Question 4.4.2: Calculate the solution of the n-degree-of-freedom system of ......
Calculate the solution of the n-degree-of-freedom system of Figure 4.8 for n=3 by modal analysis. Use the values m_1=m_2=m_3=4\ kg and k_1=k_2=k_3=4\ N / m, and the initial condition x_1(0)=1\ m with all other initial displacements and velocities zero.
The mass and stiffness matrices for n = 3 for the values given become
M=4 I \quad K=\left[\begin{array}{rrr} 8 & -4 & 0 \\ -4 & 8 & -4 \\ 0 & -4 & 4 \end{array}\right]
Following the steps suggested in Window 4.5 yields
1. M^{-1 / 2}=\frac{1}{2} I
\begin{aligned} & =(\lambda-2)[(\lambda-2)(\lambda-1)-1]-1[(\lambda-1)-0] \\ & =(\lambda-1)(\lambda-2)^2-(\lambda-2)-\lambda+1 \\ & =\lambda^3-5 \lambda^2+6 \lambda-1=0 \end{aligned}
The roots of this cubic equation are
\lambda_1=0.1981 \quad \lambda_2=1.5550 \quad \lambda_3=3.2470
Thus the system’s natural frequencies are
\omega_1=0.4450 \quad \omega_2=1.2470 \quad \omega_3=1.8019
To calculate the first eigenvector, substitute \lambda_1=0.1981 into (\widetilde{K}-\lambda I) v _1= 0 and solve for the vector v _1=\left[\begin{array}{lll}v_{11} & v_{12} & v_{13}\end{array}\right]^T. This yields
\left[\begin{array}{ccc} 2-0.1981 & -1 & 0 \\ -1 & 2-0.1981 & -1 \\ 0 & -1 & 1-0.1981 \end{array}\right]\left[\begin{array}{l} v_{11} \\ v_{21} \\ v_{31} \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]
Multiplying out this last expression yields three equations, only two of which are independent:
\begin{array}{r} (1.8019) v_{11}-v_{21}=0 \\ -v_{11}+(1.8019) v_{21}-v_{31}=0 \\ -v_{21}+(0.8019) v_{31}=0 \end{array}
Solving the first and third equations yields
v_{11}=0.4450 v_{31} \text { and } v_{21}=0.8019 v_{31}
The second equation is dependent and does not yield any new information. Substituting these values into the vector v _1 yields
v _1=v_{31}\left[\begin{array}{c} 0.4450 \\ 0.8019 \\ 1 \end{array}\right]
4. Normalizing the vector yields
v _1^T v _1=v_{31}^2\left[(0.4450)^2+(0.8019)^2+1^2\right]=1
Solving for v_{31} and substituting back into the expression for v _1 yields the normalized version of the eigenvector v _1 as
v _1=\left[\begin{array}{l} 0.3280 \\ 0.5910 \\ 0.7370 \end{array}\right]
Similarly, v _2 and v _3 can be calculated and normalized to be
v _2=\left[\begin{array}{r} -0.7370 \\ -0.3280 \\ 0.5910 \end{array}\right] \quad v _3=\left[\begin{array}{r} -0.5910 \\ 0.7370 \\ -0.3280 \end{array}\right]
The matrix P is then
P=\left[\begin{array}{rrr} 0.3280 & -0.7370 & -0.5910 \\ 0.5910 & -0.3280 & 0.7370 \\ 0.7370 & 0.5910 & -0.3280 \end{array}\right]
(The reader should verify that P^T P=I and P^T \widetilde{K} P=\Lambda.)
5. The matrix S=M^{-1 / 2} P=\frac{1}{2} I P or
S=\left[\begin{array}{rrr} 0.1640 & -0.3685 & -0.2955 \\ 0.2955 & -0.1640 & 0.3685 \\ 0.3685 & 0.2955 & -0.1640 \end{array}\right]
and
S^{-1}=P^T M^{1 / 2}=2 P^T I=\left[\begin{array}{rrr} 0.6560 & 1.1820 & 1.4740 \\ -1.4740 & -0.6560 & 1.1820 \\ -1.1820 & 1.4740 & -0.6560 \end{array}\right]
(Again the reader should verify that S^{-1} S=I.)
6. The initial conditions in modal coordinates become
\dot{ r }(0)=S^{-1} \dot{ x }_0=S^{-1} 0 = 0
and
r (0)=S^{-1} x _0=\left[\begin{array}{rrr} 0.6560 & 1.1820 & 1.4740 \\ -1.4740 & -0.6560 & 1.1820 \\ -1.1820 & 1.4740 & -0.6560 \end{array}\right]\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right]=\left[\begin{array}{r} 0.6560 \\ -1.4740 \\ -1.1820 \end{array}\right]
7. The modal solutions of equation (4.85) are each of the form given by equation (4.67) and can now be determined as
\begin{aligned} \ddot{r}_1(t)+\omega_1^2 r_1(t) & =0 \\ \ddot{r}_2(t)+\omega_2^2 r_2(t) & =0 \\ \vdots & \\ \ddot{r}_n(t)+\omega_n^2 r_n(t) & =0 \end{aligned} (4.85)
\begin{aligned} & r_1(t)=(0.6560) \sin \left(0.4450 t+\frac{\pi}{2}\right)=0.6560 \cos (0.4450 t) \\ & r_2(t)=(-1.4740) \sin \left(1.247 t+\frac{\pi}{2}\right)=-1.4740 \cos (1.2470 t) \\ & r_3(t)=(-1.1820) \sin \left(1.8019 t+\frac{\pi}{2}\right)=-1.1820 \cos (1.8019 t) \end{aligned}
8. The solution in physical coordinates is next calculated from
x =S r (t)=\left[\begin{array}{rrr} 0.1640 & -0.3685 & -0.2955 \\ 0.2955 & -0.1640 & 0.3685 \\ 0.3685 & 0.2955 & -0.1640 \end{array}\right]\left[\begin{array}{r} 0.6560 \cos (0.4450 t) \\ -1.4740 \cos (1.2470 t) \\ -1.1820 \cos (1.8019 t) \end{array}\right]
\left[\begin{array}{l} x_1(t) \\ x_2(t) \\ x_3(t) \end{array}\right]=\left[\begin{array}{l} 0.1075 \cos (0.4450 t)+0.5443 \cos (1.2470 t)+0.3492 \cos (1.8019 t) \\ 0.1938 \cos (0.4450 t)+0.2417 \cos (1.2470 t)-0.4355 \cos (1.8019 t) \\ 0.2417 \cos (0.4450 t)-0.4355 \cos (1.2470 t)+0.1935 \cos (1.8019 t) \end{array}\right]
The calculations in this example are a bit tedious. Fortunately, they are easily made using software as done in Section 4.9 and in the Engineering Vibration Toolbox. In fact, computing the frequencies, the matrix of eigenvectors P, and subsequently the mode-shape matrix S starting with the determinant is not the recommended way to proceed. Rather, the symmetric algebraic eigenvalue problem should be solved directly, and this is best done using the software methods covered in Section 4.9. The solution in this example is plotted and compared to a numerical simulation in Section 4.10.