This document provides a summary of the design of a steel member according to Eurocode 3. It includes: - Details of the steel section being designed, including dimensions, material properties, and classification. - Checks for shear, bending, axial compression, and buckling according to Eurocode 3, ensuring the design capacities exceed the design forces in each case. - A summary of the design confirming the steel member meets all requirements for its intended loading based on the specifications in Eurocode 3.

1. Project:
Job Ref.
Steel Member Design in Biaxial Bending And Axial
Compression Example, in accordance with EN1993-11:2005 incorporating Corrigenda February 2006 and April
2009 and the recommended values
GEODOMISI Ltd. - Dr. Costas Sachpazis
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation
Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 - Mobile: (+30)
6936425722 & (+44) 7585939944, costas@sachpazis.info
Section
Sheet no./rev. 1
Civil & Geotechnical Engineering
Calc. by
Dr. C. Sachpazis
Date
Chk'd by
Date
09/02/2014
App'd by
Date
STEEL MEMBER DESIGN (EN1993-1-1:2005)
In accordance with EN1993-1-1:2005 incorporating Corrigenda February 2006
and April 2009 and the recommended values
Section details
Section type;
UKC 356x406x287 (Corus Advance)
Steel grade;
S275
From table 3.1: Nominal values of yield strength fy and ultimate tensile strength fu for hot rolled
structural steel
Nominal thickness of element;
t = max(tf, tw) = 36.5 mm
Nominal yield strength;
fy = 275 N/mm
Nominal ultimate tensile strength;
fu = 430 N/mm
Modulus of elasticity;
E = 210000 N/mm
2
2
2
1

2. Project:
Job Ref.
Steel Member Design in Biaxial Bending And Axial
Compression Example, in accordance with EN1993-11:2005 incorporating Corrigenda February 2006 and April
2009 and the recommended values
GEODOMISI Ltd. - Dr. Costas Sachpazis
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation
Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 - Mobile: (+30)
6936425722 & (+44) 7585939944, costas@sachpazis.info
Section
Sheet no./rev. 1
Civil & Geotechnical Engineering
Date
Calc. by
Dr. C. Sachpazis
Chk'd by
Date
09/02/2014
App'd by
Date
Partial factors - Section 6.1
Resistance of cross-sections;
γM0 = 1.00
Resistance of members to instability;
γM1 = 1.00
Resistance of tensile members to fracture;
γM2 = 1.25
Lateral restraint
Distance between major axis restraints;
Ly = 5000 mm
Distance between minor axis restraints;
Lz = 5000 mm
Effective length factors
Effective length factor in major axis;
Ky = 0.700
Effective length factor in minor axis;
Kz = 1.000
Effective length factor for torsion;
KLT = 1.000
Classification of cross sections - Section 5.5
2
ε = √[235 N/mm / fy] = 0.92
Internal compression parts subject to bending and compression - Table 5.2 (sheet 1 of 3)
Width of section;
c = d = 290.2 mm
α = min([h / 2 + NEd / (2 × tw × fy) - (tf+ r)] / c, 1) =
1.000
c / tw = 13.9 × ε <= 396 × ε / (13 × α - 1);
Class 1
Outstand flanges - Table 5.2 (sheet 2 of 3)
Width of section;
c = (b - tw - 2 × r) / 2 = 173 mm
c / tf = 5.1 × ε <= 9 × ε;
Class 1
Section is class 1
Check shear - Section 6.2.6
Height of web;
Shear area factor;
hw = h - 2 × tf = 320.6 mm
η = 1.000
hw / tw < 72 × ε / η
Shear buckling resistance can be ignored
Design shear force parallel to z axis;
Vz,Ed = 200 kN
Shear area - cl 6.2.6(3);
Av = max(A - 2 × b × tf + (tw + 2 × r) × tf, η × hw × tw)
2
= 9378 mm
Design shear resistance - cl 6.2.6(2);
Vc,z,Rd = Vpl,z,Rd = Av × (fy / √[3]) / γM0 = 1489 kN
PASS - Design shear resistance exceeds design shear force
Design shear force parallel to y axis;
Vy,Ed = 30 kN
Shear area - cl 6.2.6(3);
Av = max(2 × b × tf - (tw + 2 × r) × tf, A - (hw × tw)) =
29325 mm
2
Design shear resistance - cl 6.2.6(2);
Vc,y,Rd = Vpl,y,Rd = Av × (fy / √[3]) / γM0 = 4656 kN
PASS - Design shear resistance exceeds design shear force
2

3. Project:
Job Ref.
Steel Member Design in Biaxial Bending And Axial
Compression Example, in accordance with EN1993-11:2005 incorporating Corrigenda February 2006 and April
2009 and the recommended values
GEODOMISI Ltd. - Dr. Costas Sachpazis
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation
Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 - Mobile: (+30)
6936425722 & (+44) 7585939944, costas@sachpazis.info
Section
Sheet no./rev. 1
Civil & Geotechnical Engineering
Date
Calc. by
Dr. C. Sachpazis
Chk'd by
Date
09/02/2014
App'd by
Date
Check bending moment major (y-y) axis - Section 6.2.5
Design bending moment;
My,Ed = 450 kNm
Design bending resistance moment - eq 6.13;
Mc,y,Rd = Mpl,y,Rd = Wpl.y × fy / γM0 = 1598.4 kNm
Slenderness ratio for lateral torsional buckling
Correction factor - Table 6.6;
kc = 0.603
2
C1 = 1 / kc = 2.75
Curvature factor;
g = √[1 - (Iz / Iy)] = 0.783
Poissons ratio;
ν = 0.3
Shear modulus;
G = E / [2 × (1 + ν)] = 80769 N/mm
Unrestrained length;
L = 1.00 × Lz = 5000 mm
2
Elastic critical buckling moment;
2
2
2
2
Mcr = C1 × π × E × Iz / (L × g) × √[Iw / Iz + L × G × It / (π × E × Iz)] = 29413.9 kNm
Slenderness ratio for lateral torsional buckling;
λLT = √[Wpl.y × fy / Mcr] = 0.233
Limiting slenderness ratio;
λLT,0 = 0.4
λLT < λLT,0 - Lateral torsional buckling can be ignored
Design resistance for buckling - Section 6.3.2.1
Buckling curve - Table 6.5;
b
Imperfection factor - Table 6.3;
αLT = 0.34
Correction factor for rolled sections;
β = 0.75
LTB reduction determination factor;
φLT = 0.5 × [1 + αLT × (λLT -λLT,0) + β ×λLT ] =
2
0.492
LTB reduction factor - eq 6.57;
2
2
2
χLT = min(1 / [φLT + √(φLT - β ×λLT )], 1, 1 /λLT ) =
1.000
2
f = min(1 - 0.5 × (1 - kc)× [1 - 2 × (λLT - 0.8) ], 1) =
Modification factor;
0.929
Modified LTB reduction factor - eq 6.58;
χLT,mod = min(χLT / f, 1) = 1.000
Design buckling resistance moment - eq 6.55;
Mb,Rd = χLT,mod × Wpl.y × fy / γM1 = 1598.4 kNm
PASS - Design buckling resistance moment exceeds design bending moment
Check bending moment minor (z-z) axis - Section 6.2.5
Design bending moment;
Mz,Ed = 125 kNm
Design bending resistance moment - eq 6.13;
Mc,z,Rd = Mpl,z,Rd = Wpl.z × fy / γM0 = 811 kNm
PASS - Design bending resistance moment exceeds design bending moment
Check compression - Section 6.2.4
Design compression force;
NEd = 4500 kN
Design resistance of section - eq 6.10;
Nc,Rd = Npl,Rd = A × fy / γM0 = 10057 kN
Slenderness ratio for major (y-y) axis buckling
Critical buckling length;
Lcr,y = Ly × Ky = 3500 mm
Critical buckling force;
Ncr,y = π2 × ESEC3 × Iy / Lcr,y2 = 168981.8 kN
3

4. Project:
Job Ref.
Steel Member Design in Biaxial Bending And Axial
Compression Example, in accordance with EN1993-11:2005 incorporating Corrigenda February 2006 and April
2009 and the recommended values
GEODOMISI Ltd. - Dr. Costas Sachpazis
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation
Engineering & Retaining Structures.
Sheet no./rev. 1
Civil & Geotechnical Engineering
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 - Mobile: (+30)
6936425722 & (+44) 7585939944, costas@sachpazis.info
Calc. by
Dr. C. Sachpazis
Date
Chk'd by
Date
09/02/2014
Slenderness ratio for buckling - eq 6.50;
App'd by
Date
λy = √[A × fy / Ncr,y] = 0.244
Design resistance for buckling - Section 6.3.1.1
Buckling curve - Table 6.2;
b
Imperfection factor - Table 6.1;
αy = 0.34
Buckling reduction determination factor;
φy = 0.5 × [1 + αy × (λy - 0.2) + λy ] = 0.537
Buckling reduction factor - eq 6.49;
χy = min(1 / [φy + √(φy - λy )], 1) = 0.984
Design buckling resistance - eq 6.47;
Nb,y,Rd = χy × A × fy / γM1 = 9899.8 kN
2
2
2
PASS - Design buckling resistance exceeds design compression force
Slenderness ratio for minor (z-z) axis buckling
Critical buckling length;
Lcr,z = Lz × Kz = 5000 mm
Critical buckling force;
Ncr,z = π × ESEC3 × Iz / Lcr,z = 32065.3 kN
Slenderness ratio for buckling - eq 6.50;
λz = √[A × fy / Ncr,z] = 0.560
2
2
Design resistance for buckling - Section 6.3.1.1
Buckling curve - Table 6.2;
c
Imperfection factor - Table 6.1;
αz = 0.49
Buckling reduction determination factor;
φz = 0.5 × [1 + αz × (λz - 0.2) + λz ] = 0.745
2
2
√(φz
2
- λz )], 1) = 0.809
Buckling reduction factor - eq 6.49;
χz = min(1 / [φz +
Design buckling resistance - eq 6.47;
Nb,z,Rd = χz × A × fy / γM1 = 8134.2 kN
PASS - Design buckling resistance exceeds design compression force
Check torsional and torsional-flexural buckling - Section 6.3.1.4
Torsional buckling length factor;
KT = 1.00
Torsional buckling length;
Lcr,T = max(Ly, Lz) × KT = 5000 mm
Distance from shear centre to centroid in y axis;
y0 = 0.0 mm
Distance from shear centre to centroid in z axis;
z0 = 0.0 mm
Radius of gyration;
i0 = √[iy + iz ] = 194.6 mm
Elastic critical torsional buckling force;
2
2
2
2
2
Ncr,T = 1 / i0 × [G × It + π × ESEC3 × Iw / Lcr,T ] =
57695.2 kN
Torsion factor;
2
βT = 1 - (y0 / i0) = 1.000
Elastic critical torsional-flexural buckling force
2
2
Ncr,TF = Ncr,y / (2 × βT) × [1 + Ncr,T / Ncr,y - √[(1 - Ncr,T / Ncr,y) + 4 × (y0 / i0) × Ncr,T / Ncr,y]] = 57695.2 kN
Elastic critical buckling force;
Ncr = min(Ncr,T, Ncr,TF) = 57695.2 kN
Slenderness ratio for torsional buckling - eq 6.52;
λT = √[A × fy / Ncr] = 0.418
Design resistance for buckling - Section 6.3.1.1
Buckling curve - Table 6.2;
c
Imperfection factor - Table 6.1;
αT = 0.49
Buckling reduction determination factor;
φT = 0.5 × [1 + αT × (λT - 0.2) + λT ] = 0.640
2
2
2
Buckling reduction factor - eq 6.49;
χT = min(1 / [φT + √(φT - λT )], 1) = 0.888
Design buckling resistance - eq 6.47;
Nb,T,Rd = χT × A × fy / γM1 = 8930.8 kN
4

5. Project:
Job Ref.
Steel Member Design in Biaxial Bending And Axial
Compression Example, in accordance with EN1993-11:2005 incorporating Corrigenda February 2006 and April
2009 and the recommended values
GEODOMISI Ltd. - Dr. Costas Sachpazis
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation
Engineering & Retaining Structures.
Sheet no./rev. 1
Civil & Geotechnical Engineering
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 - Mobile: (+30)
6936425722 & (+44) 7585939944, costas@sachpazis.info
Calc. by
Dr. C. Sachpazis
Date
Chk'd by
Date
09/02/2014
App'd by
Date
PASS - Design buckling resistance exceeds design compression force
Combined bending and axial force - Section 6.2.9
Normal force to plastic resistance force ratio;
n = NEd / Npl,Rd = 0.45
Web area to gross area ratio;
aw = min((A - 2 × b × tf) / A, 0.5) = 0.20
Design plastic moment resistance (y-y) - eq 6.13;
Mpl,y,Rd = Wpl.y × fy / γM0 = 1598.4 kNm
Reduced plastic mnt resistance (y-y)- eq 6.36;
MN,y,Rd = Mpl,y,Rd × min((1 - n) / (1 - 0.5 × aw), 1) =
983.3 kNm
Design plastic moment resistance (z-z) - eq 6.13;
Mpl,z,Rd = Wpl.z × fy / γM0 = 811.0 kNm
Reduced plastic mnt resistance (z-z) - eq 6.38;
MN,z,Rd = Mpl,z,Rd × (1 - ((n-aw) / (1- aw)) ) = 735.0
2
kNm
Parameter introducing effect of biaxial bending;
α_bi = 2.00
Parameter introducing effect of biaxial bending;
β_bi = max(5 × n, 1) = 2.24
Interaction formula – eq (6.41);
(My,Ed / MN,y,Rd)α
_bi
+ (Mz,Ed / MN,z,Rd)β
_bi
= 0.228
PASS - Reduced bending resistance moment exceeds design bending moment
Check combined bending and compression - Section 6.3.3
Equivalent uniform moment factors - Table B.3;
Cmy = 0.400
Cmz = 0.600
CmLT = 0.400
Interaction factors kij for members susceptible to torsional deformations - Table B.2
Characteristic moment resistance;
My,Rk = Wpl.y × fy = 1598.4 kNm
Characteristic moment resistance;
Mz,Rk = Wpl.z × fy = 811 kNm
Characteristic resistance to normal force;
NRk = A × fy = 10057 kN
Interaction factors;
kyy = Cmy × [1 + min(λy - 0.2, 0.8) × NEd / (χy × NRk /
γM1)] = 0.408
kzy = 1 - 0.1 × max(1,λz) × NEd / ((CmLT - 0.25) × χz
× NRk / γM1) = 0.631
kzz = Cmz × [1 + min(2 ×λz - 0.6, 1.4) × NEd / (χz ×
NRk / γM1)] = 0.773
kyz = 0.6 × kzz = 0.464
Interaction formulae - eq 6.61 & eq 6.62;
NEd / (χy × NRk / γM1) + kyy × My,Ed / (χLT × My,Rk / γM1)
+ kyz × Mz,Ed / (Mz,Rk / γM1) = 0.641
NEd / (χz × NRk / γM1) + kzy × My,Ed / (χLT × My,Rk / γM1)
+ kzz × Mz,Ed / (Mz,Rk / γM1) = 0.850
PASS - Combined bending and compression checks are satisfied
5