Question 31.7: Find the tip resistance of the 2 ft (0.609 m) diameter pile ...

Determine the ultimate point resistance for driven piles for fine sand.

q_{ult} = 0.35c_N \times N \times D/B \ tsf (\text{fine sand}) \\ c_N = 0.77  1og 20/p

where
p = effective overburden stress at pile tip (tsf)

p = 5 × 110 + 32 × 115 = 4,230 psf = 2.11 tsf (0.202 MPa)
c_N = 0.77  1og(20/2.11) = 0.751
D = depth driven into bearing stratum = 32 ft (9.75 m)

The fill material is not considered to be a bearing stratum.

B (width or diameter of the pile) = 2 ft (0.61 m)
q_{ult} = 0.35c_N \times N \times D/B (fine sand)
q_{ult} = 0.35 × 0.751 × 25 × 32/2
= 105 tsf (10.1 MPa)
maximum allowable point resistance = 4N for sandy soils
4 × N = 4 × 25 = 100 tsf

Hence

q_{ult} = 100 tsf (9.58 MPa)
allowable point bearing capacity = 100/F.O.S.

Assume a factor of safety of 3.0.
Hence,

q_{allowable} = 33.3 tsf (3.2 MPa)
total allowable point bearing capacity = q_{allowable} × tip area
= q_{allowable} \times \pi \times (2^2)/4 \\ = 105  tons (934  kN)