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Consider a $n$-sided regular (convex) polygon and its circumscribed circle of radius $r$, centered in $(0,0)$. Fixing $(r,0)$ as the coordinate of the first vertex, the $n$ vertices of the polygon are given by:

$$P_i = (x_i,y_i)=(r\cos(2i\pi/n),r\sin(2i\pi/n))$$

The second moment of inertia w.r.t. the $x$-axis can then be derived using

$$ {\displaystyle I_{x}={\frac {1}{12}}\sum _{i=1}^{n}(y_{i}^{2}+y_{i}y_{i+1}+y_{i+1}^{2})(x_{i}y_{i+1}-x_{i+1}y_{i})}$$ leading to the result $I_x=nr^4/48\ (4\sin(2\pi/n)+\sin(4\pi/n))$.

Very interestingly, this value does not depend on the axis, or equivalently, on the rotation of the polygon: for example the two following hexagons have the same second moment of area wrt to the $x$-axis:

enter image description here

To show this, I multiplied $(x_i,y_i)$ by a rotation of matrix of angle $\theta$, leading to

$$P_i=(r\cos(2i\pi/n +\theta),r\sin(2i\pi/n + \theta)$$

Then, I computed $I_x(\theta)$ and using symbolic computation, observed that $I_x'(\theta)=0$.

Question Can the fact that $I_x$ does not depend on $\theta$ be seen easily by hand? Are there intuitive reasons for that?

Side note on mechanics This is interesting because it means the deflexion of a beam---the displacement induced by the force $F$, (see below)---with a regular-polygonial section does not depend on how the beam is placed. This is completely false for a rectangular section: the deflexion is much higher if the height of the beam corresponds the short side of the rectangle, as you can observe by holding a sheet of paper (slender rectangular section!) in a horizontal plane (huge deflection) or a vertical plane (no observable displacement).

enter image description here

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anderstood
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1 Answers1

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The invariance is due to the regularity of the polygon, or rather due to the fact that there are more than two symmetry axes.

For an arbitrarily rotated cartesian coordinate system $r=(r_1,r_2)\in A$ of the cross-section, the second moments of area are components of a symmetric degree-$2$ tensor $$A_{ij} = \int_{r\in A} r_i r_j\,\mathrm{d}^2r\tag{1}$$ The area inertia tensor depends on $((A_{ij}))$ via $$J_{ij} = (A_{11}+A_{22}) \delta_{ij} - A_{ij} \tag{2}$$ where $\delta_{ij}$ is the Kronecker delta.

If one coordinate axis is a symmetry axis, then $(1)$ implies $A_{12} = 0$, that is, $((A_{ij}))$ is diagonalized, and the diagonal entries are the principal second moments of area. Thus a symmetry axis is a principal axis.

So, if there are more than two symmetry axes, there are more than two principal axes. That implies that $((A_{ij}))$ has an eigenspace of dimension more than one, which for a mere 2D cross-section implies $$A_{ij} = J_0\delta_{ij} \quad\text{for some scalar $J_0$}$$ and therefore $((A_{ij}))$ is invariant under rotation. Via $(2)$, this carries over to the area inertia tensor.

ccorn
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  • Neat! One question: I understand that the coordinate axes are symmetry axes and principal axes. But why does it imply that any symmetry axis is a principal axis? – anderstood Jan 23 '17 at 02:04
  • @anderstood: (Orthogonal) principal axes are those that diagonalize $A_{ij}$ when used as coordinate axes. Now say the $x$-axis is a symmetry axis, and the $y$-axis is orthogonal to the $x$-axis. Then the boundary of the cross-section can be described by $y=\pm f(x)$. Then $A_{12} = \int_{x=x_{\text{min}}}^{x_{\text{max}}}\int_{y=-f(x)}^{+f(x)} xy,\mathrm{d}x,\mathrm{d}y = 0$. – ccorn Jan 23 '17 at 02:22
  • The $y=\pm f(x)$ is not fully general, e.g. holes need to be dealt with by adding more integrals with some more $y=\pm g(x),y=\pm h(x)$ etc. But all such scenarios keep $A_{12}=0$ by symmetry. – ccorn Jan 23 '17 at 02:27
  • In the above comments I take the axes of $x,y$ as something that can be arbitrarily chosen and changed (with $A_{ij}$ properly updated), as long as the axes stay orthogonal. In the answer I have written $(r_1,r_2)$ to avoid confusion with a fixed coordinate system. – ccorn Jan 23 '17 at 02:35
  • BTW, I too had been amazed about this feature of more-than-doubly-symmetric cross-sections decades ago, when I learned about it. – ccorn Jan 23 '17 at 02:41
  • Do you agree that the invariance is for i) cross-sections with 3 axes of symmetry or more, and ii) cross-sections with 2 non-orthogonal axes of symmetry? Also, may I ask where you've heard about this (out of curiosity)? – anderstood Jan 23 '17 at 16:30
  • @anderstood: I agree. Note that ii) implies i) because the mirror image of a symmetry axis over another symmetry axis is again a symmetry axis – ccorn Jan 23 '17 at 17:30
  • Is the $I_{ij}$ part really necessary? – anderstood Jan 24 '17 at 20:56
  • In short, the question was about area moments of inertia, so about $I_{ij}$. The trace expression in $(2)$ is the form used for cartesian systems, and there is no need to consider more general coordinate systems here. The same arguments would apply, these would just need to be stated in more complicated expressions. – ccorn Jan 24 '17 at 21:32
  • That does not change the beauty of the answer, but the definition of $I_{xy}$ is $\iint yx \mathrm{d}x\mathrm{d}y$. With your notations, I find this with a minus sign. Maybe it would be simpler just to remove $A_{ij}$ from the answer and keep only $I_{x}=\iint y^2 \mathrm{d}x\mathrm{d}y$, $I_y=\iint x^2 \mathrm{d}x\mathrm{d}y$ and $I_{xy}=I_{yx}=\iint xy \mathrm{d}x\mathrm{d}y$. The arguments would still hold. – anderstood Jan 24 '17 at 21:49
  • @anderstood: I use the convention where the minus sign is in use. (Otherwise you cannot get $(2)$ correspond to your $I_{xx} = \int y^2 ,\mathrm{d}x,\mathrm{d}y$ where the indices are essentially swapped.) There are two flavors of the theory where bending (or rotation) is concerned, one for bending "around axes", one for bending "in a direction"; the latter does not need sign changes and index swappings. You'll find that variance again in the theory of plates and shells. Choose the one you want, but be consistent. – ccorn Jan 24 '17 at 22:10
  • With $(2)$, if $u=(u_1,u_2)$ is a unit axis vector then the area-based moment of inertia with respect to $u$ is $I(u) = \sum_{i,j=1}^2 I_{ij} u_i u_j$. You need the minus sign for that. – ccorn Jan 24 '17 at 22:32
  • I have changed the $I_{ij}$ to $J_{ij}$. You can translate that as $J_{xx}=I_x$, $J_{yy}=I_y$, $J_{xy}=-I_{xy}$. The indexing of $J$ is thus compatible with its tensor nature. – ccorn Jan 24 '17 at 22:54
  • You seem to know about shell mechanics; would you have reading suggestions (involving the metric tensor)? – anderstood Jan 25 '17 at 00:39
  • See also MathWorld. Note that the page bottom contains formulae for regular $n$-gons, cited as Roark 3rd ed. At the time of this writing, those are wrong; $r_n$ and $R_n$ need to be swapped, that leads to the same value for both $I_r$ and $I_R$. Roark's 7th edition has it right (table A.1#28). – ccorn Jan 25 '17 at 00:43
  • Regarding shells: Try disks and plates first, to get a feeling for the form of the equations without curvature (you'll find that those can be seen as 2D-analogs of the 1D theories for rods and beams). Then proceed to membranes and shells. This needs a bit of differential geometry (Gauss's curved surface stuff, the more exercises, the better), but then you already know the disk and plate stuff, and you will easily recognize the curvature-related additions. This is also a good preparation for Riemannian geometry (and later, general relativity). – ccorn Jan 25 '17 at 00:54