A 1000-kg machine is subject to a triangular pulse of duration 0.05 s and peak of 20,000 N. What is the range of isolator stiffness for an undamped isolator such that the maximum transmitted force is less than 8000 N and the maximum displacement is less than 2.8 cm?

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Accepted Answer

The force spectrum for the triangular pulse shows that for [latex] F_{T}/F_{0}\lt 0.4, \omega _{n}t_{0}/(2\pi)\lt 0.16,[/latex] which gives

[latex]\omega_{n}\lt \frac{2\pi(0.16)}{0.05 s} =20.1 rad/s [/latex]

The lower bound on the natural frequency is obtained by trial and error, using the displacement spectrum for the triangular pulse. For a guessed value of [latex]\omega _{n},\omega _{n}t_{0}(2\pi) [/latex] is calculated and the corresponding value of the maximum nondimensional displacement is found from the displacement spectru. The maximum dimensional displacement is calculated from Equation (5.98). If the displacement is greater than the allowable displacement, the guess for the lower bound must be increased. The calculations for this example are given in Table 5.2. The lower bound is calculated as 17 rad/s. Thus the allowable stiffness range is

[latex]2.89 imes 10^{5} N/m\lt k\lt 4.04 imes 10^{5}N/m [/latex]

[latex] x_{\max}=\frac{F_{0}}{m\omega ^{2}_{n}}\left(\frac{m\omega ^{2}_{n}x\max}{F_{0}} ight) [/latex] (5.98)

TABLE 5.2
[latex]\omega _{n}\pmb{, rad/s} [/latex]	[latex] \frac{\omega _{n}\pmb{t_{0}}}{2\pi} [/latex]	[latex]\frac{\pmb{m}\omega ^{2}_{n}\pmb{x}_{\max}}{\pmb{F_{0}}} [/latex]	[latex]\pmb{x}_{max}\pmb{,cm} [/latex]
10	0.08	0.25	5.0
15	0.12	0.38	3.4
18	0.14	0.42	2.6
17	0.135	0.40	2.8

Question 5.17: A 1000-kg machine is subject to a triangular pulse of durati...

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