Talk:Second polar moment of area

Expand tag

I removed the expand tag {my removal}-- as I see the article as complete. It has the definition, describes the relation to moment of inertia and area moment of inertia and has an example. If you want to re-add the tag, please explain what it is that you think is missing. Nephron  T|C 18:41, 16 February 2007 (UTC)

Anisotropic material properties

Suppose I have a torsion bar uniform along its length, L, but with anisotropic material properties. For example, suppose I have a composite bar including steel ribbons running down its length, oriented like this:

  +y
   ↑

  \|/ 
-  ʘz - → +x
  /|\

Obviously this beam will be less stiff in torsion than if each ribbon were rotated 90° about the z axis. I feel like I should be able to compute the torsional stiffness of such a bar with a formula along the lines of

$\kappa =L\int _{A}r_{i}r_{j}C_{ij}(\mathbf {r} )\,dA$

where r is the position of each material point and where $C_{i,j}(\mathbf {r} )$ is a tensor describing the anisotropic stiffness of the ribbons (with units of newtons per square meter so that $L\cdot C_{i,j}(\mathbf {r} )$ has units of newtons per meter). I think the above integral isn't quite right; I feel like there should be a distance term in the circumferential direction.

I considered starting with the parallel axis theorem, thinking that each ribbon has its own moment and that each is just shifted out from the center, but that seems to suggest that the orientation of the ribbons (radial or circumferential) doesn't matter, which is nonsense.

(Hopefully this helps expand Wikipedia's coverage of anisotropic continuum mechanics.) —Ben FrantzDale (talk) 13:01, 6 August 2008 (UTC)

I'm starting to think it's

$\kappa =L\int _{A}(\mathbf {r} \times \mathbf {z} )_{i}(\mathbf {r} \times \mathbf {z} )_{j}C_{ij}(\mathbf {r} )\,dA$

Which would extend to higher dimensions as

$\kappa _{mn}=L\int _{A}C_{ij}\varepsilon _{mik}\varepsilon _{npj}r_{k}r_{p}\,dA$

using the Levi-Civita symbol. But I could be wrong. —Ben FrantzDale (talk) 20:01, 6 August 2008 (UTC)

Non-circular shafts

To Nephron. Every text book I've seen describing torsion and polar mmonet shows it applied to circular shafts. Some say it can't be applied to other shapes. I think this is true and we should clarify it. The first sentence of the article implies it weakly.
However the polar moment may have some other application where it is valid for different shapes. Anybody know anything else it's used for?Kallog (talk) 14:55, 15 April 2009 (UTC)

You can use polar moment of inertia to calculate the rotational inertia of an object with constant cross-section:

$I=J_{z}\rho l$

where I is the rotational inertia of the object,

$J_{z}$ is the polar moment of inertia of the cross-section of the object,

$\rho$ is the density of the material,

and l is the length of the object.

Unlike the torsion application, this is true for every cross-sectional shape, even non-circular shapes.

--68.0.124.33 (talk) 18:19, 19 March 2010 (UTC)

Just look at the formula, it applies to any kind of area, not just circular ones. Even the diagram shows J being calculated for a non-circular shape. Owen214 (talk) 02:23, 6 November 2010 (UTC)

Polar second moment of area

Just wondering why anyone has not written that polar moment of inertia is also known as polar second moment of area ? In Mechanics of engineering materials PP. Benham, they use it all the time. Just wondering. —Preceding unsigned comment added by 129.240.64.203 (talk) 09:11, 27 October 2010 (UTC) Causes much confusion - reserve the term moment of inertia for ... inertia (kg) ... call this second moment of area

Hi! Yes, I think the whole article (and articles that reference second moment of area [2MoA]), needs to be overhauled to refer it this only as "2MoA" and not "moment of inertia" (MoI). MoI deals with interia only. The confusion is due to this malapropism from some slip-up in popular use (perhaps a textbook long ago), and it is indeed confusing. The entire article should only use 2MoA, and there should be just a sentence or two (or three) explaining how it is sometimes (incorrectly, colloquially) referred to as MoI. I have been trying over the years (maybe about 5 years at this point, maybe 3 or 4 times in total) to help make this more clear in these pages without doing a total overhaul. Partly because I didn't feel like putting in the time, and partly because I feel like someone would just change it back. But, in my opinion, it DOES need to be done.
And the whole "polar" moment of inertia is further confusing; a moment is polar by nature. I think this language is not only confusing the formulas, but also making the physical concepts themselves more convoluted and confused. As far as I know, there is inertia and moment of inertia (linear and polar inertia, if you want to think of it that way), and there is planar and polar second moment of area. The former has units of kg and kg-m^2, the latter m^4.
-- deathbykindnes, 7:36, 3 Dec 2021 — Preceding unsigned comment added by Deathbykindnes (talk • contribs)

I changed it! Phew. Deathbykindnes (talk) 16:45, 14 April 2022 (UTC)

Polar and torsional constant

In my opinion, this publication is a total mess between Polar moment of inertia(area)Jz and Torsional constant Jt.

Polar moment of area is Jz = Jx + Jy, where Jx and Jy should refer to principle moments of area. For a prismatic body with thickness t and density ρ, Jz = (Jx + Jy)·t·ρ is the mass polar moment and describes the inertia resistance of the body when you apply angular acceleration about z axis.

Torsional constant Jt is related to the ability of section to resist Saint Venant torsional stress.

Both are completely different in nature and should not be confused in any case. The (hollow) circular shape is a special case where both are calculated by the same formulas, but this is just a coincidence due to the simplicity of the shape. For other shapes, it is completely different.

Assume we have a rectangular shape with dimensions b and h (b < h):

Polar moment is: Iz = Ix + Iy = b·h/12·(b^2 + h^2)

Torsional moment is: It = b·h^3·(1 – 0.630·h/b + 0.052·(h/b)^5)/3 (Roark formulas for stress and strain)

They have completely different values.

By Ned Ganchovski, MSc Struct. Eng.

Possible typo?

In the section where the formula is given for the hollow cyclinder the quantities D and d are shown inside brackets, however there is no operator between the two quantities. Perhaps there should be a subtraction symbol present?