Question 1.7.1: Consider the system of Figure 1.9 with mass and stiffness pr......
Consider the system of Figure 1.9 with mass and stiffness properties as summarized by inequality (1.84). Suppose that the system is subject to an initial velocity that is always less than 300 mm/s, and to an initial displacement of zero (i.e., x_{0} = 0, ν_{0} ≤ 300 mm/s). For this range of mass and stiffness, choose a value of the damping coefficient such that the amplitude of vibration is always less than 25 mm.
This is a design-oriented example, and hence, as is typical of design calculations, there is not a nice, clean formula to follow. Rather, the solution must be obtained using theory and parameter studies. First, note that for zero initial displacement, the response may be written from equation (1.38) as
A = \sqrt{\frac{(ν_{0} + \zeta w_{n}x_{0})² + (x_{0}w_{d})²}{w²_{d}}}, or Φ = \tan^{-1}\frac{x_{0}w_{d}}{ν_{0} + \zeta w_{n}x_{0}} (1.38)
x(t) = \frac{ν_{0}}{w_{d}} e^{-\zeta w_{n} t} \sin(w_{d}t)
Also note that the amplitude of this periodic function is
\frac{ν_{0}}{w_{d}} e^{-\zeta w_{n} t}Thus, for small w_{d}, the amplitude is larger than for larger w_{d}. Hence for the range of frequencies of interest, it appears that the worst case (largest amplitude) will occur for the smallest value of the frequency (w_{n} = 8.16 rad/s). Also, the amplitude increases with ν_{0} so that using ν_{0} = 300 mm/s will ensure that amplitude is a large as possible. Now, ν_{0} and w_{n} are fixed, so it remains to be investigated how the maximum value of x(t) varies as the damping ratio is varied. One approach is to compute the amplitude of the response at the first peak. From Figure 1.10 the largest amplitude occurs at the first time the derivative of x(t) is zero. Taking the derivative of x(t) and setting it equal to zero yields the expression for the time to the first peak:
w_{d}e^{-\zeta w_{n}t} \cos (w_{d}t) – \zeta w_{n} e^{-\zeta w_{n}t} \sin (w_{d}t) = 0
Solving this for t and denoting this value of time by T_{m} yields
T_{m} = \frac{1}{w_{d}} \tan^{-1} \left(\frac{w_{d}}{\zeta w_{n}}\right) = \frac{1}{w_{d}} \tan^{-1} \left(\frac{\sqrt{1 – \zeta²}}{\zeta}\right)
The value of the amplitude of the first (and largest) peak is calculated by substituting the value of T_{m} into x(t), resulting in
A_{m}(\zeta) = x(T_{m}) = \frac{ν_{0}}{w_{n} \sqrt{1 – \zeta²}} e^{- \frac{\zeta}{\sqrt{1 – \zeta²}} \tan^{-1} \left(\frac{\sqrt{1 – \zeta²}}{\zeta}\right)} \sin \left(\tan^{-1}\left(\frac{\sqrt{1 – \zeta²}}{\zeta}\right)\right)Simplifying yields
A_{m}(\zeta) = \frac{ν_{0}}{w_{n}} e^{-\frac{\zeta}{\sqrt{1 – \zeta²}} \tan^{-1} \left(\frac{\sqrt{1 – \zeta²}}{\zeta}\right)}For fixed initial velocity (the largest possible) and frequency (the lowest possible), this value of A_{m}(\zeta) determines the largest value that the highest peak will have as ζ varies. The exact value of ζ that will keep this peak, and hence the response, at or below 25 mm, can be determined by numerically solving A_{m}(\zeta) = 0.025 (m) for a value of ζ. This yields ζ = 0.281. Using the upper limit of the mass values (m = 3 kg) then yields the value of the required damping coefficient:
c = 2mw_{n}ζ = 2(3)(8.16)(0.281) = 13.76 kg/s
For this value of damping, the response is never larger than 25 mm. Note that if there is no damping, the same initial conditions produce a response of amplitude A = ν_{0} /w_{n} = 37 mm.
