------------------------ | :--------------: | :-----------: | :------------------: | | Imposed — domestic, office | 0.7 | 0.5 | 0.3 | | Imposed — storage | 1.0 | 0.9 | 0.8 | | Imposed — vehicle (≤ 30 kN) | 0.7 | 0.5 | 0.6 | | Wind loads | 0.5 | 0.2 | 0.0 | | Snow loads (≤ 1000 m) | 0.5 | 0.2 | 0.0 | | Temperature (non-fire) | 0.6 | 0.5 | 0.0 |

Partial Factors for Actions per UK NA

Serviceability Limit State Combinations

Characteristic (irreversible SLS): [ Gk + Q{k,1} + \sum \psi*{0,i} Q*{k,i} ]

Frequent (reversible SLS): [ Gk + \psi{1,1} Q*{k,1} + \sum \psi*{2,i} Q_{k,i} ]

Quasi-permanent (long-term SLS): [ Gk + \sum \psi{2,i} Q_{k,i} ]

Accidental and Seismic Combinations

Accidental: G_k + A_d + ψ_1,1 Q_k,1 + Σ ψ_2,i Q_k,i

Seismic (EN 1998-1 + UK NA): G_k + E_d + Σ ψ_2,i Q_k,i

Worked Example — Office Floor Beam

Given: 533×210 UB 92, office floor. G_k = 35 kN/m (slab + self-weight + SDL), Q_k,1 = 20 kN/m (imposed), Q_k,2 = 3 kN/m (partitions).

ULS (Set B): F_Ed = 1.35 × 35 + 1.5 × 20 + 1.5 × 0.7 × 3 = 47.25 + 30.0 + 3.15 = 80.4 kN/m

SLS characteristic: F_ser = 35 + 20 + 0.7 × 3 = 57.1 kN/m

SLS quasi-permanent: F_qp = 35 + 0.3 × 20 + 0.3 × 3 = 41.9 kN/m

EN 1990 Equation 6.10 — Full Walkthrough

EN 1990:2002 Clause 6.4.3.2 defines the fundamental load combination equations for persistent and transient design situations. The UK National Annex adopts Equation 6.10 (expressions 6.10a and 6.10b) for STR and GEO ultimate limit state verifications. This is the most critical load combination equation in UK structural design, and a thorough understanding of each term is essential for correct application.

The Two Expressions of Equation 6.10:

The UK NA requires that BOTH expressions be evaluated, and the more onerous governs.

Expression 6.10a (permanent-dominated):

Sum*{j>=1} gamma_G,j * Gk,j + gamma_P * P + gammaQ,1 * psi0,1 * Q_k,1 + Sum{i>1} gamma_Q,i * psi*0,i * Q_k,i

For buildings without prestress: Sum gammaG * Gk + gamma_Q * psi0,1 * Qk,1 + Sum gamma_Q * psi_0,i * Q_k,i

Expression 6.10b (variable-dominated — the standard UK combination):

Sum*{j>=1} xi * gammaG,j * Gk,j + gamma_P * P + gammaQ,1 * Q_k,1 + Sum{i>1} gamma_Q,i * psi*0,i * Q_k,i

Where xi = 0.925 is the UK NA reduction factor for permanent actions in Eq. 6.10b.

For buildings: Sum xi _ gamma_G _ Gk + gamma_Q,1 * Qk,1 + Sum gamma_Q,i * psi_0,i * Q_k,i

Why two expressions?

Eq. 6.10a governs when permanent loads dominate (heavy dead load, modest live load). Eq. 6.10b governs when the leading variable action is dominant. For Eq. 6.10a, permanent load gets the full partial factor (1.35) while the leading variable action is reduced by psi_0. For Eq. 6.10b, permanent load gets a small reduction (xi = 0.925) but the leading variable action gets its full unreduced partial factor (1.5).

Walkthrough — Multi-Storey Building Column Load

Problem: A column in a 6-storey office building supports tributary area of 48 m^2 per floor. Calculate the design axial load at the foundation level under two load cases: (a) imposed load as leading variable action, (b) wind load as leading variable action.

Given loads (characteristic values):

Partial factors and psi values (UK NA):

gamma_G = 1.35 (unfavourable permanent) gamma_Q = 1.5 (unfavourable variable) xi = 0.925 (UK NA reduction for permanent in Eq. 6.10b) psi_0,imp = 0.7 (office imposed, Category B) psi_0,wind = 0.5 (wind)

Case (a): Imposed load as leading variable action

Eq. 6.10a:

N_Ed,6.10a = 1.35 * (1152 + 288) + 1.5 * 0.7 * 864 + 1.5 * 0.5 * 120
           = 1.35 * 1440 + 1.05 * 864 + 0.75 * 120
           = 1944.0 + 907.2 + 90.0
           = 2,941.2 kN

Eq. 6.10b:

N_Ed,6.10b = 0.925 * 1.35 * 1440 + 1.5 * 864 + 1.5 * 0.5 * 120
           = 0.925 * 1944 + 1296.0 + 90.0
           = 1798.2 + 1296.0 + 90.0
           = 3,184.2 kN

Eq. 6.10b governs: N_Ed = 3,184 kN. This is typical for office buildings where imposed load is the dominant variable action — Eq. 6.10b gives approximately 8% higher load than 6.10a in this case (3,184 vs 2,941). The reason: Eq. 6.10b applies the leading variable action at full strength (1.5 x Q_k) while Eq. 6.10a reduces it by psi_0 (1.5 x 0.7 x Q_k = 1.05 x Q_k).

Case (b): Wind as leading variable action

Eq. 6.10a:

N_Ed,6.10a = 1.35 * 1440 + 1.5 * 0.5 * 120 + 1.5 * 0.7 * 864
           = 1944.0 + 90.0 + 907.2
           = 2,941.2 kN

Eq. 6.10b:

N_Ed,6.10b = 0.925 * 1.35 * 1440 + 1.5 * 120 + 1.5 * 0.7 * 864
           = 1798.2 + 180.0 + 907.2
           = 2,885.4 kN

Eq. 6.10a now governs: N_Ed = 2,941 kN (vs 2,885 from 6.10b). This illustrates the critical point: when wind is the leading action and imposed is the companion, Eq. 6.10a CAN govern because the permanent load receives its full factor (1.35) while the relatively small wind load (Q_wind = 120 kN) is reduced by psi_0 = 0.5 in Eq. 6.10b — but that reduction does not save much total load. The permanent load dominates, and Eq. 6.10a applies the higher permanent factor.

Summary of design axial loads:

The imposed-leading case (Case a, Eq. 6.10b) governs overall column design with N_Ed = 3,184 kN.

UK NA Eq. 6.10 vs EN 1990 Recommended Values

The UK NA xi = 0.925 (vs EN 1990 recommended 0.85) is notably higher. This means UK Eq. 6.10b gives higher design loads than would be obtained using the EN 1990 recommended values — a deliberate UK choice to maintain reliability levels consistent with previous British Standards (BS 5950). This higher xi ensures that columns in multi-storey buildings, where permanent load dominates, are not under-designed relative to historic UK practice.

Practical Tips for Eq. 6.10 Application

1. Always check both 6.10a and 6.10b. The governing expression depends on the ratio of permanent to variable loads. For G_k / Q_k > 3: Eq. 6.10a typically governs. For G_k / Q_k < 2: Eq. 6.10b typically governs.

2. Alternate the leading variable action. For each load case (imposed leading, wind leading, snow leading), evaluate both 6.10a and 6.10b. A multi-storey column may have a different governing case than a roof beam.

3. Wind as leading vs companion. Wind is rarely the leading variable for columns in multi-storey buildings, but for lightweight roofs, portal frames, and tall buildings, wind can be the critical leading variable.

4. Do not apply xi to Eq. 6.10a. The reduction factor xi = 0.925 ONLY applies to the permanent action term in Eq. 6.10b per EN 1990 Clause 6.4.3.2(3).

5. Accidental and seismic combinations use different rules. Eq. 6.10 is for persistent and transient design situations only. Accidental uses EN 1990 Clause 6.4.3.3, seismic uses EN 1998-1 Clause 3.2.2.5.

Design Resources

• UK Wind Load — EN 1991-1-4 + UK NA
• UK Snow Load — EN 1991-1-3 + UK NA
• UK Seismic Design — EN 1998-1 + UK NA
• UK Beam Design — Flexural design
• UK Column Design — Compression design
• All UK References

Frequently Asked Questions

What load combinations are used in UK practice?

The UK NA to EN 1990 defines three sets for STR/GEO. Set B is most common for UK buildings: 1.35G_k + 1.5Q_k,1 + Σ1.5ψ_0,iQ_k,i. For wind leading: 1.35G_k + 1.5W_k + Σ1.5ψ_0,iQ_k,i. Accidental: G_k + A_d + ψ_1,1Q_k,1 + Σψ_2,iQ_k,i.

What are ψ factors and when are they used?

ψ factors reduce the probability of simultaneous extreme loads. ψ₀ (combination) for ULS with multiple variables, ψ₁ (frequent) for SLS frequent and accidental, ψ₂ (quasi-permanent) for long-term and seismic. For office loads: ψ₀ = 0.7, ψ₁ = 0.5, ψ₂ = 0.3.

How does UK NA to EN 1990 differ from recommended values?

The UK NA adopts most EN 1990 recommended values but modifies: partial factors for Sets A, B, C, specific ψ values for UK structures, and simplified guidance for wind-imposed load combination. It also categorises buildings into consequence classes CC1, CC2, CC3.

How are serviceability deflections checked per UK NA?

Total deflection under characteristic combination is checked against span/200. Imposed load deflection uses frequent combination against span/300 (standard) or span/500 (brittle finishes). These limits per UK NA to EN 1990 ensure acceptable performance without damage to cladding and partitions.

Load Combinations

Frequently Asked Questions

What is the recommended design procedure for this structural element?

The standard design procedure follows: (1) establish design criteria including applicable code, material grade, and loading; (2) determine loads and applicable load combinations; (3) analyse the structure for internal forces; (4) check member strength for all applicable limit states; (5) verify serviceability requirements; and (6) detail connections. Computer analysis is recommended for complex structures, but hand calculations should be used for verification of critical elements.

How do different design codes compare for this calculation?

EN 1990 (Eurocode) with the UK National Annex defines the load combination framework for UK structural design. UK NA to EN 1990 adopts Equation 6.10 (both 6.10a and 6.10b) for STR and GEO ultimate limit states. Key UK NA differences from EN 1990 recommended values include: xi = 0.925 (vs recommended 0.85), ψ0,wind = 0.5 (vs recommended 0.6), and categorisation of buildings into consequence classes CC1-CC3. Engineers should verify which National Annex is adopted in their project jurisdiction.

Reference only. Verify all values against the current edition of EN 1993-1-1:2005, UK National Annex, and BS EN 1090-2. This information does not constitute professional engineering advice.