Module 11 · Lesson 11.2
Eccentric loads and pile caps
When the base lifts off at one edge, and when the member is too deep for bending theory to apply at all.
Why this matters
Two situations break the assumptions of the previous lesson. A base carrying a moment as well as an axial load may lift off at one edge, and once it does the familiar N/A ± M/Z stops being true — in the unsafe direction. And a pile cap is so deep relative to its span that plane sections do not remain plane, so bending theory does not apply to it at all. Both are ordinary situations, and both are routinely designed with the wrong model.
By the end of this lesson you should be able to
- Calculate bearing pressure under axial load and moment
- Identify when the resultant leaves the middle third, and what changes
- Explain why the elastic formula becomes unsafe once the base lifts
- Design a pile cap tie by strut and tie
- Say why pile-cap reinforcement is never curtailed
What you should already know
- Pad footings (previous lesson)
- Bending stress and the section modulus (Structural Analysis Fundamentals, Module 9)
- The variable-angle truss, as an example of strut-and-tie thinking (Module 5)
The middle third
A base carrying axial load N and moment M has an eccentricity e = M/N. The pressure under it, while the whole base is in contact with the ground, follows the familiar combined-stress formula:
p = N/A ± M/Z
The minimum pressure reaches zero when e = L/6 — the edge of the middle third. This is not a code rule; it falls directly out of the algebra, and it is the same middle-third rule that governs masonry.
Push the eccentricity beyond L/6 and the formula would predict tension between the base and the ground. The ground cannot pull. So the base lifts off over part of its length, and the pressure redistributes into a triangle over a reduced contact length:
contact length = 3(L/2 − e), and pmax = 2N/(B × contact)
The important consequence: the elastic formula understates the peak pressure once the base lifts. Continuing to use N/A + M/Z past the middle third is unconservative, which is the wrong direction for a bearing check.
What it calculates: The pressure distribution under a base carrying axial load and moment.
- e
- Eccentricity M/N (mm)
- L
- Base length in the direction of the moment (mm)
- B
- Base width (mm)
- Z
- Section modulus of the base in plan, BL²/6 (mm³)
This assumes
- The base is rigid relative to the ground — a strong assumption for a large raft
- The ground carries compression only, and no tension
- Linear pressure distribution, which is a convention rather than a soil model
In plain terms: Two formulae, and which one applies is decided by a single comparison. A designer who only knows the first will get a plausible, wrong, unsafe answer whenever the base lifts — and nothing in the arithmetic will signal it.
Worked example
A base that lifts
Given
- Base 3.0 m long × 2.5 m wide
- Axial load 1200 kN with a moment of 300 kNm, and a second case with 700 kNm
Find
The pressure distribution in each case.
Assumptions
- Rigid base
- Ground carries no tension
Pile caps: the wrong tool is bending theory
A pile cap over two piles at 1.8 m centres might be 900 mm deep. Its span-to-depth ratio is about 2. Plane sections emphatically do not remain plane in a member that stocky, so the flexural theory of Module 4 — which rests entirely on that assumption — does not apply.
What actually happens is simpler and more direct. The column load travels down two inclined compression struts, straight from the column into the piles. Those struts push outwards at the bottom, and something must hold the piles together against that push.
That something is the reinforcement, and it is a tie, not flexural steel. The distinction has three practical consequences:
- The tie force follows from the strut geometry, not from a bending moment.
- The bar must be anchored fully beyond the centre of each pile — the tie is doing its job right out to the ends.
- The reinforcement is never curtailed. A tie carries its full force along its whole length; there is no point at which it is 'no longer needed'.
What it calculates: The tension the bottom reinforcement must carry, from strut geometry.
- N
- Factored column load (N)
- l
- Pile spacing, centre to centre (mm)
- d
- Effective depth of the cap (mm)
This assumes
- The member is genuinely deep — pile spacing less than about 3d
- The load enters the cap through the column and leaves through the piles only
- The struts can develop without crushing, which needs a separate check at the node
In plain terms: Read the d in the denominator: a deeper cap has steeper struts, which push outwards less, so the tie force falls in inverse proportion. Doubling the depth halves the reinforcement. That is a much stronger effect than in a beam, where doubling the depth roughly halves the steel too but for an entirely different reason — and it is why pile caps are made deep.
Worked example
A two-pile cap
Given
- Two piles at 1800 mm centres
- Factored column load 2400 kN
- Cap 800 mm deep, giving d = 700 mm
- Grade 500 reinforcement, fyd = 434.8 N/mm²
Find
The tie force and the reinforcement required.
Assumptions
- Strut-and-tie applies: the spacing 1800 mm is less than 3d = 2100 mm
- The pile load is concentrated at the pile centre
Practice
A 2.4 m long by 2.0 m wide base carries 900 kN with a moment of 200 kNm. Is the resultant within the middle third, and what is the maximum bearing pressure, in kPa?
Practice
Two piles at 2400 mm centres carry a factored column load of 3000 kN through a cap with d = 800 mm. What is the tie force, in kN?
Practice
For that same cap, if the depth were increased so d = 1000 mm, what would the tie force become, in kN?
Check yourself
Why is pile-cap reinforcement never curtailed?
Summary
- e ≤ L/6: the whole base bears and p = N/A ± M/Z
- e > L/6: the base lifts, the pressure is triangular over 3(L/2 − e), and the elastic formula becomes unsafe
- A negative computed pressure is the signal that the wrong formula is in use
- A pile cap is a deep member: plane sections do not remain plane
- The reinforcement is a tie, with T = N l/(4d)
- Doubling the cap depth halves the tie force
- The tie is anchored beyond the pile centres and never curtailed
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