# Hollow Tube Stronger than Solid bar of same Outside Diameter (O.D.)?

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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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. –  Ted Bunn Jul 28 '11 at 21:35
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. –  Nwsue Dec 2 at 10:34

"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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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. –  Omega Centauri Jul 28 '11 at 22:29
@Omega This is one of the reasons not to use tubes as beams. –  Georg Jul 29 '11 at 11:17