Question

A circular concrete pile has a low displacement driven with a diameter of 40 cm and a length of 15 m is impeded in a sandy soil. The values of the standard penetration number are assigned for each 1 m depth from top to bottom and these numbers are: 6, 7, 9, 8, 12, 13, 12, 15, 15, 13, 18, 22, 28, 30, 27, 12, 15, 16, 30, 27. Fine the allowable pile capacity of this pile by using Meyerhof’s method. Assume FS=3

foundations engineering

i wan’t answer after 14 minuts please

Answer 1

Ans) We know, according to Meyerhoff,

Allowable pile capacity () = (Qp + Qs) / FOS

Also, Qp = Ap qp

where, Qp = point resistance of pile

Ap = pipe area = (/4)( = 0.1256

qp = 0.40 Pa N60 (L/D) 4 Pa N60

where, Pa = Atmospheric pressure = 100 kPa

N60 = average N60 value 10D above and 5D below pile tip

Tip of pile is 15 m below ground , so 10D above means 10(0.4) = 4 m above pile tip and 11 m below surface

Similarly 5D below means 5(0.4) = 2 m below pile tip or 17 m below ground surface

Hence, average SPT N60 for 11 m to 17 m = (18 + 22 + 28 + 30 + 27 + 12 + 15)/7 = 21.7 22

=>  qp = 0.40 Pa N60 (L/D) = 0.40 x 100 x 22 x (15 / 0.40) = 33000 kN/

=> Qp = Ap qp = 0.1256 x 33000 = 4144.8 kN

or

qp = 4 Pa N60

=> qp = 4 x 100 x 22 = 8800 kN/

=> Qp = Ap qp = 0.1256 x 8800 = 1105.28 kN

Final value will be lower of two values calculated above , so Qp = 1105.28 kN

Now,

Qs = p L

where, p = perimeter = D = 1.256

L = Pile length = 15 m

= frictional resistance = 0.02 Pa 60

60 = average penetration number till pile length = (6+7+ 9+8+12+13+12+15+15+13+18+22+28+30+37) / 15 = 16.33 16

=> Qs = 1.256 x 15 x 16

=> Qs = 301.44 kN

Therefore,

Allowable load carrying capacity () = (Qp + Qs) / FOS

where, FOS = factor of safety = 3

=> = (1105.28 + 301.44) / 3

=> = 468.90 kN

Hence, allowable capacity of pile is 468.90 kN