Module 11 · Lesson 11.1
Pad footings
Two calculations on the same base, at two different limit states, with two different sets of loads.
Why this matters
A foundation is the one element where a structural engineer's work is governed by someone else's number — the allowable bearing pressure. Two mistakes follow from not understanding what that number is. Sizing the base with factored loads makes every foundation in the building 35% too big. Designing the reinforcement with unfactored loads makes them all under-strength. Both are common, and the second one matters.
By the end of this lesson you should be able to
- State which limit state governs sizing and which governs the concrete design
- Size a square pad from an allowable pressure, allowing for its own weight
- Calculate the design moment and the two shear checks
- Explain why the base's self-weight cancels out of the structural design
What you should already know
- Combinations of actions, and the difference between ULS and SLS (Module 2)
- Flexural design (Module 4)
- Shear and punching shear (Module 5)
Two calculations, two limit states
A pad footing is designed twice, and the two calculations barely speak to each other.
Sizing the plan area is a geotechnical question. How big must the base be so that the ground beneath it neither fails nor settles unacceptably? The number a geotechnical engineer supplies — an allowable bearing pressure — has usually already had a factor applied to it and is usually settlement-controlled. It is therefore compared against characteristic, unfactored loads. Settlement is a serviceability matter.
Designing the concrete is a structural question. How thick must the base be, and how much reinforcement does it need, so that it does not fail in bending or shear? That is an ultimate check with factored loads.
Use factored loads to size the plan area and every foundation is about 35% too large. Use unfactored loads to design the reinforcement and every foundation is under-strength by the same margin.
The first error is expensive; the second is dangerous. Both come from not asking which question is being answered.
Worked example
A square pad footing, from ground to reinforcement
Given
- Internal column 400 mm × 400 mm
- Characteristic axial load 1400 kN; design (factored) axial load 1900 kN
- Allowable bearing pressure 200 kPa = 0.20 N/mm²
- C30/37 concrete, grade 500 reinforcement, 50 mm cover
- Trial base thickness 650 mm
Find
The plan size, the reinforcement, and whether the shear checks pass.
Assumptions
- The allowable pressure is a serviceability value, compared with characteristic loads
- Concrete at 25 kN/m³
- Two layers of H16, so d ≈ 650 − 50 − 20 = 580 mm
Practice
A column carries a characteristic load of 900 kN. The allowable bearing pressure is 150 kPa and the base is 600 mm thick (concrete at 25 kN/m³). What square base side is required, in m?
Practice
A 3.0 m square pad carries a factored column load of 2100 kN from a 450 mm square column. What is the design moment at the column face, in kNm?
Check yourself
Which loads should be used to determine the PLAN AREA of a pad footing?
Summary
- Sizing: characteristic loads plus base self-weight, against the allowable pressure
- Design: factored loads, and the self-weight cancels exactly
- Bending is taken at the column face; beam shear at d from it
- Punching is checked on a perimeter at 2d, with the enclosed pressure deducted
- For a thick compact pad the punching perimeter nearly reaches the edge, so punching rarely governs
- Minimum reinforcement usually governs a footing, because d is large and K is tiny
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