EN 1993-1-8 Base Plate Design — Column Base, T-Stub, EN 1992 Concrete Bearing

Quick Reference: This comprehensive guide covers EN 1993-1-8 base plate design for steel column bases. It combines the column base provisions of Clause 6.2.8 with the T-stub model (Cl. 6.2.4) for anchor bolt tension and the concrete bearing check per EN 1992-1-1 Section 6.7. A full worked example for a HEB 240 column base with axial compression, shear, and nominal moment demonstrates the complete design procedure. Use the free base plate calculator for instant EN 1993 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.

Scope of EN 1993-1-8 for Column Bases

EN 1993-1-8 does not provide a single self-contained clause for base plate design. Instead, the designer combines three clauses:

1. Clause 6.2.8 — Column Bases. Provisions for nominally pinned and fixed bases. Covers the distribution of compression into the foundation through the base plate.
2. Clause 6.2.4 — Equivalent T-Stub in Tension. The T-stub model used for the tension side of moment-resisting bases, anchor bolt pull-out, and flange plate yielding with prying action.
3. EN 1992-1-1 Section 6.7 — Partially Loaded Areas. Concrete bearing strength under concentrated loads, including the enhancement factor √(Ac1/Ac0) for confined bearings.

Additionally, anchor bolt design references EN 1992-4 (fasteners in concrete) for post-installed anchors.

Base Plate Configurations

Design Parameters and Partial Factors

Full Worked Example — HEB 240 Column Base

Design Data

SECTION PROPERTIES — HEB 240

Step 1: Concrete Bearing Strength

fjd = βj × αcc × fck / γc

Using UK NA values: αcc = 0.85, γc = 1.50

fjd = (2/3) × 0.85 × 30 / 1.50 = 0.667 × 25.5 / 1.50

fjd = 17.0 / 1.50 = 11.33 MPa

If the foundation dimensions allow the full Ac1/Ac0 enhancement:

Ac0 = bp × lp = 340 × 400 = 136,000 mm²

For a foundation of 800 × 800 mm: Ac1 = 800 × 800 = 640,000 mm²

kj = √(Ac1/Ac0) = √(640,000/136,000) = √(4.706) = 2.17 ≤ 3.0 ✓

Enhanced bearing strength: fjd,enh = fjd × kj / 3.0

fjd,enh = 11.33 × 2.17 / 3.0? Wait — EN 1992-1-1 §6.7 uses FRdu = Ac0 × fcd × √(Ac1/Ac0) ≤ 3 × fcd × Ac0.

Simplified: fjd ≤ 3 × fcd, so fjd = min(βj × αcc × fck/γc × √(Ac1/Ac0), 3 × βj × αcc × fck/γc)

Direct approach: fjd = 11.33 × min(√(640/136), 3.0) = 11.33 × min(2.17, 3.0)

But the enhancement is already capped: the design bearing resistance FRdu = Ac0 × fcd × √(Ac1/Ac0) ≤ 3 × Ac0 × fcd.

For this worked example, conservatively use fjd = 11.33 MPa without enhancement.

Step 2: Effective Bearing Area (Compression Base, Cl. 6.2.8.2)

The additional bearing width c:

c = tp × √(fy / (3 × fjd × γM0))

c = 25 × √(275 / (3 × 11.33 × 1.00)) = 25 × √(275 / 33.99) = 25 × √(8.09)

c = 25 × 2.845 = 71.1 mm

Effective bearing area: Assuming the column footprint fully bears — this is conservative for compression-only bases. The effective area Ac0,eff:

beff = bf + 2c = 240 + 2 × 71.1 = 382.2 mm → capped at bp = 340 mm

heff = h + 2c = 240 + 2 × 71.1 = 382.2 mm → capped at lp = 400 mm

Effective bearing width = min(340, 382.2) = 340 mm Effective bearing length = min(400, 382.2) = 382 mm

Aeff = 340 × 382 = 129,880 mm²

Note: The c value exceeded the plate overhang (340 − 240)/2 = 50 mm. When c > overhang, the effective bearing area is limited by the plate geometry. This means the 25 mm plate is thick enough that bearing stress can distribute fully across the available plate width.

Step 3: Compression Base — Bearing Check

Nc,Rd = fjd × Aeff = 11.33 × 129,880 / 1000 = 1471 kN

NEd / Nc,Rd = 850 / 1471 = 0.578 → OK ✓

This is the simplest case — a compression-only base. The plate is adequate without anchor bolt tension contribution.

Step 4: Anchor Bolt Design for Tension (Moment Side)

Given MEd = 30 kN·m about the minor axis, the moment induces a tension-compression couple across the base plate. The lever arm z between bolt lines is taken as the center-to-center distance:

z = lp − 2e = 400 − 2 × 60 = 280 mm (center-to-center of bolt rows)

Tension in the bolt group from the moment:

Ft,Ed = MEd / z = 30 / 0.280 = 107.1 kN

Distributed over 2 bolts per side: Ft,Ed,per bolt = 107.1 / 2 = 53.6 kN

Bolt Tension Resistance (per bolt, M24 Grade 8.8)

As for M24 = 353 mm², fub = 800 MPa, k2 = 0.9

Ft,Rd = k2 × fub × As / γM2 = 0.9 × 800 × 353 / 1.25 = 254,160 / 1.25 = 203.3 kN

Ft,Ed,per bolt / Ft,Rd = 53.6 / 203.3 = 0.264 → OK ✓

Step 5: T-Stub Check for Plate Bending (Mode 1 — Plate Yielding)

The relevant T-stub parameters for the bolt row:

m = (lp − h − tw) / 4 = (400 − 240 − 10) / 4 = 150/4 = 37.5 mm (approximate; depends on bolt-to-web distance)

Effective length for an individual bolt row (circular pattern):

leff,circ = 2π × m = 2π × 37.5 = 235.6 mm per bolt

For two bolts: leff,1 = 2 × 235.6 = 471.2 mm

Plastic moment resistance of the T-stub flange (i.e., the base plate):

Mpl,1,Rd = 0.25 × leff,1 × tp² × fy / γM0 = 0.25 × 471.2 × 25² × 275 / 10⁶

Mpl,1,Rd = 0.25 × 471.2 × 625 × 275 / 10⁶ = 20,240,000 / 10⁶ = 20.24 kN·m — wait, this is the moment capacity over the effective width leff.

Mode 1 resistance: FT,1,Rd = 4 × Mpl,1,Rd / m

FT,1,Rd = 4 × 20.24 / 0.0375 = 2159 kN — this is very high because the 25 mm plate and small m value produce a very stiff T-stub.

When FT,1,Rd exceeds the bolt capacity (Mode 3), the bolts govern — which is the intended behavior for a compact base plate.

Mode 3 resistance: FT,3,Rd = ΣFt,Rd = 4 × 203.3 = 813.2 kN

FT,Rd = min(FT,1,Rd, FT,2,Rd, FT,3,Rd) = min(2159, ... , 813.2) = 813.2 kN

Since FT,Rd >> Ft,Ed = 107.1 kN, the base plate design is adequate for the tension side.

Step 6: Shear Resistance

Friction capacity: The compression force transfers through bearing. The friction force available:

Vf,Rd = Cf,d × Nc,Ed = 0.20 × 850 = 170 kN

VEd = 65 kN < 170 kN → OK ✓ (shear carried by friction)

If friction alone is insufficient, anchor bolt shear capacity is checked per EN 1993-1-8 Table 3.4 (Category A):

Fv,Rd per M24 bolt = αv × fub × As / γM2 = 0.6 × 800 × 353 / 1.25 = 135.6 kN

With 4 bolts: ΣFv,Rd = 4 × 135.6 = 542.4 kN — far exceeding the demand.

However, when anchor bolt holes are oversized (standard for base plates), it is better practice to provide a shear lug for VEd > 0.15 NEd, as the bolts may not engage immediately.

Step 7: Shear-Lug Check (Optional)

If a shear lug is provided (e.g., a 100 × 20 mm flat bar welded to the underside of the base plate, embedded 60 mm into the grout pocket):

Bearing of the lug on the grout/concrete:

fjd,lug = βj × αcc × fck / γc = 0.667 × 0.85 × 30 / 1.50 = 11.33 MPa

Lug bearing area: Alug = 100 × 60 = 6,000 mm² (projected bearing area)

Vlug,Rd = fjd,lug × Alug = 11.33 × 6,000 / 1000 = 68.0 kN

Together with friction: Vtotal,Rd = 170 + 68 = 238 kN > 65 kN ✓

Step 8: Base Plate Weld to Column

The column is welded to the base plate with 8 mm continuous fillet welds around the HEB 240 profile. For a column in axial compression with moderate shear:

Weld perimeter (approximate for HEB 240): lw = 2 × (240 + 240) − 4 × 10 = 920 mm (subtract web thickness overlap)

Weld force per unit length from axial load (assume load transferred through the flange and web welds proportionally):

fv,Ed = NEd / lw = 850,000 / 920 = 924 N/mm

Fw,Rd per mm (8 mm leg, a = 5.66 mm, S275, simplified method):

Fw,Rd = 223.4 × 5.66 = 1264 N/mm

fv,Ed / Fw,Rd = 924 / 1264 = 0.731 → OK ✓

Summary of Results (HEB 240 Base Plate)

Key Takeaways

1. The additional bearing width c determines the effective area. A thick plate (large tp) increases c, allowing load spread across a larger concrete contact area. For heavily loaded bases, increasing plate thickness from 20 to 30 mm can increase effective area by 50%.
2. T-stub Mode 1 (plate yielding) governs thin base plates. For a 25 mm plate with 37.5 mm bolt-to-web distance, the T-stub is very stiff and the bolts govern (Mode 3). For a 15 mm plate, Mode 1 would likely govern, reducing capacity by 40-60%.
3. Friction should be the primary shear transfer mechanism. Anchor bolts in oversized holes provide no immediate shear engagement. A shear lug is recommended for VEd/NEd > 0.15.
4. National Annex variations in αcc can change the bearing capacity by 18% (αcc=0.85 vs 1.00). Always verify the applicable NA.

Related Pages

• EN 1993-1-8 Bolt Design Guide — Fv,Rd, Fb,Rd, Table 3.4
• EN 1993-1-8 Weld Design Guide — Directional Method, βw
• EN 1993-1-1 Column Design — HEA 200 Example
• EN 1993-1-1 Beam Design — IPE 300 Example, χLT
• Column Base Design — AISC 360-22 DG1 Comparison