FIRST MOMENT OF INERTIA 4

[GESQUICOR]

I'M TRYING TO FIGURE OUT A WAY OF OBTAINING THE INERTIA AROUND X-X (Ixx, cm4) OF A COMPLEX STEEL PROFILE.

STARTING WITH THE FIRST MOMENT OF INERTIA (Kg cm2), AND KNOWING THE DENSITY OF THE MATERIAL, IS THERE A WAY OF REACHING Ixx??

THANKS.

GESQUICOR

 

[EnglishMuffin]

As far as I am aware, there is no such term as "First Moment of Inertia". The term "Moment of Inertia" is used either in its true sense (ie having the units M.L^2), or is used to refer to the "Second Moment of Area", having the units L^4. Second moment of area is also sometimes called "Area Moment of Inertia". There is also a term "First Moment of Area", which has the units L^2. There are certainly ways to find the Second Moment of Area of any given plane area, about a given axis. But when you say you want to start with what appears to be a true moment of inertia, one needs to ask which moment of inertia you are referring to. For any solid, there are three principle moments of inertia, and in the case of a beam none of them have any special relationship to the second moments of area of any plane section, since they would be heavily dependent on the length of the beam, while the area properties of the section would not. It is possible to define what is known as the moment of inertia of a plane lamina, which is an imaginary object, being infinitely thin yet having a finite mass per unit area, This is about the only object I can think of for which there would be a true correlation between "moment of inertia " and "second moment of area".

 

[EnglishMuffin]

Sorry - I goofed - first moment of area has units L^3.

 

[EnglishMuffin]

I also wrongly spelt "principal" as "principle".

 

[dbuzz]

With units of kg.cm^2 is sounds like GESQUICOR has the mass moment of inertia, which is calculated thus: mass x length^2.

This equation has a similar form to that for the second moment of area, approximately calculated thus: area x distance from centroid ^2 (m^4).

I'm not sure that you can derive one from the other though. Is the only difference density (kg/m^2)? Dividing kg.m^2 by kg/m^2 would give you the correct units, m^4, although I'm unsure if this is the correct solution technically.

It's not that hard to calculate the second moment of area for most shapes. What do you call a "complex" shape?

 

[GESQUICOR]

Thanks for the input on this matter.


Dbuzz, I do have the mass moment of inertia, thanks for pointing that out. I don't think I can get to the second moment of area with the info I have in hand.

We just finished the installation of a Finnish machine that produces roof trusses made from galvanized steel. The truss chord is the section that I reffer to as complex.

It's a rollformed section with over twenty sides and round turns. If you want to see details about it you can visit

http://www.rosettesystems.com

I've already requested this information from the machine producer in Finland, but I really appreciate your comments. I'm a new user and I think I found the best site for engineering research, wether you have a question or you want to browse through interesting topics.

Gesquicor

 

[corus]

Try this site

http://yakpol.net/

Download Section Properties for an Excel spreadsheet for determining your Ixx values. It will mean you having to input the co-ordinates of the points defining the outline of the shape, in a continuous anti-clockwise direction.

 

[TFL]

the easiest methid is to use autocad.

1. DRAW THE SHAPE AS A CLOSED POLYLINE( TWO IF IT IS HOLLOW)

2.FROM THE DRAW MENU SELECT REGION, AND SELECT THE SHAPE(s) TO TURN INTO A REGION.

3. IF THE SECTION IS HOLLOW YOU NEED TO SUBTRACT THE INNER REGION FROM THE OUTER. FROM THE MODIFY MENU SELECT BOOLEAN>>SUBTRACT. YOU CAN THEN SUBTRACT ONE REGION FROM ANOTHER.

4. FROM THE TOOLS MENU SELECT INQUIRY>>MASS PROPERTIES. AND SELECT THE REQION YOU WANT THE PROPERTIES OF. bE CAREFULL THE PROPERTIES ARE TAKEN ABOUT 0,0,0 IN AUTOCAD. I USUALLY MOVE THE OBJECT SO THE CENTROID IS LOCATED AT 0,0,0

TRY IT ON A SIMPLE SHAPE YOU CAN CHECK FIRST
GOOD LUCK

 

[GESQUICOR]

corus , thanks for pointing out that site, that spreadsheet will be very usefull to me!!

I'll have to sharpen up my CAD to follow tfl's advice, but I'll keep that in mind too.

Thanks again guys.

Gesquicor

 

[corus]

Gesquicor,
To obtain the stresses in the section you have drawn refer to the Steel Designer's Manual, IV, which gives a little formula for determing the stresses at any point given Mxx, Myy and all the calulcated values of Inertia. You can adapt the spreadsheet to show them for each of your points by inserting a column and referring to the cells that contain the values of Ixx, Iyy, and Ixy.

 

[ajose]

Hi there,

The method described by tfl is the one i normally use.
But in the description is missing somethig.
The output of the inquiry is the inertia arroud de drawing axes, so you should move the drawing to fit your needs.

ajose

 

[TFL]

ajose,

that is what i was trying to point out in the second half of step #4.

thanks for aleast confiring someone out there can use autocad for more then drafting!

tfl

 

[OzarkMTBr]

I use autocad exactly as TLF described it. Once you get the shape created and run the MASSPROP command, It will give you coords to the centroid. I move it by those numbers until the mass properties give the centroid as being 0,0,0. Then it will give you Ixx, Iyy, Radius of Gyration. Below is an example of the output for a circle shape with a diameter of 3":

---------------- REGIONS ----------------

Area: 7.0686 sq in
Perimeter: 9.4248 in
Bounding box: X: -1.5000 -- 1.5000 in
Y: -1.5000 -- 1.5000 in
Centroid: X: 0.0000 in
Y: 0.0000 in
Moments of inertia: X: 3.9761 sq in sq in
Y: 3.9761 sq in sq in
Product of inertia: XY: 0.0000 sq in sq in
Radii of gyration: X: 0.7500 in
Y: 0.7500 in
Principal moments (sq in sq in) and X-Y directions about centroid:
I: 3.9761 along [0.7071 0.7071]
J: 3.9761 along [-0.7071 0.7071]

 

[corus]

Qzark,
Am I reading the results correctly in that Autocad says that the principal axes are at 45 degrees from the global X axis?

 

[OzarkMTBr]

"You might want to be careful about using the principal moments. If the principal axes are not parallel to the X- and Y-axes, the moments about X and Y will be different from those about I and J. For many practical problems (L-shaped members for instance) the principal axes
are rotated."

I always use the 'Moments of Inertia' because they are based on the current X,Y axes.

I am not sure though as to why it gave the axes on the circle at a 45°. The Ix and Iy are the same either way.

Here is a 3" square results:

---------------- REGIONS ----------------

Area: 9.0000 sq in
Perimeter: 12.0000 in
Bounding box: X: -1.5000 -- 1.5000 in
Y: -1.5000 -- 1.5000 in
Centroid: X: 0.0000 in
Y: 0.0000 in
Moments of inertia: X: 6.7500 sq in sq in
Y: 6.7500 sq in sq in
Product of inertia: XY: 0.0000 sq in sq in
Radii of gyration: X: 0.8660 in
Y: 0.8660 in
Principal moments (sq in sq in) and X-Y directions about centroid:
I: 6.7500 along [1.0000 0.0000]
J: 6.7500 along [0.0000 1.0000]

It gives the Principal axes along the x and y.

 

[corus]

Thanks Qzark,
A circle must just give an odd result as I checked it with another program that gave the same answer.