Re: Rod or tube...which is stronger?

Hello Ewen,

 You asked the question "For the same outer diameter, is a solid rod of 
material stronger than a hollow tube of the same material?

The simple answer is 'Yes', the solid rod will be stronger.  But then we 
have to consider- What does it mean to be "stronger"?  Let’s look at 
several scenarios:

1) Tension
If the rod is in tension (the same as a rope or a cable), then the total 
cross-sectional area of the rod will determine its strength.  In other 
words, more cross-sectional area gives more strength.

As an example, let’s take two rods: Rod 'A' is .50" diameter, solid.
Rod 'B' is hollow, with a .50 outer diameter and a .30" inner diameter 
("ID").  

Each rod is pulled in tension with 200 pounds of force.

The 'Stress'** on each rod is determined by the ‘force’ per unit area:

**(Any given material has a particular value of "stress" that it can 
resist before it will "fail" when loaded)

Area 'A' = pi  x D^2  /4  = 3.14 x .5^2 / 4 = .196 in^2

Area 'B' = pi x (D^2 - (ID)^2)/ 4 = 3.14 x (.5^2  -.3^2) / 4 = .125 in^2

Stress 'A' = 200 / .196 = 1020 psi.
Stress 'B' = 200 / .125 = 1600 psi.

Clearly, rod A will be stronger because it is under a smaller stress.

If we now look at the 'bending' resistance of the rods, it becomes a 
little more complicated:

The bending resistance of an object is determined by several factors 
which include the properties of the material, and another engineer factor 
called the "moment of inertia" (designated “I”).  I is determined by the 
geometry of the cross-section of the object (a rod in our case). Think of 
it as the "stiffness" factor due to the shape.

The I for a round, solid object is determined by the following formula: 

 I =  ¼ R^4     where R = outer radius 

( I use the “^” symbol for “to the power of” because this site does not 
accept superscript)

For a hollow rod, the formula would be:

I = ¼ (R^4 – r^4)     where r = inner radius.

For both of our rods, from above, we have the follow:

I (rod A, solid) = ¼ (.25^4) = .000976 in^4

I (rod B, hollow) = ¼ (.25^4 - .15^4) = .000850 in^4

Notice how little difference there is between the two ‘I’ values (only 
about 13%)!

The difference in stiffness between the solid and hollow rod is only 
13%.  If you calculate the weight differences (I will leave that to you), 
you will see that the hollow rod is 36% lighter than the solid.

So, you can see that for engineering design, a hollow rod is a much more 
efficient use of materials, and uses much less weight than an equivalent 
solid rod, even though the solid rod will be “stronger”.  Think about 
both of these rods spanning a large horizontal distance:  Which rod would 
sag more due to its own weight?

Thanks for an interesting question!

Best Regards,

Jay Shapiro