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Torsional moment of inertia

Torsional moment of inertia

Torsional moment of inertia

(OP)
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?

RE: Torsional moment of inertia

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

RE: Torsional moment of inertia

....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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RE: Torsional moment of inertia

(OP)
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.

RE: Torsional moment of inertia

That's correct.

RE: Torsional moment of inertia

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.

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