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How can I find the amount of point force at the end of a cantilever plastic beam that produces e.g. 45° slope at the end of the beam? Is this the right equation: $$F=\frac{2EI\theta ^2}{L^2 \sin(\theta)}$$

I derived this equation by putting $x=\frac{L}{\theta}\sin(\theta)$ and $\theta_x=\theta$ in the following equation from the "Large and small deflections of a cantilever beam" paper: $$x=\sqrt{ \frac{2EI}{F} } \left(\sqrt{\sin(\theta)}-\sqrt{\sin(\theta)-\sin(\theta_x)}\right)$$

where $\theta$ is the maximum slope at beam end and $\theta_x$ is the beam slope at x.

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  • $\begingroup$ I think that you did something wrong, since according to Wikipedia I do not see any therms with a sine. $\endgroup$
    – fibonatic
    Commented Dec 30, 2013 at 21:38
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    $\begingroup$ That's the whole point, as most of the equations are for small deflections. The equation which I have stated above is, I think, for large deflection, but I need somebody to confirm it. $\endgroup$
    – NESHOM
    Commented Dec 30, 2013 at 22:09
  • $\begingroup$ Do you measure deflection along a fixed axis, or does it follow the arc of the end of the beam? $\endgroup$
    – John Alexiou
    Commented Apr 21, 2015 at 12:37

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I'm not sure if you are still interested, but I believe the equation you are looking for is:

$$F = \frac{2\sin(θ)EI}{L^2}$$

where θ is the angle at the end of your cantilever. I base this equation #16 from the paper, "An integral approach for large deflection cantilever beams"

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