# Large deflection of cantilever beam

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.

• I think that you did something wrong, since according to Wikipedia I do not see any therms with a sine. Commented Dec 30, 2013 at 21:38
• 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. Commented Dec 30, 2013 at 22:09
• Do you measure deflection along a fixed axis, or does it follow the arc of the end of the beam? Commented Apr 21, 2015 at 12:37

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"