Module 11 · Lesson 11.1
How composite action works, and what it buys
The slab is already there. Connect it to the beam and the two act as one section — with the concrete doing what it is good at and the steel doing what it is good at.
Why this matters
Almost every steel-framed floor has a concrete slab on it. Left unconnected, the slab is a load the beam carries. Connected, it becomes the compression flange of a much deeper section — and since concrete is cheap in compression and steel is excellent in tension, the two are almost ideally matched. The gains are large enough to change what section is needed. What makes composite design demanding is not the mechanics, which is straightforward, but the bookkeeping: at several points in its life the beam is a different structure, and the checks have to follow.
By the end of this lesson you should be able to
- Say why connecting the slab changes the section so much
- Calculate the plastic moment and identify the efficient case
- Explain the two stud failure modes and which governs
- Quantify what partial shear connection delivers
What you should already know
- Plastic moment and the position of the plastic neutral axis (Module 6)
- Lateral-torsional buckling, for the construction stage (Module 8)
- Concrete in compression and the design strength fcd — the RC course covers this
Two sections, or one?
Put a slab on a beam without connecting them and they bend together but slide against each other at the interface. Each has its own neutral axis; each carries its share of the moment in proportion to its own stiffness. The lever arm available to the pair is small.
Connect them and the sliding stops. The two become one section with one neutral axis, and because the concrete is above and the steel below, that neutral axis moves a long way up — usually into or just below the slab.
That is the whole mechanism, and its consequence is worth stating carefully:
If the neutral axis is at or above the bottom of the slab, the entire steel section is in tension, at a lever arm of roughly half the beam depth plus half the slab depth.
Compare that with the bare steel beam, where half the section is in compression and half in tension, working about the beam's own centroid.
For the worked example the numbers are:
| Bare steel | Composite | |
|---|---|---|
| Moment resistance | 595 kNm | 1057 kNm |
| Second moment of area | 33 500 cm⁴ | 109 000 cm⁴ |
1.78 times the moment and 3.26 times the stiffness, from a slab that was going to be there anyway and about 86 studs.
The stiffness gain is the larger of the two, and on a floor beam it is usually the one that matters — floor beams are very often governed by deflection or by vibration rather than by strength.
Try it
The section, the studs, and what they buy
A 457 × 191 × 74 UB under a solid slab in C30/37. The bars are the bare steel beam, the full composite section, and what the studs you have actually provided deliver.
- Effective width beff
- 2250 mm
- Governed by
- shear lag
- Steel force Na
- 3464 kN
- Concrete force Nc
- 4227 kN
- Plastic neutral axis
- slab
- Bare steel Mc,Rd
- 595 kNm
- Full composite MRd
- 1057 kNm
- Moment gain
- 1.78 ×
- Stiffness gain
- 3.26 ×
- One stud PRd
- 81.7 kN
- Stud failure mode
- stud shear
- Studs for full connection
- 43 per half span
- Degree of connection η
- 1.00
- Moment as provided
- 1057 kNm
- Share of the full moment
- 100 %
The plastic neutral axis is in the slab, so the whole steel section is in tension — the efficient case. Composite action multiplies the moment by 1.78 and the stiffness by 3.26, for 43 studs per half span.
Things worth trying
- Start at the defaults: 9 m span, 3 m spacing, 130 mm slab. The moment goes from 595 to 1057 kNm and the stiffness roughly triples.
- Take the studs down to 60%. The moment falls only to about 83% of the full value, because the interpolation starts at the bare steel beam and not at zero. This is why partial connection is so often economic.
- Take them to 20%. Even then the beam keeps most of what it had — the last studs are always the least productive.
- Now widen the beam spacing from 3 m to 6 m at the 9 m span. The effective width does not move at all, because shear lag already governs it. Widening the spacing beyond about a quarter of the span buys nothing.
- Shorten the span to 4 m instead and watch the effective width collapse — the force has less length over which to spread sideways.
- Now thin the slab to 90 mm and take the SPAN down to 4 m with beams at 1.5 m. Na exceeds Nc, the neutral axis drops into the steel, and the verdict changes: part of the steel is now in compression and the section has become inefficient. Note it is the short span that does it, not the thin slab on its own — a thin slab under a long span still has a wide effective width and plenty of concrete force.
- Watch the stud failure mode as you change nothing else. At C30/37 the shank and the concrete are within a couple of percent, so it is a marginal call — and the two failures have completely different remedies.
Worked example
A 9 m composite floor beam
Given
- 457 × 191 × 74 UB in S355, 9.0 m simply supported, at 3.0 m centres
- 130 mm solid slab in C30/37
- 19 mm diameter studs, 100 mm high, fu = 450 N/mm²
- Unpropped construction
Find
The composite moment resistance, the studs needed, and what partial connection would give.
Assumptions
- Plastic design of the composite section: both materials reach their design strengths
- 0.85 fcd on the concrete — a calibrated allowance for sustained loading, from the RC course
- Section properties computed from plate geometry, so slightly below published values
Predict first
Only 60% of the studs needed for full shear connection are provided. What fraction of the full composite moment resistance survives?
Practice
A composite beam has beff limited by shear lag on a 9.0 m span, with beams at 3.0 m centres. What is beff, in mm?
Practice
A steel section of 9760 mm² in S355 acts compositely. What is Na, the force it can develop in tension, in kN?
Practice
That force must cross the interface over the half span. If one stud carries 81.7 kN, how many studs are needed?
Practice
A composite beam has MRd = 1057 kNm with full connection and Ma = 595 kNm as bare steel. With η = 0.60, what is MRd in kNm?
Check yourself
Why is it desirable for the plastic neutral axis of a composite beam to lie inside the slab?
Summary
- Connecting the slab makes ONE section with one neutral axis, high up
- So nearly the whole steel section is in tension, at a large lever arm
- The worked beam went from 595 to 1057 kNm — 1.78× — and 3.26× on stiffness
- Effective width allows for shear lag: 2250 mm here, span-governed not spacing-governed
- The efficient case has the plastic neutral axis inside the slab
- A stud fails in its shank or by crushing the concrete — here they were 2% apart
- 43 studs per half span for full connection, from an equilibrium argument
- Partial connection is strongly non-linear: 60% of the studs gave 83% of the moment
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