Design of Steel Columns for Biaxial Bending | Eurocode 3

When a column section is subjected to bending moment in the two axes in addition to a compressive axial force, the column is said to be biaxially loaded. The design of steel columns for biaxial bending involves the verification of the steel section’s capacity in bending, shear, compression, flexural buckling, and interaction of all these forces. Interaction formulas are available in EN 1993-1-1:2005 (Eurocode 3) for the design of members that are biaxially loaded.

Clause 6.2.9 of EN 1993-1-1:2005 describes the design of cross-sections subjected to combined bending and axial force (such as steel columns). Bending can occur along one or both major axes, with tensile or compressive axial forces (with no difference in treatment). Eurocode 3 provides several approaches for designing Class 1 and 2, Class 3, and Class 4 cross-sections in order to deal with the combined effects.

A basic linear interaction presented below and in equation (1) can be applied to all cross-sections (clause 6.2.1(7)). Although Class 4 cross-section resistances must be based on effective section properties. Furthermore, any additional moments arising from the resulting shift in neutral axis should be allowed for in class 4 sections. These extra moments necessitate the use of the expanded linear interaction expression.

N_Ed/N_Rd + M_y;Ed/M_y;Rd + M_z;Ed/M_z;Rd ≤ 1.0 ———— (1)

where N_Rd, M_y,Rd, and M_z,Rd are the design cross-sectional resistances, and any required reduction due to shear effects should be included (clause 6.2.8). The goal of equation (1) is to allow a designer to obtain a quick, approximate, and safe solution, possibly for initial member sizing, with the option to refine the calculations for final design.

Bi-axial bending with or without axial force (Class 1 and 2 sections)

EN 1993-1-1, like BS 5950: Part 1, treats bi-axial bending as a subset of the combined bending and axial force regulations. Clause 6.2.9.1 specifies the checks for Class 1 and 2 cross-sections subjected to bi-axial bending with or without axial forces (6). Although equation (1) shows a simple linear interaction expression, equation (2) represents a more sophisticated convex interaction expression that can result in large efficiency gains:

(M_y;Ed/M_N;y;Rd)^α + (M_z;Ed /M_N;z;Rd)^β ≤ 1.0 ———- (2)

in which α and β are constants, as defined below. Clause 6.2.9(6) allows α and β to be taken as unity, thus reverting to a conservative linear interaction.

For I- and H-sections:
α = 2 and β = 5n but β ≤ 1.0

For circular hollow sections:
α = 2 and β = 1

Rectangular hollow sections
α = β = 1.66/(1 – 1.13n²) but α = β ≤ 6.0

n = N_Ed / N_c,Rd

Worked Example

Verify the capacity of a 3500 mm tall column of UKC 254x254x89 in a commercial complex to withstand the following ultimate limit state actions;

Axial load; N_Ed = 1500 kN; (Compression)
Major axis moment at end 1 – Bottom; M_y,Ed1 = 89.0 kNm
Major axis moment at end 2 – Top; M_y,Ed2 = 77.0 kNm
Minor axis moment at end 1 – Bottom; M_z,Ed1 = 7.9 kNm
Minor axis moment at end 2 – Top; M_z,Ed2 = 2.4 kNm
Major axis shear force; V_y,Ed = 56 kN
Minor axis shear force;V_z,Ed = 14 kN

Solution

Partial factors
Resistance of cross-sections; γ_M0 = 1
Resistance of members to instability; γ_M1 = 1
Resistance of cross-sections in tension to fracture;  γ_M2 = 1.1

Column details
Column section; UKC 254x254x89
Steel grade; S275
Yield strength; f_y = 265 N/mm²
Ultimate strength;  f_u = 410 N/mm²
Modulus of elasticity;  E = 210 kN/mm²
Poisson’s ratio;  υ = 0.3
Shear modulus; G = E / [2 × (1 + υ)] = 80.8 kN/mm²

Column geometry
System length for buckling – Major axis; L_y = 3500 mm
System length for buckling – Minor axis; L_z = 3500 mm

The column is part of a sway frame in the direction of the minor axis
The column is part of a sway frame in the direction of the major axis

Column loading
Axial load; N_Ed = 1500 kN; (Compression)
Major axis moment at end 1 – Bottom; M_y,Ed1 = 89.0 kNm
Major axis moment at end 2 – Top; M_y,Ed2 = 77.0 kNm

Minor axis moment at end 1 – Bottom; M_z,Ed1 = 7.9 kNm
Minor axis moment at end 2 – Top; M_z,Ed2 = 2.4 kNm
Major axis shear force; V_y,Ed = 56 kN
Minor axis shear force;V_z,Ed = 14 kN

Buckling length for flexural buckling – Major axis
End restraint factor; K_y = 1.000
Buckling length; L_{cr_y} = L_y × K_y = 3500 mm

Buckling length for flexural buckling – Minor axis
End restraint factor;  K_z = 1.000
Buckling length;  L_{cr_z} = L_z × K_z = 3500 mm

Web section classification (Table 5.2)
f_y = 265 N/mm²
Coefficient depending on f_y; ε = √(235/ f_y) = 0.942
Depth between fillets; c_w = h – 2 × (t_f + r) = 200.3 mm
Ratio of c/t;  ratio_w = c_w / t_w = 19.45
Length of web taken by axial load; l_w = min(N_Ed / (f_y × t_w), c_w) = 200.3 mm
For class 1 & 2 proportion in compression; α = (c_w/2 + l_w/2) / c_w = 1.000

Limit for class 1 web;                                                     
Limit_1w = (396 × e) / (13 × a – 1) = 31.08
The web is class 1

Flange section classification (Table 5.2)
Outstand length; c_f = (b – t_w) / 2 – r = 110.3; mm
Ratio of c/t; ratio_f = c_f / t_f = 6.38

Conservatively assume uniform compression in flange

Limit for class 1 flange; Limit_1f = 9 × e = 8.48
Limit for class 2 flange; Limit_2f = 10 × e = 9.42
Limit for class 3 flange; Limit_3f = 14 × e = 13.18

The section is class 1

Resistance of cross section (cl. 6.2)

Shear – Major axis (cl. 6.2.6)
Design shear force; V_y,Ed = 56.0 kN
Shear area; A_vy = max((h – 2t_f) × t_w, A – 2 × b × t_f + (t_w + 2 × r) × t_f) = 3081 mm²
f_y = 265 N/mm²
Plastic shear resistance;  V_pl,y,Rd = A_vy × (f_y/√3)/ γ_M0 = 471.4 kN
V_y,Ed / V_pl,y,Rd = 0.119
PASS – Shear resistance exceeds the design shear force

V_y,Ed ≤ 0.5×V_pl,y,Rd – No reduction in f_y required for bending/axial force

Shear – Minor axis (cl. 6.2.6)
Design shear force; V_z,Ed = 13.5 kN
Shear area; A_vz = 2 × b × t_f – (t_w + 2 × r) × t_f = 8250 mm²
Plastic shear resistance; V_pl,z,Rd = A_vz × (f_y /√3) / γ_M0 = 1262.3 kN
V_z,Ed / V_pl,z,Rd = 0.011
PASS – Shear resistance exceeds the design shear force
V_z,Ed ≤ 0.5×V_pl,z,Rd – No reduction in f_y required for bending/axial force

Compression (cl. 6.2.4)
Design force; N_Ed = 1500 kN
Design resistance; N_c,Rd = N_pl,Rd = A × f_y / γ_M0 = 3003 kN
N_Ed / N_c,Rd = 0.5
PASS – The compression design resistance exceeds the design force

Bending – Major axis (cl. 6.2.5)
Design bending moment;  M_y,Ed = max(abs(M_y,Ed1), abs(M_y,Ed2)) = 89.0 kNm
Section modulus;  W_y = W_pl.y = 1223.9; cm³
Design resistance; M_c,y,Rd = W_y × f_y / γ_M0 = 324.3 kNm
M_y,Ed / M_c,y,Rd = 0.274
PASS – The bending design resistance exceeds the design moment

Bending – Major axis(cl. 6.2.5)
Design bending moment; M_z,Ed = max(abs(M_z,Ed1), abs(M_z,Ed2)) = 7.9 kNm
Section modulus; W_z = W_pl.z = 575.3; cm³
Design resistance; M_c,z,Rd = W_z × f_y / γ_M0 = 152.5 kNm
M_z,Ed / M_c,z,Rd = 0.052
PASS – The bending design resistance exceeds the design moment

Combined bending and axial force (cl. 6.2.9)
f_y = 265 N/mm²;
N_pl,Rd = A × f_y / γ_M0 = 3003 kN
Ratio design axial to design plastic resistance; n = abs(N_Ed) / N_pl,Rd = 0.500
Ratio web area to gross area; a = min(0.5, (A – 2 × b × t_f) / A) = 0.217

Bending – Major axis (cl. 6.2.9.1)
Design bending moment; M_y,Ed = max(abs(M_y,Ed1), abs(M_y,Ed2)) = 89.0 kNm
Plastic design resistance; M_pl,y,Rd = W_pl.y × f_y / γ_M0 = 324.3 kNm
Modified design resistance; M_N,y,Rd = M_pl,y,Rd × min(1, (1 – n) / (1 – 0.5 × a)) = 182.1 kNm
M_y,Ed / M_N,y,Rd = 0.489
PASS – Bending resistance in presence of axial load exceeds the design moment

Bending – Minor axis (cl. 6.2.9.1)
Design bending moment;M_z,Ed = max(abs(M_z,Ed1), abs(M_z,Ed2)) = 7.9 kNm
Plastic design resistance;  M_pl,z,Rd = W_pl.z × f_y / γ_M0 = 152.5 kNm
Modified design resistance;M_N,z,Rd = M_pl,z,Rd × [1 – ((n – a) / (1 – a))²] = 132.6; kNm
M_z,Ed / M_N,z,Rd = 0.059
PASS – Bending resistance in presence of axial load exceeds the design moment

Biaxial bending
Exponent α; α = 2.00
Exponent β;  β = max(1, 5 × n) = 2.50

Section utilisation at end 1; UR_{CS_1} = [abs(M_y,Ed1) / M_N,y,Rd] ^α + [abs(M_z,Ed1) / M_N,z,Rd] ^β = 0.240
Section utilisation at end 2; UR_{CS_2} = [abs(M_y,Ed2) / M_N,y,Rd] ^α + [abs(M_z,Ed2) / M_N,z,Rd] ^β = 0.179
PASS – The cross-section resistance is adequate

Buckling resistance (cl. 6.3)
Yield strength for buckling resistance;  f_y = 265 N/mm²

Flexural buckling – Major axis
Elastic critical buckling force; N_cr,y = π² × E × I_y / L_{cr_y}² = 24140 kN
Non-dimensional slenderness; λ_y = √(A × f_y / N_cr,y) = 0.353
Buckling curve (Table 6.2);  b
Imperfection factor (Table 6.1); α_y = 0.34
Parameter Φ;  Φ_y = 0.5 × [1 + α_y × (λ_y – 0.2) + λ_y²] = 0.588
Reduction factor;  χ_y = min(1.0, 1 / [Φ_y + √(Φ_y ² – λ_y²)]) = 0.944
Design buckling resistance; N_b,y,Rd = χ_y × A × f_y  / γ_M1 = 2835.9 kN
N_Ed / N_b,y,Rd = 0.529
PASS – The flexural buckling resistance exceeds the design axial load

Flexural buckling – Minor axis
Elastic critical buckling force; N_cr,z = π² × E × I_z / L_{cr_z}² = 8219 kN
Non-dimensional slenderness; λ_z = √(A × f_y / N_cr,z) = 0.604
Buckling curve (Table 6.2);  c
Imperfection factor (Table 6.1); α_z = 0.49
Parameter Φ; Φ_z = 0.5 × [1 + α_z × (λ_z – 0.2) + λ_z²] = 0.782
Reduction factor; χ_z = min(1.0, 1 / [Φ_z + √(Φ_z² – λ_z²)]) = 0.783
Design buckling resistance; N_b,z,Rd = χ_z × A × f_y  / γ_M1 = 2350.4 kN
N_Ed / N_b,z,Rd = 0.638
PASS – The flexural buckling resistance exceeds the design axial load

Torsional and torsional-flexural buckling (cl. 6.3.1.4)

Torsional buckling length factor; K_T = 1.00
Effective buckling length; L_{cr_T} = K_T × max(L_y, L_z) = 3500 mm
Distance from shear ctr to centroid along major axis;  y₀ = 0.0 mm
z₀ = 0 mm
Distance from shear ctr to centroid along minor axis; z₀ = 0.0 mm

i₀ = √(i_y² + i_z² + y₀² + z₀²) = 129.9 mm
b_T = 1 – (y₀ / i₀)² = 1.000

Elastic critical torsional buckling force; N_cr,T = 1 / i₀² × (G × I_t + π² × E × I_w / L_{cr_T}²) = 12085 kN
Elastic critical torsional-flexural buckling force; N_cr,TF = N_cr,y/(2 × b_T) × [1 + N_cr,T/N_cr,y – √[(1 – N_cr,T/N_cr,y)² + 4 × (y₀/i₀)² × N_cr,T/N_cr,y]]
N_cr,TF = 12085 kN

Non-dimensional slenderness; λ_T = √(A × f_y / min(N_cr,T, N_cr,TF)) = 0.498
Buckling curve (Table 6.2); c
Imperfection factor (Table 6.1); α_T = 0.49
Parameter Φ; Φ_T = 0.5 × [1 + α_T × (λ_T – 0.2) + λ_T²] = 0.697
Reduction factor; χ_T = min(1.0, 1 / [Φ_T + √(Φ_T ² – λ_T²)]) = 0.844
Design buckling resistance; N_b,T,Rd = χ_T × A × f_y  / γ_M1 = 2533.9 kN
N_Ed / N_b,T,Rd = 0.592
PASS – The torsional/torsional-flexural buckling resistance exceeds the design axial load

Minimum buckling resistance
Minimum buckling resistance; N_b,Rd = min(N_b,y,Rd, N_b,z,Rd, N_b,T,Rd) = 2350.4 kN
N_Ed / N_b,Rd = 0.638
PASS – The axial load buckling resistance exceeds the design axial load

Buckling resistance moment (cl.6.3.2.1)
Lateral torsional buckling length factor;  K_LT = 1.00
Effective buckling length; L_{cr_LT} = K_LT × L_z = 3500 mm
End moment factor; y = M_y,Ed2 / M_y,Ed1 = 0.865
Moment distribution correction factor (Table 6.6);  k_c = 1 / (1.33 – 0.33 × y) = 0.957
C₁ = 1 / k_c² = 1.091
Curvature factor;  g = √[1 – (I_z / I_y)] = 0.812
Poissons ratio;  υ = 0.3
Shear modulus;  G = E / [2 × (1 + υ)] = 80769 N/mm²

Elastic critical buckling moment; M_cr = C₁ × π² × E × I_z × √[I_w / I_z + L_{cr_LT}² × G × I_t /(π² × E × I_z)]/(L_{cr_LT}² × g)
M_cr = 1739.3 kNm

Slenderness ratio for lateral torsional buckling; λ_LT = √[W_y × f_y / M_cr] = 0.432
Limiting slenderness ratio; λ_LT,0 = 0.40
Correction factor for rolled sections; β_r = 0.75
Buckling curve (Table 6.5); b
Imperfection factor (Table 6.1); α_LT = 0.34

Parameter Φ_LT;   Φ_LT = 0.5 × [1 + α_LT × (λ_LT – λ_LT,0) + β_r × λ_LT²] = 0.575
Reduction factor; χ_LT = min(1.0, 1/λ_LT², 1 / [Φ_LT + √(Φ_LT² – β_r × λ_LT²)]) = 0.988
Modification factor;   f = min(1 – 0.5 × (1 – k_c) × [1 – 2 × (λ_LT – 0.8)²], 1) = 0.984
Modified LTB reduction factor – eq 6.58; χ_LT,mod = min(χ_LT / f, 1, 1/λ_LT²) = 1.000

Design buckling resistance moment; M_b,Rd = χ_LT,mod × W_y × f_y / γ_M1 = 324.3 kNm
Design bending moment; M_y,Ed = max(abs(M_y,Ed1), abs(M_y,Ed2)) = 89.0 kNm
M_y,Ed / M_b,Rd = 0.274
PASS – The design buckling resistance moment exceeds the maximum design moment

Combined bending and axial compression (cl. 6.3.3)
Characteristic resistance to normal force;  N_Rk = A × f_y = 3003 kN
Characteristic moment resistance – Major axis; M_y,Rk = W_pl.y × f_y = 324.3 kNm
Characteristic moment resistance – Minor axis; M_z,Rk = W_pl.z × f_y = 152.5 kNm

Moment factor – Major axis; C_my = 0.9
Moment factor – Minor axis; C_mz = 0.9

Moment distribution factor for LTB;  y_LT = M_y,Ed2 / M_y,Ed1 = 0.865
Moment factor for LTB;  C_mLT = max(0.4, 0.6 + 0.4 × y_LT) = 0.946

Interaction factor k_yy; k_yy = C_my × [1 + min(0.8, λ_y – 0.2) × N_Ed / (χ_y × N_Rk / γ_M1)] = 0.973
Interaction factor k_zy;  k_zy = 1 – min(0.1, 0.1 × λ_z) × N_Ed / ((C_mLT – 0.25) × (χ_z × N_Rk/γ_M1)) = 0.945
Interaction factor k_zz; k_zz = C_mz × [1 + min(1.4, 2 × λ_z – 0.6) × N_Ed / (χ_z × N_Rk / γ_M1)] = 1.250
Interaction factor k_yz; k_yz =  0.6 × k_zz = 0.750

Section utilisation;                       
UR_{B_1} = N_Ed / (χ_y × N_Rk / γ_M1) + k_yy × M_y,Ed / (χ_LT × M_y,Rk / γ_M1) + k_yz × M_z,Ed / (M_z,Rk / γ_M1)
UR_{B_1} = 0.838

UR_{B_2} = N_Ed / (χ_z × N_Rk / γ_M1) + k_zy × M_y,Ed / (χ_LT × M_y,Rk / γ_M1) + k_zz × M_z,Ed / (M_z,Rk / γ_M1)
UR_{B_2} = 0.965

PASS – The buckling resistance is adequate