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Plastic Section Modulus for an asymmetric cross section?q 1

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BoukenKage

Mechanical
Apr 13, 2009
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Situation at hand:

I've got a body that is experiencing enough forces that it has been bent enough that it has a permanent set. I've worked out the elastic side of the analysis and have been trying to discern elements of plastic design to figure out what has created the set that is seen.

This body is designed using plates that end up creating an asymmetrical, atypical cross section. The closest thing that I can classify it as would be is an offset T (i.e. the web is not centered).

Issue:

Elsewhere on the forum (while searching) I found a post StructuralEIT that gives an overview of the plastic cross section modulus, Z. Using AutoCAD, I was able to work out what the necessary separation for equal areas above and below the plastic neutral axis (PNA). This in turn let me calculate a value for Z=A*c1+A*c2; where A = half of the total area, c1 = distance from PNA to centroid of top area, and c2 = distance from PNA to centroid of bottom area.

1) Does method apply to any cross section or is it supposed to be for symmetric cross sections? I've been looking at some resources and haven't been able to come to a conclusion.

2) I found something disturbing while checking the value I calculated for Z. I was checking the shape factors using f = Z/S and ended up with f<1 for the top and f>1 for the bottom. I can understand the f values being asymmetric but the f<1 does not seem right.

Unfortunately, I have not been able to find any examples of asymmetric cross sections in order to check this.

Thoughts?

~Cody
 
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Finding the plastic section modulus will only provide you with that particular limit state.

You must additionally determine other limit states, including local buckling limit states (your shape cannot achieve plastic moment capacity if the elements are non-compact) and LTB (For the shape you mentioned, I assume LTB will control)
 
My best guess is that you have incorrectly calculated Z and/or the neutral axes. Plastic Neutral Axis is not always in the same location is Elastic Neutral Axis; for an assymetrical section, it probably won't be. The PNA is where there is equal area cross section above and below. ENA = Sum(A*y-bar)/Sum(A). Then Z= Sum[Area*(dist from PNA to centroid of area portion)] *Don't cross the PNA when breaking into components! S= I/y-bar.
 
Thanks for the response. Unfortunately, I'm not too familiar with plastic design as of yet. As I've found out while graduating two years ago, there are a number of topics that I never got into at school. So I'm not sure what you mean by LTB.

~Cody

"We don't know what we're going to do when we get there, but we aren't going to die without good reason." - Heero Yuy, Gundam Wing
 
LTB = lateral torsional buckling. frv was just trying to remind you that there are a bunch of different limit states which could have failed the structure, and all need to be checked (don't assume it was a global flexural failure)
 
@prex:

I refer to two f's (top and bottom), due to my cross section not being symmetrical. From what I understand as I mentioned before the shape factor, f, is found by f=Z/S. Since I have a cross section that is not symmetrical, I have two seperate S values (one for above the ENA and one for below the ENA). I assumed this would mean I have two seperate shape factors as well (two S values and 1 Z value). If this is incorrect than how is the shape factor found for an asymmetric cross section?

@stellion:

I would assume my ENA is fine as it was found in AutoCAD using region/massprop/UCS commands. My PNA is not at the same location as the ENA. As I mentioned, I found my PNA by finding the total area and then finding what seperation fo the regions would yield half of the total area on either side.

Also, it might help for me point out that this offset 'T' configuration is assumed as a cantilever setup. The plates that create the T are tapered as well. The tapers don't make them triangles, but I suppose you could view them as one half of a paper airplane.

~Cody

"We don't know what we're going to do when we get there, but we aren't going to die without good reason." - Heero Yuy, Gundam Wing
 
If the web is not centered, the plastic moment is not relevant. The Canadian Standard, CSA S16-01 requires that Class 1 sections be doubly symmetric, Class 2 sections be symmetric in the plane of loading, so you have, at best a Class 3 section. A channel is a Class 3 section and the steel handbook lists Section Modulus but not Plastic Modulus for channels.

For a laterally supported Class 3 section, the factored resisting moment Mr = [&phi;]SF[sub]y[/sub] where [&phi;] is the resistance factor = 0.90, S is the section modulus and F[sub]y[/sub] is the yield stress of the material.

For a laterally unsupported Class 3 section, it becomes a bit more complicated.

BA
 
Where is load applied with respect to the section's shear center? If it is a horizontally asymmetric cross-section, then it is most likely that torsion is being induced.
 
@abusementpark:

Hopefully this will answer your question about the load application. As I mentioned the basic setup is a cantilever beam with a tapered bottom edge up towards the free end. The loading is applied some distance from the free end and normal to the tapered edge, such that the normal force is acting up and to the left (if the supported end is at the left side). Also, I'd like to note there is a friction force acting on the tapered edge as the loading is applied from another part moving against the tapered edge. However, when analyzing the free body these forces were broken down into their components.

As I'm sure you have realized this isn't structure per se like a building, but this part of the forum had the most about plastic design.

Thanks for your help everyone. I think I can finish with this analysis now.

~Cody

"We don't know what we're going to do when we get there, but we aren't going to die without good reason." - Heero Yuy, Gundam Wing
 
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