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In a number of designs, a beam will have an underslung welded plate attached, as shown below:

enter image description here

In this sketch, a symmetrical standard steel beam/column of dimensions H1 x B1, with web and flange thickness w and t, has a wider plate welded under it of dimensions H2 x B2 (with B2 >> H2, ratio around 10:1 ~ 30:1).

For simplicity, I'm only considering stiffness in the vertical plane, not torsion/eccentric/lateral loads.

An example might be, if symmetrical static vertical point loads/UDLs were applied on the plate of such a beam, at both sides, at various points along the beam's length, and one wanted to evaluate the beam's bending moments or deflection along its length, or something like that. The usual methods for loads on a simple beam would be used, but you'd need to know its area moment of inertia first.

I'm assuming one doesn't just add the area M.o.I. (or 2nd moment of inertia - same thing) of the two together, as the value of the composite depends on the positions of the two.

What's the easiest way to calculate the value for this setup?

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    $\begingroup$ "I'm assuming one doesn't just add the area M.o.I. (or 2nd moment of inertia - same thing) of the two together, as the value of the composite depends on the positions of the two." - Correct. Learn about the parallel axis theorem, and how to calculate the position of the combined neutral axis. $\endgroup$ – alephzero Jun 6 '19 at 13:17
  • $\begingroup$ The easiest way is to use a CAD system where you can draw the beam cross section and calculate the properties. But if the OP was using such a system, there wouldn't be a question to ask. $\endgroup$ – alephzero Jun 6 '19 at 13:19
  • $\begingroup$ Thanks. No CAD system. So yes, I did mean the easiest way to calculate it formulaically in terms of the parameters, not just the easiest way to get a numerical value for some specific profile, as you clearly understood :) $\endgroup$ – Stilez Jun 6 '19 at 13:25
  • $\begingroup$ You want the Second Moment of Area and not the mass moment of inertia which is used in dynamics. Right? $\endgroup$ – John Alexiou Jul 2 at 6:13
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To calculate the second moment of area of a composite beam, we need to find its neutral axis and then use the parallel axis theorem to add individual sections moments of area. I am assuming you made a mistake and meant to say $B_2\gg H_2$, if we go by the scale of your diagram.

First the H beam. Because of symmetry, we subtract the I of the rectangular left between the 2 flanges and then add the I of the web.

$$ I_{Hbeam}= \dfrac{B_1H_1^3}{12} +\dfrac{(- B_1+W)(H_1-2T)^3}{12}$$

And the neutral axis again by symmetry is at $ H1/2$ and the area is

$$ A_{Hbeam}= B_1H_1+(-B_1+W)(H_1-2T) $$

Now we are set to find the neutral axis of the composite beam. setting the bottom of the red flange as reference axis we get.

$$Y_{neutral\ axis}=\frac{ \sum A_n \bar{Y}_n}{\Sigma A_n} $$

We denote the distance from each section's neutral axis to the reference line on the bottom of the red plate, $\bar{Y} $

And the composite I:

$$ I_{composite}=\sum (A_n d_y^2 + I_n) $$

where $d_y$ is the distance between the section's and composite section's neutral axes.

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  • $\begingroup$ Most of this makes sense to me. Can you reword or clarify "we subtract the I of the rectangular left between the 2 flanges and then add the I of the web", which is a bit too much compacted, and not yet making sense to me? $\endgroup$ – Stilez Jun 7 '19 at 0:15
  • $\begingroup$ @Stilez, in sections that are composed of partitionable geometry and are symmetrical like your H beam, we can decompose them into their subsections and calculate the I by just adding or subtracting the subsection parts' I. The key word here is these subsections should share the same neutral axis. $\endgroup$ – kamran Jun 7 '19 at 0:59
  • $\begingroup$ You don't need to calculate the properties of a standard I beam. Just look them up in the structural steel tables. That way you get the correct properties, not approximate ones assuming an ideal geometry. Real I beams don't have perfectly rectangular flanges, and they have fillets not perfectly sharp corners. $\endgroup$ – alephzero Jun 7 '19 at 9:17
  • $\begingroup$ @alephzero, I did the exact geometry which is asked by OP, not an approximate, simplified version. My goal is to invite them to look at the basic simplicity and elegance of the concept. $\endgroup$ – kamran Jun 7 '19 at 15:20

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