EN 1994-1-1 Composite Beam Design — Steel-Concrete with Headed Stud Connectors

Complete reference for EN 1994-1-1:2004 composite steel-concrete beam design. Covers the effective width of concrete flange (Clause 5.4.1.2), headed stud shear connector design (Clause 6.6), degree of shear connection (Clause 6.6.1.2), partial interaction theory, and plastic moment resistance Mpl,Rd (Clause 6.2.1). Includes a fully worked composite beam example with an HEA section and profiled steel decking.

Composite construction combines a steel beam acting compositely with a reinforced concrete slab through headed stud shear connectors. This system is the dominant floor construction method for European multi-storey steel buildings.

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Composite Beam Advantages

Material Properties

Concrete (EN 1992-1-1)

For UK and European building practice, C30/37 is typical for composite slabs on profiled decking.

Headed Stud Connectors (EN ISO 13918)

Effective Width of Concrete Flange (Clause 5.4.1.2)

[ b*{eff} = b_0 + \sum b*{ei} ]

Where (b_{ei} = \min(L_e / 8, b_i)) for each side of the beam web.

For secondary beams at 3.0 m centres in a 9.0 m grid: b1 = b2 = 1,500 mm, Le = 0.7 (\times) 9.0 = 6.3 m, bei = min(788, 1,500) = 788 mm, beff = 788 + 788 = 1,576 mm.

Headed Stud Shear Resistance (Clause 6.6.3)

[ P*{Rd} = \min\left( \frac{0.8 \times f_u \times \pi \times d^2}{4 \times \gamma_V},\quad \frac{0.29 \times \alpha \times d^2 \times \sqrt{f*{ck} \times E_{cm}}}{\gamma_V} \right) ]

Where (\alpha = 0.2 \times (h*{sc}/d + 1)) for (3 \leq h*{sc}/d \leq 4), and (\alpha = 1.0) for (h_{sc}/d > 4).

For a 19 mm stud, hsc = 100 mm (hsc/d = 5.26, (\alpha = 1.0)), C30/37 concrete:

• Stud steel failure: (P_{Rd,1} = 0.8 \times 450 \times \pi \times 19^2 / (4 \times 1.25) = 81.6) kN
• Concrete cone failure: (P_{Rd,2} = 0.29 \times 1.0 \times 19^2 \times \sqrt{30 \times 33,000} / 1.25 = 83.3) kN
• (P_{Rd} = \min(81.6, 83.3) = 81.6) kN (stud steel governs)

Reduction for Profiled Decking (Clause 6.6.4)

For deck ribs perpendicular to the beam with ComFlor 60 deck (hp = 60 mm, b0 = 152 mm), one stud per rib (Nr = 1), hsc = 100 mm:

[ kt = \frac{0.7}{\sqrt{N_r}} \times \frac{b_0}{h_p} \times \left(\frac{h{sc}}{h_p} - 1\right) = \frac{0.7}{1.0} \times \frac{152}{60} \times \left(\frac{100}{60} - 1\right) = 1.183 ]

But (kt \leq 1.0) for Nr = 1 and (k_t \leq k{t,max} = 0.85) for trapezoidal deck with Nr = 1. So (k_t = 0.85).

Reduced stud resistance: (P_{Rd,red} = 0.85 \times 81.6 = 69.4) kN per stud.

Degree of Shear Connection (Clause 6.6.1.2)

The degree of shear connection (\eta = n / n_f), where n is the number of studs provided and nf is the number for full shear connection.

For ductile connectors, the minimum degree of shear connection: [ \eta_{min} = 1 - (355/f_y) \times (0.75 - 0.03 \times L_e) \geq 0.4 ]

For S355, Le = 9.0 m: (\eta_{min} = 1 - 1.0 \times (0.75 - 0.27) = 0.52).

Reduced moment resistance with partial interaction: [ M*{Rd}(\eta) = M*{pl,a,Rd} + \eta \times (M*{pl,Rd} - M*{pl,a,Rd}) ]

Plastic Moment Resistance Mpl,Rd (Clause 6.2.1)

Case 1 — PNA in concrete flange (most common): If (N*{c,f} \geq N*{pl,a}), the plastic neutral axis is in the concrete flange: [ N*{c,f} = 0.85 \times f*{ck} \times b*{eff} \times h_c / \gamma_C ] [ N*{pl,a} = Aa \times f_y / \gamma{M0} ]

Case 2 — PNA in steel top flange or web: If (N*{c,f} < N*{pl,a}), the PNA is in the steel section.

Worked Example — HEA 320 Composite Beam

Problem: Design a composite secondary beam for an office floor. Span 9.0 m, beams at 3.0 m centres. Slab: 130 mm overall, ComFlor 60 deck (hp = 60 mm), hc = 70 mm. Concrete C30/37. Steel S355. Beam self-weight estimated at 1.00 kN/m.

Loading:

• Construction: slab + deck = 3.0 kN/m², construction LL = 0.75 kN/m²
• Composite: finishes = 1.5 kN/m² dead, imposed = 4.0 kN/m²

Step 1 — Construction stage (steel alone): Try HEA 320 (Aa = 12,440 mm², Wpl,y = 1,680 cm³, Iy = 22,930 cm⁴) (M_{c,Rd} = 1,680 \times 10^3 \times 355 / 1.0 = 596.4) kNm

(w*{Ed,c} = 1.35 \times (3.0 \times 3.0 + 1.00) + 1.5 \times (0.75 \times 3.0) = 1.35 \times 10.0 + 1.5 \times 2.25 = 16.88) kN/m (M*{Ed,c} = 16.88 \times 9.0^2 / 8 = 170.9) kNm (\ll) 596.4 kNm — OK for construction.

Step 2 — Effective width: (Le = 0.7 \times 9.0 = 6.3) m, (b{ei} = \min(6,300/8, 1,500) = 788) mm (b_{eff} = 788 + 788 = 1,576) mm

Step 3 — Composite design moment: (w*{Ed} = 1.35 \times (1.5 \times 3.0) + 1.5 \times (4.0 \times 3.0) = 6.08 + 18.0 = 24.08) kN/m (M*{Ed} = 24.08 \times 9.0^2 / 8 + 170.9 = 243.8 + 170.9 = 414.7) kNm

Step 4 — Full composite moment resistance (Case 2 — PNA in steel): (N*{c,f} = 0.85 \times 30 \times 1,576 \times 70 / 1.50 = 2,143) kN (N*{pl,a} = 12,440 \times 355 / 1.0 = 4,416) kN

Since (N*{c,f} < N*{pl,a}), PNA is in steel section. Compression in steel: (N_{ac} = (4,416 - 2,143) / 2 = 1,137) kN

Approximate Mpl,Rd: (M*{pl,Rd} \approx N*{pl,a} \times ha/2 + N{c,f} \times (ha/2 + h_c/2) - N{ac}^2 / (4 \times t_w \times f_y)) (= 4,416 \times 0.155 + 2,143 \times (0.155 + 0.035) - 1,137^2 / (4 \times 9.0 \times 355)) (= 684.5 + 407.2 - 101.2 = 990.5) kNm

Utilisation: 414.7 / 990.5 = 0.42 — OK, with significant reserve.

Step 5 — Shear connector design: Number for full connection: (nf = N{c,f} / P_{Rd,red} = 2,143 / 69.4 = 30.9 \rightarrow) 31 per half-span. With deck ribs at 300 mm centres: 9,000/300 = 30 ribs per span. One stud per rib: 30 studs per half-span (\approx) nf — full shear connection achieved. Total: 60 studs (19 mm (\times) 100 mm) for the full beam.

Step 6 — Deflection: Composite Ic approx 3.5 (\times) Iy = 80,255 cm⁴. Unfactored imposed load: 4.0 (\times) 3.0 = 12.0 kN/m. (\delta = 5 \times 12.0 \times 9,000^4 / (384 \times 210,000 \times 80,255 \times 10^4) = 60.8) mm L/360 = 25.0 mm required for office. Pre-camber of 35 mm recommended.

Design Resources

• EN 1993 Steel Grades
• European Steel Properties
• EN 1993 Beam Design
• EN 1993 Connection Design
• IPE/HEA/HEB Beam Sizes
• All European References

Frequently Asked Questions

What is the difference between full and partial shear connection in composite beams? Full shear connection means enough studs are provided (n (\geq) nf) so that Mpl,Rd is governed by section bending capacity. Partial shear connection (η = n/nf < 1.0) uses fewer studs, reducing moment resistance per M_Rd = Mpl,a,Rd + η (\times) (Mpl,Rd - Mpl,a,Rd). EN 1994-1-1 permits partial interaction for ductile connectors with η (\geq) η_min (0.4-0.52 for S355). Partial interaction is economical when the full composite capacity exceeds the design moment.

How do profiled steel decking ribs affect shear connector design? Decking ribs perpendicular to the beam reduce stud capacity through the reduction factor kt per EN 1994-1-1 Clause 6.6.4. For trapezoidal decking with one stud per rib: kt = 0.85 (maximum). For two studs per rib: kt = 0.70. The stud must project at least 2d (38 mm for 19 mm studs) above the top of the deck to ensure adequate shear cone development in the concrete.

What is the effective width of the concrete flange? The effective width beff = b0 + Le/8 each side (capped by geometric width) accounts for shear lag. For a 9.0 m span secondary beam at 3.0 m centres: beff = 1,576 mm. For a 6.0 m span: beff = 1,050 mm. EN 1994-1-1 Clause 5.4.1.2 limits the effective width so that a uniform stress block assumption is valid for section analysis.

How is the construction stage checked for composite beams? The steel beam alone must resist construction loads (wet concrete, decking, self-weight, construction LL 0.75 kN/m²) per EN 1993-1-1. Unpropped construction is preferred, meaning the steel section must be adequate for M_Ed,c without temporary supports. For long spans (>12 m), propping at midspan can halve the construction moment, allowing a lighter steel section with lower embodied carbon.

Reference only. Verify all values against the current edition of EN 1994-1-1:2004 and the applicable National Annex. Educational reference only.