EN 1994 Composite Beam Design — Shear Connectors, IPE Profile, Worked Example

Quick Reference: This guide presents a complete EN 1994-1-1 composite beam design for a simply supported IPE 400 floor beam with a 130 mm concrete slab on profiled decking. We cover effective width (Cl. 5.4.1.2), plastic moment resistance with full interaction (Cl. 6.2.1), shear connector design per Cl. 6.6.3.1, degree of shear connection for partial interaction, longitudinal shear transfer (Cl. 6.6.6), and serviceability deflection including shrinkage effects. All clauses reference EN 1994-1-1:2004.

PRELIMINARY — NOT FOR CONSTRUCTION. All calculations are illustrative educational examples. Results must be verified by a licensed Professional Engineer before use in any design project.

1. The Composite Action Principle

A composite beam combines a steel section with a concrete slab connected by shear studs. The slab resists compression, the steel beam resists tension, and the shear connectors transfer longitudinal shear at the interface. The result is a structural element 30–50% stronger and 2–3 times stiffer than the bare steel beam alone.

The key EN 1994-1-1 design clauses:

2. Worked Example Problem Statement

Given:

• Steel beam: IPE 400, S275 (h = 400 mm, b = 180 mm, tw = 8.6 mm, tf = 13.5 mm)
• Aa = 8450 mm², Iy = 231.3 × 10⁶ mm⁴, Wpl,y = 1310 × 10³ mm³
• Concrete slab: C30/37, thickness hc = 130 mm, 120 mm above deck ribs
• Profiled deck: ComFlor 60, ribs perpendicular to beam, hp = 60 mm
• Slab width each side: bi = 3.0 m, span L = 12.0 m, simply supported
• Spacing: beams at 6.0 m centres (interior beam)
• Shear connectors: 19 mm diameter headed studs, fu = 450 N/mm², hsc = 100 mm
• Partial factors: γM0 = 1.00, γM1 = 1.00, γV = 1.25, γC = 1.50
• Design loads: gk = 4.5 kN/m² dead, qk = 5.0 kN/m² imposed

3. Step 1 — Effective Width

The effective width determines how much slab concrete acts compositely with the steel beam.

beff = b0 + Σbei

Where:

• b0 = centre-to-centre stud spacing = 0 for single beam
• bei = min(Le/8, bi) for each side
• Le = equivalent span = distance between points of zero moment

For a simply supported beam: Le = L = 12.0 m. Le/8 = 12.0/8 = 1.5 m. bi = 3.0 m each side.

bei = min(1.5, 3.0) = 1.5 m per side. beff = 0 + 1.5 + 1.5 = 3.0 m.

The effective concrete flange is 3000 mm wide by 120 mm deep (above the deck ribs). The concrete in the ribs below the top of the profiled deck is excluded from the section resistance.

Ac = beff × hc,eff = 3000 × 120 = 360,000 mm²

4. Step 2 — Design Actions

Tributary width = 6.0 m beam spacing.

Design UDL per beam:

• wG = 4.5 × 6.0 = 27.0 kN/m (dead)
• wQ = 5.0 × 6.0 = 30.0 kN/m (imposed)

Design bending moment at mid-span (EN 1990 Eq. 6.10): MEd = (γG × gk + γQ × qk) × L² / 8 = (1.35 × 27.0 + 1.50 × 30.0) × 12.0² / 8 = (36.45 + 45.0) × 144 / 8 = 81.45 × 18 = 1466.1 kN·m

Design shear at support: VEd = 81.45 × 12.0 / 2 = 488.7 kN

5. Step 3 — Plastic Moment Resistance (Full Interaction)

The plastic moment capacity assumes the steel beam yields in tension and the concrete slab reaches 0.85fcd in compression. The plastic neutral axis (PNA) location determines which elements are in compression.

5.1 Material Strengths

Concrete design strength: fcd = fck / γC = 30 / 1.50 = 20 N/mm² Design compressive stress block: 0.85 fcd = 17.0 N/mm²

Steel design strength: fyd = fy / γM0 = 275 / 1.00 = 275 N/mm²

5.2 Compression Capacity of Concrete Flange

Nc,f = 0.85 × fcd × beff × hc = 0.85 × 20 × 3000 × 120 = 6120 kN

5.3 Tension Capacity of Steel Section

Na = Aa × fyd = 8450 × 275 = 2323.8 kN

Since Nc,f (6120 kN) > Na (2323.8 kN), the PNA lies within the concrete slab — the steel beam yields fully in tension before the concrete crushes. This is the most efficient case for composite beams.

5.4 Depth of PNA in Slab

zpna = Na / (0.85 × fcd × beff) = 2323.8 × 10³ / (0.85 × 20 × 3000) = 2323.8 × 10³ / 51,000 = 45.6 mm

The PNA is 45.6 mm below the top of the slab, well within the 120 mm slab thickness.

5.5 Plastic Moment Capacity

Lever arm z = ha/2 + hc − zpna/2 = 400/2 + 120 − 45.6/2 = 200 + 120 − 22.8 = 297.2 mm

Mpl,Rd = Na × z = 2323.8 × 0.2972 = 690.7 kN·m

Check: MEd (1466.1) > Mpl,Rd (690.7) → FAIL — section inadequate.

This result highlights a common finding: the bare IPE 400 in S275 is too light for a 12 m span with 6 m beam spacing. We need either a heavier section (e.g. IPE 500 or IPE 600), higher steel grade (S355), or closer beam spacing. For the remainder of this example, we will work with the IPE 500 to demonstrate the full procedure.

6. Revised with IPE 500 Steel Section

IPE 500 Properties (S275)

• h = 500 mm, b = 200 mm, tw = 10.2 mm, tf = 16.0 mm
• Aa = 11,550 mm², Iy = 482.0 × 10⁶ mm⁴, Wpl,y = 2200 × 10³ mm³

Na = 11,550 × 275 = 3176.3 kN zpna = 3176.3 × 10³ / 51,000 = 62.3 mm z = 500/2 + 120 − 62.3/2 = 250 + 120 − 31.15 = 338.9 mm

Mpl,Rd = 3176.3 × 0.3389 = 1076.4 kN·m → still less than MEd = 1466.1 kN·m.

7. Revised with IPE 600 Steel Section (Final)

IPE 600 Properties (S275)

• h = 600 mm, b = 220 mm, tw = 12.0 mm, tf = 19.0 mm
• Aa = 15,600 mm², Iy = 920.8 × 10⁶ mm⁴, Wpl,y = 3510 × 10³ mm³

Na = 15,600 × 275 = 4290.0 kN zpna = 4290.0 × 10³ / 51,000 = 84.1 mm (still < 120 mm → PNA in slab) z = 600/2 + 120 − 84.1/2 = 300 + 120 − 42.05 = 378.0 mm

Mpl,Rd = 4290.0 × 0.3780 = 1621.6 kN·m

Mpl,Rd = 1621.6 > MEd = 1466.1 → OK. Utilisation = 1466.1/1621.6 = 0.904.

8. Step 4 — Shear Connector Design

8.1 Stud Capacity PRd

For headed studs, EN 1994-1-1 Cl. 6.6.3.1:

d = 19 mm, hsc = 100 mm, fu = 450 N/mm² As = π × 19² / 4 = 283.5 mm²

Steel failure (shank fracture): PRd,s = 0.8 × fu × As / γV = 0.8 × 450 × 283.5 / 1.25 = 81,648 N = 81.6 kN

Concrete failure (cone pull-out): α = 1.0 for hsc/d = 100/19 = 5.26 > 4.0 PRd,c = 0.29 × α × d² × √(fck × Ecm) / γV = 0.29 × 1.0 × 19² × √(30 × 33,000) / 1.25 = 0.29 × 361 × √(990,000) / 1.25 = 0.29 × 361 × 994.99 / 1.25 = 83,350 N = 83.4 kN

PRd = min(81.6, 83.4) = 81.6 kN

8.2 Profiled Decking Reduction Factor kt

Deck ribs perpendicular to beam, ComFlor 60: hp = 60 mm, b0 = 150 mm (rib width) Number of studs per rib: nr = 1

kt = 0.85/√nr × (b0/hp) × (h/hp − 1) = 0.85/√1 × (150/60) × (100/60 − 1) = 0.85 × 2.50 × 0.667 = 1.417 → but kt ≤ 1.0 per code

kt = 1.0 (no reduction — the stud penetration ratio is favourable)

Maximum kt allowed is 1.0 when nr = 1. For nr = 2 (two studs per rib), kt typically reduces to 0.6–0.8.

8.3 Design Stud Capacity

PRd,design = kt × PRd = 1.0 × 81.6 = 81.6 kN

9. Step 5 — Number of Shear Connectors

9.1 Full Shear Connection

The total longitudinal shear force to transfer between the steel and concrete equals the lesser of:

• Na = 4290.0 kN (steel tensile capacity)
• Nc,f = 0.85 × fcd × beff × hc = 6120 kN (concrete compression capacity)

Vl = min(4290.0, 6120) = 4290.0 kN

Number of studs for full connection between support and mid-span (half the beam): nf = Vl / PRd = 4290.0 / 81.6 = 52.6 → 53 studs per half-span

Total for full connection: 2 × 53 = 106 studs over 12 m.

9.2 Stud Spacing

Stud spacing = 12,000 / 106 = 113 mm centre-to-centre.

EN 1994-1-1 Cl. 6.6.5.7 specifies minimum longitudinal spacing = 5d = 5 × 19 = 95 mm and maximum = min(6hc, 800 mm) = min(720, 800) = 720 mm. At 113 mm, spacing meets both criteria.

10. Step 6 — Degree of Shear Connection (Partial Interaction)

Full connection requires 106 studs. Partial connection can be economical — fewer studs at the cost of reduced moment capacity.

10.1 Minimum Degree of Connection

ηmin = max(1 − (355/fy) × (0.75 − 0.03Le), 0.4) = max(1 − (355/275) × (0.75 − 0.03×12), 0.4) = max(1 − 1.291 × (0.75 − 0.36), 0.4) = max(1 − 1.291 × 0.39, 0.4) = max(1 − 0.503, 0.4) = max(0.497, 0.4) = 0.497

Minimum 49.7% shear connection permitted → minimum 0.497 × 53 = 27 studs per half-span.

10.2 Moment Resistance with Partial Connection

For n = 30 studs per half-span (η = 30/53 = 0.566):

MRd = Ms,Rd + (Mpl,Rd − Ms,Rd) × η

Ms,Rd (steel alone) = Wpl,y × fy / γM0 = 3510 × 10³ × 275 / 1.00 = 965.3 kN·m

MRd = 965.3 + (1621.6 − 965.3) × 0.566 = 965.3 + 656.3 × 0.566 = 965.3 + 371.5 = 1336.8 kN·m

MEd = 1466.1 > MRd = 1336.8 → FAIL with 30 studs. Need more connectors.

Try n = 45 studs (η = 45/53 = 0.849): MRd = 965.3 + 656.3 × 0.849 = 965.3 + 557.3 = 1522.6 kN·m

MEd = 1466.1 < MRd = 1522.6 → OK with 45 studs per half-span.

Total studs required: 2 × 45 = 90 studs (saving 16 studs vs. full connection).

11. Step 7 — Transverse Reinforcement (Longitudinal Shear)

The concrete slab must resist longitudinal shear at the steel-concrete interface. EN 1994-1-1 Cl. 6.6.6 requires checking potential shear surfaces:

The longitudinal shear per unit length at each shear surface: vL,Ed = Vl / (2 × Lshear) where Lshear is the distance between critical sections.

• Surface a-a: through the slab thickness at the stud position → requires transverse reinforcement ≥ Asf/sf per EN 1992-1-1 Cl. 6.2.4.
• Surface b-b: through the slab around the stud group.

For our configuration, minimum transverse reinforcement = 0.2% of the concrete area in the shear plane per EN 1994-1-1 Cl. 6.6.6.2. Typically satisfied by the slab's A252 mesh or equivalent (252 mm²/m in each direction).

12. Step 8 — Serviceability Deflection

Composite beams are significantly stiffer than bare steel beams. The second moment of area for the composite section accounts for the transformed concrete area:

Modular ratio n0 = Ea / Ecm = 210,000 / 33,000 = 6.36

Transformed concrete width = beff / n0 = 3000 / 6.36 = 471.7 mm

The composite second moment of area Icomp for the uncracked section (short-term loading) is computed using the parallel axis theorem:

Icomp ≈ 920.8 × 10⁶ + 15,600 × (378.0)² + (471.7 × 120³)/12 + (471.7 × 120) × (600 + 120 − 84.1/2 − 600/2)²

For practical purposes, Icomp ≈ 3.5 × Iy = 3.5 × 920.8 × 10⁶ = 3223 × 10⁶ mm⁴

Serviceability deflection under characteristic imposed load: wQ = qk × beam spacing = 5.0 × 6.0 = 30.0 kN/m

δ = 5 × wQ × L⁴ / (384 × Ea × Icomp) = 5 × 30.0 × 12,000⁴ / (384 × 210,000 × 3223 × 10⁶) = 5 × 30.0 × 2.074 × 10¹⁶ / (384 × 210 × 3223 × 10⁹) = 3.111 × 10¹⁷ / 2.600 × 10¹⁷ = 11.97 mm

Span/deflection = 12,000/11.97 = 1003 > 360 → OK for imposed load.

Long-term deflection (including concrete creep) approximately doubles the short-term deflection → δLT ≈ 24 mm → span/δ = 500, which is within typical floor deflection limits.

13. Summary Table

Related Pages

• EN 1993-1-1 Beam Design — IPE 300 Worked Example, χLT
• EN 1993-1-1 Column Design — HEA 200 Example, Buckling Curves
• EN 1993-1-2 Fire Design — Section Factor, Critical Temperature
• EN 1993-1-8 Bolt Design Guide — Fv,Rd, Fb,Rd
• EN 1993-1-8 Weld Design — Directional Method