Module 18 · Lesson 18.2
Capacity and deflection
Which one actually sizes the pile.
Why this matters
A bridge pier that has not collapsed but has moved 60 mm at bearing level is a failed structure in every sense that matters. Lateral pile design is a serviceability problem far more often than a capacity one, and the two are calculated quite differently.
Broms' short-pile solution in clay is a moment balance. It takes the soil resistance as zero over the top 1.5 diameters — that soil is disturbed, can heave, and cannot be relied on — and as 9·su·D below that. The pile rotates about a point, and the ultimate lateral load follows from equilibrium about it.
For a 0.6 m pile, 4 m long, in clay with su = 50 kPa, loaded at ground level, the ultimate lateral load is 530 kN. Raise the load to 1.0 m above ground and it falls to 376 kN — a 29% loss for a metre of eccentricity, because the applied moment about the rotation point has grown while the resisting moment has not.
Lengthening helps a short pile substantially: at 6 m the same pile carries 789 kN, more than double the 4 m value, because the active length appears squared in the resisting moment.
Deflection is usually what governs. A pile that satisfies its ultimate lateral check by a comfortable margin may still move far more at working load than the structure will accept — Module 10's angular distortion argument, applied horizontally.
The subgrade reaction approach gives a deflection directly from kh, but its weakness is now exposed: a single spring constant cannot represent soil that is stiff at small strain and soft at large, and the answer is only as good as the kh chosen for the right movement.
The better tool is the p–y method: replace the single spring with a set of non-linear springs down the pile, each with its own load–deflection curve appropriate to its depth and soil. It is a numerical analysis rather than a formula, it is standard practice for offshore and bridge foundations, and it accommodates layering, cyclic degradation and head fixity in a way no closed-form solution can.
Check yourself
Why does raising the point of load application above ground level reduce a short pile's lateral capacity so much?
Practice
A short pile's ultimate lateral load is 530 kN when loaded at ground level and 376 kN when loaded 1.0 m above it. What percentage of its capacity did the eccentricity cost?
Try it
Rigid or flexible?
The classification comes before any calculation, because the two mechanisms are completely different.
The boundaries
- L/T below about 2 — rigid. Above about 4 — flexible.
- T takes the fourth root of EI, so doubling the section stiffness raises it by only 19%.
- Broms' short-pile solution below is valid ONLY in the rigid range.
- Second moment I
- 0.00636 m⁴
- EI
- 190852 kN·m²
- Stiffness factor T
- 2.49 m
- L/T
- 1.61
- Broms capacity at this e
- 376 kN
- Broms with load at ground level
- 530 kN
- Cost of the eccentricity
- 29 %
Rigid — rotates as a body and fails in the SOIL. Broms' short-pile solution applies, and raising the load 1.0 m above ground has cost 29% of the capacity.
Summary
- T = (EI/kh)^¼ is a length; L/T decides the mechanism and must be found first
- Below L/T ≈ 2 the pile rotates and fails in the soil; above ≈ 4 it hinges and fails in the pile
- The same section was rigid at 4 m and flexible at 10 m
- Load eccentricity of 1 m cost 29% of the short pile's capacity; lengthening 4 m to 6 m more than doubled it
- Deflection normally governs, and p–y analysis is the tool for it
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