Question 31.5: This example concerns multiple sand layers with no groundwat...

STEP 1: Compute the end bearing capacity.

Q_{ultimate} = (\sigma^{\prime}_t \times N_q \times A) + (K \times \sigma^{\prime}_v \times \tan \delta \times A_p)
End bearing term     Skin friction term
end bearing term = \sigma^{\prime}_t \times N_q \times A

where
\sigma^{\prime}_t = effective stress at the tip of the pile

\sigma^{\prime}_t=\gamma_1\times 5+\gamma_2\times 7 \\ \sigma^{\prime}_t=17.3\times 5+16.9 \times 7 \\ = 204.8\space kN/m^2(4.28 \space ksf)

Find N_q using Table 31.1. Use the friction angle of the soil where the pile tip rests. For a friction angle of 32°, N_q = 29 for driven piles. The diameter of the pile is given as 0.5 m.
The φ value of the bottom sand layer is used to find N_q , since the tip of the pile lies on the bottom sand layer.

end bearing capacity = \sigma^{\prime}_t \times N_q \times A\\=204.8\times 29\times(\pi\times d^2/4)=1,166.2\space kN(262.2 \space kip)

STEP 2: Compute the skin friction.
The skin friction of the pile needs to be done in two parts.

• Skin friction of the pile portion in sand layer 1 (A to B)
•  Skin friction of the pile portion in sand layer 2 (B to C)

Compute the skin friction of the pile portion in sand layer 1 (A to B).

skin friction term = K\times \sigma^{\prime}_p\times \tan\delta\times A_p \\ A_p=(\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}_v (average effective stress at the perimeter of the pile). Obtain the \sigma^{\prime}_v value at the midpoint of the pile in sand layer 1.

\sigma_v^{\prime}(midpoint)=2.5\times(\gamma_1)\\ = 2.5 \times 17.3\space kN/m^2 \\ =43.3 \space kN/m^2(0.9\space ksf)

Obtain the skin friction angle, δ. From Table 31.3, the skin friction angle,δ, for concrete piles is ¾ φ.

δ = 3/4 × 30° = 22.5°

This is true because the friction angle of layer 1 is 30°.

skin friction in sand layer 1 = K\times\sigma_v^{\prime}\times \tan\delta\times(\pi \times d)\times L \\ =1.25 \times 43.3 \times (\tan 22.5^{\circ})\times (\pi \times 0.5)\times 5\\ =176.1\space kN(39.6\space kip)

STEP 3: Find the skin friction of the pile portion in sand layer 2 (B to C).

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

Obtain the \sigma_v^{\prime} (effective stress at the perimeter of the pile). Obtain the \sigma_v^{\prime} value at the midpoint of the pile in sand layer 2.

\sigma_v^{\prime}(midpoint)=5\times \gamma_1+3.5\times \gamma_2 \\ =5\times 17.3+3.5\times 16.9\space kN/m^2 \\ =145.7 \space kN/m^2(3,043 \space psf)

Obtain the skin friction angle, δ.
From Table 31.3, the skin friction angle, δ, for concrete piles is ¾ φ.

 δ = 3/4 × 32° = 24°

This is true because the friction angle of layer 2 is 32°.

skin friction in sand layer 2 = K\times\sigma_p^{\prime}\times \tan\delta\times(\pi \times d)\times L \\ =1.25 \times 145.7 \times (\tan 24^{\circ})\times (\pi \times 0.5)\times 7\\ =891.6\space kN(200.4\space kip)
P_{ultimate } = end bearing capacity + skin friction in layer 1 + skin friction in layer 2
end bearing capacity = 1,166.2 kN
skin friction in sand layer 1 = 176.1 kN
skin friction in sand layer 2 = 891.6 kN
P_{ultimate } = 2,233.9 kN (502.2 kip)

Note that the bulk of the pile capacity comes from the end bearing. After that comes the skin friction in layer 2 (bottom layer). The skin friction in the top layer is very small. One reason for this is that the effective stress acting on the perimeter of the pile is very low in the top layer. This is due to the fact that effective stress is directly related to depth.
Hence

allowable bearing capacity of the pile = P_{ultimate } /F.O.S.
= 2,233.9/3.0 kN
allowable pile capacity = 744.6 kN (167.4 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).