Module 15 · Lesson 15.1
Composite action and the connection
What connecting a slab to a beam actually buys, and how the studs that do it are sized.
Why this matters
Lay a concrete slab on a steel beam and you have two members. Connect them and you have one — with roughly twice the stiffness and half again the moment capacity, from the same materials. Everything that gain depends on passes through the shear connectors, which is why composite design is really connection design with a moment calculation attached.
By the end of this lesson you should be able to
- Explain composite action in terms of the lever arm
- Determine the effective flange width and say what shear lag is
- Calculate the plastic moment with full shear connection
- Size the studs from equilibrium
What you should already know
- Composite beams and transformed sections (Structural Analysis Fundamentals, Module 12)
- The rectangular stress block and plastic analysis (Module 4)
- Bond and the idea of transferring force between materials (Module 5)
What the connection buys
Take a slab resting freely on a steel beam and load it. Each bends about its own centroid, and they share the load in proportion to their stiffnesses. The slab slides over the beam at the interface — you can watch it happen at the ends.
Now connect them so that sliding cannot occur. They must bend about a common axis, which lies far above the steel centroid and near the interface. The steel is now almost entirely below that axis, in tension, and the concrete almost entirely above it, in compression.
The gain comes from the LEVER ARM. The steel force and the concrete force are separated by roughly half the beam depth plus half the slab depth, and the moment is that separation times the force.
That is also the ideal arrangement of the two materials: concrete in compression, where it is strong, and steel in tension, where it is strong. Composite construction is efficient because the geometry and the material properties happen to agree.
Effective width
The slab does not compress uniformly across its full width. Stress is highest directly over the beam and falls away either side — shear lag, the same effect met in wide flanges in the Structural Analysis Fundamentals course.
Rather than model it, an effective width is used: a narrower strip carrying the full stress, chosen to deliver the same total force.
beff = Σ min(Le/8, bi)
Le is the distance between points of zero moment, which for a simply supported beam is the span. The Le/8 term is the shear-lag limit; the bi term is simply how much slab is available before the next beam claims it. Which governs tells you something useful: if spacing governs, the beams are close together and the whole slab works; if shear lag governs, widening the spacing buys nothing.
What it calculates: The moment of resistance when the neutral axis falls within the slab.
- Na
- Tensile resistance of the whole steel section (N)
- Nc
- Compressive resistance of the full concrete flange (N)
- x
- Depth of concrete actually needed, Na/(0.85 fcd beff) (mm)
- ha
- Steel section depth (mm)
- hs
- Slab depth (mm)
This assumes
- FULL shear connection — the connectors can transfer everything demanded
- Both materials reach their design strengths, which plastic design permits
- Na ≤ Nc, so the neutral axis is inside the slab
- The steel section is compact enough not to buckle locally before yielding
In plain terms: Plastic design makes this remarkably simple: both forces are known immediately from the material strengths, so the moment is one multiplication. Compare with the elastic transformed-section calculation of Module 12 in the Structural Analysis Fundamentals course, which needs a modular ratio and a neutral-axis solution. The plastic method is simpler and gives more capacity — which is why it is used wherever the section is compact enough to allow it.
Worked example
A composite floor beam
Given
- Steel beam, 450 mm deep, area 9500 mm², grade S355
- 130 mm concrete slab, C30/37, beams at 3.0 m centres
- Simply supported span 9.0 m
- 19 mm diameter headed studs, 100 mm high
Find
The effective width, the plastic moment, and the studs required.
Assumptions
- Full shear connection for this first calculation
- The steel section is compact, so plastic design applies
- γa = 1.0 for the steel; γV = 1.25 for the studs
Practice
A simply supported composite beam spans 12.0 m with beams at 3.5 m centres. What is the effective flange width, in mm?
Practice
A steel section of 9500 mm² in S355 acts with a slab. What is its tensile resistance Na, in kN?
Practice
That force must be transferred by studs of 81.7 kN each, over a half span. How many studs are needed for full shear connection?
Check yourself
What does connecting the slab to the steel beam actually change?
Summary
- Composite action comes from the LEVER ARM between the two forces
- Concrete in compression, steel in tension — geometry and material strengths agree
- beff = Σ min(Le/8, b) allows for shear lag
- Plastic design is simple: both forces follow from the material strengths
- The neutral axis should fall in the slab; in the steel means an inefficient section
- Compositing roughly doubles the moment capacity of the bare steel section
- Stud numbers follow from equilibrium, and full connection is often impractically tight
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