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Sectional Properties 1

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DBCox

Automotive
Apr 9, 2003
58
Hello everyone,

I have an interesting problem (I think so anyway). I am trying to determine the section modulus of a part that I have and am having some trouble. The cross-sectional shape of the part is basically a rectangular tube with a vertical dividing wall in the middle (it is actually an odd external shape with a vertical divider, but this best describes it for the purposes of this question). So, when looking at its cross section, it looks like a rectagle with 2 square through holes. I need the moment of interia so that I can calculate the section modulus...

My plan of action:
1. Due to the odd shape, draw it in autocad
2. Create profiles of each feature (perimeter and 2 holes)
3. Create regions for each profile (outer profile, and 2 inner profiles)
4. Use the massprop command to determine the Ix and Iy of each profile/region
5. Use parallel axis theorem to determine the Ix and Iy of the part.

So heres the problem. When I figure out the Ix and Iy of each of the "holes", their sum is larger than the Ix and Iy of the outer profile. My plan was to subtract these Is of the inner regions from the outer regional to get an overall Ix and Iy. This won't work if the sum of the inner regions is larger than that of the outer profile...

My ultimate goal is to determine the section modulus in the X and Y directions of this part. How can I do it?

Thanks!
 
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In Autocad, make the whole thing one continuous closed polyline. If the segments are not connected in real life, connect them with negligibly small line segments. Then use Autocad region/massprop as you were going to.
 
IFR,

Thanks for the info, but thats the problem. I cannot figure out how to make it all one continuous profile. However, what I did do was make the inner cutouts one region and left a disconnect at the bottom of the divider (to allow a single continous profile). I then found the Ix and Iy of both the inner and outer regions, and subtracted the inner from the outer using the parallel axis theorem. Is this method accurate?

Thanks!
 
Hi blakrapter

ould you give the dimensions of the shape I can calculate it by hand and we can compare.
I don't understand why your using the parallel axis theorem
to get your final values why don't you just put the x-y
axis in autocad at the point you want your moment of inertia
about.

regards

desertfox
 
There must be a mistake in your calculations: the moment of inertia of a hole cut into a plane shape cannot be higher than that of the surrounding shape (just the same holds for areas, as is quite intuitive to understand).

prex
: Online tools for structural design
: Magnetic brakes for fun rides
: Air bearing pads
 
Prex,

I agree. I am either implimenting the PA Theorem wrong, or there is an error somewhere else. The moments of inertia were not larger, however, the theorem states:

Ie=Ia+Ad^2

I don't remember the subscripts of the I's but I used Ie for the effective Inertia (different since the centriods do not overlap), and Ia as the actual inertia ignoring the different centroids. The Ad^2 was large enough to make the Ie larger than the I of the outer profile.

I think I am implimenting it wrong, or something else is screwy b/c I agree it seems very wrong.

The odd shape is a cross section of a rocker panel on a vehicle. I am trying to figure out the stiffness. I tried to upload it to my website, but for some reason it isn't working.

Thanks!
 
blakrapter: I think your approach sounds good. If I understand correctly the way you are doing it, then working this out using parallel axis theorem in reverse, and simplifying, gives

Ix = Ix1 - Ix2 - {[(A1*y1 - A2*y2)^2]/(A1-A2)},

where Ix = centroidal moment of inertia of composite cross section, Ix1 = moment of inertia of outer shape output by program, Ix2 = moment of inertia of inner shape (the holes) output by program, A1 = outer shape cross-sectional area output by program, A2 = inner shape cross-sectional area output by program, y1 = y-coordinate of outer shape centroid output by program, y2 = y-coordinate of inner shape centroid output by program. Similarly,

Iy = Iy1 - Iy2 - {[(A1*x1 - A2*x2)^2]/(A1-A2)}.
 
Vonlueke,

Thanks for the info. I believe I may be using the parallel axis theorem incorrectly, at least in my first attempt. In my first attempt, I had three bodies; the outer or main body, and two individual bodies for the two holes. After I simplified it to a single hole, the theory seemed to work better.

Please explain how you derrived the equation you reference above, I am not quite sure where the multiple A's come from. Is the "y" the distance to the true centroid of the composite cross section? That may be difficult to find given the multiple shapes. How could it be modified for 3 bodies as in my first case?

On another note, I did figure out how to make everything one profile so that I can use AutoCADs massprops and forget about the parallel axis theorem like IFRs suggested. I ended up with very similar numbers as when I used the parallel axis theorem with the single hole, so I think it is probably accurate. I created the single profile by using 2 0.005" disconnects, which I doubt will make a huge difference. It would in real life of course, but mathematically it should be sound.

Thanks!
 
You can almost always make a single continuous polyline, just connect the separate areas with line segments that are very close together 0.00001 or so so they connect the areas but do not contribute to area or inertia.
 
blakrapter: A2 is the cross-sectional area of the second region described in your second post. The Ix derivation is, summation of moments of inertia of the two shapes/regions (outer and inner) described in your second post (about an arbitrarily-located x axis), minus summation(Ai)*d^2, where d = summation(Ai*yi)/summation(Ai). This can be extended to three shapes (one outer shape and two holes), giving

Ix = Ix1 - Ix2 - Ix3 - {[(A1*y1 - A2*y2 - A3*y3)^2]/(A1-A2-A3)}.

The location of the centroid of the composite cross section is parameter d in the above paragraph, where the Ai's and yi's are described in my previous post.
 
blakrapter,

It sounds to me like you are doing some of this manually. While certainly can add and subtract these areas by hand, let me go over how I use acad to calc properties for complicated shapes:

Draw the perimeter with the two holes.

Go to the draw regions and select all three profiles. Acad should indicate that 3 regions were found. If not you may need to make these into polylines.

Once you have 3 regions, the main shape and the two holes, go to Modify --> Region --> Subtract

Select the outline and then subtract the two holes.

I always cross-section the shape to make sure I got the holes out, sometimes it doesn't work depending on how the select is made.

Then go to Tools --> Inguiry --> Region/Mass properties

Make sure that you are looking moments of inertia about the centroid and not drawing 0,0 point.

Sometimes I move the geometry from the cg to the 0,0 point just to be sure.

And this should do it..!

I hope this is what you were trying to do.

Regards,
-Mike
 
vonlueke,
just to make things right, I can't understand the derivation of those formulae.
To me it should go this way:
Ix=Ix1-Ix2+A1y12-A2y22
where y1 is the distance between the centroid of region 1 and the centroid of the composite region, and similarly for region 2.
This might be equivalent to your formulation, just don't understand the meaning of y in your definitions for y1 and y2.

prex
: Online tools for structural design
: Magnetic brakes for fun rides
: Air bearing pads
 
Go here:


and download SectProp. It's a nicely written Excel spreadsheet that will give you properties of any shape you can come up with. Follow the directions - you must draw the outline in one direction and holes in the other.
 
prex: If you run two separate regions in an analysis program, as described in blakrapter's second post, the program has no knowledge of the composite cross section centroid location. Your formula requires this composite centroid location, whereas mine doesn't. My y1 is the distance between the x axis and the region 1 centroid; i.e., my y1 is the y-coordinate of the outer shape centroid. Also, Ix1 in your formula is the centroidal moment of inertia of region 1, whereas my Ix1 is region 1 moment of inertia about the x axis.

To clarify in my first post, Ix1 = moment of inertia (about the x axis) of outer shape, output by program, Ix2 = moment of inertia (about the x axis) of inner shape (the holes), output by program.
 
what's the big deal in the centroid location (which you'll need for bending stresses in any case.

my standard section properties s/sheet is ...
b d A=b*d y Ay Ay^2 Io
b is the breadth of part of the section (say a flange width, or a web thickness)
d is the depth of part of the section (a flange thickness, a web width)
y is the part's centroid location (about a convenient datum)

then ybar = sum(Ay)/sum(A) ... the centroid of the composite section.
and I = sum(ay^2+Io) - sum(A)*(ybar)^2
as someone posted above, the parllel axes theorem in reverse.
 
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