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Relationship of torsional constant to polar moment of inertia.

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AaronWhitaker

Structural
Nov 8, 2004
1
Does anyone know the relationship of the torsional constant to the polar moment of inertia for other than circular cross sections? I am trying to compare several different cross sections using a software package that only gives me polar moment of inertia. Is it safe to say that as the polar moment increases the torsional constant increases some proportional amount?

Thanks!
Aaron W.
 
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From Roark's, for bars in pure torsion,

T = QKG/l, where Q = angle of twist, G is shear modulus of material, and l is the length of the bar

"For a circular section, K is the polar moment of inertia J; for other sections K is less than J and may only be a small fraction of J."

Primarily this (reduction in torsional stiffness) is due to the warping of plane sections that occurs when non-circular bars are twisted.
 
If I remember correctly, the torsional stiffness of a circular tube is a lot different from the torsional stiffness of a circular tube with a slit down it, although the polar moment of inertia would be about the same, so it's hard to make blanket rules.
 
Best bet is to build a plate Fe model of a beam and back calculate K. In the slit tube example, K is ~0.01*J, the possibilities for error are enormous.

There are analytical approaches which may work, the membrane analogy is one, adding up the K for each plate in an I beam is another, conservative one.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
There is a good discussion of this in Section 2.10 of Blodgett's "Design of Welded Structures". For closed sections there is not a lot of difference between the torsional constant and the polar moment of inertia. Open sections are another story. The polar moment of inertia can't be used to calculate the twist because shear stresses are not uniform in an open section.
 
Whoa there, for closed /non reentrant/ sections...

Rentrant sections, or even closed ones with slightly silly shapes, can be very poor in torsion.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
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