Читать книгу Vibroacoustic Simulation - Alexander Peiffer - Страница 28

We start with a simplified system without damping and external forces in order to get the natural frequencies of the system.

(1.74)

This equation can be rearranged so that it corresponds to the general eigenvalue problem

(1.75)

The non trivial solutions of this are given by:

(1.76)

This leads to the characteristic equation with λ=ω2

(1.77)

With ω12=ks1/m1, ω12=ks1/m1 and ωc2=ksc(m1+m2)m1m2 the solutions are:

(1.78)

The eigenvalues shall be entered into the equations to solve for {Ψi}.

(1.79)

leading to the surprisingly simple eigenvalues after some painful math

(1.80)

The results present the modes of the system or the shape of movement for this natural frequency. Let us simplify the above expression by additional conditions: ks1=ks2=ks and m1=m2=m.

So the modal frequencies read with ω02=ks/m

(1.81)

with the eigenvectors:

Thus, the first mode represents a uniform motion of both masses without any relative motion and thus no effect of the centre spring. The second mode is a symmetric resonance and the centre spring adds some extra stiffness leading to higher frequencies. See Figure 1.12 for both modes.

Figure 1.12 Mode shapes of the 2DOF example. Source: Alexander Peiffer.

When we enter numerical figures with m = 0.1 kg, ks=10 N/m and ksc=2 N/m we get the modal frequencies ωn1=10.0 s−1(fn1=1.59 Hz) and ωn2=11.83 s−1(fn2=1.88 s−1).