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In a direct shear test, the shear stress and normal stress on a dry sand sample at failure are 0.6 kg/ cm2 and 1 kg/cm2 respectively. The angle of internal friction of the sand will be nearly
- a)25º
- b)31º
- c)37º
- d)43º
Correct answer is option 'B'. Can you explain this answer?
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In a direct shear test, the shear stress and normal stress on a dry sa...
Calculation of angle of internal friction
Given data:
Shear stress (τ) = 0.6 kg/cm2
Normal stress (σ) = 1 kg/cm2
To calculate the angle of internal friction (ϕ), we can use the following equation:
τ = σ tan ϕ
Substituting the given values, we get:
0.6 = 1 x tan ϕ
tan ϕ = 0.6/1
tan ϕ = 0.6
ϕ = tan-1 (0.6)
ϕ = 31.0 degrees (approximately)
Therefore, the angle of internal friction of the sand sample is approximately 31 degrees.
Explanation
Direct shear test is a common laboratory test used to determine the shear strength parameters of soil samples. In this test, a cylindrical sample of soil is placed in a shear box apparatus and subjected to shearing forces until failure occurs. The shear stress and normal stress at failure are used to calculate the shear strength parameters of the soil.
The angle of internal friction is a measure of the resistance of soil to shear stress. It is defined as the angle between the normal force acting on a soil mass and the resultant shear force that causes failure. The higher the angle of internal friction, the greater the shear strength of the soil.
In the given problem, the shear stress and normal stress at failure are known. Using the equation τ = σ tan ϕ, we can calculate the angle of internal friction of the sand sample. The calculated angle of internal friction is approximately 31 degrees, indicating that the sand has a moderate resistance to shear stress.
Given data:
Shear stress (τ) = 0.6 kg/cm2
Normal stress (σ) = 1 kg/cm2
To calculate the angle of internal friction (ϕ), we can use the following equation:
τ = σ tan ϕ
Substituting the given values, we get:
0.6 = 1 x tan ϕ
tan ϕ = 0.6/1
tan ϕ = 0.6
ϕ = tan-1 (0.6)
ϕ = 31.0 degrees (approximately)
Therefore, the angle of internal friction of the sand sample is approximately 31 degrees.
Explanation
Direct shear test is a common laboratory test used to determine the shear strength parameters of soil samples. In this test, a cylindrical sample of soil is placed in a shear box apparatus and subjected to shearing forces until failure occurs. The shear stress and normal stress at failure are used to calculate the shear strength parameters of the soil.
The angle of internal friction is a measure of the resistance of soil to shear stress. It is defined as the angle between the normal force acting on a soil mass and the resultant shear force that causes failure. The higher the angle of internal friction, the greater the shear strength of the soil.
In the given problem, the shear stress and normal stress at failure are known. Using the equation τ = σ tan ϕ, we can calculate the angle of internal friction of the sand sample. The calculated angle of internal friction is approximately 31 degrees, indicating that the sand has a moderate resistance to shear stress.
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Community Answer
In a direct shear test, the shear stress and normal stress on a dry sa...
0.6=0+1*tan(Φ)
0.6/1=tan(Φ)
Φ=tan`(0.6)
Φ=30.963~31
0.6/1=tan(Φ)
Φ=tan`(0.6)
Φ=30.963~31
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In a direct shear test, the shear stress and normal stress on a dry sand sample at failure are 0.6 kg/ cm2 and 1 kg/cm2 respectively. The angle of internal friction of the sand will be nearlya)25ºb)31ºc)37ºd)43ºCorrect answer is option 'B'. Can you explain this answer?
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