Torsional moment of inertia

[4Boomers]

I am having difficulty figuring out how to calculate the torsional moment of inertia of a built up steel member comprised of vertical plates, horizontal plates, and angles?

Can anyone provide any guidance?

 

[Trenno]

J = Sum of (1/3)*b*t^3 for each individual plate.

 

[JAE]

....unless those plates form an enclosed space. So for a square tube the sum of the four plates = 1/3bt^3 wouldn't work. The value would be much higher.

For a square tube b x h in size with walls of thickness = t, the torsional constant, J = 2(b^2)(h^2)t^2 / (bt + ht)

What does your shape look like?

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[4Boomers]

Thanks you for the help. The section geometry varies by truss member.
I found a post on the STAAD forum that provided the torsional constant of a w-beam. When I tried to verify the Torsional constant with the formula: sum of 1/3*(b)(t)^3, I was able to get the same result when I took "b" as the horizontal length of the flanges and as the vertical length of the web. Is that correct - "b" is the longer dimension of the rectangle regardless of its orientation (vertical or horizontal)?

Thanks again.

 

[Lomarandil]

That's correct.

 

[JAE]

Yes

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[Denial]

4Boomers.[ ] Strictly speaking, that 1/3bt^3 formula applies only to very thin rectangles.[ ] So, for an extreme example, you would not apply it to a solid square component.