EN 1993-1-8 Bolted Connection — Worked Example

This worked example demonstrates the complete verification of a bolted tension splice in structural steelwork, following EN 1993-1-8:2005 with UK National Annex parameters. Educational use only — always verify by a qualified engineer.

Problem Statement

Connection: Tension splice between two S355JR plates, Category A (bearing type) per Table 3.2 Fasteners: 6 × M20 Grade 8.8 bolts in 22 mm clearance holes, double row arrangement Design action: N_Ed = 500 kN (factored ULS tension)

Geometry

         e1=40  p1=70  p1=70  e1=40
         |------|------|------|
    e2=40  o     o     o     o
          |                      |  300 mm
    e2=40  o     o     o     o
         |------|------|------|
           80    70    70    80

Section Properties

Partial Factors (UK NA to EN 1993-1-8)

Step 1 — Bolt Shear Resistance per Shear Plane (Table 3.4)

For M20 Grade 8.8 bolts in shear (threads included in the shear plane, which is the usual case for structural connections):

Fv,Rd = αv × fub × As / γ_M2

For Grade 8.8: αv = 0.6 (Table 3.4, shear through threaded portion)

Fv,Rd = 0.6 × 800 × 245 / 1.25 = 94,080 N = 94.1 kN

The connection has 6 bolts in single shear:

Total V_Rd = 6 × 94.1 = 564.6 kN

Utilization: N_Ed / total V_Rd = 500 / 564.6 = 0.89 ✓ (89%)

Bolt shear is the governing check for this connection, utilising 89% of capacity.

Step 2 — Bolt Bearing Resistance (Table 3.4)

Bearing resistance of the plate per bolt:

Fb,Rd = k1 × αb × fu × d × t / γ_M2

Edge Bolts (end bolts in the direction of load transfer)

k1 = min(2.8 × e2 / d0 - 1.7, 2.5)
   = min(2.8 × 40 / 22 - 1.7, 2.5)
   = min(3.39, 2.5) = 2.5

αb = min(e1 / (3 × d0), fub / fu, 1.0)
   = min(40 / (3 × 22), 800 / 470, 1.0)
   = min(0.606, 1.70, 1.0) = 0.606

Fb,Rd,edge = 2.5 × 0.606 × 470 × 20 × 12 / 1.25
           = 136,634 N = 136.6 kN

Inner Bolts

k1 = min(1.4 × p2 / d0 - 1.7, 2.5)
   = min(1.4 × 80 / 22 - 1.7, 2.5)
   = min(3.39, 2.5) = 2.5

αb = min(p1 / (3 × d0) - 1/4, fub / fu, 1.0)
   = min(70 / (3 × 22) - 0.25, 1.70, 1.0)
   = min(0.811, 1.70, 1.0) = 0.811

Fb,Rd,inner = 2.5 × 0.811 × 470 × 20 × 12 / 1.25
            = 182,885 N = 182.9 kN

Total Bearing Capacity

ΣFb,Rd = 2 × 136.6 + 4 × 182.9 = 273.2 + 731.6 = 1,004.8 kN

Utilization: 500 / 1,004.8 = 0.50 ✓ (50%)

Bearing is not critical. If edge and end distances were reduced to minimum values (1.2d0 = 26.4 mm), bearing would approach bolt shear capacity.

Step 3 — Net Cross-Section Tension Resistance (Cl. 6.2.3(2))

The net area of the plate at the bolt holes:

Anet = (b - n × d0) × t
     = (300 - 2 × 22) × 12
     = 256 × 12 = 3,072 mm²

Note: For staggered holes, the net section would account for alternate load paths per Cl. 6.2.3(3), but with a simple double-row arrangement, the straight line across all holes governs.

Nu,Rd = Anet × fu / γ_M2
      = 3,072 × 470 / 1.25
      = 1,155 × 10³ N
      = 1,155 kN

Utilization: 500 / 1,155 = 0.43 ✓ (43%)

Step 4 — Gross Section Yielding (Cl. 6.2.3(1))

The gross cross-section must also be checked for yielding:

Npl,Rd = A × fy / γ_M0
       = (300 × 12) × 355 / 1.00
       = 3,600 × 355
       = 1,278 × 10³ N
       = 1,278 kN

Utilization: 500 / 1,278 = 0.39 ✓ (39%)

Step 5 — Block Tearing (Cl. 3.10.2)

Block tearing considers a segment of the plate tearing out around a group of bolts.

For the symmetric bolt group under concentric tension:

Ant = (e2 - d0 / 2) × t = (40 - 11) × 12 = 348 mm²  (net tension area per side)
Anv = (e1 + (n1 - 1) × p1 - (n1 - 0.5) × d0) × t × 2 sides
    = (40 + 2 × 70 - 2.5 × 22) × 12 × 2
    = (40 + 140 - 55) × 24
    = 125 × 24 = 3,000 mm²

Veff,1,Rd = (fu × Ant) / γ_M2 + (fy × Anv) / (√3 × γ_M0)
          = (470 × 348) / 1.25 + (355 × 3,000) / (1.732 × 1.00)
          = 130,848 + 614,780
          = 745,628 N
          = 745.6 kN

Utilization: 500 / 745.6 = 0.67 ✓ (67%)

Step 6 — Summary

Conclusion: 6 × M20 Grade 8.8 bolts in a 300×12 mm S355JR plate are adequate for a factored tension load of 500 kN, governed by bolt shear at 89% utilisation. The bolt shear check is typically the governing failure mode for bearing-type connections when edge distances and spacing meet the standard minimums. Consider M24 bolts or Grade 10.9 if higher capacity is required.

This worked example is for educational purposes. All designs must be verified by a qualified engineer. Use the EN 1993 bolted connection calculator to check other bolt sizes or configurations.

See Also

• Bolted Connections Calculator
• Welded Connections Calculator
• Bolt Torque Calculator
• Bolt Hole Sizes Reference
• Connection Design Workflow