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I'm having some work done in the UK (Eurocode) which involves a steel column to support a beam at 1st floor (US: 2nd floor?) level. The column bears a very-close-to-vertical total load of 47.3kN (including any relevant safety factors I guess, as that's the final figure in his calcs).

The engineer has specified a 152x152 23kg UC to be used for the column, with suitable end+base plates and fixings.

My issue is that the engineer has offered a low rate as its for a "charity job" (house belongs to an elderly frail relative). I've gone back to him with a set of 3 queries and got replies that are professional but curt and I don't want to go back again to ask if anything else is possible, unless I'm already privately very sure it is and can back it with calcs, as I don't want to lose his goodwill.

The only time he's said yes to any change, was when I also attached calcs to justify why I thought it should be OK. Unhelpful and not my job, I know!

In this case, I really need a slimmer column, to run flush with the 100mm partition wall. Unfortunately I don't understand column slenderness calculations well enough to check if there's scope to replace the 3m 152x152 23kg UC by, say, a 100x100x10 25kg RHS, or even a 100x180x5 21kg RHS. My tentative figures suggest that because of the RHS vs UC profile, both of these would be stiffer in any direction, than the UC and they'd fit nicely flush to the wall, so maybe they'd work.

I don't need an absolute calculation ("Does this profile work for that length and load"), but only a relative one ("Would this profile be at least as satisfactory as the one he's already said is good enough")?

What calculations are needed, to confirm if a more slender (in one dimension at least!) RHS profile would be at least as satisfactory/rigid in performance as a column, compared to the currently-proposed UC?

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    $\begingroup$ So, to be short, the engineer gave you an answer based on the original question at a special rate. Now you want to keep submitting changes and expect the engineer to work for free. If you had paid the correct rate, then the engineer may well be prepared to do some re-work but not ad-infinitum... Note that loaded beams are safety critical so investing in proper professional solutions is a good idea. $\endgroup$
    – Solar Mike
    May 3, 2019 at 9:25
  • $\begingroup$ Bloody hell, you don't make charitable assumptions do you? The hourly rate was discounted, not an "all in" flat fee. He said he'd think how to do it and if anything in his design didn't work in a practical sense to discuss it (but not really discussed). My email with queries specifically said that I appreciate it will add cost but will also be worth it. Now let's get back to the actual question. $\endgroup$
    – Stilez
    May 3, 2019 at 16:34
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    $\begingroup$ done favours on prices for people before and then had umpteen requests that follow - when it came to settling up, they had short memories.... $\endgroup$
    – Solar Mike
    May 3, 2019 at 16:38
  • $\begingroup$ Well, that's not this situation, and your bad experiences aren't being replicated here, so the aggressive assumption of automatic bad conduct was unfounded. There's an agreed hourly rate, and any requests that take extra hours come with an extra hourly rate, and with no haggling or niggling. The rate is helpful and agreeable. Let's move on to the actual question and close this distraction. $\endgroup$
    – Stilez
    May 3, 2019 at 18:25
  • $\begingroup$ So one has to ask why that engineer is not keen to do extra work... $\endgroup$
    – Solar Mike
    May 3, 2019 at 18:27

1 Answer 1

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The root problem of your question is that you're asking us to make some engineering judgement calls and calculations for a building where we don't know all the relevant details. You could however calculate the slenderness of the existing column and then propose a section that has this slenderness or less.

The slenderness parameter that is commonly used here in the US is $ \sqrt{\frac{KL}{r}} $ where $r$ is the radius of gyration $L$ is the column length, and $K$ is the effective length factor you can assume for comparison that this is $K=1$. Be sure and use consistent length units.

You can find tabulated section properties that should give you the $r_{x}$ and the $r_{y}$ value (radius of gyration) for different shapes (including the one originally specified). In general you will want to take the smaller of the $r$ values as that direction buckling will control.

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  • $\begingroup$ I did something similar to that - took the smallest moment area of inertia in both directions (stiffness of this kind seemed a sensible proxy for other forms of column bending I didn't know how to calculate), that's how I got to 100mm RHS as an option. The above is roughly the kind of answer I'm hoping for. But there seem to be multiple measures for slenderness, depending on which mode of bending would apply, and other factors, and its about there that I get lost. But yes, this is roughly what I'm after. $\endgroup$
    – Stilez
    May 3, 2019 at 21:37
  • $\begingroup$ Yes there are different buckling types. But given the same material, end connections, and height I would be pretty comfortable proposing a column with a smaller slenderness ratio. Obviously if you want to do all the calcs for him you need to get into more detail with the code. $\endgroup$
    – ShadowMan
    May 3, 2019 at 23:16
  • $\begingroup$ Yes, assumptions are good. Same material, height, end connections etc. Just different profile. So it should (probably) all come down to RoG being the same or better then? That's easy enough to check, or even calculate from 1st principles even, if the tables aren't easily found. For slenderness, I want * same or larger * value for minimum RoG as it's present as a reciprocal, right? (Smaller RoG = more prone to buckling) $\endgroup$
    – Stilez
    May 4, 2019 at 4:12
  • $\begingroup$ For buckling controlled column least radius means lower force. For you you want higher radii $\endgroup$
    – ShadowMan
    May 5, 2019 at 14:47

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