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While learning about I-beams I became curious to know if there is any relation between the flange length and web thickness of I-beams or if its arbitrary.

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    $\begingroup$ A question I've also had for a while. Theoretically, you want the web to be as thin as possible so that the section can be as tall as possible with as much mass near the top and bottom fibers. But if it's too thin you get local buckling, and I've never seen an authoritative source describing a method to get the optimal shape (or a demonstration that there's no such thing). Nor do I know if there are any relevant codes (in any country) that constrain the problem. $\endgroup$ – Wasabi Mar 19 at 12:44
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    $\begingroup$ "optimum I-beam" isn't a good specification. Are we talking strength vs weight? Minimum height? Maybe we're using it to hold a trolley. We could design for these things. But I've never heard of anyone designing a bespoke I-beam, although it certainly could be done. You just buy the one closest to your specs. $\endgroup$ – Tiger Guy Mar 19 at 16:52
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The answer is "Yes", and the direct source to find such relationship is from the technical publishing, that offer steel design tables/charts showing the most optimum beam/column sections for certain load with respect to the length of the beam column.

The image below is an example of a chart comparing moment capacity (indicated on the left axis) of beam sections with respect to the unbraced length indicated on the bottom. While there are many sections will satisfy either constrain, there is only one that will satisfies both constrains with the least "weight", which often is the focus of optimization.

You can also make your own calculation to find the most optimum cross section for the constrains you have in mind. For instance, You want a section can resist a set of loads, moment and shear, yet having the least weight. In realizing the stress formulas, $f_b = My/I$, and $f_v = V/A_w$, and the facts that $I$ (moment of inertia), and $A_w$ (effective area of the web for shear) must be kept constant to yield the constant stresses due to the given set of loads, now is the simple mater by writing the equations for $I$, $A$ and $A_w$ ($A$ is area of the entire cross section), and plug in varies combination of $b_f$, $t_f$, $d$, $t_w$ that will yield the constant $I$ and $A_w$, yet the $W = r \cdot A$ is the minimum.

Hope this helps.

enter image description here

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  • $\begingroup$ Appreciate the edit. Note that "r" is the weight density of steel. $\endgroup$ – r13 Mar 19 at 19:35

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