Structural Dynamics: Dynamic Response Flashcards
What is free and forces response?
Given the equation: Mq_dot_dot + Cq_dot + Kq = F with q(0) = q0 and q_dot(0) = q_dot0:
*Free response is the response to an initial perturbation of conditions without any forcing, F=0.
*Forces response is the response to a forcing excitation when the initial conditions are zero.
The total response (T) is equal to the homogenous solution (H) and the particular integral (P) or equal to the free response (X) and the forces response (F). If the solution is stable, the homogeneous or the free response dies out and so for t->inf the forces response becomes equal to the particular integral of the solution.
How can the dynamic response be computed?
1) Direct solution in the time domain by modal decomposition and integration.
2) Direct solution in the time domain by time marching integration.
3) Solution using Laplace Transform.
4) Solution in the frequency domain using Fourier series/transform.
The solution method can be applied to the full problem or after a coordinate reduction.
What is the Laplace transform?
The Laplace transform is an integral transformation used to solve differential equations. It converts a function of time to a function of a complex variable s. The Laplace transform takes a linear differential equation and transforms it into an algebraic equation.
What is the transfer function or transfer matrix of the system?
G(s) is defined as the ratio between the input and output.
What is the frequency response of a system?
If the input is harmonic, we can compute the particular integral by saying that also the generalised degrees of freedom will be harmoning. This is true, since the system is linear and stable.
What is the dynamic stiffness?
Kdyn is equal to (-ω^2M +jωC + K). It is called dynamic stiffness because it is the dynamic generalization of the stiffness matrix K (q = Kdyn^-1F). The result obtained is the particular integral, it corresponds to the forces response of the system if the system is stable.
We can non-dimensionalise the frequency response: H = K*Kdyn^-1 = Kq/F = elastic forces/external forces.
What is the modal computation of the response?
H(ω) = Σ φiφi’/(μi(ωi^2 + jω2ξiωi - ω^2).
The response is the sum of each mode. The final response is the summation of several effects coming from each mode.
What is the dynamic aplification factor?
H(ω) = Σ φiφi’/(μi(ωi^2 + jω2ξiωi - ω^2).
Di(ω) = 1/(1 + j2ξ(ω/ωi) - (ω/ωi)^2). This means that high frequency modes loser their relevance due to the dependancy on ω/ωi.
*If ω«ωi, the response is static.
*If ω = ωi, the contribution of the mode is close to the resonance peak.
*If ω»_space; ωi, the mode is not responding and can be neglected.
Diccuss periodical signals and Fourier series.
The representation with the Fourier series will be exact by using an infinite number of sinusoidal signals. This approach is applicable by truncating the series to a limited number of terms, but H(nΩ) must converge towards zero, while the frequency increases.
Can non-periodic signals be periodic-izable?
In a non periodic signal,a fter a certain time, both the input and the output will go to zero for a stable system. This time becomes the period of the periodical signal. By having a periodical signal, we can apply the Fourier series and find the solutions ar a sum of single sinusoids.
What is the relation between Fourier and Laplace transform?
The Fourier transform is the Laplace transform where s=jω, but there are no initial conditions, because now we consider only a forces response, so the system is considered stable and any effect generated by initial condition are damped after a while.
Discuss limited bandwidth excitation.
For most of the cases, the input signal is a band limited signal (it decays to zero after some time). This happens because it has a finite energy and above a certain frequency, the amplitude must go to zero. For the frequencies out of band, it is possible to consider a static response and for the frequencies inside the band it is possible to consider a dynamic response. For the final response, only the modes inside the bandwidth are considered to have a dynamic response.
Ultimately, a system can be represented with a finite set of degrees of freedom: q(ω) = Σφiφi’/μiωi^2 * DiFi + RF(ω), where R is the quasi-steady correction of the modes outside the bandwidth. R = K^-1 - Σ…
Why can’t the limited bandwidth approach work with rigid modes?
When the system is characterized by one or more rigid modes, this procedure won’t work because the stiffness matrix is singular and so it can’t be inverted.
What approach can be used when rigid body modes are present?
The inertia relief method can be used, where the rigid and elastic dofs are separated.
q = qrigid + qelastic = ΦRzR + ΦeZe. Substituting into the equation Mq_dot_dot + Kq = F and premultiplying with ΦT assuming a unit mass normalization, the acceleration of the rigid body modes is obtained. Substituting this expression to the original equation again, a system for the elastic modes are found by using the acceleration of the rigid modes as a forcing term.
MΦeze_dot_dot +KΦeze = (I - MΦRΦR^T)F = P^TF, where P^T is the projection of the external forces on the elastic modes.
The loads P^TF do not work for the rigid body modes. The work of the loads P^TF for the elastic modes is the same as the work of F for the same modes. This is true only if the elastic modes are orthogonal to the rigid body modes.
In order to invert K, we define a dummy constraint to remove the rigid body modes. The constraints consist of ground the unconstrained system with 6 lumped springs, so 3 linear and 3 rotational springs can be used.
K’ = K + ΦRKgΦR^T, where Kg is a symmetric positive definit matrix. Since K’ is not singular, the inverse is always defined. so that:
qe0 = K’^-1F = K’^-1P^TF = PK’^-1P^TF and the resulting qe0 is always the same no matter the choice of Kg.
What is anti-resonance?
Considering an undamped system and considering the diagonal components of the transfer matrix (harmonic response of dof k to the harmonic excitation applied to the same dof k), the amplitude will tend to go to infinity and the phase will have a 90deg shift. So it is changing sign from positive to negative.
Since everytime there is a natural frequency the function goes to plus infinity and returns to minus infinity, there is a point of the amplitude of frequency that is equal to zero and that is called anti-resonance. If any force is applied there, the system will not move (zero of a transfer function).
Anti-resonance doesn’t depend on how forces are applied, but only frequencies. Undamped system have alternating poles and zeros on the imaginary axis. Poles are resonance and zeros are anti-resonance.
