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?
 A: 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. 
A: 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. 
A: 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.
