We need to find the natural frequency of vibration for the following cases
longitudinal In this mode the system can be modeled as the following
Since both springs are in parallel, then the equivalent spring stiffness is\[ k_{eq}=k_{1}+k_{2}\] The equivalent mass is just the mass of the flywheel \(\frac{W}{g}\). Hence the overall system can now be modeled as follows
Which has the equation of motion \begin{align*} m_{eq}\ddot{y}+k_{eq}y & =0\\ \ddot{y}+\frac{k_{eq}}{m_{eq}}y & =0 \end{align*}
Therefore\[ \omega _{n}=\sqrt{\frac{k_{eq}}{m_{eq}}}\] We now just need to determine \(k_{eq}=k_{1}+k_{2}\). But from mechanics of materials we know that \(k_{1}=\frac{AE}{a}\) and \(k_{2}=\frac{AE}{b}\). Therefore the above becomes\begin{align*} \omega _{n} & =\sqrt{\frac{\frac{AE}{a}+\frac{AE}{b}}{\frac{W}{g}}}\\ & =\sqrt{\frac{gAE}{W}\left ( \frac{1}{a}+\frac{1}{b}\right ) } \end{align*}
Transverse In this mode the system can be modeled as beam with fixed ends with load \(W\) at distance \(a\) from one end and distance \(b\) from the other end. From tables, the stiffness coefficient in this case is given by \[ k_{eq}=3EI\left ( \frac{L}{ab}\right ) ^{3}\] The equivalent mass remains as before which is just the mass of the flywheel \(\frac{W}{g}\). Therefore, as above we find the natural frequency as\[ \omega _{n}=\sqrt{\frac{k_{eq}}{m_{eq}}}\]
Or
\[ \fbox{$\omega _n=\sqrt{\frac{3gEI}{W}\left ( \frac{L}{ab}\right ) ^3}$}\]
Torsional In this mode, the flywheel is twisted by some degree \(\theta \), and therefore the top part of the beam and the bottom part of the beam will resist this twist by applying moment against the twist as shown in this diagram
From mechanics of materials, there is relation between the twisting angle and resisting torque by beam which is given by\[ M=\frac{GJ}{L}\theta \] Where here \(\theta \) is the twist angle (radians) and \(M\) is the torque (\(Nm)\) and \(L\) is length of beam and \(G\) is modulus of rigidity (\(N\) per \(m^{2}\)) and \(J\) is the second moment of area of the cross section (\(m^{4}\)) about its center. Therefore total moments is\begin{align*} M_{1}+M_{2} & =\frac{GJ}{a}\theta +\frac{GJ}{a}\theta \\ & =GJ\theta \left ( \frac{1}{a}+\frac{1}{b}\right ) \end{align*}
Comparing the above to definition of stiffness which is \(F=K\Delta \) but in this problem \(\Delta \equiv \theta \) and \(F\equiv \left ( M_{1}+M_{2}\right ) \), then we see that the equivalent stiffness is\[ k_{eq}=GJ\left ( \frac{1}{a}+\frac{1}{b}\right ) \] We now need the equivalent mass. In this case it is the mass moment of inertia of flywheel. We are given that radius of gyration is \(r\), hence \[ m_{eq}=\frac{W}{g}r^{2}\] We now have all the pieces needed to find \(\omega _{n}\)\begin{align*} \omega _{n} & =\sqrt{\frac{k_{eq}}{m_{eq}}}\\ & =\sqrt{\frac{GJ\left ( \frac{1}{a}+\frac{1}{b}\right ) }{\frac{W}{g}r^{2}}}\\ & =\sqrt{\frac{gGJ}{Wr^{2}}\left ( \frac{1}{a}+\frac{1}{b}\right ) } \end{align*}
From tables, for circular bar of radius \(d\), we see that \(J=\frac{\pi }{32}d^{4}\). Hence the above becomes\[ \fbox{$\omega _n=\sqrt{\frac{gG\pi d^4}{32Wr^2}\left ( \frac{1}{a}+\frac{1}{b}\right ) }$}\]
Summary of results
| case | \(\omega _{n}\) |
| longitudinal | \(\sqrt{\frac{gAE}{W}\left ( \frac{1}{a}+\frac{1}{b}\right ) }\) |
| Transverse | \(\sqrt{\frac{g}{W}\left ( 3EI\right ) \left ( \frac{L}{ab}\right ) ^{3}}\) |
| Torsional | \(\sqrt{\frac{g}{W}\frac{G}{r^{2}}\frac{\pi d^{4}}{32}\left ( \frac{1}{a}+\frac{1}{b}\right ) }\) |
The first step is to draw the free body diagram and the kinematic diagram
Taking moments about the joint \(O\), noting that positive is anti-clockwise gives
\begin{equation} k_{t}\theta +k_{1}\left ( a\sin \theta \right ) a+k_{2}\left ( L\sin \theta \right ) L=-I_{o}\ddot{\theta } \tag{1} \end{equation}
Using parallel axis theorem, \begin{align*} I_{o} & =I_{CG}+m\left ( \frac{L}{2}\right ) ^{2}\\ & =\frac{1}{12}mL^{2}+m\frac{L^{2}}{4}\\ & =\frac{1}{3}L^{2}m \end{align*}
Hence (1) becomes
\[ \frac{1}{3}L^{2}m\ddot{\theta }+k_{t}\theta +k_{1}\left ( a\sin \theta \right ) a+k_{2}\left ( L\sin \theta \right ) L=0 \]
For small angle approximation the above becomes (we have to apply small angle approximation in order to obtain the form that allows us to determine \(\omega _{n}^{2}\), since this only works for linear equations of motion).
\begin{align*} \frac{1}{3}L^{2}m\ddot{\theta }+k_{t}\theta +k_{1}a^{2}\theta +k_{2}L^{2}\theta & =0\\ \frac{1}{3}L^{2}m\ddot{\theta }+\theta \left ( k_{t}+k_{1}a^{2}+k_{2}L^{2}\right ) & =0\\ \ddot{\theta }+\theta \frac{3\left ( k_{t}+k_{1}a^{2}+k_{2}L^{2}\right ) }{mL^{2}} & =0 \end{align*}
Comparing the above to standard form of linearized \(\ddot{\theta }+\omega _{n}^{2}\theta =0\) we see that the natural frequency (radians per second) is
\[ \fbox{$\omega _n=\sqrt{\frac{3\left ( k_t+k_1a^2+k_2L^2\right ) }{mL^2}}$}\]