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Question 11.2: In Fig. 11.10, the mass m1 and spring k1 represent a single-......

In Fig. 11.10, the mass m_{1} and spring k_{1} represent a single-DOF system with low inherent damping. Its response z_{1}, due to the applied force, F_{0}, is found to be excessive. The obvious solution of adding a damper in parallel with k_{1} is found not to be feasible in this case, and it is proposed to add a damped vibration absorber consisting of m_{2}, k_{2}  \mathrm{and}  c, where the mass of m_{2} is one-tenth that of m_{1}.
(a) Define the properties of the vibration absorber non-dimensionally, and plot the non-dimensional magnitude of the displacement, \left|\underline{z}_{1} \right| / z_{s},  \mathrm{versus}  \Omega_{1} = \omega / \omega_{1}.  Discuss the improvement compared with the original system.
(b) If the mass of m_{1} is 100 kg, and the natural frequency of the original single-DOF system is 10 Hz, define the optimum properties of the absorber dimensionally.

11.10
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Part (a):
The value of the mass ratio, \mu = m_{2}/m_{1} = 0.1.

From Eq. (11.16) the optimum value of R is

R_{\mathrm{opt}} = \frac{1}{1+\mu } = \frac{1}{1 +0.1} = 0.909                     (A)

From Eq. (11.17), with \mu = 0.1, the optimum value of the damping parameter \gamma = c/2m_{2}\omega _{1}  \mathrm{is} \\ \gamma_{\mathrm{opt}} = \sqrt{\frac{3\mu }{8(1+\mu )^{3}} } = 0.168                                     (B)

Substituting the above values of \mu , R_{\mathrm{opt}}  \mathrm{and}  \gamma_{\mathrm{opt}} into Eq. (11.15), the plot shown as a solid line in Fig. 11.12 was obtained using a standard spreadsheet. The maximum magnification of the system, with the optimized damper, is about 4.6.

\frac{\left|\underline{z} _{1}\right| }{z_{s}} = \left[\frac{(2\gamma \Omega_{1})^{2} + (\Omega^{2}_{1} – R^{2})^{2}}{(2\gamma \Omega_{1})^{2} (\Omega_{1}^{2} – 1 + \mu \Omega^{2}_{1})^{2} + \left[\mu R^{2}\Omega^{2}_{1} – (\Omega^{2}_{1} – 1)(\Omega^{2}_{1} – R^{2}) \right]^{2} } \right]^{\frac{1}{2} }                        (11.15)

An assessment of the improvement effected by adding the damper can be made by plotting the magnification of the original single-DOF system, without the added damper, for comparison. This is given by Eq. (4.12), which with the present notation is

\frac{\left|z\right| }{z_{s}} = \frac{1}{\sqrt{(1 – \Omega^{2})^{2} + (2\gamma \Omega )^{2}} }                            (4.12) \\ \frac{\left|z_{1}\right| }{z_{s}} = \frac{1}{\sqrt{(1 – \Omega^{2}_{1})^{2} + (2\underline{\gamma }\Omega _{1} )^{2}} }                                      (C)

where \underline{\gamma } is the damping coefficient of the original single-DOF system before adding the auxiliary damper. Taking \underline{\gamma } = 0.02, the original magnification curve is shown as a dashed line in Fig. 11.12. In comparing the two curves in Fig. 11.12 it should be remembered that the damping in the original single-DOF system was neglected when deriving the response for the system with the added damper, giving a small but conservative error. Ignoring this, the maximum magnification at the worst excitation frequency in each case is seen to be reduced by a factor of at least 5.

Part (b):
To define the properties of the vibration absorber dimensionally, we have

m_{1} = 100  kg;             \mu = 0.1;             m_{2} = \mu m_{1}= 10  kg\\ \omega _{1} = 2\pi \times 10 = 20 \pi  rad/s;             \omega _{2} = R_{\mathrm{opt}}\omega _{1} = 0.909 \times 20\pi \\ \omega _{2} = \sqrt{\frac{k_{2}}{m_{2}} } ;       k_{2} = m_{2}\omega ^{2}_{2} = 10(0.909 \times 20\pi )^{2} = 32620  N/m \\ \gamma_{\mathrm{opt}} = \frac{c}{2m_{2}\omega _{1}} ;       c = 2 \gamma_{\mathrm{opt}}m_{2} \omega _{1} = 2 \times 0.168 \times 10 \times 20\pi = 211  N/m/s

Thus the auxiliary damper has
Mass = m_{2} = 10 kg;
Spring stiffness = k_{2} = 32 620 N/m;
Damper constant = c = 211 N/m/s.

11.12

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