# Moment of inertia and the geometric center

• masterchiefo
In summary: Is this better...You seem to be taking the height of the mass centre of the C section from where it contacts the I section as the sole basis for computing how far the addition of the C section displaces the vertical height of the mass centre from the centre of the I section. This ignores the distance from the mass centre of the I section to that point of contact. It ought to have given you a very small result, but you also seem to have lost a decimal point (46641.7, 466417)."Correct, the height of the mass centre of the C section is not the only variable that affects the displacement of the mass centre. The distance from the mass centre of the
masterchiefo

## Homework Statement

The profiled steel I above is topped by a " profiled C ".
Determine the moments of inertia of the structure composed by the axes
central x and y through its geometric center C.

Picture representing the problem:

## Homework Equations

Profiled C and I information:

## The Attempt at a Solution

Section
All numbers are coming from the table for this first part.
Profiled I: A= 14400mm2 /y=0 /x=0 /y*A=0 /x*A=0
Profiled C: A= 2897mm2 /y=16.10 /x=0 /y*A=46641.7 /x*A=0
Total A = 17297mm2

/Y∑A = ∑/yA /Y*(17297) = 46641.7 /Y =26.9652mm

For the Profiled I :
Ix' = /Ix +A/Y2 =554*106 + 14400*26.96522
= 564.47*106

For the Profiled C:
Height of profiled I = 460, half = 230
Now to get height of mass center of I to mass center of section C I do 230-(22.8-16.10)
16.10 is the distance from mass center of section C to the top of the section, but I need mass center to bottom of that section so 22.8-16.10.

Ix' = /Ix +A*height2 = 0.949*106 +(2897)*(230-(22.8-16.10))2
=145.402*106

Total
145.402*106 + 564.47*106
= 709.872*106

Last edited:
It's hard having to reverse engineer your logic from the numbers. It would be a lot easier to check if you were to create symbolic names for all the dimensions and post your working in terms of those.
masterchiefo said:
Profiled I: A= 14400mm2 /y=0 /x=0 /y*A=0 /x*A=0
Profiled C: A= 2897mm2 /y=16.10 /x=0 /y*A=46641.7 /x*A=0
Total A = 17297mm2

/Y∑A = ∑/yA /Y*(17297) = 466417 /Y =26.9652mm
You seem to be taking the height of the mass centre of the C section from where it contacts the I section as the sole basis for computing how far the addition of the C section displaces the vertical height of the mass centre from the centre of the I section. This ignores the distance from the mass centre of the I section to that point of contact. It ought to have given you a very small result, but you also seem to have lost a decimal point (46641.7, 466417).

haruspex said:
It's hard having to reverse engineer your logic from the numbers. It would be a lot easier to check if you were to create symbolic names for all the dimensions and post your working in terms of those.

You seem to be taking the height of the mass centre of the C section from where it contacts the I section as the sole basis for computing how far the addition of the C section displaces the vertical height of the mass centre from the centre of the I section. This ignores the distance from the mass centre of the I section to that point of contact. It ought to have given you a very small result, but you also seem to have lost a decimal point (46641.7, 466417).
Most numbers are coming from the table in the picture.
I will explain some that aren't in the table in the main post.

"This ignores the distance from the mass centre of the I section to that point of contact."
Sorry I don't understand that part. My numbers for the first part are all coming from the table.

edit: I have edited the thread with some explanation.

Last edited:
masterchiefo said:
Most numbers are coming from the table in the picture.
I don't care where they come from, I don't want to have to read them then trace back to find out what they represent.
Let's just say you have an I-beam height hI, width wI, cross-sectional area AI, etc., and a cap height hC, etc. Leave numbers out of it. Post an equation for your answer in terms of these unknowns. After that we can move on to plugging in numbers to get a numerical answer.
masterchiefo said:
"This ignores the distance from the mass centre of the I section to that point of contact."
Sorry I don't understand that part. My numbers for the first part are all coming from the table.
Sure, but I presume you are trying to compute the mass centre of the composite. Even if the C cap were a point mass, it would shift the mass centre because of its distance from the mass centre of the I beam.

haruspex said:
I don't care where they come from, I don't want to have to read them then trace back to find out what they represent.
Let's just say you have an I-beam height hI, width wI, cross-sectional area AI, etc., and a cap height hC, etc. Leave numbers out of it. Post an equation for your answer in terms of these unknowns. After that we can move on to plugging in numbers to get a numerical answer.

Sure, but I presume you are trying to compute the mass centre of the composite. Even if the C cap were a point mass, it would shift the mass centre because of its distance from the mass centre of the I beam.
Is this better ? Sorry if I made it confusing. I thank you very much for your time into helping me.

Section C:
Ix' = Inertia in X-X axis +Surface Area*height2
height: mass center of section C to mass center of section I

Section I:
Ix' = Inertia in X-X axis +Surface Area*height2
height: mass center of section I to point C in the drawing.

That's not exactly what I was after.
In the tables you posted, there are 7 numbers that describe the I section. Assign a unique variable name to each.
Likewise, there are 8 numbers for the C section. Assign unique variable names to each of those.
Then show your working and results entirely in terms of those variable names. Don't substitute their numerical values until the absolute final step of the calculation.

(Now, it is a bit confusing that the C section's description has x and y swapped from the way the C section is employed in the composite. I don't mind too much how you handle that in assigning the names, as long as it is clear.)

Anyway, to respond to your post,
masterchiefo said:
Section C:
Ix' = Inertia in X-X axis +Surface Area*height2
height: mass center of section C to mass center of section I
I assume this is with X in the C section's co-ordinates, so it's vertical in the composite.
That should give the MoI of the C section within the composite about the mass centre of the I section.
But
(a) I don't see how you got 230-(22.8-16.10) for that height. The 230 is ok, if a little inaccurate, but where does the rest come from?
(b) You want the MoI about the mass centre of the composite.
masterchiefo said:
Section I:
Ix' = Inertia in X-X axis +Surface Area*height2
height: mass center of section I to point C in the drawing.
Which co-ordinate system is this in? In the OP you have Inertia in X-X axis = 554*106, but don't explain how this was obtained. (This is the sort of thing that would be perfectly clea if you were to work symbolically.)

In my first response I challenged this calculation:
masterchiefo said:
/Y∑A = ∑/yA /Y*(17297) = 46641.7 /Y =26.9652mm

haruspex said:
That's not exactly what I was after.
In the tables you posted, there are 7 numbers that describe the I section. Assign a unique variable name to each.
Likewise, there are 8 numbers for the C section. Assign unique variable names to each of those.
Then show your working and results entirely in terms of those variable names. Don't substitute their numerical values until the absolute final step of the calculation.

(Now, it is a bit confusing that the C section's description has x and y swapped from the way the C section is employed in the composite. I don't mind too much how you handle that in assigning the names, as long as it is clear.)

Anyway, to respond to your post,

I assume this is with X in the C section's co-ordinates, so it's vertical in the composite.
That should give the MoI of the C section within the composite about the mass centre of the I section.
But
(a) I don't see how you got 230-(22.8-16.10) for that height. The 230 is ok, if a little inaccurate, but where does the rest come from?
(b) You want the MoI about the mass centre of the composite.

Which co-ordinate system is this in? In the OP you have Inertia in X-X axis = 554*106, but don't explain how this was obtained. (This is the sort of thing that would be perfectly clea if you were to work symbolically.)

In my first response I challenged this calculation:

Hey man, to be honest I am not sure at all with anything I did on this problem, this is very new subject to me and very confusing at the moment, I will try to read more and figure out how moment of inertia work.

thanks again.

## 1. What is moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in rotational motion. It is similar to mass in linear motion, but instead of measuring the object's resistance to changes in linear motion, it measures the object's resistance to changes in rotational motion.

## 2. How is moment of inertia calculated?

The moment of inertia of an object can be calculated by summing the products of each infinitesimal mass element in the object and the square of its distance from the axis of rotation.

## 3. What is the relationship between moment of inertia and the geometric center?

The geometric center, also known as the center of mass, is the point where the object's mass is evenly distributed. The moment of inertia is directly related to the distribution of mass around the axis of rotation, so the geometric center plays a crucial role in calculating the moment of inertia.

## 4. How does the shape of an object affect its moment of inertia?

The shape of an object has a significant impact on its moment of inertia. Objects with a larger radius of gyration (a measure of how far the mass is distributed from the axis of rotation) have a higher moment of inertia. This means that objects with a larger mass concentrated away from the axis of rotation will have a higher moment of inertia.

## 5. Why is the moment of inertia important in physics?

The moment of inertia is important in physics because it is a crucial factor in understanding an object's rotational motion. It helps us predict how an object will behave under the influence of external forces and how it will respond to changes in its rotational speed. It is also used in many practical applications, such as designing structures and machinery that need to rotate smoothly and efficiently.

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