From what I understand most types of steel are strong both in tension and compression, yet they are still usually significantly stronger in tension. Suppose a common I-beam supported on both ends and with a load on top of it; the bottom flange experiences tension, while the top one compression, yet since steel is somewhat stronger in tension, why would the bottom flange be as thick as the top one? Is it not done because it's easier to manufacture symmetrical ones or is there some other reason?

Is it possible to use assymetrical I-beam to save weight and steel in that case or is my thinking completely off?

  • 2
    $\begingroup$ My initial thought is that in a real world situation it would be more likely for your modified beam to be installed upside down rather than a symmetrical beam failing from overload. $\endgroup$
    – Drew_J
    Feb 6, 2018 at 23:37
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    $\begingroup$ where did you get the idea steel has significantly different strength in tension vs compression? $\endgroup$
    – agentp
    Feb 7, 2018 at 5:02
  • $\begingroup$ @agentp: He's probably talking not about the material, but steel beams in general: since they have quite slender cross-sections, their effective compressive strength is reduced due to buckling (be that due to simple compression or bending). $\endgroup$
    – Wasabi
    Feb 7, 2018 at 17:55
  • $\begingroup$ @agentp From what I gathered a typical mild steel has UTS of about 400 MPa and compressive strength of about 250 MPa $\endgroup$
    – Ardath
    Feb 7, 2018 at 19:29
  • $\begingroup$ The balance of forces with respect to compression and tension is important because it will affect the behaviour of the element. For example, concrete has negligible tensile capacity. Steel is added to increase the tensile capacity and introduce ductility effects. If I am not mistaken, the codes contain factors to balance the compression and tension so that the element will fail due to tension rather than crushing of the compression face. This is to avoid the risk of a sudden catastrophic, brittle failure. $\endgroup$
    – AsymLabs
    Feb 8, 2018 at 9:45

1 Answer 1


Academically, I believe you are correct in that you could potentially optimize an I-beam to have a higher failure-load to mass ratio. You would have to play with the area moment of inertia equations to check for sure. T-beams would be the extreme of what you are proposing. They are less common, but composite steel/concrete T-beams are used extensively in parking garages, with additional steel in the lower tensile section.

It is important to note that many structures are design-limited by maximum deflection rather than structural failure. Additional area on the tensile member will reduce this defection. It would be interesting to see how deflections would compare between a symmetrical and the hypothetical failure-load optimized beam.

Symmetrical is superior for torsional loads or compound loads where the flange loading may be variable. Symmetrical is also superior for columns because it reduces eccentricity and the potential for buckling for a given mass/length.

Being able to use a beam in multiple situations is important because structural steel is a commodity. The price of a particular style of I-Beam is inversely proportional to how much it is used in industry in your region. This is often a bigger factor than the cost of the additional steel in the beam or cost of the additional mass in your design.

Additionally if your structure requires lots of trapezoidal shapes, you want to use symmetrical material so you waste less material. Right-side-up > Upside-down > Right-side-up > Upside-down...

Also, for every symmetrical beam, it saves the engineers an orientation specification in the drawings, and saves the fabricators a potential error.

Here is a good I-beam thesis to research further.


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