What quadratic moment should I choose in Solidworks?

3D Design Volume Model
1

Hello, as part of an RDM calculation, I want to extract the quadratic momentum.

So I did "rate", "section property" and I selected the slice of my profile.

Knowing that my force applies perpendicular to the vertical of my profile as you see it here, could you confirm that the quadratic moment I should use is indeed the one in bold below:

Moment of inertia of the zone, at the center of gravity: ( millimeters ^ 4 )
    Lxx = 434110.20    Lxy = 0.00    Lxz = 0.00
    Lyx = 0.00    Lyy = 294172.10    Lyz = 0.00
    Lzx = 0.00     Lzy = 0.00    Lzz = 728282.30

Polar moment of inertia of the area, at center of gravity = 728282.30 millimeters ^ 4

Angle Between Main Axes and Workpiece Axes = 90.00 degrees

Main moments of inertia of the zone, at the center of gravity: ( millimeters ^ 4 )
    ix = 294172.10
    Iy = 434110.20


Moment of inertia of the area, at the output coordinate system: ( millimeters^4 )
    LXX = 1443818306.78    LXY = 0.00    LXZ = 0.00
    LYX = 0.00    LYY = 1443316725.15    LYZ = 22844250.44
    LZX = 0.00    LZY = 22844250.44    LZZ = 1089925.84

I must admit that I am a little lost with all these values.

Thank you for your help!


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2

Good evening

I will give you three pieces of information.

1°) The definition of moments of inertia and products of inertia

2°) Moments of inertia and principal moments of inertia 

3) the most important remark of our eminent colleague @Aliende who tells us this
This confusion between Quadratic Moment and Moment of Inertia comes mainly from the vocabulary chosen in Mechanics@alliende.  

Kind regards

2 Likes
3

Good evening

I take this opportunity to ask a question:

- why use section property instead of mass property because although it goes back a long time to my mecha classes we always worked on a volume and not a surface?

Otherwise I also take the opportunity to say that you can change the coordinate system in the 2 tools, knowing that by default it takes the origin of the part.

See you soon^^

4

[Tease ON Mode]
@Fuz3D: it must have been a long time ago and Alois is starting to have an effect ;-P
[Tease Mode OFF]

In very simplified terms:
- For a flexion calculation, it is the cross-section that is used (arrow=FL^3/(48EI) with I=(bh^3)12 for a rectangular cross-section, we find the mm^4 of the quadratic momentum);
- The volume is used for dynamic calculation.

5

I must be too stiff for bending, I prefer dynamics:) Thank you Stefbeno.

1 Like
6

Hello

Thank you for your answers. I had read @Alliende 's post before asking my question, which allowed me to look in the right place for the quadratic moment (in mm4).
But in fact my question is to know what value to use in this precise calculation. Is it the value I put in bold that I should use (Iy = 434110.20)? This seemed logical to me, since I have to choose the I in the direction perpendicular to my applied force (in this case my force is applied along the x-axis, so I choose Iy).

Thank you all

P.S.: I use my quadratic moment I (Iy here so if I'm not mistaken) in the arrow formulas (-(FL^3/48EI) and constraints (Mfmax / Wel where Wel = I/v).