Can a vertical round tube safely support a cantilevered arm mounted near the top of the tube?


  • 6061 Aluminum
  • 2 inch Outside diameter
  • 0.25 inch wall thickness
  • 3 feet height


(I don’t know the weight of the interface clamp but it is aluminum)


  • The Arm is 4 feet long with a working load of 100 lbs. at the end of the arm
  • The arm itself weighs 11.4 lbs


  • The manufacturer of the articulated cantilevered arm (dectron USA) sells a Steel 2 inch OD, 0.125 inch wall thickness tube for this purpose. But their solution won't work for my application.

  • I am not worried about the strength of the arm it is rated to support a 100 lb. working load.

  • Don’t think it matters but I have two 12” sections, one 6” and one 18” all rigid
  • The only limit I want is 100 lbs. at the end of the arm other than that I want to use this thing with confidence for decades:)

  • Should I consider High-Strength 2024 Aluminum with the same dimensions instead?

  • OR could I safely use THINNER walled tubing, 0.125 instead of 0.25 inch wall thickness, 6061 Aluminum, 2 inch OD?

Thank you very much for looking at this and your help

  • $\begingroup$ What exactly are you asking? can you provide a sketch or something? $\endgroup$ Commented Mar 24, 2016 at 2:14
  • $\begingroup$ I would like an answer but if possible I would like to learn how to do the required calculations myself too. I have been searching the web on how to do this, thanks:) $\endgroup$
    – serac
    Commented Mar 24, 2016 at 2:18
  • 2
    $\begingroup$ I think my point was that your question isn't clear enough to provide someone the information required to do any calculations. You've provide good information about the individual members but not much on the assembled built structure. How long do you want the cantilever? How tall do you want the tube? How are these things attached to the supporting structure? What do S the supporting structure? Etc. A sketch would be very helpful. $\endgroup$ Commented Mar 24, 2016 at 15:43

1 Answer 1


I'm going to run with this assuming the arm looks like the following diagram (I'm ignoring the 11.4 pounds of the bar for now to make the concepts easier to explain - that can be added later by assuming all 11.4 pounds run through the center of the bar.):

enter image description here

This is a pretty simple setup. To figure out the loading on the shaft, we have to move the load from the tip of the arm to the center of the shaft. Fortunately, we can use the principle of force couples. The basis is that a force off-center can be moved to the center, but you have to add a couple, or moment, equal to the force times the distance moved. Here is what I came up with.

enter image description here

Now we can apply some basic formulas. We need to look at the cross section of the hollow tube and get some basic properties of the cross section, as well as on the material (we need yield strength $\sigma_y$, and $E$, the young's modulus). The important sectional properties are the area, $A$, the area moment of inertia, $I_{xx}$, and the Section Modulus, $S_{xx}$. Formulas for these can be described below (site also has a calculator).

enter image description here

Now we can start looking at the four methods of failure:

We can eliminate vibration immediately as a method of failure, but for reference, the critical frequency for a cantilevered beam with an end weight only is: (Ref Roark's Formulas for Stresses and Strains, Eighth Ed., Table 16.8, Case 3a) $$f = \frac{1.732}{2\pi}\sqrt{\frac{EI_{xx}g}{WL^3}}$$ $$f = {0.2757}\sqrt{\frac{10000000 \frac{lbf}{in^2}*0.537in^4*386\frac{in}{s^2}}{100lbf*(36in)^3}} = 5.811 Hz$$

Deflection in z (down the pole) is minimal. In this case then: $$\delta_z = \frac{W*L}{E*A}$$ $\delta_z = \frac{100lbf*36in}{10000000 \frac{lbf}{in^2}*1.374{in^2}} = 0.000262 in$ which is clearly negligible. Deflection in x/y (across the body) is likely not. In this case $$\delta_x = \frac{ML^2}{2EI_{xx}}$$ (Ref Roark again, Table 8.1 Case 3a). $$\delta_x = \frac{5000 in*lbf * (36in)^2}{2*10000000\frac{lbf}{in^2}*0.537in^4} = 0.603 in$$ This is fairly significant and should be considered (the column will tip over just shy of 5/8").

Stress is also a concern here. We need to look at the stress from the bending moment as well as from the downward force. In this case, the formula is: $$\sigma = {F \over A} \pm \frac {M}{S_{xx}}$$ (Ref: Wikipedia). $$\sigma = \frac{100 lbf}{1.374 in^2} \pm \frac{5000 lbf*in}{0.537in^3} = 72.78psi \pm 9310.98 psi$$

Two things to note - the bending part dominates over the downward portion, so this is the most important part to look at in a preliminary analysis. The second is that this is pretty high, but not terrible. It has a 4.26 F.O.S, which is better than 1.8 typically used in most structures - but aluminum has a fatigue life that needs to be investigated. Repeated quick loading (pushing on it and pulling within seconds) could cause break, but it should have a decent life. This, however, is a separate question entirely.

Buckling is where things get interesting - we need to account for both the moment buckling factor of safety and the downward force factor of safety separately, and combine. Read the Wikipedia article for more information about buckling, but suffice it to say that, with the exception of metal bars (like you have here), it's mostly empirical and relies on a lot of testing. I'll put the formulas here to cover the details so you can get going on further testing if needed. The formula for the downward force is: $$P' = 0.25*\frac{\pi^2*E*I}{L^2}$$ Source: Roark's again, Table 15.1, Case 1a. This leads to $P' = 40895 lbf$, or a FOS of $\frac{40895lbf}{100lbf} = 409$. Bending is $$M' = 0.72*\frac{Ert^2}{\sqrt{1-\nu}}$$ (Source is Roark's again, Table 15.2, Case 16) $\nu$ for aluminum is 0.35. This leads to $M' = 558312 in*lbf$, or FOS of $\frac{558312 in*lbf}{5000 in*lbf} = 111.6$. Combining the two is $$\frac{1}{\frac{1}{111.6} + \frac{1}{409}} = 87.7$$ Therefore, buckling is not a method of failure. (Anything over 5 is usually acceptable without further analysis). Therefore, the 2" aluminum tube is satisfactory for the given loads

Now onto the additional notes:

Should I consider High-Strength 2024 Aluminum with the same dimensions instead? You should be fine with this design

Could I safely use THINNER walled tubing, 0.125 instead of 0.25 inch wall thickness, 6061 Aluminum, 2 inch OD? This I leave as an exercise for you. It should be pretty easy to investigate with the tools given - though note the deflection was already fairly high.

  • $\begingroup$ Thank you very much Mark, William, et al! I did find two online calculators that found a very similar amount of deflection. William I saw your request for a diagram and was putting it together then saw Marks post / diagram. This is exactly what I was looking for:) I try to "pay it forward"on the web and off. $\endgroup$
    – serac
    Commented Mar 24, 2016 at 20:27

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