The formula for the area moment of inertia for a rectangular cross-section is b * h^3 / 12. But it looks like when computing the value for angle-iron in this table:


... they have assumed that the piece is bent into a flat plate, so that both sides are aligned to resist the bending force. Of course, this is completely unrealistic. Only one face is typically resisting the bending force (the vertical face in most applications) and the other face simple keeps the vertical face from buckling. As I missing something here ?

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    $\begingroup$ Using this calculator (civilengineer.webinfolist.com/str/micalcl.php) I get ~115 in^4, which ignores the rounds. The value I get is reasonably close to the 98 in^4 listed for 8x8x11/8. I also don't see where they are assuming a flat plate, neither in writing nor by diagram. Am I missing something from their page? $\endgroup$ – wwarriner Feb 5 '16 at 6:26
  • $\begingroup$ Also, bending is resisted by the entire section, not just the vertical. The vertical is primarily responsible for increasing the area moment of inertia, but it's not as if the horizontal can be ignored. In fact, if the piece is properly oriented such that the horizontal is on the face under compression in bending, the horizontal will reduce significantly the risk of buckling, as you yourself mention. This is intimately linked to the fact that it also participates in resisting the bending moment. $\endgroup$ – Wasabi Feb 5 '16 at 9:40
  • $\begingroup$ I didn't explain well. Let's take 6x6x1 angle, for simplicity. Table I linked says 35.5 in^4. But I figure only one of the two faces is resisting a typical bending load (e.g. the one oriented vertically, if supporting a load). So it's not much stronger than a 6x1 piece of plate, which is 18 in^4. The horizontal face only gives you 0.5 in^4; of course it's very important since it resists buckling, which is important (depending on the application, doesn't matter if vertical plate sandwiched between wood like a flitch plate), but I can't see how it doubles that 18 in^4 figure to 36 in^4. $\endgroup$ – RustyShackleford Feb 5 '16 at 21:40
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    $\begingroup$ You can't infer the second moment of area for more complex sections from the formula for a rectangular beam. In the general formula the distance of any area element from the neutral axis must be taken into account as elements furthest from the neutral axis see the greatest stress ie the stiffness of an angle beam is not the sum of the stiffness of a horizontal and vertical strip considered separately $\endgroup$ – Chris Johns Feb 6 '16 at 12:14
  • $\begingroup$ And recall that the parallel axis theorem states that elements at Delta Y from the centroid contribute Delta Y SQUARED to the overall Moment Of Inertia. $\endgroup$ – Andyz Smith Mar 1 '17 at 20:32

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