Edwin Pawlowski wrote:

"CL (dnoyeB) Gilbert" wrote in message

Say what? I disagree. Further a 1/2" tube should not be stiffer than a
1/2" rod.


I didn't write the laws of physics, I just use them like everyone else. Why
are airplane wings not solid?

Weight, mostly...

I'm not sure what laws you're thinking of...let's see--if we consider a
simple beam w/ uniform load w/ simple support at both ends the maximum
deflection at the center is 5/384 (W*l^3)/(EI) where

E = modulus of elasticity (material property only)
I = moment of inertia (dependent on geometry)
W = applied load
l = length

Now for a rod Irod = MR^2

where M = mass of beam and R = radius

and for a hollow tube it is Itube = M(R1^2 + R2^2) where

R1,R2 are inner/outer radii, respectively.

This superficially makes it look like ItubeIrod for R2 = R, but that
doesn't include the mass which will be less for a hollow tube than for a
solid rod.

Since we're after comparing two geometries of the same material, we can
consider the density of the two to be the same as well as the length.
On that basis, for the rod the weight/unit length is

mRrod = density*pi*R^2/4

and similarly,

mTube = density*pi*(R2^2-R1^2)

Substituting into the formulae for I the geometrical terms for each M we
get that for each the moment of inertia is proportional to

iRod ~ R^4

and

iTube ~ (R2^2 - R1^2)*(R1^2 + R2^2) = R2^4 - R1^4

Thus, it can be seen that the moment of inertia for the tube section is
always slightly smaller than that of the solid rod and since I is in the
denominator of the deflection, the larger deflection will occur for the
tube, not the rod for R2==R (the outer diameters equal).

If you figure on an equivalent weight basis, the tube will be stronger
as the same amount material will be located at a farther distance from
the neutral axis.