Moment of inertia with T shape

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  • #1
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]1. Homework Statement [/b]



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The Attempt at a Solution



On paper it looks like not much of an attempt has been made, but I have tried a few different ways to find I_y. I thought to take the entire rectangle and subtract the parts on the upper left and right sides of the upside down T. But, I can't get my disk method to forgo the part of the upside down T that is present, meaning the bottom part of the T is still there and I can't figure out how to set my limits and my equation to not include that portion of the bottom that is present.
 

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  • #2
Simon Bridge
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OK you have an inverted T on x-y axis.
The horizontal of the T goes from x=-4.5in to x=+4.5in and is 2in high.
The stem of the T is 6in high, making the T 8in tall, and 3in wide - so it strentches from x=-1.5in to x=+1.5in.

There is no information about it's shape in the z-direction.
The problem cannot be done without this information.

You want to find the moment of inertia about the y axis.
Your strategy was to find the moment of inertia due to a solid block, and subtract the moment of inertial due to the buts you have to cut away to make the T shape.

I don't understand why you feel the need to do an integration.

What's wrong with using one of this list and the perpendicular axis theorem, parralel axis theorem, and the stretch rule, as required?
 
  • #3
SteamKing
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There is no information about it's shape in the z-direction.
The problem cannot be done without this information.

I don't think the poster is trying to find the mass moment of inertia of a T-bar.

He is trying to calculate the second moment of area of the T-shaped cross section.
 
  • #4
Simon Bridge
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I don't think the poster is trying to find the mass moment of inertia of a T-bar.

He is trying to calculate the second moment of area of the T-shaped cross section.
Fair enough - requires clarification all the same.
 
  • #5
SteamKing
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]1. Homework Statement [/b]



Homework Equations





The Attempt at a Solution



On paper it looks like not much of an attempt has been made, but I have tried a few different ways to find I_y. I thought to take the entire rectangle and subtract the parts on the upper left and right sides of the upside down T. But, I can't get my disk method to forgo the part of the upside down T that is present, meaning the bottom part of the T is still there and I can't figure out how to set my limits and my equation to not include that portion of the bottom that is present.

If you are going to calculate the inertia of a rectangle, be careful you understand the formula you are using. I = (1/3)*b*h^3 gives the moment of inertia about the base of the rectangle, not about its centroid.
 
  • #6
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I solved the problem. The only thing that is tripping me up though is the fact that the distance from the y-axis with the upper part of the upside down T is irrelevant. For example, if the shape of the figure was an L, my moment of inertia around the y axis would be the same as if it were an upside down T???
 
  • #7
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I solved the problem. The only thing that is tripping me up though is the fact that the distance from the y-axis with the upper part of the upside down T is irrelevant. For example, if the shape of the figure was an L, my moment of inertia around the y axis would be the same as if it were an upside down T???
Times two.
 
  • #8
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why times two? All I did to find the moment of inertia around the y axis was find the individual moments of the two parts of the T and added them together, which only deals with the dimensions of the two rectangles
 
  • #9
SteamKing
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Without seeing your calculations, we don't know for certain if you have answered the question correctly.
In this thread,
https://www.physicsforums.com/showthread.php?t=714921

you seem confused about first and second moment of area calculations. For your work in this problem, did you determine the location of the centroidal axes? After calculating your moment of inertia, did you use the parallel axis theorem to calculate the moments of inertia about the centroidal axes of the T-section?
 
  • #10
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I will put it up when I get home
 
  • #11
Simon Bridge
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I solved the problem. The only thing that is tripping me up though is the fact that the distance from the y-axis with the upper part of the upside down T is irrelevant. For example, if the shape of the figure was an L, my moment of inertia around the y axis would be the same as if it were an upside down T???
That seems a bit odd to me. The two objects described have different mass don't they?

Do you mean moment of inertia or second moment of area?
 
  • #12
member 392791
second moment of area,

Here is my attempt 2. To be honest I watched a video on youtube and then followed his procedure (he did it with a T right side up). I don't understand why the distance d is the distance between the center of each section and the centroid. I thought d was supposed to be the distance between the axis and an axis that goes through the centroid. Either way it got me the right answer, I just don't understand how, it seems like I am not using the parallel axis theorem, but at the same time am.

Allow me to clarify, the parallel axis theorem is where you know the second moment of area around the centroid, then add Ad^2 to get the moment around a different axis (d is the distance between centroid and arbitrary axis). However, in this problem I found the second moment of area around the centroid using the formula for a rectangle (1/12 bh^3), but still used parallel axis theorem and added it together?? It really isn't making sense to me what I did
 

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  • #13
Simon Bridge
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second moment of area,

Here is my attempt 2. To be honest I watched a video on youtube and then followed his procedure (he did it with a T right side up). I don't understand why the distance d is the distance between the center of each section and the centroid.
"centroid" of what?

We cannot tell either - because you have not provided a link to the video or your working.
It is generally not a good idea to follow the methods of others without understanding them - sure you'll finish the problem, but what have you learned?
 
  • #14
SteamKing
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I checked your calculations. The good news is, your calculation for Iy is correct. The bad news is, your calculation for Ix has omitted some steps.

The calculations of the two pieces of the T-section about their own centroids is OK, as is the calculation of the location of the vertical centroid, or y-bar. Your problems start with applying correctly the parallel axis theorem.

The correct procedure is to take the moment of inertia of each rectangle and add to it the quantity A*d^2, where A is the area of the rectangle and d is the distance of the centroid of the rectangle from the original x-axis. After you have done this for each rectangle, and then added the I + A*d^2 values together, then this sum must be adjusted to transfer the moment to the centroid for the entire T-section.

So: I(T-bar) = (I + A*d^2) for rect. 1 + (I + A*d^2) for rect. 2 - (A * d^2) for the whole T-bar

Note: the d value for the whole T-bar is the y-bar value.
 
  • #15
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why times two? All I did to find the moment of inertia around the y axis was find the individual moments of the two parts of the T and added them together, which only deals with the dimensions of the two rectangles

Oh. From what you had written, I had the impression that you did it for the L, but had not included its mirror image.
 
  • #16
member 392791
Here is a link to the video that I got my information from



Why would I have to subtract the moment of the entire T bar from the individual pieces?? Why moment of inertia, I thought I was calculating second moment of area?? And if I did the I_x wrong, how is the answer correct?

It seems to me that the terms moment of inertia and second moment of area are being used interchangeably, are they the same thing??
 
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  • #17
SteamKing
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'Second moment of area' and 'moment of inertia' are terms used interchangeably.

After studying your calculation of Ix more closely, I think I understand what you did. It just so happens for this particular calculation, the centroids of the flange and the web of the T-section both happen to lie 2 inches from the centroid of the entire T, and your Ix calculation is correct. My apologies for not seeing this before.
 

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