Moment of inertia of rectangular tube with welded flat bar

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SUMMARY

The moment of inertia for a rectangular stainless steel tube measuring 4"x2"x0.25" thick with a 3"x0.75" flat bar welded on top has been calculated to be approximately 22.16 in4. The calculations confirm that the total moment of inertia at the x-x axis is accurate, with a detailed breakdown provided in a tabular format. The method involves calculating the contributions from each section and adjusting for the centroid of the combined section.

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johnmech6718
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hello,

I am trying to find the new moment of inertia total for a rectangular stainless steel 4"x2"x.25"thk with a 3"x.75" flat bar welded on top of the short side of the rectangular tube. The short side (.75") of the flat bar will be welded to the center of the rectangular tube to the 2" side, total height of 7". I am looking for the moment of inertia at the x-x axis. I get a new moment of inertia total of 22.15 in^4. Can anyone confirm that the total is correct? I have also attached my calcs for reference.
 

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johnmech6718 said:
hello,

I am trying to find the new moment of inertia total for a rectangular stainless steel 4"x2"x.25"thk with a 3"x.75" flat bar welded on top of the short side of the rectangular tube. The short side (.75") of the flat bar will be welded to the center of the rectangular tube to the 2" side, total height of 7". I am looking for the moment of inertia at the x-x axis. I get a new moment of inertia total of 22.15 in^4. Can anyone confirm that the total is correct? I have also attached my calcs for reference.

Your calculations are correct.

You can extend your tabular form calculation by adding two columns to your original table as shown below:

Code:
Item     A      y-bar     Ay        Ay^2       I
I        8.0     2.0     16.0       32.0      10.67
II      -5.25    2.0    -10.5      -21.0      -5.36
III      2.25    5.5     12.375     68.06      1.69
----------------------------------------------------
Total    5.00  3.575     17.875     79.06      7.00

The Inertia about the combined centroid of the section:

INA = 7.00 + 79.06 - 5*3.575^2

INA = 22.16 in^4

This form is a little neater and easier to check.
 

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