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.

 

[Denial]

No.[ ] You weren't roped in on this one[sup]*[/sup].[ ] I used the expression "partial proof" quite deliberately, aware of the small gap in what I put forward, and hoping to get away with it.[ ] Naïve of me, well spotted by you.[ ] It is an assumption that underpins beam-bending theory, one that 101% of the structural engineers this side of Alpha Centauri[ ] take for granted every day, but I cannot ever recall seeing it proved.

I do not have a ready proof available.[ ] I suspect that there will be at least two approaches to the proof, a simple one involving logic and a harder one involving some form of vector calculus. I'll try to put brain into gear and pencil to paper over the weekend.

[sup]*[/sup]Terminating my earlier proof one step before the end was intentional (as you had guessed).[ ] It had taken me a while to convince myself of the validity of that final step, and I wanted to "share the joy". BAretired has expressed it more clearly and concisely than I would have been able to do.

 

[BAretired]

I'm not sure what is to be proved. The elastic centroid is a single point in two dimensional space. A point can be measured in any co-ordinate system, but it's location remains the same irrespective of the co-ordinate system used.



BA

 

[KootK]

Without Denial's negative proof, we could say the same of the plastic centroid. Just because the point doesn't shift with a rotation of axes does't meant that the point actually is the centroid for that new set of axes. I'm sure that it is; I just don't know how to prove it.

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

 

[BAretired]

You cannot say the same about the plastic centroid because the plastic centroid is not always a single point. In those cases where it is a single point, it makes no difference which co-ordinate system is used, the plastic centroid remains the same.

The centroid of a cross section can be found using any co-ordinate system and, for any cross section is always a single invariant point.

BA

 

[KootK]

The centroid of a cross section can be found using any co-ordinate system and, for any cross section is always a single invariant point.

This is the supposition that we're trying to prove BA. We can't start with it as a given.

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

 

[BAretired]

Okay, you are going back to elementary theory where we consider the definition of center of gravity of an area.

Using cartesian co-ordinates, X = ∑(x.da)/∑da and Y = ∑(y.da)/∑da where X and Y are the position of the c.g. of the section, da is an elemental area and x, y are co-ordinates of each element, da.

All dimensions are taken from an arbitrary zero point. The result, X, Y represent a point in two dimensional space and can be represented by any system of co-ordinate axes.

But if you are saying that the fiber stress will always be zero at the center of gravity of a section subjected to bending about an arbitrary axis, I don't believe that is true.

BA

 

[BAretired]

On second thought, any applied moment can be resolved into two components, each parallel to each of the principle axes, so the fiber stress should be zero at the c.g. for any applied moment.

The c.g. is not necessarily within the section however, as is the case with an angle.

BA

 

[Denial]

Proof turns out to be quite straight forward, but devoid of penetrating geometrical insights.

Take an arbitrary shape and a set of X,Y Cartesian axes.[ ] Locate the axes' origin at the shape's centroid (the centroid as calculated relative to the X,Y axis system, to avoid any hint of circularity in the argument).[ ] Create a second set of axes U,V sharing the same origin but rotated by an angle t (for theta).

The rules for transformation between these two axis systems is
u = c.x - s.y[ ][ ] and [ ][ ]v = s.x + c.y
where[ ][ ] c = cos(t)[ ][ ] and [ ][ ]s = sin(t)

In what follows, I will use S[sub]A[/sub] to represent the operation of integration over the area of the shape.

By the definition of the centroid we know that
S[sub]A[ ][/sub]y.dA = 0[ ][ ] and [ ][ ]S[sub]A[ ][/sub]x.dA = 0
The u coordinate of the (perhaps different) centroid in the U,V coordinate system is defined as
S[sub]A[ ][/sub]u.dA
which, upon transformation to the X,Y system becomes
S[sub]A[ ][/sub](c.x-s.y).dA
= c.S[sub]A[ ][/sub]x.dA - s.S[sub]A[ ][/sub]y.dA
= 0 - 0

TCTP
(This Completes The Proof, to use an English FLA for a change.)

 

[KootK]

Well done Denial! It took me until just now to fully buy in but I'm now in full agreement. I think that the penetrating geometrical insight is BA's observation above. Not exactly earth shattering but, then, we're playing around in remedial mechanics of material 101 here. Thanks for the help guys.

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

 

[KootK]

I'd still like to know if there's a general formula for the location of the plastic centroid. I haven't seen one so here's my guess:

Sa (y-y_cp)/|y-y_cp| dA = 0

Any thoughts? Anybody know how to integrate and solve for y_cp?

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

 

[Denial]

While the formula you give is correct, I'm not sure it is a huge amount of help.[ ] The problem is probably going to require a strikingly different approach.

Maybe some inspiration can be found in statistics, where there is a (sort of) one-dimensional analogy.[ ] Consider the difference between finding the mean of a set of numbers (their "elastic centroid") and finding their median (their "plastic centroid").

 

[BAretired]

How can there be a general formula for the location of the plastic centroid if it is not always in the same place?

BA

 

[KootK]

It would be a general formula for the distance from a given axis to the location of a parallel plastic neutral axis. On an axis by axis basis, it would be similar to the elastic centroid.

My suspicion is that I haven't seen a closed form solution because there isn't one. But I'd love fire someone to prove me wrong which is why I posted my formula. A seek and find solution like median wouldn't let me implement a surveying formula style spreadsheet solutions a I had hoped.

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