Module 6 · Lesson 6.1
The effective area method
Shrink the footing until the load is in the middle.
Why this matters
Almost no foundation is loaded exactly through its centroid. Columns carry moments, retaining walls lean, machine bases have offset loads. The device engineers use to handle this is so simple it looks like a cheat — and it works well enough that it is in every design code.
By the end of this lesson you should be able to
- Compute the effective area of an eccentrically loaded footing
- Apply the middle-third rule and explain its physical meaning
- Say why the method is conservative
Meyerhof's idea: if the load acts at an eccentricity e from the centre, imagine trimming the footing down until the load is central on what is left. To move the centroid by e you must remove 2e from the width — e from the far side twice over, in effect — so the effective width is
B′ = B − 2e
Then calculate the bearing capacity for a footing of width B′, and check it against the load spread over the effective area only. The trimmed-off strip is assumed to contribute nothing.
That last assumption is why the method is conservative. The discarded soil is still there and still carries something; the method simply declines to count it.
What it calculates: The reduced footing area used for an eccentric load
- B′
- Effective width (m)
- e
- Eccentricity of the resultant from the centroid (m)
- A′
- Effective area, used for both capacity and applied pressure (m²)
This assumes
- The resultant acts within the footing — beyond that there is no effective area at all
- Eccentricity in each direction is treated independently
In plain terms: Eccentricity costs twice its own value in width. A 0.4 m eccentricity on a 3 m footing removes 0.8 m — over a quarter of it.
Worked example
A pad with a column moment
Given
- Square pad, B = L = 3.0 m, founded at 1.5 m
- Vertical load V = 900 kN
- Column moment M = 360 kNm about one axis
Find
The effective area, and whether contact is lost at an edge
The middle-third rule falls out of elastic stress distribution. Under combined axial load and moment the edge pressures are
q = V/A ± M·c/I
and the minimum goes to zero exactly when e = B/6. Beyond that the arithmetic gives a negative pressure at one edge — tension between the footing and the soil.
Soil cannot pull. So beyond the middle third the footing lifts off over part of its base, the contact area shrinks, and the pressure on the remaining contact rises faster than the eccentricity does.
Check yourself
A footing's resultant load falls outside the middle third. What physically happens?
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