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What is Euler buckling load of a beam in axial compression when the axes of loading pins at top and bottom are skew, instead of being parallel?

Should the ends be considered fixed because skewed pins restrict lateral bending?

Sketch of a tube with non-parallel ( skewed ) colored pins introducing compressive loads shown.

Skewed Pins

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  • $\begingroup$ Not clear what you mean. Buckling is caused by axial compression. So what do you mean by "axis of loading pins at top and bottom are perpendicular"? I diagram would be of help. $\endgroup$
    – Bob D
    Jan 7 '20 at 13:37
  • $\begingroup$ The pin axes should be parallel for load $P_{crit}= \pi^2 EI/l^2$ buckling with rotations of opposite sense at either end. But what if they are skewed? $\endgroup$
    – Narasimham
    Jan 7 '20 at 13:53
  • $\begingroup$ Sorry but I'm still having difficulty picturing this. Can't you provide a diagram? A picture is still worth a thousand words. $\endgroup$
    – Bob D
    Jan 7 '20 at 13:56
  • $\begingroup$ "Rotations of opposite sense"? If there is rotational stress involved, that's not axial compression... $\endgroup$
    – probably_someone
    Jan 7 '20 at 22:38
  • $\begingroup$ This was in the edited out question describing standard pinned hinges. The two hinges are smooth and rotate without resistance.in opposite sense (ccw and cw). $\endgroup$
    – Narasimham
    Jan 8 '20 at 9:42
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There's no simple exact formula as for both ends pinned/fixed. The answer will be somewhere in between.

The wikipedia article on Euler's Critical Load gives the general formula $$P_{cr}=\frac{\pi^2 EI}{(KL)^2}$$ where $K$ is the column effective length factor.

The table in Figure 1 gives values of $K$ for the 6 combinations of pinned, fixed and free ends. So for your combination the recommended design value lies between 1.0 and 0.8. Clearly the larger value is the safer option.

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