.5 t/m2Correct answer is option 'B'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A dense dry sand sample under tri-axial test has the confining pressure of 20 t/m2. The angle of internal! friction is 350. The sample will fail at normal pressure of about:a)30.5 t/m2b)31.5 t/m2c)32.5 t/m2d)33.5 t/m2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A dense dry sand sample under tri-axial test has the confining pressure of 20 t/m2. The angle of internal! friction is 350. The sample will fail at normal pressure of about:a)30.5 t/m2b)31.5 t/m2c)32.5 t/m2d)33.5 t/m2Correct answer is option 'B'. Can you explain this answer?.
A dense dry sand sample under tri-axial test has the confining pressure of 20 t/m2. The angle of internal! friction is 350. The sample will fail at normal pressure of about:a)30.5 t/m2b)31.5 t/m2c)32.5 t/m2d)33.5 t/m2Correct answer is option 'B'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A dense dry sand sample under tri-axial test has the confining pressure of 20 t/m2. The angle of internal! friction is 350. The sample will fail at normal pressure of about:a)30.5 t/m2b)31.5 t/m2c)32.5 t/m2d)33.5 t/m2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A dense dry sand sample under tri-axial test has the confining pressure of 20 t/m2. The angle of internal! friction is 350. The sample will fail at normal pressure of about:a)30.5 t/m2b)31.5 t/m2c)32.5 t/m2d)33.5 t/m2Correct answer is option 'B'. Can you explain this answer?.
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A dense dry sand sample under tri-axial test has the confining pressure of 20 t/m2. The angle of internal! friction is 350. The sample will fail at normal pressure of about:a)30.5 t/m2b)31.5 t/m2c)32.5 t/m2d)33.5 t/m2Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A dense dry sand sample under tri-axial test has the confining pressure of 20 t/m2. The angle of internal! friction is 350. The sample will fail at normal pressure of about:a)30.5 t/m2b)31.5 t/m2c)32.5 t/m2d)33.5 t/m2Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A dense dry sand sample under tri-axial test has the confining pressure of 20 t/m2. The angle of internal! friction is 350. The sample will fail at normal pressure of about:a)30.5 t/m2b)31.5 t/m2c)32.5 t/m2d)33.5 t/m2Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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