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I recall being told once that hollow rods/shafts tend to resist torsion more than solid rods/shafts...but I wasn't told why this is the case.

Now that I'm a little older, running this "fact" through my mind, my intuition tells me that a solid rod should be able to resist torsion better than a hollow one (assuming the wall of the hollow one is thick enough to withstand being "dimpled" in the process).

But I need this verified.

A quick spot of Googling didn't lead to any answers (maybe it did, but I couldn't comprehend most of the stuff out there...way off-topic for a high-school student).


Physics.SE has this post that's similar to my question; however, that post requires a comparison between a hollow shaft's and a solid shaft's resistance to bending, whereas I want to know about their resistance to torsion.

I (vaguely) understood the ideas expressed in the answers there (still trying to wrap my head around the "second moment of inertia")...but I'm not sure if those ideas apply to torsion as well as bending of the shafts.


Can someone tell me (in a simplified way...not too simplified though, I don't want to miss out on all the good stuff) why a hollow shaft tends to resist torsion better than a solid one?

(Since it was mentioned in the question I linked)

The two possible cases that arise are : $1)$ Solid and hollow rods of same diameter, $2)$ Solid and hollow rods of same mass.

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    $\begingroup$ Are you sure that the claim is that they're more resistant rather than more resistant per unit mass? $\endgroup$ – Emilio Pisanty Sep 3 '17 at 10:58
  • $\begingroup$ @Emilio Yes, it was a pretty vague claim...which is also why I posted this here wanting to know how accurate it is. The two cases of interest (as I mentioned in the post) seem to be between rods having same mass, or rods having same outer diameter. $\endgroup$ – Alan Sep 3 '17 at 11:05
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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.

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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.

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  • $\begingroup$ Quick note that I didn't catch earlier: The $\tau$ you have as torsion should be a $T$ for torque. $\endgroup$ – JMac Sep 3 '17 at 17:05
  • $\begingroup$ Aka, his intuition is right, but that would waste material in applications which do not need to resist forces in weird general directions. Besides weighting more. $\endgroup$ – Vendetta Sep 6 '17 at 20:05
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The stiffness depends on the concentration of the material present near center of gravity. If the material has high concentration of material near center of gravity then it is less stiffer( solid shaft).But in hollow shaft the material concentration is lesser near the center of gravity, so it is more stiffer.But when I told this answer to my Professor he asked me a simple answer... If any one know answer it.....

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The case of same mass can be understood by comparing the torque per unit twist for both the cylinders (https://i.stack.imgur.com/JS1no.jpg)

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