# 统计代写|主成分分析代写Principal Component Analysis代考|Mathematical Notes: Natural Frequencies and Modal Amplitude Vectors in a 3-DOF System

The passage above explains the process of finding the natural frequencies and modal amplitude vectors for a 3 degrees-of-freedom system composed of springs and masses. The system is analyzed using the finite element method and the equations of motion follow the form:

[M]U¨+[K]U=0[M]{\ddot{U}} + [K]{U} = {0}[M] U ¨ +[K]U=0

Here, [M][M][M] is the mass matrix and [K][K][K] is the stiffness matrix. When assuming a sinusoidal response for each degree of freedom Ui(t)=Aisin⁡(Ωt+ϕi)U_i(t) = A_i \sin(\Omega t + \phi_i)U i ​ (t)=A i ​ sin(Ωt+ϕ i ​ ), the frequency equation is derived from the eigenvalue problem:

∣[K]−Ω2[M]∣=0|[K] – \Omega^2 [M]| = 0∣[K]−Ω 2 [M]∣=0

For the specific 3 degrees-of-freedom system described, the stiffness and mass matrices were assembled, taking into account the constraint U1=0U_1 = 0U 1 ​ =0:

[K]=[k−k00−k3k−2k00−2k3k−k00−kk],[M]=[00000m0000m00002m][K] = \begin{bmatrix} k & -k & 0 & 0 -k & 3k & -2k & 0 0 & -2k & 3k & -k 0 & 0 & -k & k \end{bmatrix}, [M] = \begin{bmatrix} 0 & 0 & 0 & 0 0 & m & 0 & 0 0 & 0 & m & 0 0 & 0 & 0 & 2m \end{bmatrix}[K]= ​

k −k 0 0 ​

−k 3k −2k 0 ​

0 −2k 3k −k ​

0 0 −k k ​

​ ,[M]= ​

0 0 0 0 ​

0 m 0 0 ​

0 0 m 0 ​

0 0 0 2m ​

After reducing the system due to the constraint, the simplified frequency equation is derived as a cubic equation in Ω2\Omega^2Ω 2 . Upon solving it, three natural circular frequencies are found:

Ω1=0.3914km,Ω2=1.1363km,Ω3=2.2485km\Omega_1 = 0.3914\sqrt{\frac{k}{m}}, \quad \Omega_2 = 1.1363\sqrt{\frac{k}{m}}, \quad \Omega_3 = 2.2485\sqrt{\frac{k}{m}}Ω 1 ​ =0.3914 m k ​

​ ,Ω 2 ​ =1.1363 m k ​

​ ,Ω 3 ​ =2.2485 m k ​

To find the modal amplitude vectors (eigenvectors), the frequencies are individually substituted back into the reduced equations of motion, setting Ai,2=1A_{i,2} = 1A i,2 ​ =1 arbitrarily. This results in a set of homogeneous equations that are solved for the amplitudes Ai,3A_{i,3}A i,3 ​ and Ai,4A_{i,4}A i,4 ​ . For instance, for the fundamental mode Ω1\Omega_1Ω 1 ​ , the amplitude vector is:

1 1.4235 2.0511 \end{bmatrix} \] Similar calculations yield the amplitude vectors for the second and third modes. These amplitude vectors represent the relative magnitudes of displacement for each degree of freedom in each mode of vibration. Thus, every N degree-of-freedom system has N natural frequencies and corresponding N modal amplitude vectors (mode shapes). In the context of structural engineering, finite element models provide a way to compute these quantities accurately and utilize them to analyze the system’s response under various external loads.

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