Harmonic forced vibration of a cantilever beam

[FEAnalyst]

TL;DR

How to use the formula for stress in cantilever subjected to harmonic force at the free end?

Hi,
in the book titled "Formulas for Dynamics, Acoustics and Vibrations" by R.D. Blevins, I've found a formula that can be used to calculate the bending stress in a cantilever beam subjected to harmonic force applied at the free end. The formula looks like this: $$\sigma=\frac{F_{0}Ec}{m \omega_{1}^{2}L} \frac{\pi^{2}}{4L^{2}} \frac{2.1^{2}(1.78)(1-cos{\pi x_{0}}/2L)}{\sqrt{(1-f^{2}/f_{i}^{2})^{2}+(2 \xi_{i} f / f_{i})^{2}}} cos{\frac{\pi x}{2L}} cos{(\omega t - \varphi)}$$ Quite complex but I managed to calculate the first part of this equation (let's call it $S$) and I'm left with just the last term: $$\sigma=S \cdot cos{(\omega t - \phi)}$$ where: $\omega$ - frequency, $t$ - time and $\varphi$ - phase angle. I think that I can calculate $\omega$ using the frequency of excitation ($\omega=2 \pi f$). The thing is that I don't know what to substitute for time - this problem is supposed to be steady state and I want to use the formula to verify the numerical analysis done in frequency, not time, domain. I also don't know what to do with the phase angle. In the numerical simulation I only define the force magnitude. I do realize that loads in such analyses are given by $F= F_{mag} \cdot sin(\omega t+ \varphi)$ but for the simulation I don't have to specify $\varphi$.

Does anyone know how to use this formula (what to do with the last term)?

 

[Dr.D]

The formula you found is for the time domain response. If you only want the amplitude, take the first part (your S) and that is the amplitude. The steady state forced response will necessarily occur at the forcing frequency, so S will give you the amplitude at that one frequency.

 

[hutchphd]

Notice that the phase angle is really determined by what you choose to call t=0, and so probably can be chosen arbitrarily for your purposes.

 

[FEAnalyst]

Thanks for replies.

If you only want the amplitude, take the first part (your S) and that is the amplitude.

That’s what I wanted to do but the result doesn’t match the numerical value and I thought that maybe the remaining part of the equation is always necessary.

 

[Dr.D]

I think I'd go back to Blevins' book (which I do not have) to see if you have correctly understood (a) what it is intended to evaluate, and (be) what the meaning of all the symbols is.

 

[FEAnalyst]

All symbols are explained in the book as follows:

$F_{0}$ - force amplitude, $E$ - Young’s modulus, $c$ - distance in cross-section from neutral axis to stress point on edge of section, $m$ - mass per unit length, $\omega_{i}=2 \pi f_{i}$ - natural frequency (rad/s) for i-th mode, $L$ - beam length, $x_{0}$ - distance of the force from the fixed end of the beam, $f$ - frequency of excitation, $f_{i}$ - natural frequency (Hz) for i-th mode, $\xi_{i}$ - damping factor for i-th mode, $x$ - axial coordinate of the point where stress is calculated

In my case the beam is $1 \ m$ long, has square cross-section ($0.04$ x $0.04 \ m$) and $5000 \ N$ force applied at the free end. It’s made of steel (Young’s modulus: $210 \ GPa$, density: $7850 \ kg/m^{3}$) and I assume damping factor of $0.03$. The frequency of excitation is $40 \ Hz$ and I use the first mode which happens to be at $33.3908 \ Hz$.

Thus the data I use for the equation is:
$F_{0}=5000 \ N$, $E=210 000 000 000 \ Pa$, $c=0.02 \ m$, $m=12.56 \ kg/m$, $\omega_{1}=209.801 \ rad/s$, $L=1 \ m$, $x_{0}=1 \ m$, $f=40 \ Hz$ , $f_{1}=33.3908 \ Hz$, $\xi_{1}=0.03$, $x=0 \ m$

And the result I got is: $S=1.67128 \cdot 10^{9} \ Pa$ while the numerical value is: $S=1.141 \cdot 10^{9} \ Pa$

Maybe you will notice some mistake. It’s important for me to solve this exemplary case as I want to be able to verify such analyses using this book.

 

[hutchphd]

I note that the ratio of your result to the correct result is close to $\sqrt2$ ...could this be an rms vs peak issue? I would check your math but that would be foolish as mine is atrocious.

 

[FEAnalyst]

That’s an interesting idea. I also think that this difference in results is not random. However, from what I know, the numerical result is not RMS as the software that I use calculates RMS stress output only for random vibration analyses (and it has different symbol). By the way, this value is real component of a complex number.
In the book it’s only mentioned that $\sigma$ is axial bending stress.

I solved this equation once again using computer algebra system (CAS) to make sure that the math is correct. The result was the same as before so the problem lies somewhere else.