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A fully submerged infinite sandy slope has an inclination of 30° with the horizontal. The saturated unit weight and effective angle of internal friction of sand are 18 kN/m3 and 38°, respectively. The unit weight of water is 10 kN/m3. Assume that the seepage is parallel to the slope. Against shear failure of the slope, the factor of safety (round off to two decimal places) is _______.
Correct answer is '0.60'. Can you explain this answer?
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A fully submerged infinite sandy slope has an inclination of 30° w...
= 0.601
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A fully submerged infinite sandy slope has an inclination of 30° w...
Degrees with the horizontal. The density of sand is 1600 kg/m³ and the angle of internal friction is 35 degrees. What is the height of the slope at which the sand starts to slide?
To determine the height at which the sand starts to slide, we need to calculate the critical height based on the angle of internal friction.
The angle of internal friction (φ) is given as 35 degrees, which means the tangent of φ is equal to the coefficient of friction (μ).
μ = tan(φ) = tan(35) ≈ 0.7002
The weight of the sand column above a certain height h is given by the formula:
W = γ * A * h
Where:
W = weight of the sand column
γ = density of sand = 1600 kg/m³
A = cross-sectional area of the sand column
h = height of the sand column
The critical height is the height at which the weight of the sand column is equal to the resisting force due to friction.
Weight of the sand column (W) = Resisting force due to friction
W = μ * N
Where:
N = Normal force acting perpendicular to the slope
The normal force (N) can be calculated as:
N = W * cos(θ)
Where:
θ = inclination of the slope = 30 degrees
Substituting the value of N in the equation for W:
W = μ * N = μ * W * cos(θ)
Simplifying the equation:
1 = μ * cos(θ)
Substituting the value of μ:
1 = 0.7002 * cos(30)
Solving for cos(30):
cos(30) = 0.866
0.7002 * 0.866 = 0.606
Therefore, the critical height at which the sand starts to slide is approximately 0.606 times the height of the sand column.
Let's assume the height of the sand column is H.
Critical height = 0.606 * H
The inclination of the slope is 30 degrees. In a right-angled triangle formed by the slope, the critical height is the opposite side and the height of the sand column is the hypotenuse.
Using the trigonometric ratio:
sin(30) = opposite/hypotenuse
sin(30) = critical height/H
Solving for critical height:
critical height = sin(30) * H
critical height = 0.5 * H
Therefore, the critical height at which the sand starts to slide is half the height of the sand column.
Hence, the height of the slope at which the sand starts to slide is equal to twice the critical height.
Height of slope = 2 * critical height
Height of slope = 2 * (0.606 * H)
Height of slope = 1.212 * H
Therefore, the height of the slope at which the sand starts to slide is approximately 1.212 times the height of the sand column.
To determine the height at which the sand starts to slide, we need to calculate the critical height based on the angle of internal friction.
The angle of internal friction (φ) is given as 35 degrees, which means the tangent of φ is equal to the coefficient of friction (μ).
μ = tan(φ) = tan(35) ≈ 0.7002
The weight of the sand column above a certain height h is given by the formula:
W = γ * A * h
Where:
W = weight of the sand column
γ = density of sand = 1600 kg/m³
A = cross-sectional area of the sand column
h = height of the sand column
The critical height is the height at which the weight of the sand column is equal to the resisting force due to friction.
Weight of the sand column (W) = Resisting force due to friction
W = μ * N
Where:
N = Normal force acting perpendicular to the slope
The normal force (N) can be calculated as:
N = W * cos(θ)
Where:
θ = inclination of the slope = 30 degrees
Substituting the value of N in the equation for W:
W = μ * N = μ * W * cos(θ)
Simplifying the equation:
1 = μ * cos(θ)
Substituting the value of μ:
1 = 0.7002 * cos(30)
Solving for cos(30):
cos(30) = 0.866
0.7002 * 0.866 = 0.606
Therefore, the critical height at which the sand starts to slide is approximately 0.606 times the height of the sand column.
Let's assume the height of the sand column is H.
Critical height = 0.606 * H
The inclination of the slope is 30 degrees. In a right-angled triangle formed by the slope, the critical height is the opposite side and the height of the sand column is the hypotenuse.
Using the trigonometric ratio:
sin(30) = opposite/hypotenuse
sin(30) = critical height/H
Solving for critical height:
critical height = sin(30) * H
critical height = 0.5 * H
Therefore, the critical height at which the sand starts to slide is half the height of the sand column.
Hence, the height of the slope at which the sand starts to slide is equal to twice the critical height.
Height of slope = 2 * critical height
Height of slope = 2 * (0.606 * H)
Height of slope = 1.212 * H
Therefore, the height of the slope at which the sand starts to slide is approximately 1.212 times the height of the sand column.
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A fully submerged infinite sandy slope has an inclination of 30° with the horizontal. The saturated unit weight and effective angle of internal friction of sand are 18 kN/m3 and 38°, respectively. The unit weight of water is 10 kN/m3. Assume that the seepage is parallel to the slope. Against shear failure of the slope, the factor of safety (round off to two decimal places) is _______.Correct answer is '0.60'. Can you explain this answer?
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A fully submerged infinite sandy slope has an inclination of 30° with the horizontal. The saturated unit weight and effective angle of internal friction of sand are 18 kN/m3 and 38°, respectively. The unit weight of water is 10 kN/m3. Assume that the seepage is parallel to the slope. Against shear failure of the slope, the factor of safety (round off to two decimal places) is _______.Correct answer is '0.60'. 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 A fully submerged infinite sandy slope has an inclination of 30° with the horizontal. The saturated unit weight and effective angle of internal friction of sand are 18 kN/m3 and 38°, respectively. The unit weight of water is 10 kN/m3. Assume that the seepage is parallel to the slope. Against shear failure of the slope, the factor of safety (round off to two decimal places) is _______.Correct answer is '0.60'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A fully submerged infinite sandy slope has an inclination of 30° with the horizontal. The saturated unit weight and effective angle of internal friction of sand are 18 kN/m3 and 38°, respectively. The unit weight of water is 10 kN/m3. Assume that the seepage is parallel to the slope. Against shear failure of the slope, the factor of safety (round off to two decimal places) is _______.Correct answer is '0.60'. Can you explain this answer?.
A fully submerged infinite sandy slope has an inclination of 30° with the horizontal. The saturated unit weight and effective angle of internal friction of sand are 18 kN/m3 and 38°, respectively. The unit weight of water is 10 kN/m3. Assume that the seepage is parallel to the slope. Against shear failure of the slope, the factor of safety (round off to two decimal places) is _______.Correct answer is '0.60'. 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 A fully submerged infinite sandy slope has an inclination of 30° with the horizontal. The saturated unit weight and effective angle of internal friction of sand are 18 kN/m3 and 38°, respectively. The unit weight of water is 10 kN/m3. Assume that the seepage is parallel to the slope. Against shear failure of the slope, the factor of safety (round off to two decimal places) is _______.Correct answer is '0.60'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A fully submerged infinite sandy slope has an inclination of 30° with the horizontal. The saturated unit weight and effective angle of internal friction of sand are 18 kN/m3 and 38°, respectively. The unit weight of water is 10 kN/m3. Assume that the seepage is parallel to the slope. Against shear failure of the slope, the factor of safety (round off to two decimal places) is _______.Correct answer is '0.60'. Can you explain this answer?.
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A fully submerged infinite sandy slope has an inclination of 30° with the horizontal. The saturated unit weight and effective angle of internal friction of sand are 18 kN/m3 and 38°, respectively. The unit weight of water is 10 kN/m3. Assume that the seepage is parallel to the slope. Against shear failure of the slope, the factor of safety (round off to two decimal places) is _______.Correct answer is '0.60'. Can you explain this answer?, a detailed solution for A fully submerged infinite sandy slope has an inclination of 30° with the horizontal. The saturated unit weight and effective angle of internal friction of sand are 18 kN/m3 and 38°, respectively. The unit weight of water is 10 kN/m3. Assume that the seepage is parallel to the slope. Against shear failure of the slope, the factor of safety (round off to two decimal places) is _______.Correct answer is '0.60'. Can you explain this answer? has been provided alongside types of A fully submerged infinite sandy slope has an inclination of 30° with the horizontal. The saturated unit weight and effective angle of internal friction of sand are 18 kN/m3 and 38°, respectively. The unit weight of water is 10 kN/m3. Assume that the seepage is parallel to the slope. Against shear failure of the slope, the factor of safety (round off to two decimal places) is _______.Correct answer is '0.60'. Can you explain this answer? theory, EduRev gives you an
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