Pile Capacity: Tip and Skin

Q_{u l t} = q_{p} A_{p} + \sum f_{s} A_{s} Q_{a l l} = Q_{u l t} / FS

Ultimate pile resistance is end bearing qp Ap plus skin friction fs along the shaft. For clay, qp ~ 9c (Skempton) and fs = α c. For sand, qp = Nq σ'v (limited) and fs = K σ'v tanδ. Apply FS (or LRFD φ) for allowable.

$q_{p}$: Unit tip bearing
$f_{s}$: Unit skin friction
$A_{p}, A_{s}$: Tip and shaft areas

Step-by-step solved example

Clay c=80 kPa, D=0.40 m circular, L=12 m, α=0.5, qp=9c, FS=2.5. Qall? Neglect weight.

Pile: tip bearing + skin friction along the shaft.

1. Tip
Ap=0.126 m²; qp=720 kPa; Qp=90.5 kN.
2. Skin
As=πDL=15.08 m²; fs=40 kPa; Qs=603 kN. Qult=694 kN; Qall=277 kN.

Answer: Qall ≈ 280 kN

10 practice questions

0/10 correct

1.Skempton qp for clay is on the order of

2.Skin in clay often uses

3.A larger diameter increases

4.FS of 2–3 on Qult is typical for

5.Negative skin friction (downdrag) occurs when

6.Group efficiency is often < 1 because

7.Sand tip Nq increases with

8.Dynamic pile formulas (ENR etc.) use

9.Point bearing dominates in

10.Qult is a force; qp is a