According to BS5950, a beam section can be classified as plastic, semi-compact, compact or slender. For the same section area, a H-beam can take axial compression (without buckling) better than an I-beam, and as such, uses a different strut curve in the code:

enter image description here

Now, I understand that a H-beam has a wider flange compared to an I-beam, but at what point, precisely, does this transition from I- to H- occurs? For example, is a 400x300 (depth x width) beam considered a H- or an I-beam?


Extracted from BS5950 guide, the following table shows H-beams (also known as universal columns, some of which with depth greater than width. This is the reason why I don't believe the differentiation is so straight forward.

Section Property Table

  • $\begingroup$ I'm probably wrong, but I'd say the only difference between these two is orientation at which they should be used... $\endgroup$
    – Slovakov
    Commented Feb 21, 2015 at 11:36
  • $\begingroup$ @AndyT, please refer to Table 23 above to understand why this H- or I-beam distinction needs to be clear. Also, from the SCI guide to BS5950, it is quite clear that Universal Column refers to H-beam and Universal Beam refers to I-beam. $\endgroup$ Commented Feb 28, 2015 at 12:31

1 Answer 1


BS5950-1:2000 Clause 1.3.23 defines an H-section as having "an overall depth not greater than 1.2 times its overall width", and Clause 1.3.25 defines an I section as having "an overall depth greater than 1.2 times its overall width".

Note that at exactly a ratio of 1.2, it would be an H section not an I section.

  • $\begingroup$ Wow, I can't believe it is actually right at the front. Thank you! $\endgroup$ Commented Mar 3, 2015 at 15:12
  • 3
    $\begingroup$ lol. Don't worry, yours isn't the only question on here where the answer is "read the design code you quoted" - I had one too! $\endgroup$
    – AndyT
    Commented Mar 3, 2015 at 16:51
  • $\begingroup$ Is this a case of RTFS where S represents 'Standard'? :-) $\endgroup$
    – Paul Uszak
    Commented Apr 30, 2017 at 21:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.