Universidade Federal do Rio Grande do Sul (UFRGS)
Programa de Pós-Graduação em Engenharia Civil (PPGEC)

PEC00025: Introduction to Vibration Theory¶

Class 05 - Forced vibration in time domain¶

1. Introduction
2. Impulse response
3. General response to forced vibration
4. Short transient loads
4.1. Impulse response approximation
4.2. Dynamic factor approach
5. Assignment

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Prof. Marcelo M. Rocha, Dr.techn. (ORCID)
Porto Alegre, RS, Brazil

1. Introduction ¶

So far we have calculated the system response in damped free vibration, which depends solely on the initial conditions with $F(t) = 0$.

We call forced vibration the system response to any force function, $F(t)$, which is the solution of the equation:

$$m\ddot{u} + c\dot{u} + ku = F(t)$$

In this class we shall present the closed-form solution for any given force function, starting with the very fundamental one: the unit impulse function.

2. Impulse response ¶

We start by re-writing the equilibrium equation, which after division by the system mass is:

$$\ddot{u} + 2\eta \dot{u} + \omega_{\rm n}^2 u = F(t)/m$$

Now we take the force function as $F(t) = m\delta(t)$ (the Dirac's delta function at time origin), then we apply Laplace transform on this equation disregarding the initial conditions ($u_0 = 0$ and $v_0 = 0$). It gives:

$$\mathscr{L}\left\{ \ddot{u} + 2 \eta \dot{u} + \omega_{\rm n}^2 u \right\} = \mathscr{L}\left\{ \delta(t) \right\}$$ $$s^2 \bar{u}(s) + 2 \eta s \bar{u}(s) + \omega_{\rm n}^2 \bar{u}(s) = 1$$

what solving for $\bar{u}(s)$ results in:

$$\bar{u}(s) = \left[\frac{\omega_{\rm D}}{(s + \eta)^2 + \omega_{\rm D}^2}\right] \left(\frac{1}{\omega_{\rm D}} \right)$$

where we have used the same definition we made for free damped vibration: $\omega_{\rm D}^2 = \omega_{\rm n}^2 - \eta^2$.

Recognizing the sine function transform and using the translation theorem gives a special response called unit impulse response, $g(t)$:

$$g(t) = \frac{1}{\omega_{\rm D}} \; e^{- \zeta \omega_{\rm n} t} \sin \omega_{\rm D}t$$

This response corresponds to the free vibration response for $u_0 = 0$ and $v_0 = 1$ (m/s), which means that the unit impulse force $m\delta(t)$ at time origin causes the same response as a unit initial velocity.

From the practical viewpoint, the system response to an impulsive force allow us to know all the system mechanical properties by fitting the theoretical function $g(t)$ to the measured dynamic response.

3. General response to forced vibration ¶

Now let us apply Laplace transform to the general equilibrium equation:

$$\mathscr{L}\left\{ \ddot{u} + 2 \eta \dot{u} + \omega_{\rm n}^2 u \right\} = \mathscr{L}\left\{ F(t)/m \right\}$$

where we define $\mathscr{L}\left\{ F(t)/m \right\} = \bar{f}(s)$. Recalling that the Laplace transform of the unit impulse response, $g(t)$, is:

$$\bar{g}(s) = \frac{1}{(s + \eta)^2 + \omega_{\rm D}^2}$$

results in:

$$\bar{u}(s) = \bar{g}(s)\bar{f}(s) + \bar{u}_0(s)$$

where:

$$\bar{u}_0(s) = \bar{g}(s) \left[ u_0 s + \left( v_0 + 2 \eta u_0 \right) \right]$$

is the response to the initial conditions, $u_0$ and $v_0$, while the term $\bar{g}(s)\bar{f}(s)$ is the response to the force function, $F(t)/m$.

With the convolution theorem, the inverse Laplace transform finally leads to:

$$u(t) = u_0(t) + \frac{1}{m} \int_0^t {g(t - \tau) F(\tau) \, d\tau}$$

which is the general solution of the dynamic equilibrium equation for a sdof system. It implies that the response, $u(t)$, for any given force function, $F(t)/m$, can be obtained by calculating its convolution with the impulse response, $g(t)$, superposed to the response for initial conditions, $u_0(t)$.

The numerical approach for directly solving the convolution integral is not recommended. In the next class we shall be presented to a more efficient numerical technique called the Duhamel integral.

4. Short transient loads ¶

In this section we present a more practical approach to evaluate the system response to a force function with short duration. To illustrate these approaches, we make use of a numerical example, calculated with MRPy. The numerical method will be explained in details next class. By now we are only interested in the results provided by the Python class.

4.1. Impulse response approximation¶

Let us start assuming a short transient load consisting in a constant force, $F_0$, applied at $t = 0$ and removed at $t = t_{\rm d}$.

Now we use MRPy to calculate the response of a system with $m = 1$kg, natural vibration frequency $f_{\rm n} = 1$Hz, and damping ratio of critical $\zeta = 1$%.

We see that MRPy has calculated the peak response amplitude as 15.4mm. Now we will perform the same calculation by assuming that the applied force function can be approximated by an impulse function at time origin.

The impulse function equivalent to the load defined above is given by:

$$F(t) = F_0 t_{\rm d} \; \delta(t) = I \delta(t)$$

where $F_0 t_{\rm d}$ is the total impulse (integral of the Dirac's delta), which is not unitary in the example. As previously stated, the equivalent initial velocity, $v_0$, is approximatelly:

$$v_0 = \frac{I}{m} = \frac{F_0 t_{\rm d}}{m} = 1{\rm m/s}$$

where $I$ is the given impulse. The response amplitude to this initial condition is:

$$u_{\rm max} \approx \frac{v_0}{\omega_{\rm D}} \approx \frac{v_0}{\omega_{\rm n}} \approx \frac{1.0}{2 \pi \cdot 1.0} \approx 159{\rm mm}$$

which is quite close to the numerical result (the discretization $N$ was intentionally chosen to give the required accuracy).

4.2. Dynamic factor approach¶

As a rule of thumb, the approximation of a short transient load as an impulse function at time origin provides a good approximation up to $t_{\rm d} \leq T_{\rm n}/4$ (where $T_{\rm n}$ is the system natural period of vibration). For instance, by setting $t_{\rm d} = 0.25$s in the previous example, the numerical result would be $u_{\rm max} = 35.3$mm, while the impulse function approximation would give $u_{\rm max} = 39.8$mm, overestimating the correct result (try it by yourself!).

To overcome this approximation error, numerically computed results may be provided as design curves, as the one reproduced below (Clough & Penzien, 1995):

The curves provide a dynamic amplification factor, to be multiplied by the system static response to the force function maximum amplitude:

$$u_{\rm max} = A \frac{F_0}{k} = A \frac{F_0}{\omega_{\rm n}^2 m}$$

In the previous example, with $t_{\rm d} = 0.25$s, the dynamic amplification taken from the respective curve (rectangular function shape) is $A \approx 1.4$. Hence the peak response amplitude is:

$$u_{\rm max} = 1.4 \frac{4.0}{(2 \pi \cdot 1.0)^2 \cdot 1.0} \approx 142{\rm mm}$$

which is much closer to the correct solution.

5. Assignments ¶

1. Utilizar o registro de vibração livre obtido no trabalho anterior.
2. Uma vez identificados frequência e amortecimento, calcular: (a) Máximo deslocamento para uma carga transiente de curta duração, através de aproximação por resposta impulsiva. (b) Mesmo cálculo feito por MRPy.sdof_Duhamel().
3. Relatório com descrição do objeto, gráficos e resultados (nome do arquivo T3_xxxxxxxx.ipynb).

Prazo: 20 de maio de 2020.