Hollow shaft vs. Solid shaft: Which one's more resistant to torsion?
A solid rod will be stiffer (both in torsion and in bending) than a hollow rod of the same diameter. But it won't be much stiffer because almost all the stiffness comes from the outer layers of the rod (this is what the whole second-moment-of-area thing is about). So if you have a certain amount of material to use (a certain mass, or a certain mass per unit length), then rather than making a rod out of it, you get a stiffer structure by making a tube with a larger diameter. The tradeoff between diameter and wall-thickness is complicated: larger diameter tubes with thinner walls tend to be stiffer, but much more fragile and have nastier (more abrupt) failure modes, and as the walls get very thin I think they effectively get less stiff again as they become so fragile that they collapse under ordinary loads.
But what people mean by saying tubes are stiffer is that they are stiffer for a given amount of material, not stiffer for a given diameter.
(This whole area is something engineers spend a lot of time thinking about and there is a lot of literature. I don't have any pointers to it, sadly, as what little I know about this stuff comes from speaking to people about bike frames & car chassis design.)
An earlier version of this answer mistakenly used the term 'strength' to mean 'stiffness'. Here's the difference, as I understand it (disclaimer: not an engineer):
- strength tells you at what point something will fail -- either completely or by deforming in some way from which it does not then recover (passing its elastic limit in other words);
- stiffness tells you the how the deformation of the object goes as the stress on it -- it's basically the slope of the linear section of the strain-stress curve before the proportional limit is reached.
I think the definition of strength I've used is slightly dependent on the regime: some engineering components are designed to deform permanently in use (think of crumple zones in cars, for instance), and for those you'd need a more complicated definition of strength.
Finally it should be clear that stiffness and strength are not the same thing: something can be very stiff but can have a rather low maximum stress.
One can quantify tfb's answer through the second moment of the shaft's cross sectional area. Let the $z$ axis lie along the shaft, and suppose the latter is put under torsion $\tau$ about the $z$ axis. Then, if the shaft's length is $\ell$, the angular twist $\varphi$ along the shaft's length is given by:
$$\varphi = \frac{\ell\,\tau}{G\,J_z}$$
where $G$ is the shear modulus (measuring the material's resistance to torsion) and $J_z$ the second moment of area. The bigger $J_z$ is for the cross section, the less pronounced is the twist.
The second moment of area of a disk of radisu $r$ about an axis normal to it and through its center is:
$$J_z = \pi \frac{r^4}{2}$$
$J_z$ is additive for regions, so for a ring shapen region of inner and outer radius $r_i$ and $r_o$ is $\pi \frac{r_o^4-r_i^4}{2}$.
You should be able to see readily, and if not you can crunch some numbers to do so - the removal of most of the inner part of the cross section makes very little difference to $J_z$ and thus little difference to the shaft's torsional stiffness.