Base Plate in Compression — EN 1993-1-8 Clause 6.2.6.5

The compression resistance of a base plate connection:

N_j,Rd = f_jd × b_eff × l_eff

Where:

• f_jd = design bearing strength of concrete under the base plate = 2/3 × f_cd × sqrt(A_c1 / A_c0) ≤ 3.0 × f_cd
• f_cd = design concrete compressive strength = alpha_cc × f_ck / gamma_C
• A_c0 = area of base plate (b_p × l_p)
• A_c1 = maximum spread area of concrete foundation (geometrically similar to A_c0)
• b_eff, l_eff = effective width and length of T-stub compression zone

For C30/37 concrete with UK NA: alpha_cc = 0.85, gamma_C = 1.5. f_cd = 0.85 × 30 / 1.5 = 17.0 MPa.

For square foundation (800 × 800 mm) with 350 × 350 base plate: A_c0 = 122,500 mm², A_c1 = 640,000 mm². sqrt(A_c1/A_c0) = sqrt(5.22) = 2.29 ≤ 3.0. f_jd = 2/3 × 17.0 × 2.29 = 25.9 MPa.

The 2/3 factor in f_jd accounts for the flexible base plate not distributing stress uniformly — it is a joint coefficient, not a material factor. The UK National Annex does not modify this value.

Base Plate Bending — EN 1993-1-8 Clause 6.2.6.9

The base plate bending resistance is checked using the T-stub model. The effective length of the T-stub flange depends on the bolt layout pattern. For a base plate with bolts outside the column flanges:

l_eff,1 = 4m_x + 1.25e_x (circular yielding) or l_eff,2 = 2πm_x (non-circular yielding)

Where m_x is the distance from bolt centre to the column flange/web face (minus 0.8 × a√2 for weld leg).

For M24 bolts at 60 mm from HEB 240 flange face, flange thickness weld a = 6 mm: m = 60 - 0.8 × 6 × √2 = 60 - 6.8 = 53.2 mm. e = 50 mm (edge distance). l_eff,1 = 4 × 53.2 + 1.25 × 50 = 212.8 + 62.5 = 275.3 mm.

Base plate plastic moment resistance: m_pl,Rd = t_p² × f_yp / (4 × gamma_M0). For 25 mm plate S275: m_pl,Rd = 25² × 275 / (4 × 1.00) = 42,969 N·mm/mm = 43.0 kN·m/m.

T-stub tension resistance (Mode 1 — complete yielding): F_T,1,Rd = 4 × M_pl,1,Rd / m = 4 × 43.0 × 275.3 / 53.2 / 1000 = 890 kN per bolt row.

Anchor Bolt Tension — EN 1993-1-8 Table 3.4

Bolt tension resistance: F_t,Rd = 0.9 × f_ub × A_s / gamma_M2

For M24 Grade 8.8 (f_ub = 800 MPa, A_s = 353 mm², gamma_M2 = 1.25): F_t,Rd = 0.9 × 800 × 353 / 1.25 = 203.3 kN.

Check: bolt tension governs over plate bending (203 kN < 890 kN). The connection is bolt-limited.

Grout Joint Requirements — EN 1090-2

Per EN 1090-2 (Execution of Steel Structures), the grout joint under base plates must satisfy:

• Grout thickness: 30-50 mm for cementitious grout. The minimum 30 mm ensures adequate flow under the plate; the maximum 50 mm prevents excessive grout shrinkage.
• Grout strength class: Minimum C30/37, matching or exceeding the foundation concrete (typically C25/30 or C30/37).
• Execution class: EXC2 for buildings, EXC3 for bridges and structures with high consequences of failure.
• Surface preparation: The foundation surface must be scabbled to expose coarse aggregate (amplitude 3-5 mm), cleaned of laitance and loose material, and pre-wetted for 24 hours before grouting.
• Grout holes: Provide 50 mm diameter vent holes for base plates larger than 600 × 600 mm. Locate holes symmetrically to ensure complete grout coverage.

Worked Example 1 — HEB 240 Column Base Plate (Fixed Base)

Problem: HEB 240 column (S355) with N_Ed = 1200 kN compression and M_Ed = 55 kN·m bending. Design base plate and anchor bolts.

Step 1 — Base Plate Dimensions:

Base plate 350 × 350 × 25 mm, S275. Four M24 Grade 8.8 bolts at 240 mm gauge, 60 mm from column face, 50 mm edge distance.

Step 2 — Concrete Bearing Check:

N_Ed / (b_p × l_p × f_jd) = 1200 × 10³ / (350 × 350 × 25.9) = 1200 / 3173 = 0.378. OK (38% utilisation).

With moment: assume triangular stress block. Effective bearing width depends on neutral axis depth. For e = M/N = 55/1200 = 46 mm, the base is almost entirely in compression. No tension in bolts for this load combination.

Step 3 — Check Uplift (Wind):

N_Ed = 80 kN (compression after uplift), M_Ed = 65 kN·m. e = 65/80 = 813 mm — base is in partial compression.

Tension force from bolts: T_Ed = (M_Ed - N_Ed × a_c) / z.

Where a_c = (350/2) - (x/3) and z ≈ 280 mm (bolt centre to compression centroid).

Approximate: T_Ed = (65 - 80 × 0.14) / 0.28 = (65 - 11.2) / 0.28 = 192 kN — shared between two bolts. F_t per bolt = 96 kN < 203 kN. OK.

Step 4 — Shear Transfer:

Horizontal shear V_Ed = 45 kN. Transfer via anchor bolts (bearing on base plate) or shear key. Without shear key: bolt shear F_v,Rd per bolt = alpha_v × f_ub × A / gamma_M2 = 0.6 × 800 × 353 / 1.25 = 135.6 kN.

Total shear resistance (4 bolts) = 543 kN > 45 kN. OK.

Selected: Base plate 350×350×25 mm S275, 4×M24 Grade 8.8, 300 mm embedment, grout pad 30 mm min.

Worked Example 2 — IPE 300 Column Base (Pinned Base with Shear Key)

Problem: IPE 300 column (S355) in a braced frame, N_Ed = 350 kN compression, V_Ed = 160 kN shear (wind bracing), M_Ed = 0 (nominally pinned). C25/30 foundation. Design a pinned base with shear key.

Step 1 — Base Plate Sizing:

Pinned base — 2-M20 Grade 5.6 bolts inside the flanges. Base plate 250 × 220 × 15 mm S235. Foundation 600 × 600 mm.

Step 2 — Concrete Bearing:

f_cd = 0.85 × 25 / 1.5 = 14.2 MPa. sqrt(A_c1/A_c0) = sqrt(360,000/55,000) = sqrt(6.55) = 2.56, capped at 3.0. f_jd = 2/3 × 14.2 × 2.56 = 24.2 MPa. N_j,Rd = 24.2 × 250 × 220 / 1000 = 1331 kN > 350 kN. OK.

Step 3 — Shear Key Design:

Friction capacity (mu = 0.20 for painted base plate on grout per EN 1993-1-8 Clause 6.2.2): V_friction = 0.20 × 350 = 70 kN < 160 kN — shear key required.

Shear key: 80 × 15 mm flat S355, 150 mm long, welded to plate underside. Concrete bearing: V_key,Rd = f_cd × b_key × d_key = 14.2 × 80 × 80 / 1000 = 90.9 kN per key face. Total = 181.8 kN > 160 kN. OK.

Key weld: 6 mm FW both sides, a = 4.2 mm, length = 2 × 150 = 300 mm. F_w,Rd = f_u / (√3 × β_w × γ_M2) = 360 / (1.732 × 0.80 × 1.25) = 207.8 MPa (S235 plate). V_weld,Rd = 4.2 × 300 × 207.8 / 1000 = 262 kN >> 160 kN. OK.

Step 4 — Anchor Bolts:

Bolt tension is zero for pinned base. Verify bolt shear is not required (shear key carries all shear). 2-M20 Grade 5.6 bolts for location only.

Selected: 250×220×15 mm base plate S235, 2-M20 Grade 5.6 bolts, 80×15×150 mm shear key S355, 30 mm grout.

Frequently Asked Questions

How is the design bearing strength f_jd calculated per EN 1993-1-8? f_jd = 2/3 × f_cd × sqrt(A_c1/A_c0) ≤ 3.0 × f_cd, where f_cd = alpha_cc × f_ck / gamma_C. The 2/3 factor is a joint coefficient accounting for the flexible base plate distributing bearing stress non-uniformly, as opposed to the rigid bearing assumed in EN 1992-1-1. For standard C30/37 concrete with a UK foundation, typical f_jd ranges from 15-30 MPa depending on the foundation-to-plate area ratio.

When is a pinned base vs fixed base classification used in Eurocode design? A column base is classified as pinned if its rotational stiffness S_j,ini ≤ 0.5 × E × I_c / L_c (braced frames) per EN 1993-1-8 Clause 5.2.2. For typical building columns, a base plate thickness of 15-20 mm with 2 bolts inside the flanges produces a nominally pinned base. A base plate thickness of 25-40 mm with 4 bolts outside the flanges produces a fixed base. The classification affects the frame's buckling length (K-factor) and the need for a second-order global analysis.

What is the difference between EN 1993-1-8 and EN 1992-1-1 concrete bearing formulas? EN 1993-1-8 Clause 6.2.6.5 uses a reduced bearing strength (2/3 factor) compared to EN 1992-1-1 Clause 6.7. This is because the base plate is flexible and does not distribute bearing stress uniformly — the 2/3 factor converts the rigid bearing assumption of EN 1992 to the flexible joint behaviour of EN 1993. For very thick base plates (t_p > 50 mm), the 2/3 factor may be unconservative, and a finite element bearing stress analysis is recommended.

How deep should anchor bolts be embedded in European practice? Anchor bolt embedment is governed by concrete breakout per EN 1992-1-1 Section 8 and EN 1992-4 (fastenings). For headed anchors in C30/37 concrete, minimum embedment h_ef ≥ 8d (bolt diameter). Typical UK values are 250-400 mm for M20-M30 bolts. Post-installed anchors must have European Technical Assessment (ETA) approval. The embedment check is separate from the steel bolt check — both must be satisfied independently.

Related Pages

• EN 1993 Connection Design — Eurocode Overview
• EN 1993 Column Design — Worked Example
• EN 1993 Moment Connection — Bolted End Plate
• EN 1993 Beam Design — Flexural Members
• EN 1993 Bolt Capacity — Bolt Tables
• EN 1993 Bolt Grades — 4.6, 5.6, 8.8, 10.9
• European Steel Grades — S235, S275, S355
• Base Plate & Anchor Calculator — Free Tool
• Column Capacity Calculator — EN 1993-1-1

Educational reference only. Base plate design per EN 1993-1-8:2005 and EN 1992-1-1. Verify against current Eurocodes and UK National Annex values. Results are PRELIMINARY — NOT FOR CONSTRUCTION without independent Chartered Engineer verification.