Tube Strength Notes

# Use at your own risk.

Verify the ideas below from principles on the links provided or other sources. These are notes, not necessarily facts, and you use them at your own risk. Please correct or argue with errors you find.

## Strength and stiffness

In changing the shape or dimensions of a piece of given material (tube or beam), anything that would make the shape stiffer will make it stronger, because the same load distributions that cause deflection are the ones that cause failure. If you are comparing different materials, this does not apply. This idea also does not apply to complex structures with different materials (like an entire flexible composite wing engineered for a sailplane), where some elements of one material might flex, allowing the load to be distributed to other elements of other materials at high flex, making the structure both flexible and stronger.

## Strength of tubes or cables under tension

This principle would likely apply to short support tubes whose failure mode would be ripping at joints, whether the tube itself was under tension or compression. Note that increasing the size of a steel bolt attaching two aluminum tubes can decrease the joint strength if the aluminum tube rips before the steel bolt shears off, because it reduces the effective cross sectional area of the tube that supports the bolt (you would want the bolt and tube to fail at the same time for maximum strength).

The strength is proportional to the tube's (cable's) cross sectional area. So to double the strength, you would have to double the cross sectional area, and hence double the weight of that piece. For a thin walled tube $w=(D-d)/2$ is much less than D (where D and d are the outside and inside diameters), the cross sectional area is $A \approx 2\pi dw$ so doubling the diameter or wall thickness will have the same effect (even if it's torque at the joint that caused the failure, since doubling either the thickness or the diameter would double the torque the tube could apply at max strength.)

## Strength of tubes against folding

This would likely apply to long tubes such as wing spars, struts, etc which would tend to fail due to bending/folding in the lengths between supports.

Cross sectional shape
The bending stiffness and strength are proportional to the section modulus of the tube, which for a pipe is its second area moment divided by its radius, or

(1)
\begin{align} S = \frac{\pi}{32} \frac{D^4 - d^4} { D} \end{align}

where D and d are the outside and inside diameters. If it's a thin walled tube $w=(D-d)/2$ is much less than D,

(2)
\begin{align} S \approx \frac{\pi}{4} D^2 w \end{align}

So here's the point: for thin-walled tubes, the strength of the tube increases much faster with D (square relationship) than with thickness w (linear relationship). Doubling the diameter will give four times the strength. Doubling the thickness will only double the strength. Both methods will double the weight. (Drag on the tube will increase linearly with D)

But you can't increase D indefinitely and let $w$ get too small. One expert said that if $w$ is less than 6% of D, it becomes unstable against failure first by "crinkling", rather than by inherent material failure.

Length
For a beam of length $L$ with a uniformly spread load $W$, the maximum stress at the center is

(3)
\begin{align} \sigma = \frac{WL}{8S}. \end{align}

Similarly for a cantelevered beam with a uniform load the stress at the fixed end is

(4)
\begin{align} \sigma = \frac{WL}{2S}. \end{align}

Here's the takehome lesson: the maximum stress grows linearly with $L$, so the strength is inversely proportional to the length of the tube. If you put a support in the middle of a beam or wing, you actually increase the strength by four: you decrease the unsupported length $L$ by two and you decrease the load $W$ on the unsupported region by two.

page revision: 63, last edited: 31 Dec 2009 01:02