I'm trying to figure out what size of round tube to use for a tensegrity structure that'll probably have people climbing on it.

The rigid elements are 21' steel tubes, and in normal cases they undergo ~800 lbf of compression.

Assuming it's horizontal and pinned at both ends, and is under ~4000 lbf of compression (design * safety factor 5) -- I want to know how much load it can withstand at the center, given the inner and outer diameters.

Approximate answers are fine, but I'm not sure how to approach this.

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    $\begingroup$ This is a real safety concern, and you should get an analysis by an engineer to support your design at the very least. This will be necessary for insurance purposes as well. $\endgroup$ – Solar Mike Dec 4 '18 at 5:50

Just as an illustration, without any intent for real application, we assume the following.

There is no restriction on deflection, however we will control for buckling.

The tube will fail under compression stress.

lets assume R1 and R2 as outer and inner radiuses.

We limit the compressive stress to 10% so that we do not need to deal with column design charts.

We also limit the radius of gyration to r<20 and r for a hollow tube is approximated to r= R2

Now we can say $$ \sigma_{axial} = \dfrac{4000}{\pi(R1^2- R2^2)}<\dfrac{F_{yield }}{10}$$

And we know the bending moment under concentrated load in the middle is $m =\dfrac{\omega \times l}{4}$, therefore

$$\dfrac{\omega \times l}{4} = \ 0.9\times \sigma \times \dfrac{\pi\left(R_{1}^4-R_{2}^4\right)}{32R_{1}}$$

And by plugging in R1 and R2 and the allowable sigma we find the $\omega$.

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  • 1
    $\begingroup$ This is just a static analysis to see if the yield stress is exceeded. On top of that, global and local buckling need to be checked. Besides that, fatigue and dynamic amplification need to be considered. $\endgroup$ – Orbit Dec 4 '18 at 13:22

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