A pile of 0.50 m diameter and length 10 m is embedded in a deposit of ...
Calculating Skin Friction Capacity of Pile in Clay Deposit
Given Parameters:
- Diameter of Pile (D) = 0.50 m
- Length of Pile (L) = 10 m
- Cohesion of Clay (c) = 60 kN/m2
- Angle of Internal Friction of Clay (ϕ) = 0
- Adhesion Factor (α) = 0.6
Calculating Skin Friction Capacity:The skin friction capacity of the pile in the clay deposit is given by the following equation:
$$Q_s = \alpha \times c \times A_s$$
where Qs is the skin friction capacity, α is the adhesion factor, c is the cohesion of the clay, and As is the surface area of the pile in contact with the clay.
To calculate the surface area of the pile, we need to first calculate the circumference of the pile:
$$C = \pi \times D = 3.14 \times 0.50 = 1.57 m$$
The surface area of the pile in contact with the clay is then given by:
$$A_s = C \times L = 1.57 \times 10 = 15.7 m^2$$
Substituting the values in the skin friction capacity equation, we get:
$$Q_s = 0.6 \times 60 \times 15.7 = 566.4 kN$$
Therefore, the skin friction capacity of the pile in the clay deposit is 566.4 kN.