EN 1993-1-8 Bolt Design Guide — Shear, Bearing per Table 3.4, γM2=1.25

Quick Reference: This guide covers EN 1993-1-8 bolt design in detail: shear resistance Fv,Rd per Table 3.4, bearing resistance Fb,Rd with αb and k1 factors, tension resistance Ft,Rd, combined shear-tension interaction, and slip-resistant bolt categories (B and C). A complete worked example for a fin plate connection with 4 × M20 Grade 8.8 bolts demonstrates the entire design procedure. Use the free bolted connection calculator for instant EN 1993-1-8 checks.

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.

Bolt Grades and Material Properties

EN 1993-1-8 Table 3.1 classifies structural bolts by grade:

The bolt grade nomenclature: first number × 100 = fub (MPa), product of numbers × 10 = fyb (MPa). Example: Grade 8.8 → 8×100 = 800 MPa (fub), 8×8×10 = 640 MPa (fyb).

Partial factor for bolts: γM2 = 1.25 (EN 1993-1-8 §2.2, recommended value; confirmed in UK NA).

Bolt Shear Resistance Fv,Rd (Table 3.4)

The design shear resistance per bolt per shear plane depends on whether the shear plane passes through the threaded or unthreaded portion:

Shear Plane Through Threaded Portion

Fv,Rd = αv × fub × As / γM2

Shear Plane Through Unthreaded Shank

Fv,Rd = 0.6 × fub × A / γM2

where A = π × d²/4 (gross cross-sectional area). This applies when the shear plane is specifically detailed to pass through the unthreaded shank, which requires careful bolt length specification.

Worked Example — M20 Grade 8.8 Single Shear

Bolt properties (M20, Grade 8.8):

• d = 20 mm (nominal diameter)
• As = 245 mm² (tensile stress area per ISO 898-1)
• fub = 800 MPa
• αv = 0.6
• γM2 = 1.25

Fv,Rd = 0.6 × 800 × 245 / 1.25

Fv,Rd = 480 × 245 / 1.25 = 117,600 / 1.25

Fv,Rd = 94,080 N = 94.1 kN per shear plane

For double shear: 2 × 94.1 = 188.2 kN per bolt

Comparison across bolt sizes (Grade 8.8, single shear):

Bolt Tension Resistance Ft,Rd (Table 3.4)

Ft,Rd = k2 × fub × As / γM2

where k2 = 0.9 for standard bolts (countersunk bolts: k2 = 0.63).

For M20 Grade 8.8:

Ft,Rd = 0.9 × 800 × 245 / 1.25 = 176,400 / 1.25 = 141.1 kN

Bearing Resistance Fb,Rd (Table 3.4)

The bearing resistance of a bolt against the connected plate is:

Fb,Rd = k1 × αb × fu × d × t / γM2

where:

• d = bolt diameter (mm)
• t = plate thickness (mm)
• fu = plate ultimate tensile strength (MPa)
• αb = smallest of {αd, fub/fu, 1.0}
• k1 = smallest of {edge factor, spacing factor, 2.5}

αb Factor (Bearing Direction)

For end bolts: αd = e1 / (3 × d0)

For inner bolts: αd = p1 / (3 × d0) − 0.25

where e1 = end distance, p1 = bolt pitch, d0 = hole diameter (d + 2 mm for standard clearance).

k1 Factor (Perpendicular to Bearing Direction)

For edge bolts: k1 = min(2.8 × e2 / d0 − 1.7, 2.5)

For inner bolts: k1 = min(1.4 × p2 / d0 − 1.7, 2.5)

where e2 = edge distance, p2 = transverse bolt pitch.

Edge Distance and Spacing Requirements

Combined Shear and Tension (Table 3.4)

When bolts are subjected to combined shear and tension:

Fv,Ed/Fv,Rd + Ft,Ed/(1.4 × Ft,Rd) ≤ 1.0

The 1.4 reduction factor on tension capacity reflects the triaxial stress state interaction. In AISC 360, the equivalent check uses a non-linear exponent-based interaction that can be slightly less conservative for low shear ratios.

Full Worked Example — Fin Plate Connection

Design Data

Bolt Group Geometry Check

Minimum end distance: 1.2 × d0 = 1.2 × 22 = 26.4 mm

e1 = 40 mm > 26.4 mm ✓

Minimum edge distance: 1.2 × d0 = 26.4 mm

e2 = 35 mm > 26.4 mm ✓

Minimum pitch: 2.2 × d0 = 48.4 mm

p1 = 70 mm > 48.4 mm ✓

Bolt Shear Check

Design shear per bolt (assume equal distribution — elastic method):

Fv,Ed = VEd / n = 180 / 4 = 45.0 kN per bolt

M20 Grade 8.8 single shear capacity from earlier:

Fv,Rd = 94.1 kN

Fv,Ed/Fv,Rd = 45.0/94.1 = 0.478 → OK ✓

Bolt Bearing on Fin Plate (t=10 mm, S275, fu=430 MPa)

αd (end bolt): e1/(3d0) = 40/(3×22) = 0.606

αd (inner bolt): p1/(3d0) − 0.25 = 70/(3×22) − 0.25 = 1.061 − 0.25 = 0.811

Use αb = min(αd, fub/fu, 1.0) = min(0.606, 800/430, 1.0) = min(0.606, 1.860, 1.0) = 0.606

k1 (edge bolt, single column → treat all as edge bolts):

k1 = min(2.8e2/d0 − 1.7, 2.5) = min(2.8×35/22 − 1.7, 2.5) = min(4.455 − 1.7, 2.5) = min(2.755, 2.5) = 2.5

Fb,Rd = k1 × αb × fu × d × t / γM2 = 2.5 × 0.606 × 430 × 20 × 10 / 1.25

Fb,Rd = 2.5 × 0.606 × 430 × 200 / 1.25 = 130,290 / 1.25 = 104,232 N = 104.2 kN

Fv,Ed/Fb,Rd = 45.0/104.2 = 0.432 → OK ✓

Bolt Bearing on Beam Web (tw=7.1 mm, S275, fu=430 MPa)

Using same αb = 0.606 and k1 = 2.5 (conservative):

Fb,Rd = 2.5 × 0.606 × 430 × 20 × 7.1 / 1.25 = 92,635 / 1.25 = 74.1 kN

Fv,Ed/Fb,Rd = 45.0/74.1 = 0.607 → OK ✓

Block Tearing Check (EN 1993-1-8 §3.10.2)

The bolt group end region is checked for block tearing (a combination of shear yield/tension fracture or tension yield/shear fracture). For a symmetric bolt group with edge distance e2 = 35 mm and pitch p1 = 70 mm:

Ant = (e2 − d0/2) × t_fin = (35 − 11) × 10 = 240 mm² (net area in tension)

Anv = (e1 + p1 − 1.5d0) × t_fin × 2 sides = (40 + 70 − 33) × 10 × 2 = 1540 mm² (net area in shear)

Veff,1,Rd = fu × Ant / γM2 + (fy/√3) × Anv / γM0

Veff,1,Rd = 430 × 240 / 1.25 + (275/1.732) × 1540 / 1.00

Veff,1,Rd = 82,560 + 158.8 × 1540 / 1000 = 82.6 + 244.6 = 327.2 kN

VEd = 180 kN < 327.2 kN → OK ✓

Slip-Resistant Connections (Categories B and C)

For connections where slip must be prevented, the preload force Fp,C per bolt is:

Fp,C = 0.7 × fub × As (Grade 8.8 and 10.9 bolts)

For M20 Grade 8.8: Fp,C = 0.7 × 800 × 245 = 137.2 kN

Slip resistance at SLS (Category B):

Fs,Rd,ser = ks × n × μ × Fp,C / γM3,ser

Slip resistance at ULS (Category C):

Fs,Rd = ks × n × μ × Fp,C / γM3

where ks = 1.0 for standard clearance holes, n = number of friction interfaces (1 for single-lap), μ = 0.30 for blast-cleaned surfaces (Class A), and γM3 = 1.25 (ULS) or γM3,ser = 1.10 (SLS).

For M20 Grade 8.8, single interface, blast-cleaned:

Fs,Rd = 1.0 × 1 × 0.30 × 137.2 / 1.25 = 32.9 kN

This is approximately 35% of the bearing shear capacity — demonstrating that slip-critical design can govern connection sizing.

Key Takeaways

1. Bearing resistance often governs over shear for thin connected plies. The αb and k1 factors can reduce bearing capacity below Fv,Rd for small edge distances or close bolt spacing.
2. γM2 = 1.25 is applied to all bolt limit states — shear, bearing, and tension. This is 25% more conservative than the cross-section partial factor γM0 = 1.00.
3. Tensile stress area As, not gross area A, controls shear when the shear plane passes through the threads. For M20, As/A = 245/314 = 0.78 — threads reduce the effective shear area by 22%.
4. Combined shear+tension interaction uses 1.4 × Ft,Rd in the denominator, which means moderate tension demand (up to ~30% of Ft,Rd) has minimal impact on shear capacity.

Related Pages

• EN 1993-1-8 Weld Design Guide — Directional Method, βw
• EN 1993-1-8 Base Plate Design — T-Stub, Concrete Bearing
• EN 1993-1-1 Beam Design — IPE 300 Worked Example
• EN 1993-1-1 Column Design — HEA 200 Example, Curves a₀-d
• Bolt Capacity Tables — AISC 360-22 & EN 1993 Comparison