Steel Column Design Example

Civil Guide

Design of Steel Column

Steel Column Design Example: Analysis of Steel Column under Axial Load

Columns are a fundamental construction element that is usually subjected to Axial load only (additional moment can be caused around the perimeter due to wind loading).

An example calculation for a column subjected to axial load is below:

Design a UC 254x254x89 in S355 steel to resist a design axial compression force of 500 kN. (Column is pinned at both ends and is 10m high).

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Section Properties

From tata steel online blue book:

  • Area of section, A = 113 cm2
  • Thickness of flange, tf = 17.3mm
  • Radius of gyration about the major axis (y-y), iy = 112mm
  • Radius of gyration abiout minor axis (z-z), iz= 65.5 mm
  • fy = 355 N/mm2
  • Web thickness, tw = 10.3mm
  • h = 260.3 mm
  • b = 256.3 mm

Section Classification

Part subject to compression:

\(\epsilon\) = (235/fy)0.5 = (235/355)0.5 = 0.81

c/tf =88/17.3=5.1<9ε=9×0.81=7.29

C = (b-tw – 2r)/2 = (256.3 – 10 – 2 x 12.7)/2 = 220.9/2 = 110.5 mm

Part subject to bending:

c/t_w = d/t = 260.3/10.3 = 25.3<72ε=72×0.81=58.32

Therefore, from Table 5.2 of Eurocode 3, section UC 254x254x89 is class 1.

Resistance of Cross-Section

Design resistance for uniform compression, Nc,RD for class 1 is defined in the equation below:

Nc,Rd = Afy/ γMO =  11300 x  355 / 1.0 = 4012 x 103 N = 4012 kN > NED = 1750 kN   ……. OKAY

Buckling Resistance of Column

Effective length of column about both axes is determined below:

Lcr = Lcr,y = Lcr,z = 1.0L = 1.0 x 10000 = 10,000mm.

The column will buckle about it’s weak axis (z-z). Slenderness ratio can be determined below:

λ1=π√(E/fy) = π√(210 x 103/355) = 76.4

Slenderness ratio about z-z axis (λz) is

λz = √(Afy/Ncr) = Lcr/iz x 1/ λ1 = 10,000/65.5 x 1/76.4 = 2.0

h/b = 260.3 / 256.3 = 1.02 < 1.2 and tf < 100mm. Therefore, the buckling curve around z-z axis is curve c, α = 0.49.

φ = 0.5[1 + α(λz – 0.2) + λ2z] = 0.5 [1+ 0.49(2.0-0.2)+ 2.02] = 2.941

χ = 1/ (φ + √(φ2-λ2)) = 1/(1.286 + √(1.2862 – 1.072)) = 0.2

Therefore, design buckling resistance, Nb,Rd is determine by the following:

Nb,Rd = χAfy/ϒM1

=0.2 x 11300 x 355 x 10-3/1.0 = 802 kN > 500 kN

Therefore, the section is adequate.

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