stiffness of a tube Vs. a rod of the same material in a simple cantilever application.

If the tubes dimensions are .024" outer diameter and .019" inner diameter.  
The diameter of the rod is .019".  
This give you a cross sectional area of 1.69 x 10^-4 sqin. for the tube and 2.84 x 10^-4 sqin. for the rod.  

I am looking, which is stiffer the tube or the rod and lets say that the lengths are 12" long for both tube and rod and both weigh the same 1-lbm.

I calc. the mass moment of inertia of the tube and the rod and found that the tube and rod's mass moment of inertia are the same, ( 12-lbm-in^2). But this is not a function of stiffness. I think that we need to look at the flexural stresses/ the extreme fiber stresses in this situation.

Equation for calculating the flexural stresses/ the extreme fiber stresses in the tube or the rod. Sigma = [(Bending Moment) X ( radius to the outer fiber of the tube or rod)] / Mass moment of inertia  


So this would give you a smaller flexural stress in the rod. The smaller flexural stress means a stiffer member. So the rod is stiffer in a simple cantilever application for flexural stress and in the mechanical stiffness application the rod is still stiffer. But in a rotational or torsional application the tube is stiffer. It depends on what you are going to do with the tube or the rod.

Flexural stiffness is a function of the moment of inertia, I, of the cross section.  For the solid rod,I = PI x .25 x R^4 = PI x .25 x 0.095^4 = .00006397 inches^4

For the tube, it is PI x .25 x (R^4 - r^4)= PI x .25 x (.012^4 - .095^4)= .00006395 inches^4

The rod is infinitesimally stiffer than the tube.

You felt into a big and very dangerous misunderstanding: stiffness and resistance is not the same.
Stiffness is proportional to the product EJ, and if the material is the same, is proportional to J, the geometric moment of inertia about the neutral axis.
Resistance depends also on the allowable stress (or yield limit), and, for same materials, is proportional to J/r.
To be clear: higher stiffness means lower deflection under the same load, higher resistance means you can accept a higher load.
All the above was for pure bending: if you add normal stresses or torsion, then things change and a separate treatment is needed.

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Firstly, the areas as your calculations show, are different for both rod and tube. Therefore, the rod and tube cannot possibly have the same weight and the same mass moment of inertia. However, you are right in looking at flexural stresses in a cantilever application and that the rod in this situation is the stiffer of the two.
A simple way of looking at the stiffness in bending (cantilever application)would be to consider the "second moment of area" of both tube and the rod. These can be calculated as follows:
I(rod) = Pi x D^4 /64
I(tube) = Pi x (Do^4-Di^4)/64
Where Do and Di are the external and internal diameters of the tube.
Again, for torsional stiffness the "polar second moment of area" should be considered ie.
J(rod)= Pi D^4/32
J(tube) = Pi (Do^4-Di^4)/32
Thus the tube in this case would be the stiffer in torsion also.

It would appear I made a mistake, I thought the rod and tube were the same outside diameter ie 0.024",using the
formula I suggested earlier its shows the tube to be stiffer in bending (cantilever case) than the solid rod.

You're all correct - of course tubes are less stiff than rods of the same OD but a lot lighter - so for stiffness for weight always go for a tube.

But be careful of failure modes! Tubes buckle when overloaded whether in bending, torsion or compression - and they fail a lot more suddenly and catastrophically than rods. Surface flaws may precipitate bucking failures at lower levels than you would expect from the analysis of a 'perfect' tube.

Don't know your application but this may be relevant.