In: Civil Engineering
Solution –
Pile driven in clay
Given Data –
Length = 20 m
Cross-section = 0.3 m x 0.3 m
Undrained shear strength (Su) = 100 kPa
Adhesion factor (α) for stiff clay = 0.3
Nc for clay = 9
Now,
Pile load capacity (Qu) = point bearing capacity (Qpu) + frictional capacity (Qf)
Point bearing capacity (Qpu) = qp * Ab
Where,
qp = Su * Nc
= 100 * 9
= 900 kPa
Ab =area of base
= 0.3 * 0.3
= 0.09 m2
Qpu = 900 * 0.09
= 81 KN
Frictional capacity (Qf) = fs * As
Where,
fs = α * Su
= 0.3 * 100
= 30 kPa
As = Area of side
= 4 * 0.3 *20
= 24 m2
Qf = 30 * 24
= 720 KN
So,
Pile load capacity (Qu) = 81 + 720
= 801 KN
So, the pile load capacity is 801 KN.