Question 4.4.4: Compute the solution of the unrestrained system given in Fig......

Summing forces in the horizontal direction on each of the free-body diagrams given in Figure 4.11 yields

\begin{aligned} & m_1 \ddot{x}_1=k\left(x_2-x_1\right) \\ & m_2 \ddot{x}_2=-k\left(x_2-x_1\right) \end{aligned}

Bringing all the forces to the left side and writing in matrix form yields

\left[\begin{array}{ll} 1 & 0 \\ 0 & 4 \end{array}\right]\left[\begin{array}{l} \ddot{x}_1 \\ \ddot{x}_2 \end{array}\right]+4\left[\begin{array}{rr} 1 & -1 \\ -1 & 1 \end{array}\right]\left[\begin{array}{l} x_1 \\ x_2 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \end{array}\right]

Note that the determinant of the stiffness matrix K is zero, indicating that it is singular and hence has a zero eigenvalue (see Appendix C). Following the steps of Window 4.5 and substituting the values for M and K yields the following:

1. M^{-1 / 2}=\left[\begin{array}{cc}1 & 0 \\ 0 & \frac{1}{2}\end{array}\right]
2. \widetilde{K}=M^{-1 / 2} K M^{-1 / 2}=4\left[\begin{array}{cc}1 & 0 \\ 0 & \frac{1}{2}\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ 0 & \frac{1}{2}\end{array}\right]=\left[\begin{array}{cc}4 & -2 \\ -2 & 1\end{array}\right]
3. Calculating the eigenvalue problem for 2 yields

\operatorname{det}(\widetilde{K}-\lambda I)=\operatorname{det}\left(\left[\begin{array}{cc} 4-\lambda & -2 \\ -2 & 1-\lambda \end{array}\right]\right)=\left(\lambda^2-5 \lambda\right)=0

This has solutions \lambda_1=0 and \lambda_2=5, so that \omega_1=0 and \omega_2=\sqrt{5}= 2.236\ rad / s. Note the zero eigenvalue/frequency. However, the eigenvector for \lambda_1 is not zero (eigenvectors are never zero) as the following calculation for the eigenvector for \lambda_1=0 yields

\left[\begin{array}{cc} 4-0 & -2 \\ -2 & 1-0 \end{array}\right]\left[\begin{array}{l} v_{11} \\ v_{21} \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \end{array}\right] \text { or } 4 v_{11}-2 v_{21}=0

Thus 2 v_{11}=v_{21}, or v _1=\left[\begin{array}{ll}1 & 2\end{array}\right]^T. Repeating the procedure for \lambda_2 yields v _2= \left[\begin{array}{ll}2 & -1\end{array}\right]^T.
4. Normalizing both eigenvectors yields

v _1=\left[\begin{array}{l} 0.4472 \\ 0.8944 \end{array}\right] \text { and } v _2=\left[\begin{array}{r} -0.8944 \\ 0.4472 \end{array}\right]

Note that the eigenvector v _1 associated with the eigenvalue \lambda_1=0 is not zero. Combining these to form the matrix of eigenvectors yields

P=\left[\begin{array}{rr} 0.4472 & -0.8944 \\ 0.8944 & 0.4472 \end{array}\right]

As a check note that

P^T P=I \text { and } P^T \widetilde{K} P=\operatorname{diag}\left[\begin{array}{ll} 0 & 5 \end{array}\right]

5. Calculating the matrix of mode shapes yields

6. Calculating the modal initial conditions yields

\begin{aligned} r (0)=S^{-1} x _0 & =\left[\begin{array}{rr} 0.4472 & 1.7889 \\ -0.8944 & 0.8944 \end{array}\right]\left[\begin{array}{l} 1 \\ 0 \end{array}\right] \\ & =\left[\begin{array}{r} 0.4472 \\ -0.8944 \end{array}\right], \quad \dot{ r }(0)=S^{-1} v _0= 0 \end{aligned}

7. Here is where the zero eigenvalue makes a difference as equations (4.66) and (4.67) only apply for the second natural frequency. The modal equation for the first mode becomes \ddot{r}_1=0, which has solution r_1(t)=a+b t. Here a and b are the constants of integration to be determined from the modal initial conditions. Applying the modal initial conditions yields the two equations

\begin{aligned} & r_1(0)=a=0.4472 \\ & \dot{r}_1(0)=b=0.0 \end{aligned}

Thus the first modal equation has solution

r_1(t)=0.4472

a constant. The solution for the second mode follows directly from equations (4.66) and (4.67) as before and yields

r_2(t)=-0.894 \cos (\sqrt{5} t)

Thus the modal response vector is

r (t)=\left[\begin{array}{c} 0.447 \\ -0.894 \cos (\sqrt{5} t) \end{array}\right]

8. Transforming back into the physical coordinates yields the solution

Note that each of the two physical coordinates moves a constant distance 0.2 units and then oscillates at the second natural frequency.

Next consider the effect of the zero frequency in using the mode summation method. In this case, equation (4.90) becomes

q _i(t)=\left(a_i e^{-\sqrt{\lambda_i} j t}+b_i e^{\sqrt{\lambda_i} j t}\right) v _i            (4.90)

q _1(t)=(a+b t) v _1

and equations (4.93) and (4.94) become

q (0)=\sum\limits_{i=1}^n d_i \sin \phi_i v _i        (4.93)

\dot{ q }(0)=\sum\limits_{i=1}^n \omega_i d_i \cos \phi_i v _i              (4.94)

q (0)=(a+b 0) v _1+\sum\limits_{i=2}^n d_i \sin \phi_i v _i \quad \text { and } \quad \ddot{ q }(0)=b v _i+\sum\limits_{i=2}^n \omega_i d_i \cos \phi_i v _i

Following the same steps as before and using orthogonality yields

a= v _1^T q (0) \text { and } b= v _1^T \dot{ q }(0)

which replace the modal constants of integration d_i and \phi_i for the zero valued mode. Computing d_2 and \phi_2 following equations (4.97) and (4.98), and combining the modes according to equation (4.103), again yields the solution

\phi_i=\tan ^{-1} \frac{\omega_i v _i^T q (0)}{ v _i^T \dot{ q }(0)} \quad i=1,2, \ldots n            (4.97)

d_i=\frac{ v _i^T q (0)}{\sin \phi_i} \quad i=1,2, \ldots n          (4.98)

x (t)=\sum\limits_{i=1}^n d_i \sin \left(\omega_i t+\phi_i\right) u _i           (4.103)

x (t)=a u _1+d_2 \cos (\sqrt{5} t) u _2=\left[\begin{array}{c} 0.2+0.8 \cos (\sqrt{5} t) \\ 0.2-0.2 \cos (\sqrt{5} t) \end{array}\right]

The solutions are plotted in Figure 4.12.