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

PEC00025: Introduction to Vibration Theory¶

Class 04 - Free vibration in time domain¶

1. Free vibration of undamped sdof systems
2. Energy dissipation
3. Free vibration of damped sdof systems
4. Damping ratio estimation
5. Assignment

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

1. Free vibration of undamped sdof systems ¶

The figure below shows an idealization of a single degree of freedom (sdof) mechanical system with its basic elements:

1. The total system mass, $m$,
2. A restitutive force element, ideally a linear spring with stiffness coefficient $k$,
3. A dissipative force element, ideally a viscous damper with damping coefficient $c$,
4. The applied external force, $F(t)$, and
5. The system response, the displacement $u(t)$.

Although the idealization presented above may resemble some feasible device, the basic elements may also be seen as conceptual and represent a much broader concept. For instance, the restitutive force element could be the gravity acting upon a pendulum. Or the system response could also be a rotation angle instead of a displacement.

The dissipative force will be introduced in the next section and are initially disregarded. If we also disregard the external force, $F(t) = 0$, any system response will be called free vibration and that is our first case for analysis.

By applying the equilibrium equation and accounting for the inertial force as stated by Newton's law gives:

$$m \ddot{u} + k u = 0$$

Applying Laplace transform over this equation gives:

$$\mathscr{L}\left\{ m \ddot{u} + k u \right\} = m \left[ -s u(0) - \dot{u}(0) + s^2 \bar{u}(s) \right] + k \bar{u}(s) = 0$$

Recognizing that $u(0) = u_0$ is the system initial position and $\dot{u}(0) = v_0$ is the system initial velocity and solving for $\bar{u}(s)$ gives:

$$\bar{u}(s) = \left(\frac{s}{s^2 + \omega_{\rm n}^2}\right) u_0 + \left(\frac{\omega_{\rm n}}{s^2 + \omega_{\rm n}^2}\right) \frac{v_0}{\omega_{\rm n}}$$

where we have defined $\omega_{\rm n}^2 = k/m$. The inverse transform is promptly recognized from the sine and cosine transforms previously presented:

$$u(t) = u_0 \cos \omega_{\rm n}t + \frac{v_0}{\omega_{\rm n}} \sin \omega_{\rm n}t$$

It can be seen that the free vibration occurs at the natural vibration frequency $\omega_{\rm n}$, and the movement amplitude depends on the initial displacement and initial velocity.

Let us plot this function with some Python lines:

This is surely not an interesting result. Try changing the values for $u_0$ and $v_0$ to see what happens. Indeed, the solution above can also be given in the form:

$$u(t) = u_{\rm max} \cos ( \omega_{\rm n}t - \theta )$$

with:

$$\begin{align*} u_{\rm max} &= \sqrt{u_0^2 + \left(\frac{v_0}{\omega_{\rm n}}\right)^2} \\ \\ \theta &= \arctan \left(\frac{v_0}{u_0 \omega_{\rm n}} \right) \end{align*}$$

The most important observation so far is the natural vibration frequency, that holds a sort of universal law for mechanical vibrations:

$$\omega_{\rm n} = \sqrt{\frac{k}{m}}$$

Hence, the larger the system stiffness the higher the frequency, while the larger the system mass the lower the frequency, and vice-versa. This special frequency can be also obtained by energy conservation, if one recognizes that there are two kinds of energy present in the system:

1. Potential elastic energy: $$V = k u^2(t)/2$$
2. Kinectic energy: $$T = m \dot{u}^2(t)/2$$

Assuming that if the maximum displacement of the system is $u_{\rm max}$, then the maximum velocity will be $\omega_{\rm n} u_{\rm max}$. When the displacement attains its maximum the potential energy is maximum and the kinectic energy is minimun, and vice-versa. By energy conservation:

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

what leads to the equation for $\omega_{\rm n}$.

The natural frequency is also expressed in cycles per second, or Hertz, related to the frequency in radians per second with:

$$f_{\rm n} =\frac{\omega_{\rm n}}{2 \pi}$$

Alternatively, the natural period of vibration:

$$T_{\rm n} = \frac{1}{f_{\rm n}} = \frac{2 \pi } {\omega_{\rm n}}$$

can be used. These relations are illustrated in the figure below.

2. Energy dissipation ¶

Besides the potential and kinectic energy, there is always some energy dissipation mechanism taking place during the system vibration (otherwise a perpetual motion would be possible). The most common causes for this dissipation are:

1. Material internal hysteresis,
   

2. Shear deformation of embedding fluid,
   

3. Friction at supports or connections,
   

among others.

Each one of these phenomena must be properly modelled to enter the equilibrium equation, but the most common approach is to use the so-called newtonian or viscous damping model for the dissipative force, $c \dot{u}$, where $c$ is the viscous damping constant. The great advantage of this model is that the equilibrium equation preserves its linearity and solution easiness, but one must keep in mind that it is only a model and the definition of constant $c$ may be quite challenging.

Most commonly the damping constant is presented as a ratio to critical damping, $\zeta$, defined as:

$$\zeta = \frac{c}{2 m \omega_{\rm n}}$$

which is non-dimensional and for this reason more easily quantified. It can be shown that $\zeta = 1$ is the smallest damping prenventing the system from having a single oscillation (one can have an idea of this condition by releasing a pendulum under water).

The Chapter 9 of the Brazilian Code NBR6123-Forças devidas o vento nas edificações brings a table (Tabela 2) of suggested values for $\zeta$ according to the type of structural system undergoing wind induced vibration:

Structure type	$\zeta$
Reinforced concrete buildings without stiffning walls	0.020
Reinforced concrete buildings with stiffning walls	0.015
Reinforced concrete towers and chimneys, tappered	0.015
Reinforced concrete towers and chimneys, constant section	0.010
Moment resisting steel buildings with welded connections	0.010
Steel towers and chimneys, constant section	0.008
Timber structures	0.030

It can be observed that for Civil Engineering the range of values are not too wide, and a typical value $\zeta = 0.01 = 1\%$ is usually assumed when no further information is available.

In the next section it will be shown that despite of all uncertainty about energy dissipation, it is of utmost importance for the safety and good performance of mechanical systems.

3. Free vibration of damped sdof systems ¶

Assuming now the viscous model for energy dissipation gives the free vibration equilibrium:

$$m \ddot{u} + c \dot{u} + k u = 0$$

Applying Laplace transform over this equation gives:

$$\mathscr{L}\left\{ m \ddot{u} + c \dot{u} + k u \right\} = m \left[ -s u(0) - \dot{u}(0) + s^2 \bar{u}(s) \right] + c \left[ -u(0) + s \bar{u}(s) \right] + k \bar{u}(s) = 0$$

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

$$\bar{u}(s) = \left[\frac{(s + \eta)}{(s + \eta)^2 + \omega_{\rm D}^2}\right] u_0 + \left[\frac{\omega_{\rm D}}{(s + \eta)^2 + \omega_{\rm D}^2}\right] \left(\frac{v_0 + \eta u_0}{\omega_{\rm D}} \right)$$

where we have additionally defined:

$$\begin{align*} \omega_{\rm D}^2 &= \omega_{\rm n}^2 - \eta^2 = \omega_{\rm n}^2 \left( 1 - \zeta^2 \right ) \\ \\ \eta &= \frac{c}{2m} = \zeta \omega_{\rm n} \end{align*}$$

The inverse transform can be obtained with the well known sine and cosine transforms and the translation theorem in frequency domain:

$$u(t) = e^{- \zeta \omega_{\rm n} t} \left[ u_0 \cos \omega_{\rm D}t + \left( \frac{v_0 + \zeta \omega_{\rm n} u_0} {\omega_{\rm D}} \right) \sin \omega_{\rm D}t \right]$$

It can be seen that the free vibration now occurs at the natural damped vibration frequency $\omega_{\rm D}$, and the movement amplitude depends on the initial displacement and initial velocity.

Although the natural frequency has changed, by recalling that the damping ration $\zeta$ usually is in the order of 1% implies, for instance, that:

$$\omega_{\rm D} = \omega_{\rm n} \sqrt{1 - 0.01^2} \approx 0.99995 \omega_{\rm n} \approx \omega_{\rm n}$$

what means that, beyond the whole uncertainty about the actual damping, for usual damping values the natural frequency change may be disregarded.

Let us take a look at the free vibration damped response using the same time discretization from the previous example.

It can be observed that the exponential envelope causes the undamped response to have its amplitude decreased along time. Try changing the damping ratio in the script above to see how this affect the response (but it must always be within the range $0 < \zeta < 1$).

4. Damping ratio estimation ¶

As we did for the undamped response, the damped response in free vibration can be put in the form:

$$u(t) = u_{\rm max} e^{- \zeta \omega_{\rm n} t} \cos ( \omega_{\rm D}t - \theta )$$

with:

$$\begin{align*} u_{\rm max} &= \sqrt{u_0^2 + \left( \frac{v_0 + \zeta \omega_{\rm n} u_0} {\omega_{\rm D}} \right) ^2} \\ \\ \theta &= \arctan \left( \frac{v_0 + \zeta \omega_{\rm n} u_0}{\omega_{\rm D} u_0 }\right) \end{align*}$$

Now we calculate the amplitude ratio at two time instants where the cosine function reaches its maximum, which are $N$ vibration cycles apart, what gives:

$$\frac{u(t)}{u(t + N T_{\rm n})} = \frac{\exp \left[ - \zeta \omega_{\rm n} t \right]} {\exp \left[ - \zeta \omega_{\rm n} (t + N T_{\rm n}) \right]} = \exp \left( - \zeta \omega_{\rm n} N T_{\rm n} \right) = \exp \left( - 2 \pi \zeta N \right)$$

Making $u(t) = u_i$, $u(t + N T_{\rm n}) = u_{i+N}$, taking natural logarithms and isolating $\zeta$ finally gives:

$$\zeta = \frac {\log (u_i \, / \, u_{i+N})}{2 \pi N}$$

This formula is exemplified in the figure below with the output of a real accelerometer. It is important to emphasize that the estimator presented above is valid no matter which kinemactic parameter is used for the amplitude ratio (displacement, velocity or acceleration), for they are related by a constant that will cancel upon division.

Applying the formula gives:

$$\zeta = \frac{\ln (0.52 \, / \, 0.31)}{2 \pi \cdot 8} \approx 1\%$$

A usefull rule for estimating $\zeta$ without much effort is to count the number of cycles necessary for the amplitude to be reduced to half. In this case:

$$\zeta = \frac{\ln (2 \, / \, 1)}{2 \pi \,N} \approx \frac{1}{9N}$$

The MRPy module provides a more sophisticated tool for estimating all parameters of a free damped response function available as a time series. This is exemplifyed below using the same the $u(t)$ function from the previous example.

The fitting result is usually not so accurate when the original function comes from real measurements. In some cases the fitting algorithm (which uses scipy module) may even fail in presenting any usefull result.

5. Assignments ¶

1. Fazer um registro com o celular de uma superfície em vibração livre (lenta)
2. Importar para dentro de um Jupyter notebook e analisar: (a) Razão de amortecimento por decremento logarítmico. (b) Frequência e razão de amortecimento por MRPy.fit_decay().
3. Relatório com descrição do objeto, gráficos e resultados (nome do arquivo T1_xxxxxxxx.ipynb, onde xxxxxxxx é o número do cartão do aluno, zipado com o arquivo gerado pelo celular).

Prazo: 30 de março de 2022.