Question 35.3: Find the tip resistance of the 4 ft diameter caisson shown i...

Pile capacity comes from tip resistance and skin friction. In this example, only the tip resistance is calculated.
Find the ultimate point resistance for driven piles for fine sand.

q_{\text {tip }}=0.12 c_{ N } \times N \times D / B \text { tsf (for fine sand) }
c_{ N }=0.77 \log 20 / p
p = effective overburden stress at pile tip (tsf)
p = 5 × 110 + 22 × 115 = 3,080 psf = 1.54 tsf (147 kPa)
c_N = 0.77 log(20/1.54) = 0.86
D = depth into bearing stratum = 22 ft (6.7 m)

Fill material is not considered to be a bearing stratum.

B = 4 ft (width or diameter of the pile)
q_{ tip }=0.12 C_{ N } \times N \times D / B(\text { fine sand }) \\q_{\text {tip }}=0.12 \times 0.86 \times 15 \times 22 / 4=8.5 tsf (814 kPa )
maximum allowable point resistance =4N tsf for sandy soils
4 × N = 4 × 15 = 60 tsf

Hence

q_{tip}=8.5  tsf
allowable point bearing capacity = 8.5/F.O.S.

Assume a factor of safety of 3.0. Hence, the total allowable point bearing capacity can be determined.

q_{\text {tipallowable }}=2.84 tsf (272 kPa ) \\ Q_{\text {tipallowable }} \times \text { tip area }=q_{\text {tipallowable }} \times \pi \times\left(4^{2}\right) / 4 \\ = 36  ton (320  kN)

Only the tip resistance was computed in this example.