A square tube is slid into a larger square tube. If a load is applied to the end of the inner tube, I would think that inner tube would mainly contact the outer tube in two locations as the load is applied since the clearance between the tubes is not very tight. Is the best way to solve for the max bending/shear stress a shear/bending moment diagram based on the fbd? Or, should I assume the inner tube acts as a fixed-free cantilever? Would prefer the more conservative approach.



If you are only interested in approximation, the inner tube can be considered as a clamped cantilever after getting in contact with the walls of the outer tube, thus the applied load is resisted by the forces at the points in contact. If you want to go one step further, the clamped inner tube is resembling a beam simply supported on two hinges with an overhang. At the support point, there is a vertical reaction as usual, and in addition, there is a horizontal force due to friction.

However, if you want more accurate results, you need to resort to advanced calculations involving deformation and rotation of the outer tube, or using FEM, since the problem isn't that simple.

Note, how tight or loose is the connection influences the outcomes the most.

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Aside from @r13's answer, a real-world alternative is that the inner tube after tilting a bit locks into the outer tube and becomes one with it.

If the fit is close enough, we can assume the inner tube's tilt is negligible and consider the combined two parts as one cantilevered beam with variable I.

Then the I, the second moment of area, of each tube will be

$$ I_{n}= \frac{bh^3 - b_1 h_1`^3}{12}$$

  • b = base of the tube

  • b1 = base of the opening

  • h = height of the tube

  • h1 = height of the opening

And the second-moment area of the double-thick part will be the sum of the two I's.

And of course, if we could attach the two parts together with fasteners strong enough to resist the shear flow the thick part will be much stronger and act as an integrated piece with a thickness equal to the sum of the two tubes which is much stronger than adding their sperate I,s.

No matter which way we go the moment diagram will be triangular with the maximum moment at midspan support and the Max bending stress will occur at the top and bottom surface of the inner tube at that support. The max shear is at the midspan but max shear stress is any point along the inner tube cross section.



  • $\begingroup$ @bigboi, if you modify your answer, adding more details, i can better help you. dimensions in the hollow tube size and thickness, the magnitude of forces and reactions?? $\endgroup$
    – kamran
    Sep 11 at 23:20

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