Find Centroids: Determine Ix'y' for Area - a=120mm, b=290mm, d=30mm, h=390mm

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Homework Help Overview

The original poster attempts to determine the product of inertia of a specified area with respect to the centroidal x′ and y′ axes, given dimensions for a beam's cross-section. The problem involves understanding the implications of symmetry and applying the parallel axis theorem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of the area in question and the implications of symmetry on the product of inertia. There are attempts to clarify how to calculate the centroid and the product of inertia using the parallel axis theorem. Questions arise about the meaning of certain variables and how to apply them in calculations.

Discussion Status

Some participants have provided guidance on the calculations involved, including the use of tables to organize area information and moments. There is an ongoing exploration of how to relate the calculated values to the product of inertia, with various interpretations being discussed.

Contextual Notes

Participants note that the area in question may not have an axis of symmetry, which complicates the calculation of the product of inertia. There is also mention of previous problems that may provide context for the current calculations, but specific details are not fully resolved.

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Homework Statement


Determine the product of inertia of this area with respect to the centroidal x′ and y′ axes.
Locate the centroid x bar of the beam's cross-sectional area. Set a = 120mm , b = 290mm , d = 30mm , h = 390mm .

Homework Equations


?

The Attempt at a Solution


So I can't find anything online that would help with finding the Ix'y' value except maybe Ixy = Ix'y' + ACxCy. But I don't have anything to help find Ixy, so I'm pretty much stuck in an inertia loop. Any explanation would be great as to how the answer is -1.21e8 mm^4!
 
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It's not clear what 'this area' refers to. Is there a picture which didn't get attached?
 
Ah, right. There is indeed!
 

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The product of inertia Ixy of any area which has an axis of symmetry is equal to zero. After that, apply the parallel axis theorem to determine Ixy about the centroid of the entire area.
 
So Ixy is 0 in this case because each piece is a rectangle (aka symmetrical)?

So it's just -ACxCy?

If so, how do I find Cx and Cy? Is that the center of each piece of the object in the x and y directions respectively?
 
Bluestribute said:
So Ixy is 0 in this case because each piece is a rectangle (aka symmetrical)?

No, I don't think you understand. The Ixy of each rectangle forming the cross section is zero about its own centroid.

The section in the figure does not have an axis of symmetry because the areas are not distributed evenly side to side or top to bottom.

So it's just -ACxCy?

Just like you find the centroid of the whole section by calculating moments about a reference axis, so you'll have to calculate Ixy about this same axis using the PAT. Once all moments of area and moments of inertia have been computed about this reference axis, then you use the PAT again to find the MOI about the centroidal axes.

If so, how do I find Cx and Cy? Is that the center of each piece of the object in the x and y directions respectively?

Yes. In these types of calculations, it's easy to set up a table of are information for all the individual pieces and make the calculations.
 
But the total Ixy is just all the Ixy's added together . . . so wouldn't that be 0?

And in the previous problems, I found X bar and Y bar of the shape, if that helps, since I'm lost as to the whole parallel axis thing. I know you can change your axis. Don't know why you do, don't know how you apply it to real life, and don't know how to appropriately use it to calculate Ix'y' unfortunately =\, but I know it exists.
 
Bluestribute said:
But the total Ixy is just all the Ixy's added together . . . so wouldn't that be 0?

Nope, you're ignoring the distance between the centroid of each rectangle and the centroid of the section as a whole. Because the centroids of each rectangle do not coincide with the centroid of the figure as a whole, this is why applying the PAT is necessary.

And in the previous problems, I found X bar and Y bar of the shape, if that helps, since I'm lost as to the whole parallel axis thing. I know you can change your axis. Don't know why you do, don't know how you apply it to real life, and don't know how to appropriately use it to calculate Ix'y' unfortunately =\, but I know it exists.

This slide show may illustrate why the PAT makes these calculations easier:

http://www.ce.memphis.edu/3322/Pdfs/PaulsPDFs/Centroids and Moment of Inertia Calculation.pdf
 
X bar = 73.2 mm
Y bar = 154 mm

(73.2 + (30 * 120 * 60^2)) + (73.2 + (30 * 390 * 15^2)) + (73.2 + (30 * 290 * 145^2)) = 1.985e8 = Ix
(154 + (30 * 120 * 15^2)) + (154 + 30 * 390 * 195^2)) + (154 + (30 * 290 * 15^2)) = 4.477e8 = Iy

Right track in relating that to Ixy (which is the next point I'm stuck at)?
 
  • #10
Bluestribute said:
X bar = 73.2 mm
Y bar = 154 mm

(73.2 + (30 * 120 * 60^2)) + (73.2 + (30 * 390 * 15^2)) + (73.2 + (30 * 290 * 145^2)) = 1.985e8 = Ix
(154 + (30 * 120 * 15^2)) + (154 + 30 * 390 * 195^2)) + (154 + (30 * 290 * 15^2)) = 4.477e8 = Iy

Right track in relating that to Ixy (which is the next point I'm stuck at)?

This is why I suggested you put your calculations into a tabular form. It makes the calculations much easier to check.

I'm going to show you how to calculate Ix about the x-axis with a tabular form, but I am going to use the letter dimensions rather than actual numbers.

Code: 
ITEM                AREA    y-bar     A*y-bar        A*y-bar^2          Io
Lower piece          b*d     d/2      b*d*d/2       b*d*(d/2)^2      b*d^3/12
Vertical piece  (h-2d)*d     h/2   - Fill in the numbers here -  (d/12)*(h-2d)^3
Upper piece          a*d    h-d/2    a*d*(h-d/2)   a*d*(h-d/2)^2     a*d^3/12
------------------------------------------------------------------------------
Totals              Σ A    y'-bar     Σ A*y-bar    Σ A*(y-bar)^2       Σ Io
You can find the Total Area and the moment of Area about the x-axis by adding the appropriate columns in the table.
The moment of Inertia about the x-axis is the Σ Io + Σ A*y-bar^2.

y'-bar = (Σ A*y-bar) / (Σ A)

To find the moment of inertia about the centroidal x'-axis, you would apply the PAT to the totals above like so:

Ix' = Σ Io + Σ A*(y-bar)2 - ΣA*y'-bar2,

where y'-bar is the vertical distance from the x-axis to the x'-axis.

A separate table can be set up to calculate Iy and Ixy, or the table shown above can be extended by adding columns to cover the additional properties desired.
 
  • #11
Ok, so this definitely helps when I have to calculate the Ix and Iy (which I'll be doing tomorrow while studying). How exactly do you relate this information to Ixy though?
 
  • #12
Bluestribute said:
Ok, so this definitely helps when I have to calculate the Ix and Iy (which I'll be doing tomorrow while studying). How exactly do you relate this information to Ixy though?

The same principle applies to calculating Ixy as it does to calculating Ix or Iy. The Ixyo values for each rectangle are all zero (because the rectangle has an axis of symmetry), and you can use the table to compile the values of A*x*y for the transfer quantities of each rectangle about the x-axis and the y-axis. Once the location of the centroidal x'-y' axes have been established relative to the x-y axis, then the PAT can be applied to determine Ixy for the entire section relative to these centroidal axes, thus

Ixy = Σ Ixyo + Σ A*x*y - Σ A*(x-bar)*(y-bar)

In this case, Σ Ixyo = 0, leaving

Ixy = Σ A*x*y - Σ A*(x-bar)*(y-bar)

where x-bar and y-bar are the location of the centroid of the entire cross section relative to the x-y axis.
 
  • #13
Right right. What I mean though, is like Y bar is just half the height of each piece (in this particular case), I0 is bd3/12, etc.

But what's Ixy in terms of math (since I know physically it isn't anything)?
 
  • #14
Bluestribute said:
Right right. What I mean though, is like Y bar is just half the height of each piece (in this particular case), I0 is bd3/12, etc.

But what's Ixy in terms of math (since I know physically it isn't anything)?

Ixy is just as physically real as Ix or Iy, it's just not dealt with often unless the cross section doesn't have an axis of symmetry.

Maybe this article can better illustrate what I've been trying to explain:

http://m-sudo.sakura.ne.jp/soft_data/kikaisekkei_data/mo-ru eN_Kansei_Momennto5.pdf
 
  • #15
Ok. So basically, unless you're applying a force at the centroid of the object, you'll essentially be bending on a different axis? Is there a simple equation for Ixy? Using what I think I know about the other two, is it just XYA (x bar, y bar, and area)?
 
  • #16
Think of Ixy as a measure of how evenly area or mass is distributed about the centroidal axes.

If Ixy = 0, then there is at least one axis of symmetry.

Unlike Ix and Iy , it is possible for Ixy to take on negative values.

There is no simple formula for calculating Ixy except those I have tried to explain.

Ixy does not have any influence in the bending of a beam.
 
  • #17
Ah, so I just had false hope with that XY part. Dammit! Oh well, I think I can apply the rest (but we'll find out!)
 
  • #18
Ok, so I want to see if I'm messing up somewhere or if the answer key is wrong. I've done this multiple times, I've used a table, and I'm getting 2.65 inches or .942 inches. The key says 2.25 inches . . .

TOP: Y bar = 0.5 | Area = 1.3
MID: Y bar = 1.9 | Area = 1.9
BOT: Y bar = .25 | Area = 1.8

I'm summing AY as my numerator and just A as my denominator, so my fraction becomes: 4.71/5
Summing up Y bars only, which gets me closer to the key, is the 2.65. That's not right, but it's closer . . .
 

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  • #19
Bluestribute said:
Ok, so I want to see if I'm messing up somewhere or if the answer key is wrong. I've done this multiple times, I've used a table, and I'm getting 2.65 inches or .942 inches. The key says 2.25 inches . . .

TOP: Y bar = 0.5 | Area = 1.3
MID: Y bar = 1.9 | Area = 1.9
BOT: Y bar = .25 | Area = 1.8

I'm summing AY as my numerator and just A as my denominator, so my fraction becomes: 4.71/5
Summing up Y bars only, which gets me closer to the key, is the 2.65. That's not right, but it's closer . . .

Your answer isn't coming out right because you're not measuring the y-bar values for each piece from the same reference.

For instance, either pick to top or the bottom of the section and measure all of the y-bar values from that reference.

For example, the bottom piece is 0.5/2 = 0.25" from the bottom of the section, the middle piece is 3.8/2 + 0.5 = 2.4" from the bottom,
and the top piece is 1.0/2 + 3.8 + 0.5 = 4.8" from the bottom. These are the y-bar values you put into the table in order to calculate the centroid of the entire section.
 
  • #20
You know, some people feel stupid because the right method was, so to speak, on the "tip of their brain".

I'm more of just plain stupid since I was looking at the axis in the center of the picture going "How do I find how high up these axis's are? Everything needs to be from these axis's" haha. Oh man, yes, yes you are correct and I got the answer.
 
  • #21
Great.
 

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