Module 14 · Lesson 14.1
The base plate, and the concrete under it
The concrete is stronger than you expect and the plate is weaker. Both facts come from the same place: how far a plate can reach before it runs out of bending capacity.
Why this matters
A column base looks like the simplest connection in a steel frame and is one of the least well understood. Two things surprise people. The concrete beneath it is roughly twice as strong as its design cylinder strength, because it is confined. And the plate on top of it is usually working over less than half its own area, because a plate can only spread pressure as far as it can span in bending — which is a much shorter distance than most base plates are wide. Both follow from short pieces of mechanics, and together they explain why base plates look the size they do.
By the end of this lesson you should be able to
- Say why confinement raises the bearing strength, and by how much
- Derive the effective projection of a base plate
- Read what fraction of a plate is actually working
- Say what actually decides a base plate's size
What you should already know
- The plastic and elastic section modulus of a rectangle (Module 6)
- Concrete in compression and fcd — the RC course covers this
- The T-stub, for the next lesson (Module 13)
The concrete is stronger than its cylinder strength
Put a small steel plate on a large concrete pad and press. The concrete directly under the plate cannot spread sideways, because the concrete around it is in the way — it is held in triaxial compression, and concrete in triaxial compression is far stronger than an unconfined cylinder.
That is captured by a concentration factor, essentially the square root of the area ratio, capped because the confinement cannot be relied on indefinitely. Against that, the grout the plate actually sits on is weaker than the foundation, so a coefficient βj reduces it again.
For the base in this module — a 500 mm plate on a 1500 mm pad in C30/37:
| fcd = 30/1.5 | 20.0 N/mm² |
| Concentration factor √(1500²/500²) = 3.0, at its cap | × 3.00 |
| Grout coefficient βj | × 0.667 |
| fjd | 40.0 N/mm² |
Exactly twice the plain design strength. Both factors are code-calibrated and unverified in this course, but the mechanism behind the first one is real and worth understanding: it is why base plates are far smaller than an unconfined calculation would give.
From first principles
How far a base plate can spread the pressure
We want to show: Show that a base plate can only distribute bearing pressure a distance c = t√(fy/3fjd γM0) beyond the column outline, and see what that implies about plate sizes.
The plate is a cantilever sticking out beyond the column, with the bearing pressure pushing up on it. The further it sticks out, the more moment that pressure puts into it at the column face. There is a distance beyond which the plate simply does not have the bending capacity — so the pressure cannot get out there at all. The plate is not choosing to work over a smaller area; it is unable to reach any further.
Try it
How much of the plate is actually working
A 254 UC on a square base plate, drawn to scale. The outline is the plate; the shaded regions are the area that can actually deliver pressure to the concrete.
- Concrete fcd
- 20.0 N/mm²
- Concentration factor kj
- 3.00
- Bearing strength fjd
- 40.0 N/mm²
- As a multiple of fcd
- 2.00 ×
- Effective projection c
- 51.6 mm
- Effective area
- 101 × 10³ mm²
- Gross plate area
- 250 × 10³ mm²
- Fraction of the plate working
- 40 %
- Resistance NRd
- 4022 kN
- What the gross plate would suggest
- 10000 kN
- Enough for the column?
- yes
- Thickness to use the whole plate
- 70.8 mm
A 30 mm plate can spread the pressure 52 mm beyond the column outline — no further, because past that it would need more bending moment than it has. The effective area is 101 × 10³ mm², which is 40% of the plate. The rest of the plate is carrying nothing: it is there for the bolts, or for nothing at all.
Things worth trying
- Start at 30 mm on a 500 mm plate. The shaded area is 40% of the plate and carries 4022 kN — just enough for the column's 3973 kN squash load.
- Drop to 25 mm. The projection falls from 51.6 to 43.0 mm, the resistance to 3367 kN, and the base no longer delivers what the column can bring.
- Compare the resistance with the 'gross plate would suggest' figure. At 30 mm the gross calculation is 2.5 times too high — and too high in the unsafe direction.
- Now try to make the whole plate work. The readout says what thickness that needs: about 70.8 mm on a 500 mm plate. Nobody specifies that, which is the point.
- Widen the plate to 700 mm on the same 1500 mm pad. Two things happen at once, and neither is the obvious one: the concentration factor falls from 3.00 to 2.14, so fjd falls from 40 to 28.6 — and because c goes with 1/√fjd, the projection LENGTHENS from 51.6 to 61.1 mm. The utilisation still collapses, from 40% to 24%, because the plate grew faster than the reach did.
- Raise the concrete strength from 30 to 50. The bearing strength rises and the projection FALLS, because a stronger concrete puts more moment into the plate at the same reach. The resistance still rises, by pressure rather than by reach.
- Shrink the foundation pad towards the plate size. The concentration factor collapses towards 1, and with βj = 0.667 the bearing strength ends up BELOW fcd. Confinement was doing all the work.
Worked example
A base plate for a 254 UC
Given
- 254 × 254 × 89 UC in S355 — A = 111.9 cm², squash load 3973 kN
- 500 × 500 mm base plate in S355 on a 1500 × 1500 mm pad
- Concrete C30/37, γC = 1.5, sand–cement grout bedding
- γM0 = 1.0
Find
The plate thickness needed, and how much of the plate is doing work.
Assumptions
- kj and βj are calibrated and unverified here
- The plate stays elastic — the elastic modulus t²/6 is used
- Uniform bearing pressure over the effective area
Practice
A 500 mm square plate sits on a 1500 mm square pad. What is the uncapped concentration factor √(Ac1/Ac0)?
Practice
With fck = 30 N/mm², γC = 1.5, kj = 3.0 and βj = 0.667, what is fjd in N/mm²?
Practice
A 25 mm base plate in S355 bears at fjd = 40 N/mm². How far can it spread the pressure beyond the column outline, in mm? Take γM0 = 1.0.
Practice
That plate's effective area is 84 × 10³ mm². What compression can the base deliver, in kN?
Check yourself
Why does the effective projection c FALL as the concrete bearing strength rises?
Summary
- Confinement raises the bearing strength: fjd = 40 N/mm² against fcd = 20
- A plate spreads pressure only c = t√(fy/3fjd γM0) — DERIVED from a cantilever balance
- The elastic modulus t²/6 is used, because a base plate is not designed to hinge
- c is linear in thickness and falls with the square root of the bearing strength
- A 30 mm plate carried 4022 kN — the column's full squash load of 3973
- But it used only 40% of the plate area
- Reaching the plate edge would have needed a 70.8 mm plate
- So the plan size is a BOLT decision and the thickness is a bearing decision
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