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

PEC00025: Introduction to Vibration Theory¶

Class 08 - Vibration analysis in frequency domain¶

1. Frequency preservation theorem
2. Reological models
2.1. General linear model
2.2. The Kevin-Voigt model
2.3. The Maxwell model
2.4. The standard model
3. Equilibrium in frequency domain
4. Assignment

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

1. Frequency preservation theorem ¶

Let us assume now that our idealized single degree of freedom system is subjected to a harmonic (sinusoidal) loading with frequency $f_0$:

$$F(t) = F_{\rm max} \sin(2\pi f_0 t + \theta_F)$$

with $F_{\rm max}$ being the force function amplitude and $\theta$ some phase angle.

Recalling the convolution theorem for Fourier Transform presented in last class we anticipate that, in the same way as with Laplace Transform, the system response can calculated as:

$$u(t) = f(t) * h(t) = \int_{-\infty}^{\infty} {f(t - \tau}) h(\tau) \, \; d\tau$$

with $f(t) = F(t)/m$. Deriving now the displacement twice we got the acceleration:

$$\ddot{u}(t) = \int_{-\infty}^{\infty} {\ddot{f}(t - \tau}) h(\tau) \, \; d\tau$$

but replacing the sinusoidal function we get:

$$\ddot{u}(t) = -(2\pi f_0)^2 \int_{-\infty}^{\infty} {f(t - \tau}) h(\tau) \, \; d\tau$$

Replacing now the integral in the expression above by $u(t)$ we arrive at:

$$\ddot{u}(t) + (2\pi f_0)^2 u(t) = 0$$

what is a homogeneous differential equation with solution:

$$u(t) = u_{\rm max} \sin(2\pi f_0 t + \theta_u)$$

This solution states that a linear system responds to a harmonic excitation preserving the same excitation frequency. On the other hand, whenever a system is subjected to a harmonic excitation and the excitation frequency is not exclusivelly preserved in the system response, this will be a strong indication that the system is not linear.

This theorem is demonstrated below with a excitation frequency $f_0 = 1$Hz applied to a linear system with natural frequency $f_{\rm n} = 2$Hz. The system equation is solved with the MRPy method sdof_Fourier() which assumes that the loading is periodic and hence no initial conditions are required. How this method works will be explained in the following sections.

2. Reological models ¶

2.1. The general linear model ¶

The damped spring-mass system depicted at the beginning of this notebook, with a single spring in parallel with a single damper, is one amongst many possible models for the reological behavior of linear systems. In general, the equilibrium equation may be written as:

$$m \ddot{u} + r(u, t) = F(t)$$

where $r(u, t)$ is a time function abridging both restitutive and reactive forces, which obviously depend on the system response itself, $u(t)$. In general, it can be stated that a linear reological model is the solution of the equation:

$$\left(a_0 + a_1\frac{\partial}{\partial t} + a_2\frac{\partial^2}{\partial t^2} + \dots \right) r(t) = \left(b_0 + b_1\frac{\partial}{\partial t} + b_2\frac{\partial^2}{\partial t^2} + \dots \right) u(t)$$

If we apply a Fourier Transform to this equation, with $\mathscr{F} \left\{ r(t) \right\} = R(\omega)$ and $\mathscr{F} \left\{ u(t) \right\} = U(\omega)$, after some algebraic manipulation we arrive at the following general relation:

$$R(\omega) = K(\omega) \left[ 1 + i\mu(\omega) \right] U(\omega)$$

what in short means that there will be an in phase and an out of phase force to displacement response. Each reological models will have its own $K(\omega)$ and $\mu(\omega)$ functions. It is important to observe the importance of the Fourier Transform approach here, which makes possible the solution of equilibrium equation in frequency domain without the need of solving an integral-differential equation in time domain.

2.2. The Kevin-Voigt model ¶

The parallel spring-damper system, commonly used for elastic structures in Civil Engineering, is known as Kelvin-Voigt model.

In time domain the model is formulated as:

$$r(t) = c\dot{u}(t) + ku(t)$$

while its properties are:

$$\begin{align*} K(\omega) &= k \\ \mu(\omega) &= -\frac{c\omega}{k} \end{align*}$$

We will be always using this model, unless otherwise explicitly stated.

2.3. The Maxwell model ¶

The series spring-damper system, the most basic representation of a viscous material, is known as Maxwell model.

In time domain the model is stated as:

$$r(t) + \frac{c}{k} \; \dot{r}(t) = c\dot{u}(t)$$

while its properties are:

$$\begin{align*} K(\omega) &= \frac{c^2 k}{c^2 \omega^2 + k^2} \omega^2 \\ \mu(\omega) &= -\frac{k}{c\omega} \end{align*}$$

Observe that this model implies the possibility (almost a certainty) of a system non-zero final response. Furthermore, for zero excitation frequency (what means a static load) the displacement will become infinity.

2.3. The Zener (standard) model ¶

A more complex reological model is known as the Zener (or standard) model. There are two versions for these model, as illustrated below.

The Zener model (Maxwell version)	The Zener model (Kelvin version)

The time domain equation for the Maxwell version of Zener model is:

$$r(t) + \frac{c}{k_2} \; \dot{r}(t) = k_1 u(t) + \frac{c(k_1 + k_2)}{k_2} \; \dot{u}(t)$$

while the equation for the Kelvin version is:

$$r(t) + \frac{c}{k_1 + k_2} \; \dot{r}(t) = \frac{k_1 k_2}{k_1 + k_2} \; u(t) + \frac{c k_1}{k_1 + k_2} \; \dot{u}(t)$$

As an exercise, it is suggested the derivation of frequency domain functions, $K(\omega)$ and $\mu(\omega)$ for these two Zener model versions.

3. Equilibrium in frequency domain ¶

After choosing a suitable reological model, we may now go back to the linear dynamic equilibrium equation:

$$m \ddot{u} + r(u, t) = F(t)$$

and apply a Fourier transform to both sides of equation:

$$-\omega^2 U(\omega) + \frac{1}{m} \; R(\omega) = F(\omega)$$

where $\mathscr{F} \left\{ F(t)/m \right\} = F(\omega)$.

After replacing the expression for the choosen $R(\omega)$, the system response $U(\omega)$ can be calculated as a function of excitation $F(\omega)$, and transformed back to time domain (if required). One must be aware that this is complex algebra and the solution will have both an in phase and an out of phase components.

In the following we will do the math for the most used Kevin-Voigt model. The equilibrium in frequency domain is:

$$-\omega^2 U(\omega) + \frac{k}{m} \; \left( 1 - i\frac{c\omega}{k} \right) U(\omega) = F(\omega)$$

Now we isolate $U(\omega)$ and recall that $\omega_{\rm n}^2 = k/m$ and that $\zeta = c/(2m\omega_{\rm n})$, what gives:

$$U(\omega) = H(\omega) F(\omega)$$

where:

$$H(\omega) = \frac{1}{\omega_{\rm n}^2} \; \left[ \frac{1}{(1 - \beta^2) - i(2\zeta\beta)} \right]$$

is called the system mechanical admittance, with $\beta = \omega / \omega_{\rm n}$ being a nondimensional frequency.

These expressions allow a straightforward solution of the equilibrium equation in time domain (although its complex algebra). The mechanical admittance can be understood as a frequency dependent flexibility, for as the excitation frequency goes to zero (static condition) the admittance goes to $1/\omega_{\rm n}^2$, which is the flexibility coefficient (inverse of the stiffness coefficient, $\omega_{\rm n}^2$) with unity mass.

In the expression above it can also be observed that for undamped systems, $\zeta = 0$, the admittance, and consequently the system response, will rise to infinity when the excitation frequency equals the system natural frequency, $\omega = \omega_{\rm n}$. This condition is called resonance, and must be always avoided in structural systems design.

The class MRPy has a method sdof_Fourier() that is an implementation of this solution in frequency domain (for Kevin-Voigt model). It requires no inicial condition, for in Fourier analysis the numerical approach (where signals must have a finite duration) assumes the signal is always periodic.

The example below compares the solutions with sdof_Fourier() and sdof_Duhamel() for a linear system subject to a harmonic load. Observe the difference due to the initial conditions in time domain approach. The two responses become the same after some acceleration time in the solution with Duhamel. This difference between the two solution methods is exactly the system response to some initial conditions.

There are many areas in engineering analysis where design codes define dynamic loads as power spectra. For instance, surface irregularities in pavements, earthquake accelerations, wind speeds, etc. In these cases, the equation:

$$U(\omega) = H(\omega) F(\omega)$$

which preserve phase information, is replaced by:

$$S_U(\omega) = \lvert H(\omega) \rvert^2 S_F(\omega)$$

where, instead of the complex Fourier transforms, the (real) spectral densities are used. In this new equation, the absolute squared mechanical admittance becomes:

$$\lvert H(\omega) \rvert^2 = \frac{1}{\omega_{\rm n}^4} \; \left[ \frac{1}{(1 - \beta^2)^2 + (2\zeta\beta)^2} \right]$$

with $\beta = \omega/\omega_{\rm n}$. The square root of the expression between brackets is also called dynamic amplification factor, $A(\beta, \zeta)$:

$$A(\beta, \zeta) = \sqrt{\frac{1}{(1 - \beta^2)^2 + (2\zeta\beta)^2}}$$

plotted below for some typical values of the damping ratio.

The most important features of the dynamic amplification factor are:

• The factor represents the dynamic response increase with respect to the static response to a harmonic loading:
  
  $$F(t) = F_{\rm max} \sin(\omega t + \theta)$$
  such that
  $$u_{\rm max} = A(\beta, \zeta) \; \frac{F_{\rm max}}{m \omega_{\rm n}^2}$$
  where we recall that $m \omega_{\rm n}^2$ is the stiffness coefficient and the phase information is lost.
• For very low frequencies the amplification becomes $A(\beta, \zeta) = 1$, what means static response.
• For $\beta \approx 1$ the amplification attains its maximum, which is approximatelly $A(1, \zeta) = 1/(2\zeta)$. This means, for instance, that for a typical damping ratio, $\zeta = 1$%, the dynamic response is amplified approximately 50 times with respect to the static response.

Lets take a look at this theory by running an example.

4. Example of application ¶

The aerodynamic force over a bluff body due to wind speed turbulence can be calculated as:

$$F(z, t) = \frac{1}{2} \rho V^2(z, t) C_{\rm D} A$$

where $z$ is the height above ground, $\rho$ is the air density, $V(z,t)$ is the wind speed at height $z$, and the product $C_{\rm D} A$ refers to the aerodynamic characteristics of the body. It can be shown that the spectral density of this aerodynamic force is related to the spectral density of the (fluctuating part) of the wind speed through:

$$S_F(z, f) = \left [ \frac{2\bar{F}(z)}{\bar{V}(z)} \right ]^2 S_v(z, f)$$

where $f$ is the frequency, $\bar{F}(z)$ is the mean force at height $z$, and $\bar{V}(z)$ is the mean wind speed at height $z$.

The wind speed turbulence, $v(t)$, may be modelled according to Harris' spectral density, $S_V(f)$:

$$\frac{f S_V(f)}{\sigma_V^2} = \frac{0.6 Y}{\left( 2 + Y^2 \right)^{5/6}}$$

with:

$$\begin{align*} Y &= 1800 f \;/\; \bar{V}_{10} \\ \sigma_V^2 &= 6.66 \; c_{\rm as} \; \bar{V}_{10}^2 \end{align*}$$

where $\sigma_V^2$ is the wind speed variance, $c_{\rm as}$ is the surface drag coefficient and $\bar{V}_{10}$ is the mean wind speed at 10m height.

5. Assignment ¶

1. Calcular as funções $K(\omega)$ e $\mu(\omega)$ para o modelo reológico de Zener (escolha uma das duas versões, Maxwell ou Kelvin).
2. Aplique as funções acima na equação de equilíbrio dinâmico no domínio da frequência e deduza a correspondente função de admitância mecânica.
3. Plote esta admitância em função da frequência adimensionalizada.
4. Apresente a expressão da frequência natural de vibração livre para este modelo.
5. Relatório com deduções, gráficos e resultados (nome do arquivo T6_xxxxxxxx.ipynb).

Prazo: 03 de junho de 2020.