Question 31.3: This example covers a single pile in a uniform sand layer fo...

STEP 1: Compute the end bearing capacity.

Q_{ultimate} = (\sigma^{\prime}_t \times N_q \times A) + [K \times \sigma^{\prime}_p \times \tan \delta \times (\pi \times d) \times L]
End bearing term            Skin friction term

where
end bearing term = \sigma^{\prime}_t \times N_q \times A

\sigma^{\prime}_t = effective stress at the tip of the pile
\sigma^{\prime}_t = γ × depth to the tip of the pile
= 17.3 × 10 = 173 kN/m² (3,610 psf)

Find N_q using Table 31.1. For a friction angle of 30°, N_q = 21 for driven piles.

end bearing capacity = \sigma^{\prime}_t \times N_q \times A
= 173 × 21 × (π × d²/4) = 713.3 kN (160 kip)

STEP 2: Compute the skin friction.

skin friction term = K \times \sigma_p^{\prime} \times \tan\delta \times (\pi \times d) \times L

Obtain the K value.

From Table 31.4, for driven round piles, the K value lies between 1.0 and 1.5. Hence, assume K = 1.25. Obtain the \sigma^{\prime}_p (effective stress at the perimeter of the pile). The effective stress along the perimeter of the pile varies with the depth. Hence, obtain the \sigma^{\prime}_p value at the midpoint of the pile. The pile is 10 m long. Hence, use the effective stress at 5 m below the ground surface.

\sigma^{\prime}_p(midpoint) = 5 \times \gamma = 5 \times 17.3 = 86.5  kN/m^2(1.8  ksf)

Obtain the skin friction angle, δ. From Table 31.3, the skin friction angle for steel piles is 20°. Find the skin friction of the pile.

skin friction = K \times \sigma_p^{\prime} \times \tan\delta \times (\pi \times d) \times L \\ = 1.25 \times 86.5 \times (\tan 20^{\circ})\times (\pi \times 0.5)\times 10 \\ =618.2 \space kN \space (139 \space kip)

STEP 3: Compute the ultimate bearing capacity of the pile.

Q_{ultimate} = ultimate bearing capacity of the pile
Q_{ultimate} = end bearing capacity + skin friction
Q_{ultimate} = 713.3 + 618.2 = 1,331.4 kN (300 kip)

Assume a factor of safety of 3.0. Hence, the allowable bearing capacity of the pile can be calculated as

Q_{ultimate} /F.O.S. = 1,331.4/3.0
allowable pile capacity = 443.8 kN (99.8 kip)

Note that 1 kN is equal to 0.225 kip. Hence, the allowable capacity of the pile is 99.8 kip.

Table 31.1
Friction angle vs. N_q

Source: NAVFAC DM 7.2 (1984).

Table 31.3
Pile type and pile skin friction angle

Source: NAVFAC DM 7.2 (1984).

Table 31.4
Pile type and lateral earth pressure coefficient 

Source: NAVFAC DM 7.2 (1984).