Question 31.6: This example concerns multiple sand layers with groundwater ...

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 3+(\gamma_1-\gamma_w)\times 2+(\gamma_2-\gamma_w)\times 10 \\ =17.3 \times 3+(17.3-9.8)\times 2+(16.9-9.8)\times 10 \\ =137.9 \space kN/m^2(2,880\space psf)

Find N_q using Table 31.1. For a friction angle of 32°, N_q= 29 for driven piles.
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 \\ = 137.9 \times 29 \times (\pi \times d^2/4)=785.2\space kN(176.5\space kip)

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

• Skin friction of the pile portion in sand layer 1 above groundwater (A to B).
• Skin friction of the pile portion in sand layer 1 below groundwater (B to C).
• Skin friction of the pile portion in sand layer 2 below groundwater (C to D).

STEP 3: Compute the skin friction of the pile portion in sand layer 1 above the groundwater level (A to B).

skin friction term = K\times \sigma^{\prime }_v \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 }_v (effective stress at the perimeter of the pile). Obtain the \sigma^{\prime }_v value at the midpoint of the pile in sand layer 1, above the groundwater level (A to B).

\sigma^{\prime }_v(midpoint) = 1.5 \times (\gamma_1)=1.5 \times 17.3  kN/m^2 \\ =26  kN/m^2(543  psf)

Obtain the skin friction angle, δ. From Table 31.3, the skin friction angle, δ, for concrete piles is 3/4 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 (A to B) = K\times \sigma^{\prime }_v \times \tan\delta \times (\pi \times d)\times L \\ = 1.25 \times (26) \times (\tan 22.5^{\circ}) \times (\pi \times 0.5)\times 3 \\ =63.4   kN(14.3  kip)

STEP 4: Find the skin friction of the pile portion in sand layer 1 below the groundwater level (B to C).

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

Obtain the \sigma^{\prime }_v (effective stress at the perimeter of the pile).

\sigma^{\prime }_v (midpoint) =3\times \gamma_1+1.0(\gamma_1-\gamma_w) \\ =3\times17.3+1.0\times(17.3-9.8)kN/m^2 \\ =59.4\space kN/m^2(1,241\space psf)

skin friction in sand layer 1 (B to C) = K\times \sigma^{\prime }_p \times \tan\delta \times (\pi \times d)\times L\\ = 1.25 \times 59.4 \times (\tan 22.5^{\circ})\times (\pi \times 0.5)\times 2 \\ =96.6  kN(21.7  kip)

STEP 5: Find the skin friction in sand layer 2 below the groundwater level (C to D).

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

Obtain the \sigma^{\prime }_p (effective stress at the midpoint of the pile):

\sigma^{\prime }_v(midpoint) = 3\times \gamma_1 +2\times (\gamma_1-\gamma_w)+5\times (\gamma_2-\gamma_w) \\ =3\times 17.3+2.0\times (17.3-9.8)+5\times (16.9-9.8)  kN/m^2 \\ =102.4  kN/m^2(2.14  ksf)

From Table 31.3, the skin friction angle, δ, for concrete piles is 3/4 φ.

δ = 3/4 × 32° = 24°

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

skin friction in sand layer 2 (C to D) = K\times \sigma^{\prime }_v \times \tan\delta \times (\pi \times d)\times L\\ = 1.25 \times 102.4 \times (\tan 24^{\circ}) \times (\pi \times 0.5) \times 10 \\ = 895.2 kN(201.2 kip)

P_{ultimate} = end bearing capacity
+ skin friction in layer 1 (above the groundwater level)
+ skin friction in layer 1 (below the groundwater level)
+ skin friction in layer 2 (below the groundwater level)
end bearing capacity = 785.2 kN
skin friction in layer 1 (above the groundwater level)(A to B) = 63.4 kN
skin friction in layer 1 (below the groundwater level)(B to C) = 96.6 kN
skin friction in layer 2 (below the groundwater level)(C to D) = 895.2 kN
total = 1,840 kN
P_{ultimate} = 1,840.4 kN (413.7 kip)

In this case, the bulk of the pile capacity comes from end bearing and the skin friction in the bottom layer. Hence, the allowable bearing capacity of the pile can be determined by

P_{ultimate}/F.O.S. = 1,840.4/3.0
allowable pile capacity = 613.5 kN (137.9 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).