Question 3.5: Use an exact piecewise-linear method to find the displacemen......
Use an exact piecewise-linear method to find the displacement response, z, of the system represented by Eq. (3.2):
\ddot{z} + 2\gamma \omega _{n}\dot{z} + \omega ^{2}_{n}z = F/m (A)
from t = 0 to t = 0.5 s when F consists of the force pulse shown in Fig. 3.3(a), and the other numerical values are as follows: mass, m = 100 kg; damping coefficient \gamma = 0.05; the undamped natural frequency, f_{n} = 10 Hz (ω_{n} =20π rad/s). Assume z = \dot{z} = 0 at t = 0.
As shown in Fig. 3.3(b), the required input function can be formed by the superposition of:
(1) a positive ramp of slope 4000 N/s starting at t = 0;
(2) a similar negative ramp starting at t = 0.05 s; and
(3) a negative step of 200 N starting at t = 0.2 s
Once started, these functions all continue to the end of the computation.
From Example 3.4, the exact response to a ramp of slope b is
and the exact response to a step of magnitude a is
where
\omega _{d} = \omega _{n}\sqrt{1-\gamma ^{2}}Inserting numerical values, the response to the first ramp, z_{1}, which is required for 0 < t < 0.5, is
z_{1} = 0.01013\left\{ t – 0.001591 [1 – \exp(- 3.1415 t) \cos 62.753 t] – 0.01585 \exp (- 3.1415 t) \sin 62.753 t\right\} (D)
The response, z_{2}, to the second ramp, which is required for 0.05 < t < 0.5, is
z_{2} = -0.01013\left\{(t – 0.05) – 0.001591[1 – \exp(- 3.1415 (t – 0.05))\cos 62.753(t – 0.05)] – 0.01585\exp(- 3.1415(t – 0.05))\sin 62.753(t – 0.05)\right\} (E)
and the response, z_{3}, to the negative step, which is required for 0.2 < t < 0.5, is
z_{3} = -0.0005066\left\{1 – \exp[- 3.1415(t – 0.2)] × [\cos 62.753(t – 0.2) + 0.05006 \sin(t – 0.2)]\right\} (F)
The response, plotted from a spreadsheet, is shown in Fig. 3.4.
