Question 9.4.3: Estimating Mass, Stiffness, and Damping from the Step Respon...
Estimating Mass, Stiffness, and Damping from the Step Response
Figure 9.4.3 shows the response of a system to a step input of magnitude 6×10³ N. The equation of motion is
m\ddot{x} + c\dot{x} + kx = f (t)Estimate the values of m, c, and k.
From the graph we see that the steady-state response is x_{ss} = 6 cm. At steady state, x_{ss} = f_{ss}/k , and thus k = 6 × 10^{3}/6 × 10^{−2} = 10^{5} N/m .
The peak value from the plot is x = 8.1 cm, so the maximum percent overshoot is M% =[(8.1 − 6)/6]100 = 35%. From Table 9.3.2, we can compute the damping ratio as follows:
| Table 9.3.2 | Step response specifications for the underdamped model m\ddot{x} + c\dot{x} + kx = f . |
| Maximum percent overshoot | M% = 100e^{−πζ/\sqrt{1−ζ^{2}}} |
| ζ = \frac{R}{\sqrt{π^{2} + R^{2}}} , R = \ln \frac{100}{M\%} | |
| Peak time | t_{p} = \frac{π}{ω_{n} \sqrt{1 − ζ^{2}}} |
| Delay time | t_{d} \approx \frac{1 + 0.7ζ}{ω_{n}} |
| 100% rise time | t_{r} = \frac{2π − \phi}{ω_{n} \sqrt{1 − ζ^{2}}} |
| \phi = tan^{−1} \left(\frac{\sqrt{1 − ζ^{2}}}{ζ} \right) + π |
R = \ln \frac{100}{35} = 1.0498 ζ = \frac{R}{\sqrt{\pi^{2} + R^{2}}} = 0.32
The peak occurs at t_{p} = 0.32 s. From Table 9.3.2,
t_{p} = 0.32 = \frac{\pi}{ω_{n} \sqrt{1 − ζ^{2}}} = \frac{3.316}{ω_{n}}
Thus, ω^{2}_{n} = 107 and
m = \frac{k}{ω^{2}_{n}} = \frac{10^{5}}{107} = 930 kg
From the expression for the damping ratio,
ζ = \frac{c}{2 \sqrt{mk}} = \frac{c}{2 \sqrt{930(10^{5})}} = 0.32
Thus c = 6170 N · s/m, and the model is
930 \ddot{x} + 6170 \dot{x} + 10^{5}x = f (t)