Question

Sand Layer 1 r = 16.5 kN/m Ysat = 18.5 kN/m c' = 0 kPa O' = 30° 20 m Sand Layer 2 y = 18.5 kN/m Ysat = 20.0 kN/m c' = 0 kPa O' = 40° Fig. 3

Base on the soil profile above (Fig.3), it was designed to use driven pile with dimension 500mm x 500mm to carry out the following design load (use factor of safety = 2):
i. 200 kN
ii. 1000 kN
iii. 3000 kN
Determine the required depth of driven pile (assume no defective of pile during installation process)

Answer 1

Answer:

I Sand layer 1 20m 8 = 16.5 kN lm3 18.5kN/m3 c = okPa dsat ' = 30° = 18.5kN/m3 = 20 kN lm3 Sand layer a 8 8 sat c=0kPa 0l = 40° >> Neglecting end bearing copacity of pile For concretepile, Assume Kzl for tot layer k=2 for and layer. 8 = 3d for concrete pile, then, for 1st dayes, Se st 3 x 30 = 22.5° layer, X40=30 load carrying capacity of tot layer, Q = k xtans, As (16.5 x 20) x tan(22.5J x(0.5²20 for and = 3 Sa Average o 1 x Q=34l. 726 KN.

2 =) Factor of safety, load cooying capacity for which it should be designed » 2000 2 = LOOKN -) 1000 X2 = 2000 KN 3000 X2 = 6000KN. 20 Н 16.5X20 +18.5x4 Qui - 341.726+ kg Favorge x tan & XAS H00 = 341.726+ 2 (06-5 420 4,18-5 34416.5*20]tan3o xosten By solving, weget, t = 0.6 m. Total depth = 20+ 0.6=20:6m. -) 9000 = 341-726+2 •(16.5420x2 +85** )xtenzóxo sakura 2000 = 341.726 +95.25USH+2-6742 By solving, weget, it = 12.81 m Total length = 20 +12.81 32.81m

3 8000 = 34110726 +2 16.5 X 20 X2 +18.5XH ( 160 -5XH) xtanzóxosten 2 6000 = 341.72 6 + 95.2545 H +2.6742 ka By solving weget, t = 31.65m Total length = 20 +31.65 51.65 m.