Parallel Axis Theorem for Z ? 1

[271828]

I need to calculate the plastic section modulus for some moderately complicated shapes.

If I were computing the elastic section modulus (via the moment of inertia), I'd have a parallel axis theorem to help me.

Does something like a parallel axis theorem exist for plastic section modulus calcs? I've never seen something like this and I looked through my books already.

A reference to something like this would be greatly appreciated.
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[PMR06]

In calculating Z, there is no I/c, so no parallel axis theorem tricks. Z is all about setting area above neutral axis equal to area below neutral axis. Then it's area times distance between centroids of areas. It's fairly straight forward to set up on spreadsheet.

 

[NS4U]

Similar to what PMR06 said... if you assume the entire section yields a stress SigmaYield, you can set the area in tension = area in compression.

Based on that find, the moment due to the tension/compression couple (similar to elastic design) which is your Mp and you know Mp=Z*SigmaYield; Z=Mp/SigmaYield

I think programs like SAP2000 let you input custom section dimensions, then it calculates the I, Z, S etc... I could be mistaken though.

 

[271828]

Yeah, that's what I was afraid of, LOL.

Thanks guys.

 

[rb1957]

plastic bending analysis in our business means "Cozzone", a simplified approach to plastic stresses in bending.

 

[Lion06]

As PMR06 points out, once you understand exactly what Z is, it is actually a little easier to calc than S. There is no need to calc the elastic NA, and the PNA is located where there is an equal amount of AREA above and below the axis. From there, it is just (area above (or below - they will be equal) PNA)*(distance from PNA to centroid of area above PNA + distance from PNA to centroid of area below PNA).

 

[271828]

Thanks everybody.

BTW, I've calculated Z for years and understand exactly what Z and S are!

Save your keystrokes with regard to the basic stuff, LOL.

 

[rb1957]

you structures guys (no condescention implied, meant, or otherwise) are describing a simplified plastic analysis, where the entire section is working at yield (in tension and in compression), hence equal areas.

this is reasonable and conservative (which is a good thing). if you need to squeeze a little more out of something "Cozzone" gives you a methodology for allowing the remote fibers to be at a higher stress (exceed yield).

 

[271828]

rb1957:

It is an interesting idea and I'd like to research it. My analysis books don't have anything on it, so I'll have to google it or (gasp) use the library! Do you have a paper on the subject that I could download?

The yield plateau for mild steel is very long, so I don't know if we get enough enough rotation in the plastic hinge to get into strain hardening before eventual local buckling.

Our b/t limits for I-shapes are not stringent enough to actually allow squashing of the flanges. They're set to preclude local buckling up to some rotation. AISC calls it a "rotation capacity of 3" but I forgot how that's defined.

 

[rb1957]

check Bruhn "analysis of flight vehicle structures". the method is based on setting the remote fiber strain on one side of the beam, then determining the strain on the other side (plane sections remain plane), and simplifying the stress distribution (as a trapezoid).

I-beams don't gain very much from this approach (the whole idea is that the material near the neutral axis can absorb more load after the remote fibers yield); if there isn't much material there (as in an I beam) there isn't much benefit. i'd stick with your assumption of constant yield stress; much simpler and conservative.

 

[Lion06]

My understanding is that in order to account for strain hardening the section must yield almost uncontrollably. Not to mention that this methodology is not endorsed by AISC. According to AISC, in no circumstance can Mn be greater than Mp, which =Fy*Zx.
If you consider a simple (2) column with (1) beam framing between them framing system - assume shearwalls so no lateral loads present. Once the plastic hinge forms in the center of the beam, the framing has failed. You can not get additional capacity out of it. The amount of deformation necessary to get to that strain hardening will cause the beam to act like (2) cables just pulling the columns over.
It would certainly make for interesting reading, but I would not ever consider using it for normal building applications.

 

[Kwan]

Strain hardening??? Like material strain hardening???? I always understood plastic bending as yield extreme fiber and take advantage of the material near the neutral axis. Perhaps I have it wrong???

 

[271828]

Kwan, I don't think you have it wrong. rb1957 was referring to some other method that goes beyond just assuming that the stress is the yield stress throughout the section. I don't think it applies here because mild steel has a tremendous yield plateau and, like StructuralEIT typed, AISC pretty much holds us to Fy throughout the section and that's it.

 

[Lion06]

Kwan-
You are exactly right. Plastic bending is simply Zx*Fy. rb1957 is suggesting using a value greater than Fy at the extreme fibers. That only occurs with strain hardening - which requires a great deal of deformation.

 

[Denial]

Getting back to the original post...

This is not a problem I have ever given much thought to before, but what happens if the cross section is not symmetrical? You can no longer assume that the neutral axis will be parallel to the axis about which the applied moment is acting, so there is an infinite number of lines which divide the cross section into equal areas above and below that line.

 

[271828]

Denial, LOL, you're conjuring up nightmarish memories of Adv. Mech. of Materials.

Let's say monosymmetric and bent about one of the principal axes to keep it simple.

 

[UcfSE]

That is the case for every cross section, not just an unsymmetric section. We're accustomed to bending about a principal axis or geometric axis just because that's what we do most often.

 

[Boiler8019]

If you need a reference to show you how to work with the areas to find the PNA and the Plastic Moment, the Structrural Engineering reference manual 3rd ed has an example on page 4-4.

I wouldn't recommend buying this book though, I dont think it's all that good.

 

[swivel63]

i don't think there's a parallel axis theorem for plastic sections. i would presume that you just multiply Q from the CG of the section by 2.

 

[swivel63]

by the time we would get to strain hardening, i would think that we're WAAAYY past the point of servicability requirements.