Worked Examples To Eurocode 2 Volume 2 May 2026

REPORT

Step 5: Torsion link reinforcement

[ \fracA_sws = \fracT_Ed2 A_k f_ywd \cot\theta = \frac45e62 \times 105,306 \times (500/1.15) \times 2.5 ] ( f_ywd = 435 \text MPa ) [ \fracA_sws = \frac45e62 \times 105,306 \times 435 \times 2.5 = \frac45e6228.9e6 \approx 0.197 \text mm^2/\textmm ] For 2 legs (torsion acts on all walls) → per leg: ( A_sw/s = 0.098 ) Use ( \varnothing 10 ) (78.5 mm²) → spacing ( s = 78.5 / 0.098 \approx 800 \text mm ) – too large; torsion rarely governs spacing; shear likely demands tighter. worked examples to eurocode 2 volume 2

Volume 2 often includes a failed example—showing what happens if you ignore the maximum shear stress limit (v_Rd,max). REPORT Step 5: Torsion link reinforcement [ \fracA_sws

• Design Situation: A 250mm flat slab supported on 400x400mm columns.
• Step 1: Determine control perimeters (at 2d from column face).
• Step 2: Calculate the applied shear stress (v_Ed).
• Step 3: Verify shear resistance without reinforcement (v_Rd,c).
• Step 4: If failed, design punching shear reinforcement (shear rails or link legs).
• Step 5: Check the perimeter beyond which reinforcement is no longer required.

Appendices

• Appendix A: Full material property tables for C20/25 to C90/105.
• Appendix B: Printed spreadsheets (QR codes) for moment redistribution & crack control.
• Appendix C: Comparison of National Annexes (UK, Ireland, Germany, Sweden).