# Simple steel column buckling calc - where am I going wrong? (Eurocode)

I'm trying to follow an example of this, but I get a wildly different answer than expected and I can't see where I'm going wrong.

The example I've got, is for a 2.8 m steel column (RHS 150 x 100 x 5mm) fixed both ends with a simple static vertical central load. Couldn't be easier for a beginner. My calculation is:

Relevant properties (section properties taken from [here]) normalising length to metres:

• Steel section: RHS 150 x 100 x 5mm. (unspecified hot or cold)
• L: Free length = 2.8 m
• E: 205 x 109 = 2.05 x 1011
• I (table): Least 2nd moment of area = 3.923 x 106 mm4 = 3.923 x 10-6 m4
• A (table): Area = 2373 mm2 = 2.373 x 10-3 m2
• r (table): Least radius of gyration (as double check): 40.7mm = 0.0407 m
• K: Effective length factor = 1 (fixed-fixed)
• s: Slenderness = L/r = 2.8/0.0407 = 68.8 (intermediate)

EUler buckling force, calculated using I,K,L:

• π2 x (E.I) / (KL)2
= (3.1422 x (2.05 x 1011) x (3.923 x 10-6) / (1 x 2.8)2
= 1.012 MN

Euler buckling force, calculated using A,s (as double check):

• π2 x (E.A) / (s2)
= π2 x (2.05 x 1011) x (2.373 x 10-3) / (68.8)2
= 1.014 MN

So the two calculations agree within the limits of the data table figures.

The problem is, the correct answer is apparently Pc (compressive strength) = 124 N/mm2 = 294 kN.

The answer does confirm that slenderness is 68.8, and that the least radius of gyration is 40.7 mm, so I know I'm on the right track.

But there's no actual full calculation, and the error isn't an obvious factor that might suggest the problem, so I don't actually understand where Ive gone wrong.

(The actual question is part of a longer worked example, so if I get stuck anywhere else I might need to update this. But for now, that's the point I'm stuck at... )

• Does Eurocode include a safety factor? A factor of about 3.5 would not be particularly conservative. Some people might prefer using 10 instead! – alephzero Nov 16 '19 at 0:41
• No, this calc doesn't involve further safety factors, at least as far as I can see. Its pretty basic, e.g. K=1. In any case it would almost certainly involve some specific factor from a table or calculated from the above information, which is all one's told (as opposed to say an engineers personal preference in a given real-world situation), and a factor of 3.3 ~ 3.4x just doesn't seem to match anything obvious. Otherwise a " correct" answer would become "anything under 427, pick a number you like", which might be fair in real situations, but isn't likely here with a very specific answer = 124 – Stilez Nov 16 '19 at 0:46
• You title says "Eurocode" not "Euler buckling." I'm a mech engineer not civil, but a quick google search brought up this eurocodes.jrc.ec.europa.eu/doc/2014_07_WS_Steel/presentations/… which implies that the Eurocode does specify safety factors to be applied to the standard Euler formulas. See the end of Example 1. – alephzero Nov 16 '19 at 2:30

• Ohh! There's no mention of factors in the question or explanatory text. But the answer does definitely say Pc, and the value is then compared to the design stress (static load/area). The only other info is that it's an exercise from the UK (where I am), so it'll be UK specific data if anything, and also described as "the column is considered fixed top and bottom", and "is an end support for a fixed 200mm wide beam having a UDL with a reaction of 41.5 kN at each end". Combined with the OP info, is that enough information to know which factors to include and come to the correct answer? – Stilez Nov 16 '19 at 10:17