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I was listening to some co-workers talking about problems meeting stiffness requirements. Someone said that even with a solid metal rod (instead of the existing tube) we could not meet stiffness requirements.

I started daydreaming... and went back in time over a quarter of a century to some class I was taking in college. Things are hazy when you go back that far; but I am sure that someone with a Ph.D., or some other letters after his name, said that you could actually add stiffness to a solid rod by drilling out the center (and maybe by appropriately treating it, I forget). The reason, if I recall correctly, had to do with the added tensile strength of the inner surface. The reason I remember this from so long ago was that it was so counter-intuitive: I was stunned.

I'm not a stress/structures guy; so I asked a co-worker about it, and he said that the solid rod would be stiffer, because it had the greater (bending) moment of inertia. I agree with the latter part of the statement, but my hazy daydream keeps me from agreeing with the preceding conclusion. My money is still on a series of concentric tubes, appropriately processed and internally supported, being stiffer than a solid rod of the same O.D.

So, my question: Does anyone know of any references to this little structural trick (or engineering wives' tale). If so, can you quantify how much stiffness is gained?

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    $\begingroup$ A guess: Somewhere along a chain of communication, the (true) statement that a hollow tube can be stronger than a solid one of the same mass got changed into the statement that a hollow tube can be stronger than a solid one of the same diameter. $\endgroup$
    – Ted Bunn
    Jul 28, 2011 at 21:35
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    $\begingroup$ A discussion I heard regarding this topic, was related to bird leg bones. The professor was saying that bird leg bones were hollow and were stronger than if they were solid. I have always wondered about the accuracy of that statement too. $\endgroup$
    – user35091
    Dec 2, 2013 at 10:34

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http://www.physicsforums.com/archive/index.php/t-37701.html says

"Most of the strength of a cylinder comes from the outer portions. I think the contribution goes like the cube of the radial position. So, if you took a solid rod and drilled out a half the volume from the center, you do not lose half the strength. Strength to weight ratio is better for a hollow pipe than a solid rod."

The definition for the second moment of inertia $I_c$ for a filled and hollow cylinder can be found on http://en.wikipedia.org/wiki/Second_moment_of_area: $$I_c=\int\!\!\!\!\!\int_A y^2 \textrm{d}x\textrm{d}y=\int_0^R\!\!\!\!\!\int_0^{2\pi} r^2 \textrm{d}\phi\ r\textrm{d}r=\frac{\pi r^4}{4}$$

The surface area of the filled cylinder is: $$A=\pi r^2$$

Compare filled and hollow cylinder of equal mass: $$s_c=\frac{I_c}{A}=\frac{r^2}{4}$$, cylinder with fractional internal radius $r_i=xr_o$ and $x<1$: $$s_h=\frac{I_h}{A}=\frac{r^4(1-x^4)}{4r^2(1-x^2)}=\frac{r^2(1-x^2)(1+x^2)}{4(1-x^2)}=\frac{r^2(1+x^2)}{4}>s_c$$.

This means a hollow cylinder is stronger than a rod of equal mass and the same material. A hollow cylinder with a bigger inside diameter is better. In the limit $x\rightarrow 1$ the hollow cylinder is twice as strong. Note that this limit isn't physically viable as it would be an cylinder with infinite radius and infinitesimally thin wall. However it is useful to define the upper limit of the second moment of inertia. I didn't expect the increase in strength only a factor of two.

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    $\begingroup$ There will be a limit to stifness per unit weight, and also a limit to how thing you can make the wall before someother failure mode crops up. Make the cyllinder walls too thin and you will run into a (cyllinder wall) buckling limit. $\endgroup$
    – Omega Centauri
    Jul 28, 2011 at 22:29
  • $\begingroup$ @Omega This is one of the reasons not to use tubes as beams. $\endgroup$
    – Georg
    Jul 29, 2011 at 11:17
  • $\begingroup$ Using the above, if you make the comparison keeping the outer diameter fixed rather than the mass, the ratio of bending moments (hollow to solid) goes like 1-x^4 (essentially 1 in the range 0<x<0.5, and changing rapidly to 0 at x=1). Thus, for the solid cylinder, the region inside r0/2, which contains 1/4 of the total mass, gains you weight but basically no rigidity. $\endgroup$
    – ninemileskid
    Jun 24, 2016 at 22:53
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    $\begingroup$ @whoslip I don't understand what's your point? that the hollow cylinder will be stronger? the question asks about drilling the core of a rod, so the mass cannot be equal and your calculations say that the mass is equal which is irrelevant to the question and very confusing $\endgroup$
    – ergon
    Aug 10, 2016 at 12:20
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Though I generally agree with whoplisp's answer, it is worth to note that obtaining the (lower) limit of the thickness is rather tricky as long as it is defined by stability under 'strong' deformations. Where 'strong' is compared with the tube wall thickness.

Which is obvious from common sense point of view: thin rod is easier to deform than the hollow tube (of the same mass), but as long as the hollow tube starts to deform it is easier to break.

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A hollow tube of the same diameter as the solid round bar would have a larger radius of gyration hence your hazy recollection. This translates to a solid bar being more slender than a hollow one. Therefore a solid round bar of the same diameter will buckle easier than a hollow one of the same diameter, this also means that a solid round bar cannot span the same distance as a hollow pipe would. Of course reason has to be applied as well, a hollow pipe with a wall the thickness of a hair would buckle faster due to buckling of its wall.

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