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 sectionâ€™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} â‰¤ 1.0 â€”â€”â€”â€” (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})^{Î±} + (M_{z;Ed} /M_{N;z;Rd})^{Î²} â‰¤ 1.0 â€”â€”â€”- (2)

in which Î± and Î² are constants, as defined below. Clause 6.2.9(6) allows Î± and Î² to be taken as unity, thus reverting to a conservative linear interaction.

For I- and H-sections:

Î± = 2 and Î² = 5n but Î² â‰¤ 1.0

For circular hollow sections:

Î± = 2 and Î² = 1

Rectangular hollow sections

Î± = Î² = 1.66/(1 â€“ 1.13n^{2}) but Î± = Î² â‰¤ 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 â€“ Bottom; M_{y,Ed1} = 89.0 kNm

Major axis moment at end 2 â€“ Top; M_{y,Ed2} = 77.0 kNm

Minor axis moment at end 1 â€“ Bottom;Â M_{z,Ed1} = 7.9 kNm

Minor axis moment at end 2 â€“ 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; Î³_{M0} = 1

Resistance of members to instability; Î³_{M1} = 1

Resistance of cross-sections in tension to fracture;Â Î³_{M2} = 1.1

**Column details**

Column section; UKC 254x254x89

Steel grade; S275

Yield strength;Â f_{y} = 265 N/mm^{2}

Ultimate strength;Â f_{u} = 410 N/mm^{2}

Modulus of elasticity;Â E = 210 kN/mm^{2}

Poissonâ€™s ratio;Â Ï… = 0.3

Shear modulus;Â G = E / [2 Ã— (1 + Ï…)] = 80.8 kN/mm^{2}

**Column geometry**

System length for buckling â€“ Major axis;Â L_{y} = 3500Â mm

System length for buckling â€“ Minor axis; L_{z} = 3500Â 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 â€“ Bottom; M_{y,Ed1} = 89.0 kNm

Major axis moment at end 2 â€“ Top; M_{y,Ed2} = 77.0 kNm

Minor axis moment at end 1 â€“ Bottom;Â M_{z,Ed1} = 7.9 kNm

Minor axis moment at end 2 â€“ 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 â€“ Major axis**

End restraint factor; K_{y} = 1.000

Buckling length;Â L_{cr_y} = L_{y} Ã— K_{y} = 3500 mm

**Buckling length for flexural buckling â€“ Minor axis**

End restraint factor;Â K_{z} = 1.000

Buckling length;Â L_{cr_z} = L_{z} Ã— K_{z} = 3500 mm

**Web section classification (Table 5.2)**

f_{y} = 265 N/mm^{2}

Coefficient depending on f_{y}; Îµ = âˆš(235/ f_{y}) = 0.942

Depth between fillets;Â c_{w} = h â€“ 2 Ã— (t_{f} + r) = 200.3 mm

Ratio of c/t;Â ratio_{w} = c_{w} / t_{w} = 19.45

Length of web taken by axial load; l_{w} = min(N_{Ed} / (f_{y} Ã— t_{w}), c_{w}) = 200.3 mm

For class 1 & 2 proportion in compression; Î± = (c_{w}/2 + l_{w}/2) / c_{w} = 1.000

Limit for class 1 web;Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Limit_{1w} = (396 Ã— e) / (13 Ã— a â€“ 1) = 31.08

The web is class 1

**Flange section classification (Table 5.2)**

Outstand length; c_{f} = (b â€“ t_{w}) / 2 â€“ 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;Â Limit_{1f} = 9 Ã— e = 8.48

Limit for class 2 flange; Limit_{2f} = 10 Ã— e = 9.42

Limit for class 3 flange;Â Limit_{3f} = 14 Ã— e = 13.18

The section is class 1

**Resistance of cross section (cl. 6.2)**

**Shear â€“ Major axis (cl. 6.2.6)**

Design shear force; V_{y,Ed} = 56.0 kN

Shear area; A_{vy} = max((h â€“ 2t_{f}) Ã— t_{w}, A â€“ 2 Ã— b Ã— t_{f} + (t_{w} + 2 Ã— r) Ã— t_{f}) = 3081 mm^{2}

f_{y} = 265 N/mm^{2}

Plastic shear resistance;Â V_{pl,y,Rd} = A_{vy} Ã— (f_{y}/âˆš3)/ Î³_{M0} = 471.4 kN

V_{y,Ed} / V_{pl,y,Rd} = 0.119**PASS** â€“ Shear resistance exceeds the design shear force

V_{y,Ed} â‰¤ 0.5Ã—V_{pl,y,Rd} â€“ No reduction in f_{y} required for bending/axial force

**Shear â€“ Minor axis (cl. 6.2.6)**

Design shear force;Â V_{z,Ed} = 13.5 kN

Shear area;Â A_{vz} = 2 Ã— b Ã— t_{f} â€“ (t_{w} + 2 Ã— r) Ã— t_{f} = 8250 mm^{2}

Plastic shear resistance; V_{pl,z,Rd} = A_{vz} Ã— (f_{y} /âˆš3) / Î³_{M0} = 1262.3 kN

V_{z,Ed} / V_{pl,z,Rd} = 0.011

PASS â€“ Shear resistance exceeds the design shear force

V_{z,Ed} â‰¤ 0.5Ã—V_{pl,z,Rd} â€“ No reduction in f_{y} required for bending/axial force

**Compression (cl. 6.2.4)**

Design force;Â N_{Ed} = 1500 kN

Design resistance; N_{c,Rd} = N_{pl,Rd} = A Ã— f_{y} / Î³_{M0} = 3003 kN

N_{Ed} / N_{c,Rd} = 0.5

PASS â€“ The compression design resistance exceeds the design force

**Bending â€“ Major axis (cl. 6.2.5)**

Design bending moment;Â M_{y,Ed} = max(abs(M_{y,Ed1}), abs(M_{y,Ed2})) = 89.0 kNm

Section modulus;Â Â W_{y} = W_{pl.y} = 1223.9; cm^{3}

Design resistance;Â M_{c,y,Rd} = W_{y} Ã— f_{y} / Î³_{M0} = 324.3 kNm

M_{y,Ed} / M_{c,y,Rd} = 0.274

PASS â€“ The bending design resistance exceeds the design moment

**Bending â€“ Major axis(cl. 6.2.5)**

Design bending moment;Â M_{z,Ed} = max(abs(M_{z,Ed1}), abs(M_{z,Ed2})) = 7.9 kNm

Section modulus; W_{z} = W_{pl.z} = 575.3; cm^{3}

Design resistance; M_{c,z,Rd} = W_{z} Ã— f_{y} / Î³_{M0} = 152.5 kNm

M_{z,Ed} / M_{c,z,Rd} = 0.052

PASS â€“ The bending design resistance exceeds the design moment

**Combined bending and axial force (cl. 6.2.9)**

f_{y} = 265 N/mm^{2};

N_{pl,Rd} = A Ã— f_{y} / Î³_{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 â€“ 2 Ã— b Ã— t_{f}) / A) = 0.217

**Bending â€“ 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;Â M_{pl,y,Rd} = W_{pl.y} Ã— f_{y} / Î³_{M0} = 324.3 kNm

Modified design resistance; M_{N,y,Rd} = M_{pl,y,Rd} Ã— min(1, (1 â€“ n) / (1 â€“ 0.5 Ã— a)) = 182.1 kNm

M_{y,Ed} / M_{N,y,Rd} = 0.489

PASS â€“ Bending resistance in presence of axial load exceeds the design moment

**Bending â€“ 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;Â M_{pl,z,Rd} = W_{pl.z} Ã— f_{y} / Î³_{M0} = 152.5 kNm

Modified design resistance;M_{N,z,Rd} = M_{pl,z,Rd} Ã— [1 â€“ ((n â€“ a) / (1 â€“ a))^{2}] = 132.6; kNm

M_{z,Ed} / M_{N,z,Rd} = 0.059

PASS â€“ Bending resistance in presence of axial load exceeds the design moment

**Biaxial bending**

Exponent Î±; Î± = 2.00

Exponent Î²;Â Î² = max(1, 5 Ã— n) = 2.50

Section utilisation at end 1;Â UR_{CS_1} = [abs(M_{y,Ed1}) / M_{N,y,Rd}] ^{Î±} + [abs(M_{z,Ed1}) / M_{N,z,Rd}] ^{Î²} = 0.240

Section utilisation at end 2; UR_{CS_2} = [abs(M_{y,Ed2}) / M_{N,y,Rd}] ^{Î±} + [abs(M_{z,Ed2}) / M_{N,z,Rd}] ^{Î²} = 0.179

PASS â€“ The cross-section resistance is adequate

**Buckling resistance (cl. 6.3)**

Yield strength for buckling resistance;Â f_{y} = **265** N/mm^{2}

**Flexural buckling â€“ Major axis**

Elastic critical buckling force;Â N_{cr,y} = Ï€^{2} Ã— E Ã— I_{y} / L_{cr_y}^{2} = 24140 kN

Non-dimensional slenderness; Î»_{y} = âˆš(A Ã— f_{y} / N_{cr,y}) = 0.353

Buckling curve (Table 6.2);Â b

Imperfection factor (Table 6.1);Â Î±_{y} = 0.34

Parameter Î¦;Â Î¦_{y} = 0.5 Ã— [1 + Î±_{y} Ã— (Î»_{y} â€“ 0.2) + Î»_{y}^{2}] = 0.588

Reduction factor;Â Ï‡_{y} = min(1.0, 1 / [Î¦_{y} + âˆš(Î¦_{y} ^{2} â€“ Î»_{y}^{2})]) = 0.944

Design buckling resistance; N_{b,y,Rd} = Ï‡_{y} Ã— A Ã— f_{y}Â / Î³_{M1} = 2835.9 kN

N_{Ed} / N_{b,y,Rd} = 0.529

PASS â€“ The flexural buckling resistance exceeds the design axial load

**Flexural buckling â€“ Minor axis**

Elastic critical buckling force;Â N_{cr,z} = Ï€^{2} Ã— E Ã— I_{z} / L_{cr_z}^{2} = 8219 kN

Non-dimensional slenderness; Î»_{z} = âˆš(A Ã— f_{y} / N_{cr,z}) = 0.604

Buckling curve (Table 6.2);Â c

Imperfection factor (Table 6.1); Î±_{z} = 0.49

Parameter Î¦; Î¦_{z} = 0.5 Ã— [1 + Î±_{z} Ã— (Î»_{z} â€“ 0.2) + Î»_{z}^{2}] = 0.782

Reduction factor; Ï‡_{z} = min(1.0, 1 / [Î¦_{z} + âˆš(Î¦_{z}^{2} â€“ Î»_{z}^{2})]) = 0.783

Design buckling resistance; N_{b,z,Rd} = Ï‡_{z} Ã— A Ã— f_{y}Â / Î³_{M1} = 2350.4 kN

N_{Ed} / N_{b,z,Rd} = 0.638

PASS â€“ 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;Â L_{cr_T} = K_{T} Ã— max(L_{y}, L_{z}) = 3500 mm

Distance from shear ctr to centroid along major axis;Â y_{0} = 0.0 mm

z_{0} = 0 mm

Distance from shear ctr to centroid along minor axis; z_{0} = 0.0 mm

i_{0} = âˆš(i_{y}^{2} + i_{z}^{2} + y_{0}^{2} + z_{0}^{2}) = 129.9 mm

b_{T} = 1 â€“ (y_{0} / i_{0})^{2} = 1.000

Elastic critical torsional buckling force;Â N_{cr,T} = 1 / i_{0}^{2} Ã— (G Ã— I_{t} + Ï€^{2} Ã— E Ã— I_{w} / L_{cr_T}^{2}) = 12085 kN

Elastic critical torsional-flexural buckling force;Â N_{cr,TF} = N_{cr,y}/(2 Ã— b_{T}) Ã— [1 + N_{cr,T}/N_{cr,y} â€“ âˆš[(1 â€“ N_{cr,T}/N_{cr,y})^{2} + 4 Ã— (y_{0}/i_{0})^{2} Ã— N_{cr,T}/N_{cr,y}]]

N_{cr,TF} = **12085** kN

Non-dimensional slenderness;Â Î»_{T} = âˆš(A Ã— f_{y} / min(N_{cr,T}, N_{cr,TF})) = 0.498

Buckling curve (Table 6.2);Â c

Imperfection factor (Table 6.1); Î±_{T} = 0.49

Parameter Î¦; Î¦_{T} = 0.5 Ã— [1 + Î±_{T} Ã— (Î»_{T} â€“ 0.2) + Î»_{T}^{2}] = 0.697

Reduction factor; Ï‡_{T} = min(1.0, 1 / [Î¦_{T} + âˆš(Î¦_{T} ^{2} â€“ Î»_{T}^{2})]) = 0.844

Design buckling resistance;Â N_{b,T,Rd} = Ï‡_{T} Ã— A Ã— f_{y}Â / Î³_{M1} = 2533.9 kN

N_{Ed} / N_{b,T,Rd} = 0.592

PASS â€“ 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 â€“ The axial load buckling resistance exceeds the design axial load

**Buckling resistance moment (cl.6.3.2.1)**

Lateral torsional buckling length factor;Â K_{LT} = 1.00

Effective buckling length; L_{cr_LT} = K_{LT} Ã— L_{z} = 3500 mm

End moment factor;Â y = M_{y,Ed2} / M_{y,Ed1} = 0.865

Moment distribution correction factor (Table 6.6);Â k_{c} = 1 / (1.33 â€“ 0.33 Ã— y) = 0.957

C_{1} = 1 / k_{c}^{2} = 1.091

Curvature factor;Â g = âˆš[1 â€“ (I_{z} / I_{y})] = 0.812

Poissons ratio;Â Ï… = 0.3

Shear modulus;Â G = E / [2 Ã— (1 + Ï…)] = 80769 N/mm^{2}

Elastic critical buckling moment; M_{cr} = C_{1} Ã— Ï€^{2} Ã— E Ã— I_{z} Ã— âˆš[I_{w} / I_{z} + L_{cr_LT}^{2} Ã— G Ã— I_{t} /(Ï€^{2} Ã— E Ã— I_{z})]/(L_{cr_LT}^{2} Ã— g)

M_{cr} = 1739.3 kNm

Slenderness ratio for lateral torsional buckling; Î»_{LT} = âˆš[W_{y} Ã— f_{y} / M_{cr}] = 0.432

Limiting slenderness ratio;Â Î»_{LT,0} = 0.40

Correction factor for rolled sections; Î²_{r} = 0.75

Buckling curve (Table 6.5);Â b

Imperfection factor (Table 6.1);Â Î±_{LT} = 0.34

Parameter Î¦_{LT};Â Â Î¦_{LT} = 0.5 Ã— [1 + Î±_{LT} Ã— (Î»_{LT} â€“ Î»_{LT,0}) + Î²_{r} Ã— Î»_{LT}^{2}] = 0.575

Reduction factor;Â Ï‡_{LT} = min(1.0, 1/Î»_{LT}^{2}, 1 / [Î¦_{LT} + âˆš(Î¦_{LT}^{2} â€“ Î²_{r} Ã— Î»_{LT}^{2})]) = 0.988

Modification factor;Â Â f = min(1 â€“ 0.5 Ã— (1 â€“ k_{c}) Ã— [1 â€“ 2 Ã— (Î»_{LT} â€“ 0.8)^{2}], 1) = 0.984

Modified LTB reduction factor â€“ eq 6.58;Â Ï‡_{LT,mod} = min(Ï‡_{LT} / f, 1, 1/Î»_{LT}^{2}) = 1.000

Design buckling resistance moment; M_{b,Rd} = Ï‡_{LT,mod} Ã— W_{y} Ã— f_{y} / Î³_{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 â€“ The design buckling resistance moment exceeds the maximum design moment

**Combined bending and axial compression (cl. 6.3.3)**

Characteristic resistance to normal force;Â N_{Rk} = A Ã— f_{y} = 3003 kN

Characteristic moment resistance â€“ Major axis;Â M_{y,Rk} = W_{pl.y} Ã— f_{y} = 324.3 kNm

Characteristic moment resistance â€“ Minor axis; M_{z,Rk} = W_{pl.z} Ã— f_{y} = 152.5 kNm

Moment factor â€“ Major axis;Â C_{my} = 0.9

Moment factor â€“ Minor axis;Â C_{mz} = 0.9

Moment distribution factor for LTB;Â y_{LT} = M_{y,Ed2} / M_{y,Ed1} = 0.865

Moment factor for LTB;Â C_{mLT} = max(0.4, 0.6 + 0.4 Ã— y_{LT}) = 0.946

Interaction factor k_{yy};Â k_{yy} = C_{my} Ã— [1 + min(0.8, Î»_{y} â€“ 0.2) Ã— N_{Ed} / (Ï‡_{y} Ã— N_{Rk} / Î³_{M1})] = 0.973

Interaction factor k_{zy};Â k_{zy} = 1 â€“ min(0.1, 0.1 Ã— Î»_{z}) Ã— N_{Ed} / ((C_{mLT} â€“ 0.25) Ã— (Ï‡_{z} Ã— N_{Rk}/Î³_{M1})) = 0.945

Interaction factor k_{zz};Â k_{zz} = C_{mz} Ã— [1 + min(1.4, 2 Ã— Î»_{z} â€“ 0.6) Ã— N_{Ed} / (Ï‡_{z} Ã— N_{Rk} / Î³_{M1})] = 1.250

Interaction factor k_{yz};Â k_{yz} =Â 0.6 Ã— k_{zz} = 0.750

Section utilisation;Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

UR_{B_1} = N_{Ed} / (Ï‡_{y} Ã— N_{Rk} / Î³_{M1}) + k_{yy} Ã— M_{y,Ed} / (Ï‡_{LT} Ã— M_{y,Rk} / Î³_{M1}) + k_{yz} Ã— M_{z,Ed} / (M_{z,Rk} / Î³_{M1})

UR_{B_1} = 0.838

UR_{B_2} = N_{Ed} / (Ï‡_{z} Ã— N_{Rk} / Î³_{M1}) + k_{zy} Ã— M_{y,Ed} / (Ï‡_{LT} Ã— M_{y,Rk} / Î³_{M1}) + k_{zz} Ã— M_{z,Ed} / (M_{z,Rk} / Î³_{M1})

UR_{B_2} = 0.965

PASS â€“ The buckling resistance is adequate

Great job Engr Ubani

Professional post!