Polar Moment of Inertia for extruded rectangular tubing

[whotmewory]

Can anyone remind me the equation or steer me to an immediate resource for the Polar Moment of Inertia of extruded retangular tubes for figuring total shear loading at the Neutral Axis for a rectangular hollow tube with a torsional load?

Cheers!

 

[wes616]

Iz = (intergal S) (x^2+y^2)dA



Wes C.
------------------------------
When they broke open molecules, they found they were only stuffed with atoms. But when they broke open atoms, they found them stuffed with explosions...

 

[GregLocock]

It is in Roark

back transcribing from a spreadsheet, ie there may be errors

2*t^2(a-t)^2(b-t)^2/(at+bt-2*t^2)





Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[whotmewory]

Greg:

I don't have Roark's - it "took a walk" on me.

Ah, can I assume "a" = height above the Neutral Axis, "b" = width across the Neutral Axis, and "t" is the material wall thickness?

Chris

 

[rb1957]

polar moment = Ixx+Iyy
Ixx = b*a^3/12-(b-2t)*(a-2t)^3/12
= (2a^2t(3b+a)-12at^2(a+b)+8t^3(b+3a-2t))/12
Iyy = a*b^3/12-(a-2t)*(b-2t)^3/12
= (2b^2t(3a+b)-12bt^2(a+b)+8t^3(a+3b-2t))/12

 

[rb1957]

btw, a is the full depth of the section, b is the full width

simplify polar moment = [(a+b-6t)*(a+b)^2+16t^2*(a+b-t)]*t/6

 

[rb1957]

an approximate answer (maybe where greg is coming from) is ...
t/6*[3*(a+b-2t)(a-t)(b-t)+(a-t)^3+(b-t)^3]

 

[Denial]

If WhotMeWory is interested in the torsion of a rectangular hollow tube, it may well be that the use of the polar moment of inertia is completely inappropriate.

 

[GregLocock]

...or it may be completely appropriate. The spreadsheet version correlates well with my FEA model. It would not work in a structure that was not sensibly designed for torsion, (ie a,b are too dissimilar) and the efficiency of the joints between beams is an interesting exercise.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[rb1957]

gez, i wish i could read ...

shear stress of a closed section due to torque is T/(2[A]t), where [A] is the enclosed area of the tube (a-t)(b-t).

tho' the reference to "neutral axis" is abit confusing

 

[GregLocock]

That's a very dangerous formula as well. In real thin walled structures it provides an upper bound to the achievable torsional stiffness.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[rb1957]

g'day greg,

comparing your formula and mine yours has extra terms that amount to (a-t)(b-t)/(a+b-2t). for the couple of tests i did this is greater than 1 (and got bigger as the section got thinner), something like 2. this would give lower stresses, no ?

 

[GregLocock]

For a given rotation, yes, the more reality you add, the lower the stresses get, as the section softens.

A reasonable rule of thumb is that you'll only see 60% of the theoretical stiffness near the joint (say 10*a) if you use squarish tubes and butt joints. Good joint design can take that up to 90-95%.

It is almost impossible to manually estimate the stiffness of sections that are re-entrant or generally weird, such as are used in cars. A good example of that is using a light foam to stiffen a steel structure. At first sight it is ridiculous, but the foam stabilises the flat surfaces in the section and provides a measurable improvement in performance.





Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.