Are hollow pipes less likely to bend under weight

[Nicholas]

I have recently heard a rumor that a hollow pipe can endure a greater load without bending than a solid rod.
Is that true, and why?

Thanks.

[jamesrc]

No that is not true (at least not as you have stated your question). Here's why:

First thing: Both pipes will bend since neither pipe will have infinite stiffness; under a given load, both pipes will bend some finite amount.

Second thing: The character of the bending depends on the materials used and the geomoetry of the pipes. So it is possible to design a hollow pipe that is stiffer than some solid pipe. However, since you didn't state specifics, I have to assume that both pipes are made of the same material and both pipes have the same outer diameter, d. (I'm also assuming they're round.)

Now we can look more closely at the problem:

The bending stress in the beam (pipe) is given by the following:

\sigma(x) = \frac{M(x)c}{I}

where M(x) is the external moment as a function of x (same for both since loading conditions are the same), c is the distance from the neutral axis (will equal d/2 in this problem when finding the maximum bending stress) and I is the area moment of inertia (aka second moment of area) which is a function of the cross section geometry.

(If you look at the Euler beam bending formula, you will find the same dependence of deflection on moment of inertia, which is the key to this problem.)

The moment of inertia of the solid pipe:

I_s = \frac{\pi d^4} {64}

The moment of inertia of the hollow pipe (same outer diameter, d, along with inner diameter of di):

I_h = \frac{\pi \left(d^4 - d_i^4\right)}{64}

Since the amount of bending is inversely proportional to the moment of inertia, the pipe with the lower moment of inertia will bend more. Looking at the equations you should see that the hollow pipe will bend more.

[jamesrc]

A little addendum:

I just realized you may have meant that the loading was solely due to the weight of the pipe. If so, consider the expression for the maximum deflection in a beam under a uniform load (you can get these too if you work out the Euler beam bending equation):

if it's cantilevered:
\delta_{max} = \frac{qL^4}{8EI}
if it's simply supported at both ends (max deflection occurs at center of beam here):
\delta_{max} = \frac{5qL^4}{384EI}

(If you are unfamiliar with the definitions of any of these terms, please ask.) In either case (and assuming both beams are made of the same material (E)), the maximum deflection is proportional to the term q/I.

For the solid beam:

\frac q I = \frac{\frac{\rho L \pi d^2 g}{4L} }{\frac{\pi d^4}{64}} = \frac{16\rho g}{d^2}

For the hollow beam:

\frac q I = \frac{\frac{\rho L \pi \left(d^2-d^2_i\right) g}{4L} }{\frac{\pi \left(d^4-d^4_i\right)}{64}} = \frac{16\rho g}{d^2+d_i^2}

Assuming I haven't messed up my math here, the bending for the hollow beam is less than the bending for the solid beam. The solid beam is still more stiff, but it also undergoes more loading (because the solid beam weighs more). I hope that helps.

[enigma]

What you may have heard is this, Nicholas:

Between a solid pipe and a hollow pipe of the same weight, the hollow pipe will be more resistant to bending. It will also have a much larger radius.