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A pile of 0.50 m diameter and of length 10 m is embedded in a deposit of clay. The undrained strength parameters of the clay are cohension= 60 kN/m2 and the angle of internal friction = 0. The skin friction capacity (kN) of the pile for an adhesion factor of 0.6, is
- a)671
- b)565
- c)283
- d)106
Correct answer is option 'B'. Can you explain this answer?
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A pile of 0.50 m diameter and of length 10 m is embedded in a deposit ...
Skin friction capacity of clay is given by
Qs = αCAs
where, As = effective area in developing friction
= π × d × 1
Qs = αCAs
where, As = effective area in developing friction
= π × d × 1
Most Upvoted Answer
A pile of 0.50 m diameter and of length 10 m is embedded in a deposit ...
Given:
Diameter of the pile (D) = 0.50 m
Length of the pile (L) = 10 m
Cohesion of the clay (c) = 60 kN/m²
Angle of internal friction of the clay (φ) = 0°
Adhesion factor (α) = 0.6
To find:
Skin friction capacity of the pile
Solution:
The skin friction capacity of a pile can be calculated using the equation:
Qs = πDLc + 2πLαDσtan(φ)
Where,
Qs = Skin friction capacity of the pile
D = Diameter of the pile
L = Length of the pile
c = Cohesion of the clay
α = Adhesion factor
σ = Effective vertical stress at the depth of the pile
φ = Angle of internal friction of the clay
Step 1: Calculate the effective vertical stress (σ)
Since the pile is embedded in a deposit of clay, the effective vertical stress can be calculated as:
σ = γH
Where,
γ = Unit weight of the clay
H = Depth of the pile
Step 2: Calculate the skin friction capacity (Qs)
Using the values given, we can substitute them in the equation to find Qs:
Qs = πDLc + 2πLαDσtan(φ)
Step 3: Calculate the effective vertical stress (σ)
The effective vertical stress can be calculated as:
σ = γH
Step 4: Substitute the values in the equation
Now substitute the values of D, L, c, α, σ, and φ in the equation to calculate Qs.
Qs = πDLc + 2πLαDσtan(φ)
Step 5: Calculate the skin friction capacity (Qs)
Solve the equation to find the value of Qs.
Qs = πDLc + 2πLαDσtan(φ)
By substituting the values and solving the equation, we get:
Qs = 3.14 * 0.50 * 10 * 60 + 2 * 3.14 * 10 * 0.6 * 0.50 * 0 * tan(0)
Qs = 942 + 0
Qs = 942 kN
Therefore, the skin friction capacity of the pile is 942 kN.
Diameter of the pile (D) = 0.50 m
Length of the pile (L) = 10 m
Cohesion of the clay (c) = 60 kN/m²
Angle of internal friction of the clay (φ) = 0°
Adhesion factor (α) = 0.6
To find:
Skin friction capacity of the pile
Solution:
The skin friction capacity of a pile can be calculated using the equation:
Qs = πDLc + 2πLαDσtan(φ)
Where,
Qs = Skin friction capacity of the pile
D = Diameter of the pile
L = Length of the pile
c = Cohesion of the clay
α = Adhesion factor
σ = Effective vertical stress at the depth of the pile
φ = Angle of internal friction of the clay
Step 1: Calculate the effective vertical stress (σ)
Since the pile is embedded in a deposit of clay, the effective vertical stress can be calculated as:
σ = γH
Where,
γ = Unit weight of the clay
H = Depth of the pile
Step 2: Calculate the skin friction capacity (Qs)
Using the values given, we can substitute them in the equation to find Qs:
Qs = πDLc + 2πLαDσtan(φ)
Step 3: Calculate the effective vertical stress (σ)
The effective vertical stress can be calculated as:
σ = γH
Step 4: Substitute the values in the equation
Now substitute the values of D, L, c, α, σ, and φ in the equation to calculate Qs.
Qs = πDLc + 2πLαDσtan(φ)
Step 5: Calculate the skin friction capacity (Qs)
Solve the equation to find the value of Qs.
Qs = πDLc + 2πLαDσtan(φ)
By substituting the values and solving the equation, we get:
Qs = 3.14 * 0.50 * 10 * 60 + 2 * 3.14 * 10 * 0.6 * 0.50 * 0 * tan(0)
Qs = 942 + 0
Qs = 942 kN
Therefore, the skin friction capacity of the pile is 942 kN.
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A pile of 0.50 m diameter and of length 10 m is embedded in a deposit of clay. The undrained strength parameters of the clay are cohension= 60 kN/m2 and the angle of internal friction = 0. The skin friction capacity (kN) of the pile for an adhesion factor of 0.6, isa)671b)565c)283d)106Correct answer is option 'B'. Can you explain this answer?
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A pile of 0.50 m diameter and of length 10 m is embedded in a deposit of clay. The undrained strength parameters of the clay are cohension= 60 kN/m2 and the angle of internal friction = 0. The skin friction capacity (kN) of the pile for an adhesion factor of 0.6, isa)671b)565c)283d)106Correct answer is option 'B'. 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 pile of 0.50 m diameter and of length 10 m is embedded in a deposit of clay. The undrained strength parameters of the clay are cohension= 60 kN/m2 and the angle of internal friction = 0. The skin friction capacity (kN) of the pile for an adhesion factor of 0.6, isa)671b)565c)283d)106Correct answer is option 'B'. 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 pile of 0.50 m diameter and of length 10 m is embedded in a deposit of clay. The undrained strength parameters of the clay are cohension= 60 kN/m2 and the angle of internal friction = 0. The skin friction capacity (kN) of the pile for an adhesion factor of 0.6, isa)671b)565c)283d)106Correct answer is option 'B'. Can you explain this answer?.
A pile of 0.50 m diameter and of length 10 m is embedded in a deposit of clay. The undrained strength parameters of the clay are cohension= 60 kN/m2 and the angle of internal friction = 0. The skin friction capacity (kN) of the pile for an adhesion factor of 0.6, isa)671b)565c)283d)106Correct answer is option 'B'. 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 pile of 0.50 m diameter and of length 10 m is embedded in a deposit of clay. The undrained strength parameters of the clay are cohension= 60 kN/m2 and the angle of internal friction = 0. The skin friction capacity (kN) of the pile for an adhesion factor of 0.6, isa)671b)565c)283d)106Correct answer is option 'B'. 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 pile of 0.50 m diameter and of length 10 m is embedded in a deposit of clay. The undrained strength parameters of the clay are cohension= 60 kN/m2 and the angle of internal friction = 0. The skin friction capacity (kN) of the pile for an adhesion factor of 0.6, isa)671b)565c)283d)106Correct answer is option 'B'. Can you explain this answer?.
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