Module 13 · Lesson 13.1

The two resistances

Where the load actually goes.

Why this matters

A pile is usually drawn as a column standing on firm ground, which suggests the load travels down and out through the toe. For most piles in most ground that picture is almost exactly wrong: the great majority of the load never reaches the base at all.

By the end of this lesson you should be able to

Compute shaft and base resistance separately
Say which dominates, and why it depends on the ground
Account for the pile's own weight correctly

Ultimate pile capacity

Q_{u l t} = Q_{s} + Q_{b} = \sum τ_{s} π D Δ L + q_{b} A_{b}

What it calculates: The total resistance a pile can offer

Q_s: Shaft resistance, summed over the embedded length (kN)
$τ_{s}$: Unit shaft resistance in a layer (kPa)
D: Shaft diameter (m)
q_b: Unit base resistance (kPa)
A_b: Base area (m²)

This assumes

Both resistances are available simultaneously — which the second lesson shows is not true
τ_s and q_b come from correlations; only the addition is mechanics

In plain terms: The structure of this equation is equilibrium and beyond dispute. Everything difficult about pile design lives inside τ_s and q_b, and Modules 14 to 16 are about nothing else.

Worked example

Where the load goes in a bored pile in clay

Given

Bored pile, D = 0.6 m, embedded length 15 m
Firm clay throughout, s_u = 60 kPa
Adhesion factor α = 0.5
Base bearing factor N_c = 9

Find

The split between shaft and base, and the pile's net capacity

Practice

A pile has 848 kN of shaft resistance and 153 kN of base resistance. What percentage of the ultimate capacity comes from the shaft?

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This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint