Question

Homework 7: Undamped, 2-DOF System 1. A system with two masses of which the origins are at the SEPs is shown in Figure 1. The mass of m2 is acted by the external force of f(t). Assume that the cable between the two springs, k2 and k3 is not stretchable. Solve the following problems (a) Draw free-body diagrams for the two masses and derive their EOMs (b) Represent the EOMs in a matrix fornm (c) Find the undamped, natural frequencies and the corresponding mode shapes of the system for the given system parameters of k1 k2-k3-100 kN/m, and m1 = 3m2-12 kg. (d) Normalize the eigenvectors such that ATMA = 1 and confirm that your normalized eigenvectors satisfies ATKA where Λ is a diagonal matrix with the square of the natural angular frequencies as the diagonal elements (e) State the modal differential equations (including the definition of modal forces) and the coordinate transformation from the modal coordinate to physical coordinates Determine the steady-state vibration responses of the two bodies with the harmonic force of f(t) = f0 sin(wt) where fo = 2.5 kN and ω = 120 rad/sec. Plot the responses from 0 to 0.2 sec (f) (g) Find the undamped, free vibration responses with the initial conditions that the mass of m2 is pulled down by 0.05 m with zero velocity and the mass of m is held at the SEP. Plot the responses from 0 to 0.2 sec. Here, the initial conditions should be applied to the complete solution (i.e., the homogeneous solution plus the particular solution) or ki m2 Figure 1: 2-DOF System

Answer 1

a) Please refer the diagram below:

判( m1

The equations of motion are

$m_1\ddot x_1=(k_2+k_3)(x_2-x_1)-k_1x_1$

$m_2\ddot x_2=-(k_2+k_3)(x_2-x_1)+f$

b) In matrix form

$\begin{bmatrix} m_1 & 0\\ 0 & m_2\end{bmatrix}+\begin{bmatrix} (k_1+k_2+k_3) &-(k_2+k_3) \\ -(k_2+k_3) & k_2+k_3 \end{bmatrix}=\begin{bmatrix} 0\\ f \end{bmatrix}$

c) With

The matrix equation is

$\begin{bmatrix} 12 & 0\\ 0 & 4\end{bmatrix}+\begin{bmatrix} 300 &-200 \\ -200 & 200\end{bmatrix}=\begin{bmatrix} 0\\ f \end{bmatrix}$

The natural frequencies are given by solutions of the following equation

$\begin{bmatrix} 300-12\omega^2 &-200 \\ -200& 200-4\omega^2 \end{bmatrix}=0$

Solving the above in Matlab eig function:

$\omega_1=\sqrt{6.0424}=2.458\: rad/s$

$\omega_2=\sqrt{68.9576}=8.304\: rad/s$

d) The eigenvectors are given as

V =

-0.2413 -0.1585
-0.2745 0.4179

First normalisation is done by forcing the largest component of eigenvector to +1

Hence normalised eigenvectors are

$\phi_1=\begin{bmatrix} 0.8791\\ 1 \end{bmatrix},\phi_2=\begin{bmatrix} -0.3793\\ 1\\ \end{bmatrix}$

Eigenvector mass normalisation

We assume mass normalised eigenvedtors as

$\bar \phi_1=c_1\phi_1$

Hence

$\bar\phi_1^TM\bar\phi_1=1$

Hence

$c_1^2(\phi_1^TM\phi_2)=1$

Solving in Matlab

$c_1=0.2745$

Similarly

$\bar \phi_2=c_2\phi_2$

$c_2^2(\phi_2^TM\phi_2)=1$

$c_2=0.4179$

Hence mass normalised eigenvectors are

$\bar \phi_1=c_1\phi_1=\begin{bmatrix}0.2413 \\0.2745 \end{bmatrix}$

$\bar \phi_2=c_2\phi_2=\begin{bmatrix} -0.1585\\ 0.4179 \end{bmatrix}$

Check for diagonal matrix

$A=\begin{bmatrix} 0.2413 &-0.1585 \\ 0.2745 & 0.4179 \end{bmatrix}$

$A^TMA=\begin{bmatrix} 6.0430 &0 \\ 0& 68.9596 \end{bmatrix}=\begin{bmatrix} \omega_1^2 & 0\\ 0& \omega_2^2 \end{bmatrix}$

The Matlab result screenshot is given below:

Command Window K-[300-200-200 200] phil-.8791 1]'; a-phil' M*phil; c1-1/a.5 phi2-[-.3793 1]'; c2-1/b.5 A-[.2413 -.1585: .2745 .4179] -0.2413 -0.1585 -0.2745 0.4179 6.0424 0 0 68.9576 0.2745 0.4179 ans - 6.0430 0.0027 68.9596 0.0027 fx. >>

(Please note that 4 subquestions need to be answered as per Chegg guidelines)