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I am looking at a simple cantilever beam deflection:

enter image description here

I understand the general expression for deflection/force would be:

$$y_s = \frac{Fx_s^3}{3EI}$$

$$F_p = \frac{3y_sEI}{x_s^3}$$

If you were going to add viscous damping to the bending of the beam, would it be as simple as:

$$F = \frac{3y_sEI}{x_s^3} - cEI\theta_t$$

Where the equation for the angle of deflection is $\theta = \dfrac{FL^2}{2EI}$?

I have seen some suggestions that simple damping of cantilever beams is done by applying viscosity to the rate of angle change with respect to time. Is that generally correct?

I have had some strange behaviors trying this so I'm not sure what the ideal simple solution is.

Thanks for any help or answers/ideas for either question. It is appreciated.

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The suggestion that "simple damping of cantelever beams is done by applying viscosity to the rate of angle change with respect to time" is correct.

However, the expression you provided

$F = \frac{3y_sEI}{x_s^3} - cEIθ_t$

where: $θ = \frac{FL^2}{2EI}$

is not appropriate.

The reasons are:

  1. you use angle not angle rate .
  2. for different points in the beam the angle and the corresponding rate will be different, so you'd have to use calculus to get the correct expression.
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    $\begingroup$ Thanks NMech. By $θ_t$ I meant the derivative of $θ$ by time which would be the angle rate, right? If I'm only applying pressure to one point on the beam at $x_s$, and using the angle rate at this point, would this then be valid? Or would it still be a poor method? I figure one way to solve this would be to instead solve the beam as a finite difference model using something like $F = ρAy_{tt} = -EI(y_{xxxx} + c*y_{xxxxt})$ but I was hoping for a simpler solution just based on the fact that I only need one point to apply pressure to and measure from. Is there an easier approach? $\endgroup$
    – mike
    Feb 1 at 4:47
  • $\begingroup$ @mike On the way to find a feasible solution, you need to define the physical model (which you did), and the goal you wanted it to achieve (which I don't think it is clear). I am quite confused on what the "damper" is expected to do in such setup. $\endgroup$
    – r13
    Mar 3 at 16:42

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