Mechanical Vibrations

Mechanical Vibrations

Degrees of Freedom: the number of independent variables needed to completely describe the system dynamics.

NVH: Noise, Vibrations and Harshness: Vehicle Controlability

• Base: a body vibrates when there is an oscillation
• The diapason produces exactly a vibration at 440 Hz, it is used to calibrate

The effective value of a vibration is related to its energy and maybe to its destructive potential (Hertzquake)

• Source: where the vibration comes from, where the vibration excites
• Factor: Path in how vibration is transferred
• Receiver: Quantity of sound, vibration get the receiver

1 DOF Systems

Generally the vibration is induced by a pertubation from the equilibrium position, it consists in an oscillation around the equilibrium position

In the simplest case of a mass connected to a spring the equilibrium position is given by the spring elongation Δ = mg/k

If a force is applied to the mass that moves from equilibrium position and starts to oscillate subjected to the elastic force Fe = -kx as considering the D'Alembert approach we get:

This is a linear ode of second order, we have to define the Cauchy's problem in order to get a solution:

• Initial position and initial velocity must be known

The same equation can be written as:

Defining ω_n:

The general solution is given by a linear combination of sines and cosines functions:

where A = x₀, B = x˙₀ can be calculated imposing the initial conditions.

Damper: A device which dissipates energy when stretched.

If x(t) is the relative position of the piston, then the forces of the damper can be formally written as:

F = Cx_d dt + ... = Cx_d

C = Damping coefficient [ML^-1]

Consider again the system formed by a mass connected to the ground by a linear spring and a damper (see figure above), the equation is:

m x¨(t) + Cx˙(t) + κ(t) = 0 Unforced system

mx_d + κx(t) = -x_dx_s

We search the solution of the equation using exponential function because the exponential function can convert fractional with respect to the derivative operator, that means if we derive an exponential equation the result is an exponential equation:

x(t) = A eˣᵗ

If we derive x(t) and substitute in the equation we get (multiplying A):

L²m^eˣᵗ + Ceˣᵗ + κ eˣᵗ = 0 => αL² + C/m κL² = 0 =>

α² + 2 Εμ α + ω_n² = 0

When: ω_n = √^κ/_m Natural Circular Frequency

Δ = -2/κ

Resolving the 2^nd grade equation above we get (homogeneous equation):

α_1,2 = ω_n(-Ε ± √^Δ2 - 1)

If the solutions are distinct (Ε ± 1) the general solution of the equation is:

α(t) = Ae^α₁t + Be^α₂t

The type of response of the system (and the number of roots) depends on the value of Ε:

• Aperiodic non-oscillating motion: |E| > 1 => α₁, α₂ are both real and negative, the system approaches to zero without oscillating.

x(t)

For example:

∫₀^2π f(t) cos² t dt = A_r ∫₀^2π cos² t dt + ∑_r=0^∞ (A_r ∫₀^2π cos 2t cos rt dt) + ∑₀^2π (b_r cos rt sin rt dt)

Choosing t' = ∞ τ it is possible to transform the equations in real time.

Multiplying both sides of Fourier expansion for cos(w₀t) or sin(w₀t) and using orthogonality properties it is easy to obtain:

∫₀^T f(t) R_r(w₀t) cos(w₀st) dt = A_r ∞ ₀ R(t) cos t dt r =0, ..., ∞

With the same method base using sinus function we get:

b_r = 2/T ∫₀^T (f(t) sin rt dt) r = 1, ..., ∞

Any periodical function can be expressed as a sum of its harmonic functions;

The Fourier series represents the importance of each trigonometric function (each harmonic) in representing the periodic function f(t); the term a₀ represents the mean value of the function.

Each term is independent because of the orthogonality of the trigonometric functions.

The Fourier series analysis and converge if the Dirichlet Conditions are satisfied:

• f(t) is continuous or presents a finite number of discontinuities within the period (see piecewise concept)
• f(t) has a finite number of maxima or minima in the period
• f(t) is absolutely integrable in the period; ∫₀^T|f(t)| dt < ∞

The force F(t) can be obtained by an integration of dF(t) along all the interval.

Each strip is a Dirac delta instantaneous impulsive F(t)dτ as the response of the oscillator to this portion of excitation is the response of the system to the unitary impulse h(t) with the amplitude F(t)dτ shifted in time to t = τ.

The response is:

dα(t - τ) = h(t - τ)dF(t)dτ

Due to the linearity of the problem the superposition principle holds, so the response to a general force is given by the summation of all the infinitesimal contributions:

α(t) = ∫₀^t h(t - τ)dF(τ)dτ

In case of nonhomogeneous initial condition (because the linearity of the system) it's safe to use the superposition principle.

Which is the function I have to translate between h and f? => h(t - τ) or f(t, c)? It's the same, let's show:

α(t) = ∫₀^t h(t - τ)dF(τ)dτ => Introducing a transformation of variables =>

t - τ = z => dτ = -dz and {ξ = 0 -> z = t; τ = t -> z = 0}

We get => α(t) = ∫₀^t h(t - τ)dF(τ)dτ = ∫_t⁰ h(z)dF(t - z)(-dz) =

= ∫₀^t h(z)dF(t - z)dz => z and τ are both dummy variables of integration so their two forms of integral are the same

We have shown that the convolution integral is symmetric in the excitation and in the impulse response, in the sense that the result is the same regardless of which of the two functions is shifted and folded.

The convolution integral can be written in a compact form as:

α(t) = h(t) * F(t) = F(t) * h(t) => Symmetry Property

x(t) = ₁/^2π (e^jΩ₁t + e^-jΩ₁t) = ₁/² cos (Ω₁t)

X(ω)

X(ω) = j ∫ [f(w - Ω₁) - f(w + Ω₁)]∞^−∞

x(t) ∫ f(w-Ω₁) e^jΩ du - ∫_-∞^∞ ₁/^2π (f(w-Ω₁) e^jΩ du

= ₁/^2π (e^jΩ₁t - e^-jΩ₁t) = ±₁/^π sin (Ω₁t)

So, having this demonstracion, we can say that:

x(t) = A cos Ω₁t ⇒ its spectrum is given by:

wherever amplitude of the spectrum is _A/^π

x(t) = A cos Ω₁t + B cos Ω₂t

Spectrum of multi frequency function

Working in frequency domain instead of circular frequency and keeping in mind that ω = 2π f for the same function we get that the amplitude is

X(f)