Question 31.9: Find the skin friction and end bearing capacity of the pile ...
Find the skin friction and end bearing capacity of the pile shown in Fig. 31.17. Assume that the critical depth is achieved at 20 ft into the bearing layer (NAVFAC DM 7.2, 1984). The pile diameter is 1 ft, and other soil parameters are as shown in Fig. 31.17.
The skin friction is calculated in the overburden. In this case, skin friction is calculated in the soft clay. Then the skin friction is calculated in the bearing layer (medium sand) assuming the skin friction attains a limiting value after 20 diameters (critical depth). See 31.18.
STEP 1: Find the skin friction from A to B.
skin friction in soft clay = α × c × perimeter surface area
= 0.4 × 700 × π × d × L
=0.4 × 700 × π × 1 × 12
= 10,560 lb (46.9 kN)
STEP 2: Find the skin friction from B to C.
Critical depth is at point C, 20 ft below the soft clay
Skin friction in sandy soils = S = K × \sigma^{\prime}_v \times \tan \delta \times A_p
S = skin friction of the pile
\sigma^{\prime}_v = average effective stress along the pile shaft
Average effective stress along pile shaft from B to C =(\sigma_B+\sigma_C)/2
\sigma_B = effective stress at B
\sigma_C = effective stress at C
To obtain the average effective stress from B to C, find the effective stresses at B and C and obtain the average of those two values.
\sigma_B = 100 × 4 + (100 – 62.4) × 8
= 700.8 lb/ft² (33.6 kPa)
\sigma_C = 100 × 4 + (100 – 62.4) × 8 + (110 – 62.4) × 20
= 1,452.8 lb/ft² (69.5 kPa)
average effective stress along pile shaft from B to C = (\sigma_B+ \sigma_C)/2 \\ = (700.8 + 1452.8)/2 = 1076.8 lb/ft^2
skin friction from B to C = K \times \sigma^{\prime}_v \times \tan \delta \times A_p \\ = 0.9 \times 1,076.8 \times \tan(25^{\circ}) \times (\pi \times 1 \times 20) \\ = 28,407 lb
STEP 3: Find the skin friction from C to D.
Skin friπction reaches a constant value at point C, 20 diameters into the bearing layer.
skin friction at point C = K \times \sigma^{\prime}_v \times \tan \delta \times A_p \\ \sigma^{\prime}_v at point C = 100 \times 4 + (100 – 62.4) 8 + (110- 62.4) \times 20 \\ = 1,452.8 lb/ft^2
unit skin friction at point C = 0.9 \times 1,452.8 \times \tan 25^{\circ} \\ = 609.7 lb/ft^2 (29 kPa)
The unit skin friction is constant from C to D. This is due to the fact that skin friction does not increase after the critical depth.
skin friction from C to D = 609.7 × surface perimeter area
= 609.7 × (π × 1 × 8) lb
= 15,323.4 lb (68.2 kN)
Summing up,
skin friction in soft clay (A to B) = 10,560 lb
skin friction in sand (B to C) = 28,407 lb
skin friction in sand (C to D) = 15,323 lb
total = 54,290 lb (241 kN)
STEP 4: Compute the end bearing capacity.
The end bearing capacity also reaches a constant value below the critical depth.
end bearing capacity = q \times N_q \times A
where
q = effective stress at pile tip
N_q = bearing capacity factor (given as 15)
A = cross-sectional area of the pile
If the pile tip is below the critical depth, q should be taken at critical depth. In this example, the pile tip is below the critical depth, which is 20 diameters into the bearing layer. Hence, q is equal to the effective stress at the critical depth (point C).
effective stress at point C = 100 × 4 + (100 – 62.4) × 8
+ (110- 62.4) × 20
= 1,452.8 lb/ft²
end bearing capacity = q × N_q × A
= 1,452.8 × 15 × (π × d²/4) lb
= 17,115 lb
total ultimate capacity of the pile = total skin friction + end bearing
= 54,290 + 17,115 = 71,405 lb
= 35.7 tons (317.6 kN)
