UK Column Design Example -- Complete EN 1993-1-1 Design Walkthrough for a Multi-Storey Column

This worked example demonstrates the complete design procedure for a steel column in a UK multi-storey braced frame building, from initial load estimation through to final section verification. The design follows BS EN 1993-1-1:2005 with the UK National Annex and references the SCI Blue Book for section properties. The example covers load determination per EN 1991, section classification per Clause 5.5, flexural buckling per Clause 6.3.1, and the final check against the design axial force.

Project Description

A six-storey commercial office building in Manchester, UK. The structural arrangement is a steel braced frame with composite floors (steel beams with concrete slab on profiled decking). Columns are arranged on a 7.5 m x 7.5 m grid. The column under consideration is an internal column accumulating load from six floors plus the roof. Lateral stability is provided by a concrete core at each end of the building, so the steel columns are classified as braced (non-sway).

Step 1 -- Load Determination per EN 1991

Floor Construction

• Composite slab: 130 mm lightweight concrete on ComFlor 60 decking
• Services and ceiling: 0.50 kN/m^2
• Raised floor: 0.30 kN/m^2
• Steel beams: 0.30 kN/m^2 (allowance)
• Total permanent (dead) load per floor: G_k = 3.50 kN/m^2

Imposed Loads

• Office areas (Category B1 per EN 1991-1-1 Table 6.2): Q_k = 3.0 kN/m^2 (UK NA value, increased from the recommended 2.5 kN/m^2 for Category B)

Roof Loads

• Roof construction: 1.20 kN/m^2 (decking, insulation, waterproofing)
• Services: 0.30 kN/m^2
• Imposed roof load (Category H, accessible for maintenance): Q_k,roof = 0.75 kN/m^2

Column Tributary Area

Tributary area per floor = 7.5 m x 7.5 m = 56.25 m^2.

Column Axial Load at Ground Floor

Permanent: G_k_total = 5 floors x 3.50 kN/m^2 x 56.25 m^2 + 1 roof x 1.50 kN/m^2 x 56.25 m^2 = 5 x 196.9 + 1 x 84.4 = 984.5 + 84.4 = 1,069 kN

Imposed: Q_k_total = 5 floors x 3.0 kN/m^2 x 56.25 m^2 + 1 roof x 0.75 kN/m^2 x 56.25 m^2 = 5 x 168.75 + 1 x 42.2 = 843.8 + 42.2 = 886 kN

The imposed load on five floors may be reduced by the area reduction factor alpha_A per EN 1991-1-1 Clause 6.3.1.2: alpha_A = 0.5 + A_0 / A with A_0 = 50 m^2 (UK NA value, confirmed as 50 for office floors) alpha_A = 0.5 + 50 / (6 x 56.25) = 0.5 + 0.148 = 0.648

Reduced imposed load: Q_k_reduced = 0.648 x 886 = 574 kN.

ULS Design Axial Force per EN 1990

Equation 6.10 (UK NA -- use Expression 6.10): N_Ed = 1.35 x G_k_total + 1.5 x Q_k_reduced = 1.35 x 1,069 + 1.5 x 574 = 1,443 + 861 = 2,304 kN

The column at ground floor level must resist 2,304 kN in axial compression.

Step 2 -- Initial Section Selection

The column effective length for buckling is the storey height. The ground floor storey height is 4.5 m. In a braced frame, with nominally pinned connections at both ends, the effective length factor k = 1.0.

Lcr = 1.0 x 4.5 = 4.5 m.

For a first estimate, assume lambda_bar_z approximately 0.8, giving chi_z approximately 0.70 for curve b (alpha = 0.34).

Required area: A_req = N_Ed x gamma_M1 / (chi_z x fy) = 2,304 x 10^3 x 1.0 / (0.70 x 355) = 2,304,000 / 248.5 = 9,270 mm^2 = 92.7 cm^2

Try 305 x 305 x 97 UC in S355.

Step 3 -- Section Properties (from SCI Blue Book)

305 x 305 x 97 UC, S355J0:

• h = 307.9 mm, b = 305.3 mm
• tw = 9.9 mm, tf = 15.4 mm
• A = 123 cm^2 = 12,300 mm^2
• i_y = 13.4 cm = 134 mm
• i_z = 7.74 cm = 77.4 mm
• W_pl,y = 1,460 cm^3
• tf = 15.4 mm <= 40 mm

Step 4 -- Section Classification per Clause 5.5

Flange (outstand in compression): c = (b - tw)/2 - r = (305.3 - 9.9)/2 - 15.2 = 147.7 - 15.2 = 132.5 mm c/tf = 132.5 / 15.4 = 8.60 Class 1 limit: 9 x epsilon = 9 x 0.814 = 7.33 8.60 > 7.33, but Class 2 limit: 10 x epsilon = 8.14 8.60 > 8.14, Class 3 limit: 14 x epsilon = 11.39 8.60 <= 11.39

Flange classification: Class 3 for S355. The flange is not compact enough for plastic design. Moment resistance uses the elastic section modulus.

Web (internal, pure compression): cw = h - 2tf - 2r = 307.9 - 2 x 15.4 - 2 x 15.2 = 307.9 - 30.8 - 30.4 = 246.7 mm cw/tw = 246.7 / 9.9 = 24.9 Class 1 limit (pure compression, alpha = 1.0): 33 x epsilon = 33 x 0.814 = 26.9 24.9 <= 26.9

Web classification: Class 1.

Overall section: Class 3 (governed by flange). For axial compression, Class 1-3 all use A x fy, so the classification does not affect the column axial resistance check in this case. However, for combined axial and bending (if present), the elastic section modulus Wel would govern.

Step 5 -- Flexural Buckling Check

Non-dimensional slenderness: lambda_1 = 93.9 x sqrt(235/355) = 76.4

lambda_bar_y = (4,500 / 134) / 76.4 = 33.6 / 76.4 = 0.440 lambda_bar_z = (4,500 / 77.4) / 76.4 = 58.1 / 76.4 = 0.761

Buckling curve selection: y-y axis: curve a (alpha = 0.21) z-z axis: curve b (alpha = 0.34)

Reduction factors: y-y: Phi_y = 0.5 x [1 + 0.21 x (0.440 - 0.2) + 0.440^2] = 0.5 x [1 + 0.050 + 0.194] = 0.622 chi_y = 1 / [0.622 + sqrt(0.622^2 - 0.440^2)] = 1 / [0.622 + 0.439] = 0.942

z-z: Phi_z = 0.5 x [1 + 0.34 x (0.761 - 0.2) + 0.761^2] = 0.5 x [1 + 0.191 + 0.579] = 0.885 chi_z = 1 / [0.885 + sqrt(0.885^2 - 0.761^2)] = 1 / [0.885 + 0.452] = 0.748

Buckling resistance: N_b,Rd,y = 0.942 x 12,300 x 355 / 1.0 = 4,118 kN N_b,Rd,z = 0.748 x 12,300 x 355 / 1.0 = 3,272 kN (governs)

Utilisation: N_Ed / N_b,Rd,z = 2,304 / 3,272 = 0.704. OK, 70% utilised.

The 305 x 305 x 97 UC in S355 is adequate. A lighter section (e.g., 305 x 305 x 79 UC) could be considered for the upper storeys where the axial load is lower.

Step 6 -- Sensitivity Check

What if the imposed load reduction is not taken (conservative)?

N_Ed = 1.35 x 1,069 + 1.5 x 886 = 1,443 + 1,329 = 2,772 kN Utilisation = 2,772 / 3,272 = 0.847. Still OK.

What if the column length is 5.0 m (taller ground floor)?

lambda_bar_z = (5,000/77.4) / 76.4 = 64.6/76.4 = 0.846 Phi_z = 0.5 x [1 + 0.34 x (0.846 - 0.2) + 0.846^2] = 0.5 x [1 + 0.220 + 0.716] = 0.968 chi_z = 1 / [0.968 + sqrt(0.968^2 - 0.846^2)] = 1 / [0.968 + 0.471] = 0.695 N_b,Rd,z = 0.695 x 12,300 x 355 / 1.0 = 3,038 kN > 2,304 kN. Still adequate.

Step 7 -- Section Variation up the Building

For optimal material use, the column section should be varied up the height. A typical UK approach for this six-storey building:

The upper storey utilisation ratios are low because the available UC sections have minimum sizes dictated by practical considerations (transport, handling, minimum bolt group geometry). The section reduction is typically limited to one or two changes per building height.

UK National Annex Provisions

This worked example has applied the following UK NA values:

• gamma_M0 = 1.00, gamma_M1 = 1.00, gamma_M2 = 1.25
• Imposed load for office areas: 3.0 kN/m^2 (UK NA to EN 1991-1-1)
• Area reduction factor A_0 = 50 m^2 (UK NA to EN 1991-1-1)
• Load combination Expression 6.10 with UK NA psi factors
• Buckling curves per Table 6.2, adopted without modification

Design Resources

• UK Column Buckling Reference -- Buckling curves a0-d
• UK Effective Length Factors -- Annex E k factor method
• UK Steel Grades Reference -- EN 10025-2 grades
• UK UC and UB Section Properties -- Section tables
• UK Combined Loading Design -- Beam-column interaction
• All UK Steel Design References -- complete library

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Frequently Asked Questions

What UC section is appropriate for a six-storey internal column at 7.5 m grid?

A 305 x 305 x 97 UC in S355 is appropriate for the lower storeys, supporting approximately 2,300 kN at ground floor level with adequate reserve (70% utilisation at 4.5 m storey height). Upper storeys can be reduced to 305 x 305 x 79 UC or 254 x 254 x 73 UC depending on the accumulated load and storey height. The design is typically governed by weak-axis buckling (curve b, alpha = 0.34).

How is imposed load reduction applied per EN 1991-1-1?

For multi-storey columns, the imposed load may be reduced by the factor alpha_n = (2 + (n-2) x psi_0) / n, where n is the number of storeys above the loaded element and psi_0 = 0.7 for office areas (Category B). Additionally, the area reduction factor alpha_A = 0.5 + A_0/A (with A_0 = 50 m^2 per UK NA) may be applied where the loaded area exceeds 50 m^2. The UK NA to BS EN 1991-1-1 permits both reductions to be combined.

Why does the z-z axis buckling always govern for UC sections?

The minor axis radius of gyration i_z is approximately half of i_y for UC sections (e.g., 305 x 305 x 97 UC: i_y = 13.4 cm vs i_z = 7.74 cm). The resulting slenderness lambda_bar_z is correspondingly higher. Combined with the higher imperfection factor alpha for the z-z axis (curve b, alpha = 0.34) compared with y-y (curve a, alpha = 0.21), the z-z buckling resistance is always lower. Columns should be oriented with the major axis resisting the larger bending moments where present.

Educational reference only. All design values are per BS EN 1993-1-1:2005 + UK National Annex and BS EN 1991-1-1:2002 + UK NA. Verify all values against the current editions of the standards and the applicable National Annex for your project jurisdiction. Designs must be independently verified by a Chartered Structural Engineer registered with the Institution of Structural Engineers (IStructE) or the Institution of Civil Engineers (ICE). Results are PRELIMINARY -- NOT FOR CONSTRUCTION without independent professional verification.