The Chartership Forum

Structural Steel Design / Column bases and anchorages

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.

Lesson review: expert verification remains open

This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

Corrections checked in this pass

  • Original default formula already subtracted flange/web overlap: that particular audit allegation was false. Large projections and plate clipping were genuine defects, now handled by exact clipped rectangle union and matching SVG. Stated fy = 345 gives 3965.0 kN for the 30 mm illustrative plate at assumed fjd = 40.

Checks needed for project use

  • A steel-design reviewer must check the first-generation calculation profiles, second-order analysis assumptions, connection rules and execution references against the applicable standards and amendments.
  • Concrete supporting volume and dispersal geometry, genuine loaded area, grout, αcc convention, enhancement restrictions, plate yield by thickness, welds and actual column actions. fjd40 is an assumed separately verified input, not established by a pad-plan ratio.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.

Read the lesson register and outstanding review notes

Why this matters

A column base transfers forces through the plate, grout, concrete, anchors and foundation. The effective-area method credits only part of the plate for compression resistance. Its shaded area is an idealisation, not a measured pressure distribution. This lesson derives the plate-strip projection and keeps the bearing-strength assumptions separate.

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)

Joint bearing strength

A supported loaded area may have enhanced concrete bearing resistance, subject to the applicable distribution geometry and limits. A pad’s plan area alone does not establish the supporting distribution area. Grout properties, thickness and bedding conditions also matter.

Write fcd = αcc fck/γC and fjd = βj kj fcd with the factors appropriate to the selected standard and project. For illustration, fck = 30, γC = 1.5, αcc = 0.85, kj = 3 and βj = 2/3 give fcd = 17 and fjd = 34 N/mm². The worked effective-area example below instead takes fjd = 40 N/mm² as a separately assumed verified input; it does not claim to derive that value from this pad.

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.

A uniformly loaded cantilever strip gives a useful resistance idealisation for the plate beyond the column outline. Balancing its moment demand with the chosen strip resistance gives an effective projection. The resulting area receives resistance credit; actual contact pressure need not stop at that line.

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.

30 mm
500 mm
30 N/mm² cylinder
1,500 mm
Base plate outline with the effective bearing area shadedorange = column · blue = credited area · outline = plateprojection 51 mm · working 40% of the plateresistance 3965 kN · the column can bring 3973 kN
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
50.9 mm
Effective area
99 × 10³ mm²
Gross plate area
250 × 10³ mm²
Fraction of the plate working
40 %
Resistance NRd
3965 kN
What the gross plate would suggest
10000 kN
Model scope
Illustrative bearing comparison only
Thickness to use the whole plate
74.0 mm

The model credits 99.1 × 10³ mm², or 39.7% of the plate, with projection c = 50.9 mm. Shading shows resistance-credit area, not the actual contact-pressure distribution. Verify the assumed bearing input, plate material and complete base detail.

Things worth trying

  • Vary plate thickness and inspect the credited area. It is a resistance idealisation, not a contact-pressure map.
  • At large projection, the expanded flange and web footprints overlap. The drawing uses their clipped union, so overlaps do not count twice.
  • Concrete bearing enhancement requires a valid supporting distribution geometry; a pad plan area alone does not establish it.
  • Thickness-dependent material strength must be used. The full-coverage comparison is not a plate specification.

Worked example

A base plate for a 254 UC

Given

  • Idealised column outline: h = 260.3 mm, b = 256.3 mm, tf = 17.3 mm, tw = 10.3 mm. Design compression NEd = 3800 kN.
  • 500 × 500 mm base plate; use fy = 345 N/mm² for the 25–30 mm S355 plate in this example.
  • Assume a separately verified joint bearing strength fjd = 40 N/mm². The pad geometry, concrete strength and grout checks establishing this input are outside this numerical example.
  • 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

    For an assumed geometrically valid supporting distribution area Ac1 = 1500² mm² and loaded area Ac0 = 500² mm², calculate √(Ac1/Ac0). The pad plan dimensions alone do not establish Ac1.

    Practice

    For an illustrative calculation with αcc = 1.0 explicitly assumed, fck = 30 N/mm², γC = 1.5, kj = 3.0 and βj = 0.667, calculate fjd. This is not the UK αcc selection.

    Practice

    A 25 mm plate has assumed fy = 355 N/mm² and fjd = 40 N/mm². Calculate its effective projection c for γM0 = 1.0. This is an arithmetic exercise with an explicit strength input, separate from the S355 worked example.

    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

    • Bearing enhancement requires a verified supporting distribution area and applicable concrete/grout factors.
    • A plate spreads pressure only c = t√(fy/3fjd γM0) — DERIVED from a cantilever balance
    • The derivation uses an elastic strip modulus; plastic base components require their own applicable model.
    • c is linear in thickness and falls with the square root of the bearing strength
    • Using assumed fjd = 40 and plate fy = 345, the 30 mm plate gives 3965 kN against NEd = 3800 kN.
    • But it used only 40% of the plate area
    • Whole-area coverage requires a geometric union check; a thicker plate also needs its own material strength.
    • Coordinate plate size, thickness, anchors, bearing and erection access.