Geometric Properties of Complex Shapes 1

[Engineering05]

Hi Everyone,

I am a graduate structural engineer who uses a wide array of software day-to-day in the design of structures and structural elements. Some designs require me to design members of non-standard shapes e.g. parallel flange channel with unequal flange lengths. In the aforementioned example, the principle axes become rotated relative to the global axes and hence when calculating geometric properties such as the plastic section modulus, locating the plastic centroid of said shape becomes a lot more involved. My question is twofold:

1. My only guess in attempting to solve for the plastic centroid for a non-symmetric shape about a single axis is to firstly determine the orientation of the principal axes and then incrementally guess where the plastic centroid may lie vertically and horizontally. Once areas above and below the rotated neutral axis (relative to the global axes) are equated we have a solution as to where the plastic centroid lies. Can anyone offer an insight as to whether or not what I just described is indeed a sound method or algorithm in attempting to calculate the plastic centroid of any non-symmetric shape?

2. Calculating additional properties such as the Torsion and Warping constants appear to be far more involved and, admittedly, go right over my head. Where there are no published solutions for torsion and warping constants for obscure shapes (Roark's) I turn to programs like Strand7, IES Shapebuilder or other general packages that calculate the geometric properties for me. I first and foremost acknowledge that I may be treading into the realm of computational programming but I thought I would ask those in the loop how these programs are able to do this. In the examples of software I list above they both create a mesh within the outline of the shape comprised of either small triangles or quadrilaterals. How does meshing within a region allow a computer to calculate a) simple properties like area, second moment of area, elastic and plastic section moduli and b) more complex properties like torsion and warping constants?

As always, I turn to these forums as I have not been able to find the answer to these questions on my own by either researching my online university database, google searches or other asking other, more experienced engineers. If anyone knows of thesis papers, textbooks or other sources that discuss these topics in either structural mechanics or computational programming feel free to include a link or reference that I can follow up on.

Thank you in advance.

 

[BAretired]

I have never used plastic centroids, but it is important to define what you mean by a specific property. The attached link defines plastic centroid in a way which does not appear to take into account a change in direction of the principle axes. Does that agree with your definition?

http://www.egr.msu.edu/~harichan/classes/ce405/chap2.pdf

BA

 

[KootK]

A1) Your method sounds reasonable so long as you actually know how to calculate the plastic principle axes. I don't. Another option might be to do the same thing but as a two step process, using the geometrical axes (or whichever orthogonal axes are convenient). Scan the y-axis to get the vertical location and then the x-axis to get the horizontal.

A2) It's often more than just a mesh; it's a true finite element analysis run. I stumbled into this when attempting to write a spreadsheet that would find the shear center of an arbitrary cross section, including muti-cellular sections. I even purchased an arcane book on torsion from Germany by a fellow named Kollbrunner in order to find my solution. It turns our that there is no closed form solution. FEM is the only way to tackle it. Basically, the FEM solution solves the differential equations that are at the heart of the Prandtl membrane analogy.

I'm not sure which section properties do and do not require FEM for their determination. FEM is probably over kill for things like area and centroid with any section. However, if you're attempting to create a general solver for arbitrary sections that may be open, composite, multi-cellular -- or all three -- things get complicated pretty fast. You start having to code in complex ray casting functions etc just so that your algorithm can "know" when an element is inside or outside of the cross section.

There's a fellow on this forum known by the handle IDS. He's a programming wiz and is interested in a lot of things like this. He's even got a website set up where you can download a bunch of his spreadsheets. I'd recommend searching this site for some of his threads. Last we spoke, he was working on a general solver for bi-axial bending + axial in prestressed cellular bridge decks. Yeah.

Here are some documents from my adventures that might interest you:

1) My Mathcad spreadsheet that was meant to be a general purpose section calculator (Link). It's based on surveying style formulas that calculate easily and take corner point coordinates for input. That part of it is pretty slick. Ironically, I stopped at working out the plastic section properties. I expected that to be easier than elastic properties but it wasn't at all.

2) The surveying style formulas for area and centroid: Link

3) The supervening style formulas for moment of inertia: Link




The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[IDS]

Coincidentally, I have just updated my section properties spreadsheet, which can be downloaded from:

http://newtonexcelbach.wordpress.com/2014/11/02/section-properties-update/

This won't do much to help with your queries though, as it deals only with elastic section properties.

As KootK mentioned above, I have also posted on finding the neutral axis for a reinforced (or prestressed) concrete section under biaxial bending:

http://newtonexcelbach.wordpress.com/2014/08/10/faster-biaxial-bending/

The calculation is for the ultimate bending strength of a concrete section, which uses a rectangular stress block for the concrete, and bi-linear for the steel. It should be fairly simple to modify the code to make it suitable for steel design. I'll have a look at it when I have time, but that probably won't be for some time. Feel free to ask questions if you want to have a go yourself. The main difference from steel design is that the concrete in tension is ignored, and the compression block does not extend to the NA.

Regarding torsion of irregular sections, I don't think there is any universal simple approach. I have also used Strand7 for this purpose, and I think it does a good job. I think the easiest way to automate the process would be to use Strand7 with the API (or any FEA program with equivalent facilities), and you could then set up data entry and output on a spreadsheet.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[BAretired]

Is plastic centroid unique to a particular shape? Where is the plastic centroid of an equal leg angle? Of an unequal leg angle? Of an unsymmetrical Z section?

The attached article shows calculations for the strength of an equal leg angle under biaxial bending, but the term "plastic centroid" was not used anywhere in the article.

http://www.academia.edu/7854300/Ful...ctions_under_biaxial_bending_and_normal_force

BA

 

[BAretired]

Maybe, for a steel shape, the plastic centroid is the same as the centroid, i.e. the c.g. of the area. That would certainly simplify calculations.

One thing for sure, if you can't define what you mean by the plastic centroid, you'll never be able to find where it is.

BA

 

[IDS]

BA - the definition of the plastic centroid in your first link looks correct, it's the point that has an equal area either side of the two perpendicular axes through the point. In general it won't be the same point as the elastic centroid. That seems to be the definition used in the OP as well.

I couldn't open your second link.

The procedure used in my RC spreadsheet (which should work for steel as well, with appropriate adjustment of material properties) is:

- Assume a Neutral Axis angle
- Find the neutral axis position such that the resultant force is equal to the applied axial load. For steel with zero axial load this could be reduced to finding the position such that the section area above and below the axis are equal.
- Take moments about the global axes (or whatever axes the applied moments are defined for)
- Adjust the neutral axis angle and repeat until the reaction moments are close enough to the applied moments.

So you don't actually need the plastic centroid position, just the Neutral Axis position and angle.


Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[Trenno]

Do you mean the plastic neutral axis, where the area in tension and compression is the same? Thus allowing you to calculate the plastic section modulus?

I can see how this would get complicated for a doubly-unsymmetric section...

 

[BAretired]

Doug, when you said you couldn't open my second link, I thought maybe I had made a mistake in spelling or something, but I tried it again and it works fine for me. Is anyone else having trouble opening my second link?

Doug, in your program for biaxial bending in a concrete column, the neutral axis location is not strictly a property of the shape. It is also a property of the loading. That is why I am wondering if the plastic centroid of a steel section is a unique property of the steel shape or is variable depending on the loading.

BA

 

[IDS]

BA - The 2nd download worked this time. I'll have a read and respond later.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[BAretired]

The attached article shows calculations for the strength of an equal leg angle under biaxial bending, but the term "plastic centroid" was not used anywhere in the article.

http://www.academia.edu/7854300/Full_plastic_capac...

I indicated earlier that the plastic centroid was not mentioned in the article. Actually, it is...in the Appendix (see A13 for the expression for an equal leg angle). So this suggests that the plastic centroid is a property of the section, irrespective of the loading, contrary to what I was suggesting in an earlier post.

BA

 

[IDS]

The plastic centroid (as defined in the paper) is the intersection of perpendicular axes dividing the section into four equal areas, so it is a property of the section. For a section subject to bending about a principal axis with zero axial load the neutral axis will be coincident with the plastic principal axis, but for any other loading it may not be, but this is the same as for elastic analysis.

I think a point worth making is that in general for bending about a non-principal axis the neutral axis will not be parallel to the axis of bending. As far as I know for a non-symmetrical section the only way to find the NA position and angle is by iterative calculation of the reaction force and moments until they are in equilibrium with the applied loads.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[KootK]

Maybe you know this IDS: if one rotated the axes of consideration arbitrarily, would the location of the plastic centroid always remain the same?

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[Denial]

KootK.[ ] I cannot prove it readily, but I very strongly suspect that the answer to your question is "no".[ ] With considerably less confidence, I suspect that for the location of the plastic centroid to be the same regardless of the orientation of the reference axes the section must be doubly symmetric.

 

[Denial]

Managed to get a few minutes to return to the question of the "axis independency" of the location of the plastic centroid.[ ] All I need to disprove the independency is one example, and there is a simple example readily to hand:[ ] an equilateral triangle.[ ] Take one side as the "base".[ ] The height of the plastic centroid above the base does not equal the height of the conventional centroid above the base.[ ] QED.

This result is not all that surprising, as the plastic centroid has a few other oddities.[ ] Start with a standard I-beam.[ ] Reduce the web thickness until it is vanishingly thin.[ ] Where is the plastic centroid now?[ ] It is undefined, as it can be anywhere along the line where the web used to be.

 

[KootK]

Thanks for helping me out with this Denial. However, your first paragraph doesn't prove that the location of the plastic centroid is axis independent. It only proves that the elastic and plastic centroids are not, in general, the same. Am I missing something?

Your point with the vanishing web is an interesting one. Perhaps one has to restrict the conversation to contiguous sections in order for it to have meaning.

This is an embarrassing question to have to ask but, is the elastic centroid axis independent? I've always assumed so but I really don't know how to prove or even rationalize that.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[BAretired]

KootK,

If h is the height of the equilateral triangle and b is the base, the plastic centroid occurs at h/√2 down from the apex when the axis considered is perpendicular to the base.

When we shift 60 degrees, the new location for the plastic centroid is h/√2 down from the apex but it's a different apex, so not the same point. In order for it to be the same point, it would have to be at the centroid of the area, namely 2h/3 down from the top.

BA

 

[KootK]

@BA: I get it now, thanks for the push.

@Denial: your example is brilliant. Sadly, you left me with one deductive leap to make and I failed.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[Denial]

KootK.[ ] Thanks for the compliment, but it took me an embarrassingly long time to come up with that simple example.[ ] I was exhausting my brain playing around with complex shapes, when I should have homed in much faster on the simple one.

You ask about axis-independence for the elastic centroid. The answer is "yes".[ ] By way of partial proof I offer the following.[ ] If you take any arbitrary section defined relative to any arbitrary set of Cartesian axes, you can calculate a centroid relative to those axes.[ ] If you then do what any structural engineer would do and transform things so that everything is relative to the principal axes, the centroid is in the same position.[ ] If that applies to one set of arbitrary starting axes it applies to all starting axes.[ ] QED again.

 

[KootK]

Precise language, deductive reasoning, Spartan level of detail, Latin TLA... You must have some training / interest in logic. CS major turned CE?

I also believe that the answer is yes. However, at the risk of being the dope who was roped twice this afternoon, I'm going to challenge your latest proof. You have stated that the centroid is in the same location for both sets of axes without any justification for why that is the case. Please revise and resubmit.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.