Design of Steel Column
Table of Contents
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).
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.