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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][1]) 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... )

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  • $\begingroup$ Does Eurocode include a safety factor? A factor of about 3.5 would not be particularly conservative. Some people might prefer using 10 instead! $\endgroup$ – alephzero Nov 16 '19 at 0:41
  • $\begingroup$ 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 $\endgroup$ – Stilez Nov 16 '19 at 0:46
  • $\begingroup$ 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. $\endgroup$ – alephzero Nov 16 '19 at 2:30
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You've calculated the Euler buckling force, not the compressive strength. The part of the code you want is EN 1993-1-1 section 6.3.1, which specifies how to account for imperfections and safety factor in order to calculate the allowable load. The size of the imperfections depend on whether the profile is cold formed or hot finished and the relevant safety factor is defined in whatever national annex is applicable to your column.

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  • $\begingroup$ 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? $\endgroup$ – Stilez Nov 16 '19 at 10:17
  • $\begingroup$ @Stilez, I had a suspicion it was British, as Pc is the terminology used in the old British standards before they were replaced by eurocodes. Eurocodes use different terms and symbols. You'll find the safety factor in the British national annex to EN 1993-1-1 and everything else in section 6.3.1 of EN 1993-1-1. Except hot/cold, so you'll just have to pick one, but there only the two options. $\endgroup$ – ingenørd Nov 16 '19 at 12:00
  • $\begingroup$ I've used bcsatools.steel-sci.org/Compression which calculates the UK factors for hot and cold. But it gives - N(b,z,Rd) = 496 kN cold, and 617 kN hot, neither at all close to the correct answer of 294 kN? Anything else I could be missing? $\endgroup$ – Stilez Nov 17 '19 at 7:55
  • $\begingroup$ @Stilez, I just punched in some numbers and I can't reach a capacity of 124MPa either, if I assume your slenderness is correct. I noticed you haven't specified a yield stress but even assuming grade S235 (which normally isn't used for RHS-sections) doesn't help. You write that the column is connected to a beam and not a slab, so is there any chance the beam only supports the column in one direction, resulting in a completely different slenderness in the other direction? That's my best guess at the moment, but using K=2 and the larger second moment of area gives a capacity much closer to 294kN. $\endgroup$ – ingenørd Nov 17 '19 at 12:20
  • $\begingroup$ Its an exercise, so the details are sketchy as one would expect. The sketch shows the top joined by an end cap, the bottom has some kind of large bolted plate, hence as it says fixed both ends, and the text also suggests no horizontal freedom ("disregard any lateral freedom and assume the beam is rigidly held in a fixed position and orientation"). As its an RHS I'm guessing one uses whichever radius of gyration/2nd moment gives the least favourable result, though? $\endgroup$ – Stilez Nov 17 '19 at 13:08

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