Structural Steel Design / Steel–concrete composite beams

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.

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

Steel self-weight, factored construction actions and failure against assumed 165 kNm resistance made explicit. Temporary-load removal produces negative elastic recovery; sustained/transient loads use separate stiffnesses. Per-plane slab shear corrected. Unsupported plastic section and interpolation boundaries are withheld.

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.
Actual temporary restraints, construction combinations, deck/stud detailing, creep/shrinkage, propping and removal chronology, connection ductility/minimum connection. Arbitrary propped chronology deliberately unsupported.

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

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 f_cd — 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.

Shear connectors transfer longitudinal force between the slab and beam. The full-interaction idealisation neglects interface slip. 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

Composite section and shear connection

A 457 × 191 × 74 UB under a solid C30/37 slab. Compare the bare-steel and full-connection section resistances. Partial-connection interpolation is illustrative unless its minimum connection and ductility conditions are established.

Span9.0 m

Beam spacing3.00 m

Slab thickness130 mm

Studs provided100 % of full connection

Effective width b_eff: 2250 mm
Governed by: shear lag
Steel force N_a: 3464 kN
Concrete force N_c: 4973 kN
Plastic neutral axis: slab
Bare steel M_c,Rd: 595 kNm
Full composite M_Rd: 1085 kNm
Moment gain: 1.82 ×
Stiffness gain: 3.26 ×
One stud P_Rd: 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: 1085 kNm
Share of the full moment: 100 %

Cross-section interpolation is within the stated connection limits; verify connector detailing and member checks. Full-connection plastic cross-section resistance with the neutral axis in the slab. Member stability and connector detailing are separate checks.

Things worth trying

Change span and spacing to see which effective-width limit governs.
Reduce the stud count to explore the illustrative interpolation. A plotted value below full connection is not usable without the applicable minimum connection, connector ductility and detailing checks.
A narrow slab can put the plastic neutral axis into the steel. This simplified section model then withholds resistance; it needs flange and web equilibrium rather than a guessed lever arm.
The stud resistance shown assumes a solid slab. Slab cover, transverse reinforcement, spacing, welding and any deck reductions are separate checks.

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, f_u = 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 f_cd 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 b_eff limited by shear lag on a 9.0 m span, with beams at 3.0 m centres. What is b_eff, in mm?

Practice

A steel section of 9760 mm² in S355 acts compositely. What is N_a, 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 M_Rd = 1057 kNm with full connection and M_a = 595 kNm as bare steel. With η = 0.60, what is M_Rd 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
The linear interpolation starts from bare-steel resistance: 60% connection gives about 83% of full composite resistance