EN 1993-1-1 Tension Member Design — Clause 6.2.3, Net Section, Angle Worked Example
Quick Reference: This guide covers the complete EN 1993-1-1 tension member design procedure. We walk through gross section yielding (Npl,Rd), net section rupture at bolt holes (Nu,Rd), staggered hole paths per Cl. 6.2.2.2, eccentricity reduction for single-angle members per Cl. 6.2.3(5), and a full worked example using an L80×80×8 angle with three M20 bolts. All partial factors follow EN 1993-1-1:2005 + AC:2009 recommended values.
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. Design Philosophy — Two Failure Modes
Tension member design under EN 1993 is straightforward compared to compression — no buckling instability, no interaction equations. But there are two distinct limit states, and the lower capacity governs:
| Failure Mode | Formula | Partial Factor | Character |
|---|---|---|---|
| Gross section yield | Npl,Rd = A × fy / γM0 | γM0 = 1.00 | Ductile — yielding spreads across full section |
| Net section rupture | Nu,Rd = 0.9 × Anet × fu / γM2 | γM2 = 1.25 | Brittle — sudden fracture at bolt hole |
The 0.9 factor in Nu,Rd reflects the notch sensitivity of steel at net sections. Even ductile structural steels (S235, S275, S355) can exhibit reduced elongation at bolt holes due to triaxial stress states. The higher γM2 (1.25 vs 1.00) accounts for the lower reliability of fracture-governed failure modes.
The design tension resistance: Nt,Rd = min(Npl,Rd, Nu,Rd)
At ambient temperature, gross section yielding typically governs for members with few or no bolt holes. Net section rupture controls when the bolt hole area exceeds roughly 15–20% of the gross area — common in light angles and channels.
2. Gross Section Yielding — Npl,Rd
The simplest check. The entire cross-section reaches the yield stress fy before the member fails by excessive elongation.
Npl,Rd = A × fy / γM0
Where:
- A = gross cross-sectional area (mm²)
- fy = nominal yield strength (N/mm²), from EN 10025-2
- γM0 = 1.00
For S275 steel: fy = 275 N/mm² for t ≤ 40 mm.
Example for L80×80×8 angle (A = 1230 mm²): Npl,Rd = 1230 × 275 / 1.00 = 338.3 kN
No reduction for holes — yielding can redistribute around holes. The gross area is used directly.
3. Net Section Rupture — Nu,Rd
When bolt holes are present, the cross-section is reduced. The failure plane passes through the holes where stress concentration peaks.
Nu,Rd = 0.9 × Anet × fu / γM2
Where:
- Anet = net cross-sectional area (mm²)
- fu = ultimate tensile strength (N/mm²)
- γM2 = 1.25
3.1 Computing Net Area Anet
For a single row of holes aligned perpendicular to the load:
Anet = A − n × d0 × t
Where:
- n = number of holes in the transverse line
- d0 = hole diameter = bolt diameter + 2 mm (standard clearance per EN 1090-2 for M12–M30 bolts)
- t = member thickness (mm)
M20 bolt in 8 mm angle leg: d0 = 20 + 2 = 22 mm Single hole: ΔA = 22 × 8 = 176 mm²
For our L80×80×8 with one transverse hole: Anet = 1230 − 176 = 1054 mm² Nu,Rd = 0.9 × 1054 × 430 / 1.25 = 326.4 kN
For S275 steel: fu = 430 N/mm² (EN 10025-2, t ≤ 40 mm).
Comparing: Npl,Rd = 338.3 kN, Nu,Rd = 326.4 kN → net section rupture governs. Nt,Rd = 326.4 kN.
3.2 Staggered Holes — Zigzag Failure Paths
When bolt holes are staggered, the shortest failure path may not be a straight transverse line but a zigzag line connecting adjacent holes. EN 1993-1-1 Cl. 6.2.2.2(4) provides the formula for the effective net width:
bnet = bgross − Σd0 + Σ(s² / 4p)
Where:
- s = staggered pitch (centre-to-centre spacing parallel to the force)
- p = gauge distance (centre-to-centre spacing perpendicular to the force)
The s²/4p term accounts for the longer inclined path length. Consider two staggered M20 holes (d0 = 22 mm) with s = 60 mm, p = 50 mm:
s²/4p = 60² / (4 × 50) = 3600 / 200 = 18 mm
For a flat bar 120 mm wide, 10 mm thick: Path through both holes: bnet = 120 − 2×22 + 18 = 94 mm Anet = 94 × 10 = 940 mm²
Always check all candidate failure paths — straight across one hole, zigzag through two, zigzag through three — and take the minimum Anet.
4. Single-Angle Tension Members — Eccentricity Reduction
This is where many designers make mistakes. A single angle connected by one leg is loaded eccentrically. The centroid of the bolt group in the connected leg is offset from the angle's centroidal axis by the distance ey (see section property tables for each angle size). This eccentricity ey × N produces a bending moment My that reduces the effective tension capacity.
EN 1993-1-1 Cl. 6.2.3(5) and Cl. 6.2.2.2(8) provide specific reduction factors based on the number of bolts:
4.1 One Bolt
Not treated as a tension member. Design as a simple shear connection — the bolt is checked for shear and bearing. The angle leg is checked for block tearing (EN 1993-1-8 Cl. 3.10.2). No tension resistance calculation applies.
4.2 Two Bolts
Nu,Rd = 2.0 × (e2 − 0.5 d0) × t × fu / γM2
Where e2 is the edge distance from the hole centre to the free edge measured perpendicular to the load. This is a block-tearing-based expression because with only two bolts the failure mode is essentially tearing out of the bolt group rather than net section fracture across the full leg.
For e2 = 30 mm, d0 = 22 mm, t = 8 mm, fu = 430 N/mm²: Nu,Rd = 2.0 × (30 − 0.5×22) × 8 × 430 / 1.25 = 2.0 × 19 × 8 × 430 / 1.25 = 104.6 kN
4.3 Three or More Bolts
Nu,Rd = β3 × Anet × fu / γM2
Where β3 is a reduction factor based on bolt pitch p1 (edge-to-edge spacing parallel to load):
| p1 | β3 |
|---|---|
| ≤ 2.5 d0 | 0.5 |
| 2.5 d0 < p1 ≤ 5.0 d0 | 0.7 |
| ≥ 5.0 d0 | 1.0 |
For our M20 example: d0 = 22 mm → 2.5 d0 = 55 mm, 5.0 d0 = 110 mm. With p1 = 70 mm, β3 = 0.7.
Nu,Rd = 0.7 × 1054 × 430 / 1.25 = 253.9 kN
This is significantly lower than the full Nu,Rd of 326.4 kN — a 22% reduction due to eccentricity alone. For this reason, symmetrical connections (angles connected by both legs, or back-to-back angles) are strongly preferred for primary tension members.
5. Worked Example — L80×80×8 Angle with 3 M20 Bolts
Given
- Angle: L80×80×8, S275 steel (fy = 275 N/mm², fu = 430 N/mm²)
- A = 1230 mm² (from section tables)
- Connected leg: one leg only (eccentric loading applies)
- Bolts: 3 × M20 Grade 8.8, d0 = 22 mm
- Pitch p1 = 70 mm, edge distance e1 = 35 mm, e2 = 30 mm
- Load: design tension NEd = 180 kN
Step 1 — Gross Section Yield
Npl,Rd = 1230 × 275 / 1.00 = 338.3 kN
Utilisation: 180 / 338.3 = 0.53 → OK (53% capacity)
Step 2 — Net Section at Bolt Holes
Anet = 1230 − 1 × 22 × 8 = 1054 mm² Nu,Rd (full) = 0.9 × 1054 × 430 / 1.25 = 326.4 kN
Step 3 — Eccentricity Reduction (Single Angle)
p1 = 70 mm, d0 = 22 mm → 2.5d0 = 55, 5d0 = 110 70 lies between 55 and 110 → β3 = 0.7
Nu,Rd (reduced) = 0.7 × 1054 × 430 / 1.25 = 253.9 kN
Step 4 — Design Tension Resistance
Nt,Rd = min(338.3, 253.9) = 253.9 kN Utilisation: 180 / 253.9 = 0.71 → OK (71% capacity)
Step 5 — Block Tearing (EN 1993-1-8 Cl. 3.10.2)
Block tearing checks the tear-out of a rectangular block of material around the bolt group. For the angle connected leg:
VeRd,1 = (Ant × fu) / γM2 + (1/√3) × (Anv × fy) / γM0
Ant = (e2 − 0.5 d0) × t = (30 − 11) × 8 = 152 mm² Anv = (e1 + 2p1 − 2.5 d0) × t = (35 + 140 − 55) × 8 = 960 mm²
VeRd,1 = (152 × 430) / 1.25 + 0.577 × (960 × 275) / 1.00 = 52.3 + 152.4 = 204.7 kN
Block tearing capacity > NEd = 180 kN → OK.
Summary of Results
| Check | Capacity (kN) | Utilisation | Status |
|---|---|---|---|
| Gross yield Npl,Rd | 338.3 | 0.53 | PASS |
| Net rupture Nu,Rd (β3) | 253.9 | 0.71 | PASS |
| Block tearing | 204.7 | 0.88 | PASS |
The angle is adequate for NEd = 180 kN, with block tearing being the governing limit state at 88% utilisation. For higher loads, consider a larger angle, back-to-back double angles connected by both legs, or increasing the bolt pitch to push β3 to 1.0.
6. Partial Factors and National Annex Variations
| Factor | EN 1993-1-1 Recommended | UK NA | German DIN NA |
|---|---|---|---|
| γM0 (yield) | 1.00 | 1.00 | 1.00 |
| γM1 (buckling) | 1.00 | 1.00 | 1.10 |
| γM2 (rupture) | 1.25 | 1.25 | 1.25 |
Tension design is relatively consistent across National Annexes — only γM2 appears, and all major NAs retain 1.25. However, the eccentricity reduction factors for angles (Cl. 6.2.3(5)) are only given as "recommended" in EN 1993-1-1. Some National Annexes provide alternative methods — the UK NA directs designers to SCI P358 (Green Book) for angle design, which gives slightly more favourable β factors for back marks ≥ 55 mm.
7. Design Tips for Tension Members
- Symmetrical connections eliminate eccentricity. Wherever possible, connect tension members through both legs or specify back-to-back angles with stitch bolts at third points.
- Minimise bolt holes in the critical section. If net rupture governs, stagger bolts to increase the effective net area through the s²/4p term.
- Check block tearing for short connections. When e1 is tight (less than 2.0 d0) or the bolt group is compact, block tearing often governs over net rupture.
- For angles, the connected leg thickness drives β3. Thicker legs increase bolt bearing capacity but do not change β3 — eccentricity reduction depends only on bolt layout, not section size.
- Fatigue-sensitive members need additional checks. Where tension varies cyclically (wind bracing, crane support structures), run a fatigue assessment per EN 1993-1-9 alongside the static check.