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A cylindrical specimen of saturated soil failed under an axial vertical stress of 100 kN/m2 when it was laterally unconfined. The failure plane was inclined to the horizontal plane at an angle of 45°. The values of cohesion and angle of internal friction for the soil are respectively.
- a)0,5 N/mm2 and 30°
- b)0.05 N/mm2 and 0°
- c)0.2 N/mm2 and 0°
- d)0.05 N/mm2 and 45°
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A cylindrical specimen of saturated soil failed under an axial vertica...
For an unconfined compression test:
S = C = 100/2 = 50kN/m2
= 0.05 N/mm2
A laterally unconfined compression test is conducted on soils which can stand without confinement. Also, for clay angle of internal friction = φ = 0°
S = C = 100/2 = 50kN/m2
= 0.05 N/mm2
A laterally unconfined compression test is conducted on soils which can stand without confinement. Also, for clay angle of internal friction = φ = 0°
Most Upvoted Answer
A cylindrical specimen of saturated soil failed under an axial vertica...
Degrees. The effective angle of internal friction of the soil is 30 degrees. Calculate the shear strength parameters of the soil.
Solution:
Given:
Axial vertical stress (σ1) = 100 kN/m2
Angle of inclination (φ) = 45 degrees
Effective angle of internal friction (ϕ') = 30 degrees
We know that the Mohr-Coulomb failure criterion is given by:
τ = σ' * tan(ϕ') + c'
where,
τ = shear stress
σ' = effective normal stress
ϕ' = effective angle of internal friction
c' = effective cohesion
For a cylindrical specimen, the failure occurs on a plane inclined at an angle of 45 degrees to the horizontal plane. Therefore, the normal stress on the failure plane is given by:
σf = σ1 * sin(45) = 70.7 kN/m2
The effective normal stress is given by:
σ' = σf - u
where u is the pore water pressure. As the soil is saturated, u = 0.
Therefore, σ' = σf = 70.7 kN/m2
Substituting the values in the Mohr-Coulomb failure criterion, we get:
τ = 70.7 * tan(30) + c'
At failure, the shear stress (τ) is equal to the axial vertical stress (σ1).
Therefore, 100 = 70.7 * tan(30) + c'
Solving for c', we get:
c' = 27.3 kN/m2
Therefore, the shear strength parameters of the soil are:
Effective angle of internal friction (ϕ') = 30 degrees
Effective cohesion (c') = 27.3 kN/m2
Solution:
Given:
Axial vertical stress (σ1) = 100 kN/m2
Angle of inclination (φ) = 45 degrees
Effective angle of internal friction (ϕ') = 30 degrees
We know that the Mohr-Coulomb failure criterion is given by:
τ = σ' * tan(ϕ') + c'
where,
τ = shear stress
σ' = effective normal stress
ϕ' = effective angle of internal friction
c' = effective cohesion
For a cylindrical specimen, the failure occurs on a plane inclined at an angle of 45 degrees to the horizontal plane. Therefore, the normal stress on the failure plane is given by:
σf = σ1 * sin(45) = 70.7 kN/m2
The effective normal stress is given by:
σ' = σf - u
where u is the pore water pressure. As the soil is saturated, u = 0.
Therefore, σ' = σf = 70.7 kN/m2
Substituting the values in the Mohr-Coulomb failure criterion, we get:
τ = 70.7 * tan(30) + c'
At failure, the shear stress (τ) is equal to the axial vertical stress (σ1).
Therefore, 100 = 70.7 * tan(30) + c'
Solving for c', we get:
c' = 27.3 kN/m2
Therefore, the shear strength parameters of the soil are:
Effective angle of internal friction (ϕ') = 30 degrees
Effective cohesion (c') = 27.3 kN/m2
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A cylindrical specimen of saturated soil failed under an axial vertical stress of 100 kN/m2 when it was laterally unconfined. The failure plane was inclined to the horizontal plane at an angle of 45°. The values of cohesion and angle of internal friction for the soil are respectively.a)0,5 N/mm2 and 30°b)0.05 N/mm2 and 0°c)0.2 N/mm2 and 0°d)0.05 N/mm2 and 45°Correct 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 cylindrical specimen of saturated soil failed under an axial vertical stress of 100 kN/m2 when it was laterally unconfined. The failure plane was inclined to the horizontal plane at an angle of 45°. The values of cohesion and angle of internal friction for the soil are respectively.a)0,5 N/mm2 and 30°b)0.05 N/mm2 and 0°c)0.2 N/mm2 and 0°d)0.05 N/mm2 and 45°Correct 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 cylindrical specimen of saturated soil failed under an axial vertical stress of 100 kN/m2 when it was laterally unconfined. The failure plane was inclined to the horizontal plane at an angle of 45°. The values of cohesion and angle of internal friction for the soil are respectively.a)0,5 N/mm2 and 30°b)0.05 N/mm2 and 0°c)0.2 N/mm2 and 0°d)0.05 N/mm2 and 45°Correct answer is option 'B'. Can you explain this answer?.
A cylindrical specimen of saturated soil failed under an axial vertical stress of 100 kN/m2 when it was laterally unconfined. The failure plane was inclined to the horizontal plane at an angle of 45°. The values of cohesion and angle of internal friction for the soil are respectively.a)0,5 N/mm2 and 30°b)0.05 N/mm2 and 0°c)0.2 N/mm2 and 0°d)0.05 N/mm2 and 45°Correct 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 cylindrical specimen of saturated soil failed under an axial vertical stress of 100 kN/m2 when it was laterally unconfined. The failure plane was inclined to the horizontal plane at an angle of 45°. The values of cohesion and angle of internal friction for the soil are respectively.a)0,5 N/mm2 and 30°b)0.05 N/mm2 and 0°c)0.2 N/mm2 and 0°d)0.05 N/mm2 and 45°Correct 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 cylindrical specimen of saturated soil failed under an axial vertical stress of 100 kN/m2 when it was laterally unconfined. The failure plane was inclined to the horizontal plane at an angle of 45°. The values of cohesion and angle of internal friction for the soil are respectively.a)0,5 N/mm2 and 30°b)0.05 N/mm2 and 0°c)0.2 N/mm2 and 0°d)0.05 N/mm2 and 45°Correct answer is option 'B'. Can you explain this answer?.
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A cylindrical specimen of saturated soil failed under an axial vertical stress of 100 kN/m2 when it was laterally unconfined. The failure plane was inclined to the horizontal plane at an angle of 45°. The values of cohesion and angle of internal friction for the soil are respectively.a)0,5 N/mm2 and 30°b)0.05 N/mm2 and 0°c)0.2 N/mm2 and 0°d)0.05 N/mm2 and 45°Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A cylindrical specimen of saturated soil failed under an axial vertical stress of 100 kN/m2 when it was laterally unconfined. The failure plane was inclined to the horizontal plane at an angle of 45°. The values of cohesion and angle of internal friction for the soil are respectively.a)0,5 N/mm2 and 30°b)0.05 N/mm2 and 0°c)0.2 N/mm2 and 0°d)0.05 N/mm2 and 45°Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A cylindrical specimen of saturated soil failed under an axial vertical stress of 100 kN/m2 when it was laterally unconfined. The failure plane was inclined to the horizontal plane at an angle of 45°. The values of cohesion and angle of internal friction for the soil are respectively.a)0,5 N/mm2 and 30°b)0.05 N/mm2 and 0°c)0.2 N/mm2 and 0°d)0.05 N/mm2 and 45°Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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