# X1 f1 k2 x2

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Unformatted text preview: king the square root of λr2 yields the four roots of the original quartic equation. λr ≈ ± j 0.814, ± j 3.686 ≈ ±0.902 j , ±1.920 j rad s Notice that the roots occur in complex conjugate pairs. This will always be the case for physically real systems (i.e. mass, stiffness &amp; damping are positive and real.) Which means the natural frequencies are, Ω r = ±0.902 &amp; ± 1.920 rad s Recall the original development of the general solution to the homogeneous problem. It was stated then that the solution could be written as pairs of complex conjugate terms in the following form, { x(t )} = # dof r =1 ∑ ( Q {ψ } r eλrt + Qr* {ψ }r eλr t r * * ) (Ignore for the moment the Qr term. It is simply a scaling term to account for the arbitrary scaling of {ψ }r that results from the eigenvalue solution. In essence, it is the constant of integration for the differential equation.) Now that the eigenvalues (natural frequencies) are known, it is possible to solve for the eigenvectors (mode shapes). Because the solution is a sum of complex conjugate terms, it is only necessary to solve for the positive frequency vectors. Again, starting with the above homogeneous equation, 2 0 2 3 + 2 −2 X 1 0 = s + −2 2 X 2 0 0 1 Substitute the roots ( λr ) for s and expand the equations, −2 X 1 + ( λr2 + 2 ) X 2 = 0 Since the values of λr are chosen to make the equation set rank deficient, there is an arbitrary choice of one value (either X 1 or X 2 .) For this example, choosing X 1 = 1 yields the following equation set, ( 2λ 2 r + 5) X1 − 2 X 2 = 0 Lecture Notes -61- 06/16/06 12:25 PM Mechanical Vibrations I X2 = − ( 2λr2 + 5 ) −2 −2 X2 = − ( λr2 + 2 ) Substituting the first root ( λ1 = 0.902 j ) into either equation yields X 2 = 1.686 . Substituting the second root ( λ2 = 1.920 j ) into either equation yields X 2 = −1.186 . Therefore, the natural frequencies and mode shapes for this two degree of freedom system are, λ1 = j 0.902 rad s and {ψ }1 = {ψ }2 = 1 1.686 λ2 = j1.920 rad s plus their complex conjugates. 1 −1.186 (Final note: if the complete homogeneous (free response) solution was desired, the final step would be to solve for the Qr ’s by evaluating the initial conditions { x(0)} and { x(0)} .) Lecture Notes -62- 06/16/06 12:25 PM Mechanical Vibrations I #12 – Estimating Modal Parameters – Frequency Domain There are many techniques, both simple and advanced, for estimating modal parameters (frequency, damping, mode shape, and scaling) from measured experimental data. This section will focus upon the single degree-of-freedom frequency domain technique known as quadrature. Quadrature The quadrature method is based entirely upon the characteristics of a single degree-offreedom free frequency response function. The quadrature method has many advantages over the log decrement technique presented in section five. In the frequency domain, many lightly damped, multi degree-of-freedom systems behave much like a single de...
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