I'm building a set of monkey bars for my kids and want to use aluminum tubing (6061 T6) for the rungs. The tubes will be 3 feet long and 1 inch outside diameter. Supported on both ends with a circular metal pipe bracket. I have a choice of various wall thicknesses, all the way up to a full solid round bar. How do I calculate the minimum wall thickness I need to safely support (without any permanent deformation) something like 200 pounds at the middle point of the rung?


1 Answer 1


We use this equation for a simply supported beam loaded at its center with 200lbs. M= P x L/4 =200 x 3 /4 = 150 lbs.ft
for calculating the tube stress we convert this to 150 x 12 = 1800 lbs.inch

For a hollow round tube we calculate section module which is

$Z = 0.78 (R_o^4 -R_i^4)/R_o$ assuming you have 16 gauge tube with 0.065 inch walls, Z=0.1838 .

Now we have $sigma = M/Z so 42000 (allowable aluminum stress) = M/Z $

$M = 42000 x 0,1838/4= 1929 lbs$
Usually in applications like this which is dynamic loading they use a safety factor of 3 so you can expect this tubing to be able to support approximately
$ 600 lbs$.

Then again there are many standards of aluminum. For this illustration I assume 2024-T3 which has hi strength, you need to find out what you have.
This answer is just an illustration and you should check with the manufacturer about their tests on strength and ductility. Also connections and wall construction has to be checked.

  • $\begingroup$ Thanks! I have 6061-T6 alum. According to wiki it has a yield strength of at least 35ksi. I can get relatively cheap tubing with a 0.125 wall, which seems to come out to about 530 lb. That sounds like a large enough safety margin even for an adult that I shouldn't have any problems. $\endgroup$
    – Egor
    Commented Aug 4, 2016 at 20:35
  • $\begingroup$ @Egor i would check for a source on dynamic loading on these tubes. I know by just jumping to grab one can apply 10 times their weight. We talk a totally different ballpark when we talk about play/sport structures. $\endgroup$
    – kamran
    Commented Aug 5, 2016 at 0:31
  • $\begingroup$ 42000=M/Z, solving for M, you get M=Z*42000 or .1838*42000 = 7719.6. My question is where does the division by four come from to get 1929 lbs? *I am assuming allowable aluminum stress is sigma. $\endgroup$
    – Ryan R.
    Commented Jul 21, 2023 at 2:24
  • $\begingroup$ How does the tube fail when overloaded? Does it buckle, or does it tear? $\endgroup$ Commented Apr 26 at 2:13
  • $\begingroup$ It fails by bending too much and buckling under compression on top. typical failure pattern is a sharp crumpled in around the top middle of the bar and folding the bar. $\endgroup$
    – kamran
    Commented Apr 26 at 6:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.