Module 14 · Lesson 14.1
Adhesion, not cohesion
Why the pile never gets the strength the laboratory measured.
Why this matters
The obvious approach in clay is to take the undrained shear strength and apply it over the shaft area. Do that and you will overpredict shaft resistance in stiff clay by a factor of two or three. The reason is physical, and understanding it is what makes α more than a fudge factor.
By the end of this lesson you should be able to
- Explain what happens to clay around a pile during installation
- Apply α layer by layer rather than as an average
- Compute base resistance with the deep bearing factor
What it calculates: Undrained shaft resistance, summed over the layers a pile passes through
- Adhesion factor — the fraction of su the interface actually delivers (—)
- su
- Undrained shear strength of the layer (kPa)
- D
- Shaft diameter (m)
- Thickness of the layer along the shaft (m)
This assumes
- Undrained conditions — appropriate to the short term, which usually governs in clay
- α from a correlation fitted to load tests, with substantial scatter
In plain terms: A total-stress method: it never mentions effective stress or the water table, because in the undrained case su already contains them. That is its convenience and its limitation.
Why α is below one. Installing a pile disturbs the clay immediately around it. A driven pile displaces and remoulds it; a bored pile lets it swell and soften towards the open hole, and the wet concrete adds water at the interface. Either way, the shaft is in contact with clay that is weaker than the intact material the laboratory tested.
The stiffer the clay relative to its stress level, the bigger the shortfall — a heavily overconsolidated clay loses proportionally more of its strength on remoulding than a soft normally consolidated one. That is why α is correlated against the ratio su/σ′v and falls as that ratio rises.
Some of the loss recovers with time, as excess porewater pressures dissipate and the annulus reconsolidates against the shaft. Designs do not usually claim that recovery, and a load test carried out a week after installation may well be measuring a lower capacity than the pile will eventually have.
Worked example
A bored pile through two clays
Given
- Bored pile, D = 0.6 m, embedded 18 m
- 0–8 m: soft clay, su = 40 kPa. 8–18 m: firm clay, su = 90 kPa
- γ = 19 kN/m³, water table at ground level, so γ′ = 9.19 kN/m³
- α from the strength ratio su/σ′v in each layer
Find
The shaft and base resistances, and the split
Practice
A 0.6 m bored pile passes through 10 m of firm clay with su = 90 kPa and α = 0.576. What shaft resistance does that layer contribute, in kN?
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