I stumbled across a little math that allows us to compare the strength of different sizes of tubing. I thought it was interesting, so I am doing a write up in the hope that others will find it useful as well. Please remember to use common sense, and be careful when making important decisions about critical structural components. It's not my fault if you kill yourself!

The idea behind this formula is simple. It allows us to measure the deflection at the end of a piece of tubing thats being bent by force

**F**, while the other end of the tubing is fixed. I think my simple drawing will say the most:

The formula for deflection of our sample is:

**D = (F x L^3)/(3 x E x I)**

**Note:**In the formula above,

**L^3**means "L Cubed."

Where:

**D**is deflection in inches

**F**is the force applied in pounds

**L**is the length of the sample in inches

**E**is Young's Modulus in psi

**I**is the moment of inertia for the shape

The same formula can be used for any shape. Only the value for the moment of inertia (

**I**) changes with the shape of the sample being tested.

So how do we solve for

**I**for a piece of round tubing?

We know that

**I**for a solid round bar is this:

**I = 0.7854 x R^4**

**Note:**

**R**is the radius and

**R^4**means "R to the fourth."

So,

**I**for a piece of round tubing is simply

**I**for a solid bar, minus

**I**for what is removed, leaving behind only the wall of the tubing. Using this information, and skipping the derivation,

**I**for a cylinder (our piece of tubing) is:

**I = 0.7854 x (R^4 r^4)**

**Note:**The big

**R**is the outside radius and the little "

**r**" is the inside radius. Here is what it looks like:

Lets compare the deflection of three pieces of round tubing, all 24" inches long, with a force of 300 pounds applied to the end. Our samples are:

(a) 2 by 1/4 wall, (b) 2 by 3/16 wall, (c) and 1 3/4 by 1/4 wall.

Our first sample is the most common choice for T-Bucket front tube axles, and the last is the less common Total Performance axle. I thought it would be interesting to look at the 2" by 3/16" wall tubing just for fun.

Using the value for Young's modulus

**E**in

**psi**, (that's pounds per square inch),

**E**for mild steel is approximately 30,000,000psi.

Now, lets first calculate our moment of inertia for our three samples before we calculate deflection.

**I**for 2" by 1/4" wall tubing is:

**I = 0.7854 x (1^4 - 0.75^4) = 0.53689**

**I**for 2 by 3/16" wall tubing works out to be

**0.44312**

**I**for 1 3/4" by 1/4" wall tubing works out to be

**0.34054**

(a) Substituting

**I**and all other values into our equation for the 2 by 1/4 wall piece of tubing, we get:

**D = (300 x 24^3)/(3 x 30,000,000 x 0.53689)**

**= 0.0858"**of deflection. (86 thou, not too bad!)

(b)

**D**for our 2" by 3/16" wall piece of tubing works out to be

**0.10399"**of deflection.

(c)

**D**for our 1 3/4" by 1/4" wall piece of tubing works out to be

**0.1353"**of deflection.

Now it's already interesting to note the 2 by 3/16 wall tubing sample is much closer in stiffness to the 2 by 1/4 wall tubing than our 1 3/4 by 1/4 wall sample. Something to think about.

*** Just for fun, lets look at the deflection of a 2" by 3/8" wall tube made of 6061-T6 aluminum. Using a Young's modulus of 10,000,000psi for the aluminum, the calculated deflection is 0.257" inches.***

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Now lets look at the deflection of square/rectangular tubing. Again, the only change in our equation is the moment of inertia (

**I**) because the shape of the tubing is different.

**I**for a solid rectangular bar is:

**I = (W x H^3)/12**

**Note:**

**H^3**means "H cubed."

Where:

**W**is the width and

**H**is the height of our taller side (Important!)

This means

**I**for a piece of rectangular tubing is simply

**I**for a solid rectangular bar, minus

**I**for what is removed, leaving behind only the wall of the rectangular tubing. Using this information, and skipping the derivation,

**I**for a rectangular piece of tubing would be:

**I = (1/12) x [(W x H^3) - (w x h^3)]**

**Note:**Little

**h**and little

**w**are side lengths for the inner removed part of our tubing.

This is what it looks like:

Now lets work four rectangular tubing examples that are all 48" inches long, with a 500 pound force applied to the end. Remember, our dimension

**H**is the taller of the two sides. Our samples are:

(a) 2" by 4" by 1/8" wall, (b) 2" by 3" by 3/16" wall, (c) 2" by 3" by 1/8" wall, and (d) 1.5" by 3" by 1/8" wall. Should be interesting, yes?

Lets first calculate the moment of inertia for our four samples before we calculate deflection.

**I**for our first piece of 2" by 4" by 1/8" wall rectangular tubing is:

**I = (1/12) x [(2 x 4^3) - (1.75 x 3.75^3)] = 2.97624**

**I**for 2" by 3" by 3/16" wall tubing works out to be

**2.05059**

**I**for 2" by 3" by 1/8" wall tubing works out to be

**1.46712**

**I**for 1 1/2" by 3" by 1/8" wall tubing works out to be

**1.20866**

(a) Using the original equation from before to find the deflection for our 2" by 4" by 1/8" wall tubing:

**D = (500 x 48^3)/(3 x 30,000,000 x 2.97624) = 0.2064"**of deflection at the end.

(b)

**D**for the 2" by 3" by 3/16" wall tubing works out to be

**0.2996"**of deflection.

(c)

**D**for the 2" by 3" by 1/8" wall tubing works out to be

**0.4188"**of deflection.

(d)

**D**for the 1 1/2" by 3" by 1/8" wall tubing works out to be

**0.5083"**of deflection.

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Well, that's it for now. I hope some of you find this as interesting as I do.

I order you now to go forth and create great things!

Take care all!

David