For a cylinder or tube made from a given material, and of fixed diameter, is it ever the case that a hollow cylinder is stronger than its solid counterpart? By stronger I mean more resistant to bending. If the answer is “yes”, is that because the deforming forces would only be distributed across the diameter of the solid tube, but around the circumference of the hollow tube?
Thanks.
I don’t have cites ready but for a given amount of material, a hollow cylinder wll be stronger. For a given diameter, a solid will be stronger. In the first case, the hollow cylinder has a larger diameter and so the force is spread over a larger area.
Nope, solid always wins (for a constant outside diameter and material).
Let’s do the math:
As far as bending is concerned, the governing sectional property is the moment of inertia (I). The larger the I, the more resistant the section to bending.
For a solid cylinder:
I = ([sym]p[/sym]/4)R[sup]4[/sup]
where R is the outside diameter.
For a hollow cylinder:
I = ([sym]p[/sym]/4)(R[sup]4[/sup] - R[sub]i[/sub][sup]4[/sup])
where R[sub]i[/sub] is the inside diameter.
From this you can see that for any value of R[sub]i[/sub] (except zero), the resultant I will always be smaller than a solid cylinder.
No. For two cylinders of the same material, length and outer diameter, the one with the greater wall thickness (or smaller inner diameter) will always be more resistant to bending.
An illustration: put one end of each cylinder in a vise so that the free end is suspended horizontally. Hang a weight on the free end. The cylinder is now being forced to bend. The top surface will be in tension and the bottom surface will be in compression. The equations get pretty complicated so I won’t go into them (besides, it’s been a while).
The material on the outside is under the most load, while the material at the center is hardly loaded at all. That’s why I-beams are shaped as they are: put the most material where the most load is and you have an efficient use of material.
In any case, given enough load, a hollow cylinder will eventually fail. Under that same load, the solid cylinder will not fail because it has more material to resist the load.
As a follow-up, it should be noted that hollow cylinders/tubes are better from a practical standpoint. From the equations above, you can see how you can actually take out a good deal of the material from the center of a section and not lose a lot of strength. This allows a good deal of savings from a weight perspective.
Fer instance, for a 5 inch diamter solid cylinder, it’s moment of inertia would be 30.68 in[sup]4[/sup], with a cross sectional area of 19.63 in[sup]2[/sup].
Take the same cylinder, but take out the center leaving a inside diamter of 3 inches (1 inch wall thickness). This shape has a moment of inertia of 26.7 in[sup]4[/sup], with a cross sectional area of 12.56 in[sup]2[/sup].
This means that the hollow shape maintains 87% of its “strength”, with only 63% of its mass (or to put in another way, you only lose 13% of the strength, but lose 37% of the weight).
Allright, before anyone brings this up and gets all technical on my butt, I guess there could be an exception depending on how you define the rules.
Say we took a simple cantilever beam (fixed at one end, dangling horizontally), and defined its strength to be “how much additional load can it take before it fails.” I guess you could argue that for a certain weak material and offbeat diameter/wall thickness combination, that there might be a case where the inherent deadweight of the solid cylinder is already putting the cylinder within a gnat’s heiny of failure (i.e. can just barely support its own weight), whereas the hollow one has enough weight removed to give it a margin in terms of applied load.
But if we are talking in terms of most materials & sections, overall strength, and load-resistance, solid wins.
It’s a stupid exception, but it is one.
Note that if the amount of material and length are fixed, the diameter will vary. Since I wasn’t sure which question the OP was asking I answered both possibilities. I think {b]Why A Duck**'s equations answer the question most clearly and succintly, thanks.
I don’t have a cite, but I know that I can easily bend a 5 foot length of 1" mild steel rod (one end is firmly held), while I can’t bend a similar length of mild steel tubing.
Diameters of tube? Grade of steel for both?
Thanks. That was very helpful.
donkeyoatey’s experience is similar to what I thought I too had noticed. Hence my question.
A lot of it really does matter on steel chemical composition and heat treatment.
The measure of strength when you’re talking materials is something called yield strength which is given in terms of pounds/square-inch. This is the amount of stress that the material can take before it begins to yield (i.e. plastically or permanently deform).
A lot of solid round bar, such as re-bar, really isn’t meant to be a structural steel, and as such may be pretty weak.
Whereas steel tubing is usually used in a structural application (as opposed to copper or brass tubing which is pipe), so it tends to have a higher yield strength.
So it would be easy to have a case where a hollow steel bar is stronger than a solid steel bar, it’s just that the two materials were not identical, which was once of your constraints.
I know that donkeyoatey mentioned that they were both mild steel, but that term doesn’t really have a lot of meaning. You need to know things like carbon content and manufacturing process to really compare the two.
Hold on… Moment of inertia? That’s not a strength measurement. When you’re talking about moment of inertia, you’re assuming that the body is staying rigid. I just tells you how easy it is to rotate the whole thing, not how easy it is to rotate one part relative to another (i.e., bending).
Sigh…
Source: Roark’s Formulas for Stress and Strain Sixth Edition, Chapter 7, Formula (12)
[sym]s[/sym] = Mc/I
where [sym]s[/sym] is maximum fiber stress due to bending moment, c is the distance from the neutral axis to the extreme fiber, and I is the moment of inertia. The bigger the I, the lower the [sym]s[/sym].
True, it is not a strength measurement by the strict definition of terms, but it is a measure of the resistance to stress due to the geometry. All other things (material, length, etc.) being equal, it does correlate to strength.
:: Why A Duck mutters something about dern physicists…
I’ve gotta go home, but here’s something for you to chew on…
Link.
“The Moment of Inertia (I) is a term used to describe the capacity of a cross-section to resist bending.”
Are you sure it was tubing, and not pipe? 1" standard wall pipe has a 1.315" OD, with a moment of inertia about 80% greater than a 1" OD bar.
Yeah, I meant to say pipe, cynic.
I’ve been trying to find a cite but no luck yet.
I think this is what you’re looking for,KarlGauss.
(too many equations make the baby Jesus cry)
A brittle flaw-sensitive material has a strength dependent on the statistical distribution of imperfections within the material. Depending on this statistical distribution, hollow cylinders could be less likely to break with a given bending moment than solid cylinders of the same diameter and material.
If you have an anisotropic material (i.e. it is not equally strong or stiff in all directions), a hollow cylinder can be made stronger and stiffer than a solid cylinder by orienting the material differently in the two cylinders. It is still the same material, as specified in the question.
If solid cylinders are made such that there is a residual tensile stress on the surface, drilling out the centers to make them hollow and reduce the residual stress could increase their load carrying capablity.
The question is not specific about what counts as the same material, and as mentioned by others, how a material is processed affects its properties. A cold drawn tube would have different properties than a cast solid bar, even if they contain exactly the same number of atoms of each element per unit volume.
Here is something about hollow carbon fibers that may be of interest:
http://www.carbon2001.org/agenda/papers/session%2017/17.5.pdf
Well yeah. Note the key phrase “of the same area”. In order for a hollow cylinder (A=[sym]p[/sym](r[sub]o[/sub][sup]2[/sup] - r[sub]i[/sub][sup]2[/sup]) to have the same cross-sectional area as a solid cylinder (A=[sym]p/sym, the hollow cylinder must have a larger outer diameter and thus a larger moment of inertia. Two entirly different ducks.
Good link by the way. It describes a lot of what I do for a living (design submarine internal structures). Granted a lot of what I do relies on failure of material under shock loads, but for a lot of it (i.e. decks), deflection is a key factor.
Manlob - Well sure, if you want to start talking about anisotropic materials and stacking the deck in favor of the hollow cylinder, there might be exceptions. But my interpretation of the OP was that the question was material independant and based more on a pure geometry analysis.
You guys are just making this too complicated. The answer is simple: You have a hollow tube. Now you put inside it a rod made of the same material. Will it be weaker now? I don’t think so. Ergo, everything else being equal (material, outside diameter), the solid tube will be stronger (unless you can prove the air is stronger than the rod you insert).