Question 3.4: Use Table 3.1 to find the displacement response of the syste......
Use Table 3.1 to find the displacement response of the system shown in Fig. 2.1 and represented by Eq. (3.2), when,
(a) a step force of magnitude a force units is applied at t = 0 and the initial conditions are z = 0 and \dot{z} = 0 at t = 0.
(b) a ramp force, starting from zero and growing linearly at the rate of b force units/ second, is applied at t = 0, with the same initial conditions, z = 0 and \dot{z} = 0 at t = 0.
Assume that the non-dimensional damping coefficient, γ, is less than unity.
\ddot{z} + 2\gamma \omega _{n}\dot{z} + \omega ^{2}_{n}z = F/m (3.2)
| Table 3.1 Transient Responses of the System:1/ (s^{2} + 2\gamma \omega_{n} s + \omega^{2}_{n}) (\gamma < 1) | |||||
| Input function F | |||||
| Unit impulse \delta (t) | Unit step H(t) | Unit ramp t | sin ωt | cos ωt | Response functions |
| 0 | \frac{1}{\omega _{n}^{2}} | \frac{-2\gamma }{\omega _{n}^{3}} | 0 | 0 | Unit step H(t) |
| 0 | 0 | \frac{1}{\omega _{n}^{2}} | 0 | 0 | Unit ramp t |
| 0 | 0 | 0 | \frac{\omega^{2}_{n} – \omega ^{2}}{(\omega^{2}_{n} – \omega ^{2})^{2}+ (2\gamma \omega _{n}\omega )^{2}} | \frac{2\gamma \omega _{n}\omega}{(\omega^{2}_{n} – \omega ^{2})^{2}+ (2\gamma \omega _{n}\omega )^{2}} | sin ωt |
| 0 | 0 | 0 | \frac{-2\gamma \omega _{n}\omega}{(\omega^{2}_{n} – \omega ^{2})^{2}+ (2\gamma \omega _{n}\omega )^{2}} | \frac{\omega^{2}_{n} – \omega ^{2}}{(\omega^{2}_{n} – \omega ^{2})^{2}+ (2\gamma \omega _{n}\omega )^{2}} | cos ωt |
| \frac{1}{\omega _{d}} | \frac{-\gamma }{\omega _{n} \omega _{d}} | \frac{2\gamma^{2} – 1 }{\omega _{n}^{2} \omega _{d}} | \frac{\omega [(\omega^{2} – \omega^{2}_{n})+ 2\gamma ^{2}\omega^{2}_{n}]}{\omega _{d}\left[(\omega^{2}_{n} – \omega ^{2})^{2}+ (2\gamma \omega _{n}\omega )^{2}\right] } | \frac{-\gamma \omega _{n}(\omega ^{2}_{n}+\omega ^{2})}{\omega _{d}\left[(\omega^{2}_{n} – \omega ^{2})^{2}+ (2\gamma \omega _{n}\omega )^{2}\right] } | e^{-\gamma \omega _{n}t}\sin \omega _{d}t |
| 0 | \frac{-1}{\omega _{n}^{2}} | \frac{2\gamma }{\omega _{n}^{3}} | \frac{2\gamma \omega _{n}\omega}{(\omega^{2}_{n} – \omega ^{2})^{2}+ (2\gamma \omega _{n}\omega )^{2}} | \frac{\omega^{2} – \omega ^{2}_{n}}{(\omega^{2}_{n} – \omega ^{2})^{2}+ (2\gamma \omega _{n}\omega )^{2}} | e^{-\gamma \omega _{n}t}\cos \omega _{d}t |
| \omega =forcing frequency (rad/s); \omega _n=undamped natural frequency (rad/s), \omega _n =\sqrt{k/m};\omega _d= damped natural frequency (rad/s), \omega _d =\omega_n \sqrt{1-\gamma^2} ;\gamma =non-dimensional viscous damping coefficient. | |||||
The response, z, to a unit force step, of a system with unit mass, is found by multiplying all the constants in the second column, ‘unit step’, by the corresponding response functions in the last column, and adding the products. For a step of magnitude a, and a system mass of m, the result must then be multiplied by a/m:
which can be simplified to the following, for t > 0 if the step function H(t) starts at t = 0:
In this case, the constants in the third column, ‘unit ramp’, are individually multiplied by the response function in the last column, the products added, and the whole multiplied by b/m, as follows:
which can be simplified to