# Design of Steel Columns | EN 1993-1 and BS 5950-1:2000

In structural members subjected to compressive forces (e.g. steel columns and struts), secondary bending caused by imperfections within materials during fabrication processes, inaccurate positioning of loads or asymmetry of the cross-section etc, can induce premature failure either in a part of the cross-section, such as the outstand flange of an I-section, or on the element as a whole. In such cases, the failure mode is predominantly buckling and not squashing.

The design of most compressive members is governed by their overall buckling capacity, i.e. the maximum compressive load which can be carried before failure occurs by excessive deflection in the plane of greatest slenderness.

Compression members (i.e. struts and columns) should be checked for;
(1) resistance to compression
(2) resistance to buckling

The design of struts for buckling is well covered in clause 4.7 of BS 5950-1:2000 while the buckling resistance of members is covered in clause 5.5 of EN 1993-1-1:1992 (EC3). The checks for uniform compression in EC3 is found in clause 5.4.4.

In the design of steel columns, the design compressive axial force (NEd) should not exceed the design compression resistance (Nc,Rd) such that;

NEd/NC,Rd â‰¤ 1.0
NC,Rd = (A.fy)/Î³mo

Where;
A is the gross area of the section
fy is the yield strength of the section
Î³mo is the material factor of safety.

Similarly, the design axial force should not exceed the buckling resistance of the column (Nb,Rd) such that;

NEd/Nb,Rd â‰¤ 1.0
Nb,Rd = (Ï‡A.fy)/Î³mo

Where;
Ï‡ is the buckling reduction factor
A is the gross area of the section
fy is the yield strength of the section
Î³mo is the material factor of safety.

## Design Example of Steel Columns

In the downloadable solved example, the capacity of a 4m universal column (UC 305 Ã— 305 Ã— 158) pinned at both ends was investigated for its capacity to carry an axial load of 3556 kN according to the requirements of Eurocode3 and BS 5950.

On the Shear Deformation of One Span Beams Using Virtual Work Method
Example on Analysis of Statically Determinate FramesÂ

SOLUTION
The first step is to outline the properties of the section we wish to investigate.

Properties of UC 305 Ã— 305 Ã— 158

D = 327.1 mm; B = 311.2 mm; tf = 25mm; tw = 15.8mm; r = 15.2mm; d = 246.7mm, b/tf = 6.22; d/tw = 15.6; Izz = 38800 cm4; Iyy = 12600 cm4; iy-y = 13.9cm; iz-z = 7.9 cm, A = 201 cm2

### Column Design according to EN 1993-1-1 (Eurocode 3)

Thickness of flange tf = 25mm. Since tf > 16mm, Design yield strength fy = 265 N/mm2 (Table 3.1 EC3)

Section classification
Îµ = âˆš(235/fy) = âˆš(235/265) = 0.942

We can calculate the outstand of the flange (flange under compression)
C = (b â€“ tw â€“ 2r) / (2 ) = (327.1 â€“ 15.8 -2(15.2)) / (2 ) = 140.45mm.
We can then verify that C/tf = 140.45/25 = 5.618
5.618 < 9Îµ i.e. 5.618 < 8.478.

Therefore the flange is class 1 plastic

Web (Internal compression)
d/tw = 15.6 < 33Îµ so that 15.6 < 31.088. Therefore the web is also class 1 plastic

Resistance of the member to uniform compression
NC,Rd = (A.Fy)/Î³mo = (201 Ã— 102 Ã— 265) / 1.0 = 5326500 N = 5326.5 kN

NEd/NC,Rd = 3556/5326.5 = 0.6676 < 1

Therefore section is ok for uniform compression.

Buckling resistance of member (clause 5.5 EN 1993-1-1:1992)

Since member is pinned at both ends, critical buckling length is the same for all axis Lcr = 4000mm

Slenderness ratio Â¯Î» = Lcr/(ri Î»1)
Î»1 = 93.9Îµ = 93.9 Ã— 0.942 = 88.454

In the major axis
(Â¯Î»y ) = 4000/(139 Ã— 88.454) = 0.3253
In the minor axis
(Â¯Î»z ) = 4000/(79 Ã— 88.454) = 0.5724

Check D/b ratio = 327.1/311.2 = 1.0510 < 1.2, and tf < 100 mm (Table 5.5.3 EN 1993-1-1:1992)

Therefore buckling curve b is appropriate for y-y axis, and buckling curve c for z-z axis. The imperfection factor for buckling curve b Î± = 0.34 and curve c = 0.49 (Table 5.5.1)

Î¦ = 0.5 [1 + Î±(Â¯Î» â€“ 0.2) + Â¯Î»2]

Î¦z = 0.5 [1 + 0.49 (0.5724 â€“ 0.2) + 0.57242] = 0.7551
Î¦y = 0.5 [1 + 0.34 (0.3253 â€“ 0.2) + 0.32532] = 0.5742

X = 1/(Î¦+ âˆš(Î¦2 -Â¯Î»2))
Xz = 1/( 0.7551 + âˆš(0.75512 â€“ 0.57242)) = 0.8015 < 1 ok
Xy = 1/( 0.5742+ âˆš(0.57242 â€“ 0.32532)) = 0.9548 < 1 ok

In this case, the lesser holds for the calculation of the buckling load.

Therefore Nb,Rd = (Xz A.Fy)/Î³m1 = (0.8015 Ã— 201 Ã— 102 Ã— 265) / (1.0 ) = 4269189.75 N = 4269.189 kN

NEd/Nb,Rd = 3556/4269.189 = 0.8329 < 1

Therefore section is ok for buckling.

Summarily, the section is ok to resist axial load on it.

### Column Design According to BS 5950-1:2000

From Table 9 BS 5950-1:2000
Since the thickness of the flange is > 16mm but < 40mm, Py = 265 N/mm2

Obviously, the column will buckle about the z-z axis, which is the weaker axis in terms of buckling. It is however important to realise that axes are not labelled in the same way using the two codes. The z-z axis in EC3 is the y-y axis in BS 5950, while the y-y axis in EC3 is called the x-x axis in BS 5950.

Îµ = âˆš(275/fy) = âˆš(275/265) = 1.0186
b/tf = 6.22 < 15Îµ = 15.28; d/tw = 15.6 < 40Îµ = 40.74

Hence the UC section is not slender.
Therefore, let us focus on the weaker axis (z-z) in order to verify the buckling load.
Slenderness ratio = Î» = Lcr/rz = 4000/79 = 50.633

From Table 23 of BS 5950, for H sections of thickness < 40mm, strut curve c is appropriate for calculating our compressive strength PC (N/mm2). We can go through the stress of calculating the compressive strength PC using the formulars outlined in ANNEX C of BS 5950 or simply read them from Table 24 of the code. We will use the two methods in this example.

Calculating using formula
Limiting slenderness Î»0 = 0.2 [(Ï€2E)/Py ]0.5 = 0.2 [(Ï€2 Ã— 210000) / 265]0.5 = 17.6875

Perry factor for flexural buckling under axial load Î· = [a(Î» â€“ Î»0)]/1000 where a = 5.5 for strut curve c
Î· = [5.5(50.633 â€“ 17.6875)] / 1000 = 0.1812
Euler load PE = (Ï€2E)/Î»2 = (Ï€2 Ã— 210000) / 50.6332 = 808.447 N/mm2
Î¦ = (Py + (Î·+1)PE) / 2 = [265 + (0.1812 + 1) Ã— 808.447] / 2 = 609.9688 N/mm2

Therefore, the compressive strength Pc = (PE Py) / (Î¦ + (Î¦2 â€“ PE Py)0.5)

Pc = (808.447 Ã— 265) / (609.9688 + [609.96882 â€“ (808.447 Ã— 265)]0.5) = 212.699 N/mm2

If we had decided to read from chart, strut curve C, Î» < 110, Py = 265 N/mm2

Knowing that Î» = 50.633. Interpolate between Î» = 50 and 52 to obtain Pc = 212.0505 N/mm2
So let us use PC = 212.699 N/mm2
Hence Buckling load, PX = AgPC = 201 Ã— 102 Ã— 212.699 = 4275249.9 N = 4275.2499 kN

Therefore, N/PX = 3556/4275.2499 = 0.8317 < 1.0. Hence the section is ok for buckling.

The difference in result between EC2 buckling load (Nb,Rd =4269.189 KN) and BS 5950 buckling load (PX = 4275.2499 KN) for axially loaded columns disregarding load factor is just about 0.144%.

You can download the paper HERE and compare the results from the two codes.