Engineering at Alberta Courses » Transfer matrices and periodic structures

Advanced Dynamics and Vibrations: Transfer matrices and periodic structures

Transfer Matrices – State Vectors for Vibrating Systems

For many MDOF systems, it is often more efficient to use a “shooting” technique rather than solve a boundary value problem. Essentially, we employ a series of initial value problems to find the solution to the boundary value problem. In vibrations, several numerical procedures have led to the “transfer matrix/state vector” approach. This includes the methods of Holzer and Myklestad’s for beams. This technique is an iterative technique where an estimate of the natural is used at one boundary then checked at the other and if not correct an iterative procedure is used to “hone-in-on” the answer (eg. Newton and Raphson). After the correct answer is found the mode is also known as the “states” have been calculated.

Transfer Matrices – State Vectors

In the transfer matrix approach, we follow the idea of the Holzer method and guess the eigenvalue (natural frequency). This approach can be extended to continuous systems.

We will introduce this for torsional and spring systems. We define the state of a position in the system as the force and displacement at that point.

$\{\mathfrak z\}_i = \begin{Bmatrix} x \\ F \end{Bmatrix}_i \text{ (axial)}$

$\{\mathfrak z\}_i = \begin{Bmatrix} \theta \\ T \end{Bmatrix}_i \text{ (torsional)}$

Transfer Matrix – relates the state vectors at two points in the system

$F_{i + 1} = F_i$

$x_{i+1} = x_i + \frac{F_i}{k_i}$

Therefore:

$\begin{split} \begin{Bmatrix} x \\ F \end{Bmatrix}_{i+1} &= \begin{bmatrix} 1 & \frac{1}{k_i} \\ 0 & 1 \end{bmatrix} \begin{Bmatrix} x \\ F \end{Bmatrix}_i \\&= [z] \{ \mathfrak z \}_i \end{split}$

This is called a field transfer matrix.

If we have a discontinuity of mass at a point:

$\begin{split} \sum F_x &= m_i \ddot{x}_i \\&= F_i^R - F_i^L \end{split}$

$x_i^L = x_i^R$

Assume $\ddot{x}_i = -p^2x_i$ (SHM). Therefore:

$-m_ip^2x_i = F_i^R - F_i^L$

Therefore:

$\begin{split} \begin{Bmatrix} x \\ F \end{Bmatrix}_i^R &= \begin{bmatrix} 1 & 0 \\ -m_ip^2 & 1 \end{bmatrix} \begin{Bmatrix} x \\ F \end{Bmatrix}_i^L \\&= [p] \{ \mathfrak z \}_i^L \end{split}$

Where the point transfer matrix is $[p]$ . We can combine these to generate a transfer matrix from one position to another which are at like locations.

$\begin{split} \{ \mathfrak z \}_{i+1}^R &= [p]_{i+1} \{ \mathfrak z \}_{i+1}^L \\&= [p]_{i+1} [z] \{ \mathfrak z \}_i^R \end{split}$

Therefore:

$\begin{split} \{ \mathfrak z \}_{i+1}^R &= \begin{bmatrix} 1 & 0 \\ -m_ip & 1 \end{bmatrix} \begin{bmatrix} 1 & \frac{1}{k_i} \\ 0 & 1 \end{bmatrix} \{ \mathfrak z \}_i^R \\&= \begin{bmatrix} 1 & \frac{1}{k_i} \\ -p^2m_i & 1 - \frac{m_ip^2}{k_i} \end{bmatrix} \{ \mathfrak z \}_i^R \end{split}$

Example 1

$\{ \mathfrak z \}_1^L = [Z] \{\mathfrak z\}_o = \begin{bmatrix} 1 & \frac{1}{k} \\ 0 & 1 \end{bmatrix} \{ \mathfrak z \}_o$

$\begin{split} \{ \mathfrak z \}_1^R &= \begin{bmatrix} 1 & 0 \\ -mp^2 & 1 \end{bmatrix} \{ \mathfrak z \}_1^L \\&= \begin{bmatrix} 1 & 0 \\ -mp^2 & 1 \end{bmatrix} \begin{bmatrix} 1 & \frac{1}{k} \\ 0 & 1 \end{bmatrix} \begin{Bmatrix} 0 \\ F \end{Bmatrix}_o \end{split}$

$\begin{Bmatrix} x \\ 0 \end{Bmatrix}_1^R = \begin{bmatrix} 1 & \frac{1}{k} \\ -mp^2 & 1 - \frac{mp^2}{k} \end{bmatrix} \begin{Bmatrix} 0 \\ F \end{Bmatrix}_o$

$0 = \Big(1 - \frac{mp^2}{k} \Big) F_o$

Therefore:

$1 - \frac{mp^2}{k} = 0$

$p^2 = \frac{k}{m}$

A long way to get this simple result.

Consider a torsional system:

Then the positive directions are similar to the analogous linear case. The field equations are then:

$\begin{Bmatrix} \theta \\ T \end{Bmatrix}_i = \begin{bmatrix} 1 & \frac{1}{\hat{k}_1} \\ 0 & 1 \end{bmatrix} \begin{Bmatrix} \theta \\ T \end{Bmatrix}_{i-1}$

The point matrix is:

$\begin{Bmatrix} \theta \\ T \end{Bmatrix}_i^R = \begin{bmatrix} 1 & 0 \\ -J_ip^2 & 1 \end{bmatrix} \begin{Bmatrix} \theta \\ T \end{Bmatrix}_i^L$

It is often useful to divide the torques by a $\hat{k}$ to make them non-dimesional as well.

$\begin{Bmatrix} \theta \\ T \end{Bmatrix} = \begin{bmatrix} 1 & 0 \\ \frac{-J_ip^2}{\hat{k}} & 1 \end{bmatrix} \begin{Bmatrix} \theta \\ \tilde{T} \end{Bmatrix}_i$

$\begin{Bmatrix} \theta \\ T \end{Bmatrix} = \begin{bmatrix} 1 & \frac{\hat{k}}{\hat{k}_i} \\ 0 & 1 \end{bmatrix} \begin{Bmatrix} \theta \\ \tilde{T} \end{Bmatrix}_{i-1}$

Example 2

$\hat{k} = 1 \cdot 10^6 \text{ lbs in/rad}$

$J = 100 \text{ lb in sec}^2$

$\{ \mathfrak z \}_B^R = [p]_3 [z]_{2-3} [p]_2 [z]_{1-2} [p]_1 [z]_{0-1} \{ \mathfrak z \}_o$

$\begin{Bmatrix} \theta \\ 0 \end{Bmatrix}_3^R = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \begin{Bmatrix} 0 \\ T \end{Bmatrix}_o$

Note that $a_{11}$ and $a_{21}$ are not required. It is sometimes better to non-dimensionalize the equations by dividing the torques by $\hat{k}$ . Therefore:

$\begin{bmatrix} 1 & \frac{1}{1.8} \\ 0 & 1 \end{bmatrix} = [z]_{0-1}$

$\begin{bmatrix} 1 & 0 \\ -\frac{p^2(2J)}{\hat{k}} & 1 \end{bmatrix} = [p]_1$

Now we have only to choose the value of $\frac{p^2J}{\hat{k}}$ . Assume:

$\frac{p^2J}{\hat{k}} = 0.2$

Then:

$\begin{bmatrix} 0 \\ T \end{bmatrix}_o = \mathfrak z_o$

$\begin{bmatrix} 1 & 0.5556 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ T \end{bmatrix}_o = \mathfrak z_1^L$

$\begin{bmatrix} 1 & 0.5556 \\ -0.4 & 0.7778 \end{bmatrix} \begin{bmatrix} 0 \\ T \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -0.4 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0.5556 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ T \end{bmatrix} = \mathfrak z_1^R$

$\begin{bmatrix} 0.8 & 0.9449 \\ -0.4 & 0.7778 \end{bmatrix} \begin{bmatrix} 0 \\ T \end{bmatrix} = \begin{bmatrix} 1 & 0.5 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0.5556 \\ -0.4 & 0.7778 \end{bmatrix} \begin{bmatrix} 0 \\ T \end{bmatrix}_o = \mathfrak z_2^L$

$\begin{bmatrix} 0.8 & 0.9449 \\ -0.6 & 0.5416 \end{bmatrix} \begin{bmatrix} 0 \\ T \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -0.25 & 1 \end{bmatrix} \begin{bmatrix} 0.8 & 0.9449 \\ -0.4 & 0.7778 \end{bmatrix} \begin{bmatrix} 0 \\ T \end{bmatrix} = \mathfrak z_2^R$

$\begin{bmatrix} 0.4 & 1.3060 \\ -0.6 & 0.5416 \end{bmatrix} \begin{bmatrix} 0 \\ T \end{bmatrix}_o = \begin{bmatrix} 1 & 0.6667 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0.8 & 0.9449 \\ -0.6 & 0.5416 \end{bmatrix} \begin{bmatrix} 0 \\ T \end{bmatrix}_o = \mathfrak z_3^L$

$\begin{bmatrix} 0.4 & 1.3060 \\ -0.68 & 0.2804 \end{bmatrix} \begin{bmatrix} 0 \\ T \end{bmatrix}_o = \begin{bmatrix} 1 & 0 \\ -0.2 & 1 \end{bmatrix} \begin{bmatrix} 0.4 & 1.3060 \\ -0.6 & 0.5416 \end{bmatrix} \begin{bmatrix} 0 \\ T \end{bmatrix}_o = \mathfrak z_3^R$

This is not a natural frequency as $a_{22} = 0.2804$ and it should be zero. Also, look at the mode shape:

$\begin{Bmatrix} \theta_1 \\ \theta_2 \\ \theta_3 \end{Bmatrix} = \begin{Bmatrix} 0.5556 \\ 0.9449 \\ 1.3060 \end{Bmatrix} \text{ (not correct)}$

The correct answer is $\frac{Jp^2}{k} = 0.3,\ 1.725,\ 4.175$ .

Non Dimensionalized Matrices

$\begin{bmatrix} 1 & \frac{1}{1.8} \\ 0 & 1 \end{bmatrix} = [z]_{0-1} \quad \begin{bmatrix} 1 & 0 \\ \frac{-p^2(2J)}{k} & 1 \end{bmatrix} = [p]_1$

$\begin{bmatrix} 1 & \frac{1}{2} \\ 0 & 1 \end{bmatrix} = [z]_{1-2} \quad \begin{bmatrix} 1 & 0 \\ \frac{-p^2(1.25J)}{k} & 1 \end{bmatrix} = [p]_2$

$\begin{bmatrix} 1 & \frac{1}{1.5} \\ 0 & 1 \end{bmatrix} = [z]_{2-3} \quad \begin{bmatrix} 1 & 0 \\ \frac{-p^2(J)}{k} & 1 \end{bmatrix} = [p]_3$

Assume $\frac{p^2J}{k} = 0.3$ . Then:

$\begin{bmatrix} 0 \\ \tilde{T} \end{bmatrix}_o = \{ \mathfrak z\}_o$

$\begin{bmatrix} 1 & 0.5556 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ \tilde{T} \end{bmatrix}_o = \{ \mathfrak z \}_1^L$

$\begin{bmatrix} 1 & 0.5556 \\ -0.6 & 0.6666 \end{bmatrix} \begin{bmatrix} 0 \\ \tilde{T} \end{bmatrix}_o = \begin{bmatrix} 1 & 0 \\ -0.6 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0.5556 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ \tilde{T} \end{bmatrix}_o = \{ \mathfrak z \}_1^R$

$\begin{bmatrix} 0.7 & 0.8889 \\ -0.6 & 0.6666 \end{bmatrix} \begin{bmatrix} 0 \\ \tilde{T} \end{bmatrix}_o = \begin{bmatrix} 1 & 0.5 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0.5556 \\ -0.6 & 0.6666 \end{bmatrix} \begin{bmatrix} 0 \\ \tilde{T} \end{bmatrix}_o = \{ \mathfrak z \}_2^L$

$\begin{bmatrix} 0.7 & 0.8889 \\ -0.8625 & 0.3333 \end{bmatrix} \begin{bmatrix} 0 \\ \tilde{T} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -0.375 & 1 \end{bmatrix} \begin{bmatrix} 0.7 & 0.8889 \\ -0.6 & 0.6666 \end{bmatrix} \begin{bmatrix} 0 \\ \tilde{T} \end{bmatrix} = \{ \mathfrak z \}_2^R$

$\begin{bmatrix} 0.1250 & 1.10511 \\ -0.8625 & 0.3333 \end{bmatrix} \begin{bmatrix} 0 \\ \tilde{T} \end{bmatrix} = \begin{bmatrix} 1 & 0.6667 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0.7 & 0.8829 \\ -0.8625 & 0.3333 \end{bmatrix} \begin{bmatrix} 0 \\ \tilde{T} \end{bmatrix} = \{ \mathfrak z \}_3^L$

$\begin{bmatrix} 0.1250 & 1.10511 \\ 0.9000 & 0.0177 \end{bmatrix} \begin{bmatrix} 0 \\ \tilde{T} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -0.3 & 1 \end{bmatrix} \begin{bmatrix} 0.1250 & 1.10511 \\ -0.8625 & 0.3333 \end{bmatrix} \begin{bmatrix} 0 \\ T \end{bmatrix} = \{ \mathfrak z \}_3^R$

Therefore, close at $p^2 = 0.3 \frac{k}{J}$ . Modal vector is:

$\begin{Bmatrix} 0.5556 \\ 0.8889 \\ 1.1051 \end{Bmatrix} = \begin{Bmatrix} \theta_1 \\ \theta_2 \\ \theta_3 \end{Bmatrix}$

Where to start, use Dunkerley’s technique.

$\frac{1}{p_1^2} < \frac{1}{p_{11}^2} + \frac{1}{p_{22}^2} + \frac{1}{p_{33}^2}$

$p_{11}^2 = \frac{1.8}{2} \frac{k}{J}$

$p_{22}^2 = \frac{\frac{1}{\frac{1}{1.8} + \frac{1}{2.0}}}{1.25} \frac{k}{J}$

$p_{33}^2 = \frac{\frac{1}{\frac{1}{1.8} + \frac{1}{2.0} + \frac{1}{1.5}}}{1} \frac{k}{J}$

Therefore:

$p_{11}^2 = 0.9 \frac{k}{J}$

$p_{22}^2 = 0.758 \frac{k}{J}$

$p_{33}^2 = 0.581 \frac{k}{J}$

Therefore:

$\frac{1}{p_1^2} < \Big( \frac{1}{0.9} + \frac{1}{0.758} + \frac{1}{0.581} \Big) \frac{J}{k}$

Therefore:

$p_1^2 > 0.241 \frac{k}{J}$

Release Point

Consider that there are masses at the ends of the lever $k_2$ .

Again, we wish to go from $\{ \mathfrak z \}_1^L$ to $\{ \mathfrak z \}_1^R$ .

$x_1^R = -\frac{a}{b}x_1^L$

$\alpha = \frac{a}{b}$

$\begin{split} +\circlearrowleft \sum M_o &= (m_1b^2 + m_1^1a^2) \frac{\ddot{x}_1^L} {b} \\&= -F_1^R(a) - F_1^L(b) \end{split}$

$\begin{split} F_1^R &= -F_1^L \frac{b}{a} + (m_1b^2 + m_1^1a^2) \frac{p^2x_1^L}{ab} \\&= -\frac{F_1^L}{\alpha} + \Big(\frac{m_1}{\alpha} + m_1^1\alpha \Big) p^2x_1^L \end{split}$

Therefore:

$\begin{Bmatrix} x \\ F \end{Bmatrix}_1^R = \begin{bmatrix} -\alpha & 0 \\ p^2 (\frac{m_1}{\alpha} + m_1^1\alpha) & -\frac{1}{\alpha} \end{bmatrix} \begin{Bmatrix} x \\ F \end{Bmatrix}_1^L$

Consider a branched system:

$\{ \mathfrak z\}_1^R = [?] \{ \mathfrak z \}_1^L$

The transfer matrix will include the effect of the branch.

$x_1^B = -\frac{a}{b} x_1$

$\sum M_R = F_1^{RB}(a) - F_1^{LB}(b)$

$F_1^{LB} = \frac{a}{b} F_1^{RB}$

$\{ \mathfrak z \}_3^L = \begin{bmatrix} 1 & \frac{1}{k_3} \\ 0 & 1 \end{bmatrix} \{ \mathfrak z\}_1^{RB}$

$\begin{split} \{ \mathfrak z \}_3^R &= \begin{bmatrix} 1 & 0 \\ -m_3p^2 & 1 \end{bmatrix} \{ \mathfrak z \}_3^L \\&= \begin{bmatrix} 1 & 0 \\ -m_3p^2 & 1 \end{bmatrix} \begin{bmatrix} 1 & \frac{1}{k_3} \\ 0 & 1 \end{bmatrix} \{ \mathfrak z \}_1^{RB} \\&= \begin{bmatrix} 1 & \frac{1}{k_3} \\ -m_3p^2 & 1 - \frac{m_3p^2}{k_3} \end{bmatrix} \{ \mathfrak z \}_1^{RB} \end{split}$

Therefore:

$\begin{Bmatrix} x \\ 0 \end{Bmatrix}_3^R = \begin{bmatrix} 1 & \frac{1}{k_3} \\ -m_3p^2 & 1 - \frac{m_3p^2}{k_3} \end{bmatrix} \begin{Bmatrix} x \\ F \end{Bmatrix}_1^{RB}$

$0 = -m_3p^2 x_1^{RB} + \Big(1 - \frac{m_3p^2}{k^3} \Big) F_1^{RB}$

$F_1^{RB} = +\frac{m_3p^2x_1^{RB}}{\Big(1 - \frac{m_3p^2}{k_3} \Big)}$

Therefore:

$F_1^{RB} = -\Big(\frac{a}{b}\Big) \frac{m_3 p^2}{\Big(1 - \frac{m_3p^2}{k^3} \Big)} x_1$

$F_1^{LB} = -\Big(\frac{a}{b}\Big)^2 \frac{m_3 p^2}{\Big(1 - \frac{m_3p^2}{k_3} \Big)} x_1$

$F_1^R = F_1^L + F_1^{LB}$

$\{ \mathfrak z \}_1^R = \begin{Bmatrix} x \\ F \end{Bmatrix}_1^R$

$\begin{Bmatrix} x \\ F \end{Bmatrix}_1^R = \begin{bmatrix} 1 & 0 \\ -\big(\frac{a}{b}\big)^2\frac{m_3p^2}{1-\frac{m_3p^2}{k^3}} & 1 \end{bmatrix}\begin{Bmatrix} x \\ F \end{Bmatrix}_1^L$

This now has the “effect” of the branch in it.

$\{z\}_2^R = [p]_2[z]_1^2[H]_{1L}^{1R}[z]_0^1[p]_0\{z\}_0^L$

Consider the linear system analog of the torsional one, but with the mass on the other side.

$\{\mathfrak z\}_3^{LB} = [p]_3[z]_{1-3}\{\mathfrak z \}_1^{LB}$

$\begin{split}\begin{Bmatrix}x \\ 0\end{Bmatrix}_3^{LB} &=\begin{bmatrix} 1 & 0 \\ -m_3p^2 & 1\end{bmatrix}\begin{bmatrix}1 & \frac{1}{k_3} \\ 0 & 1\end{bmatrix}\begin{Bmatrix} x \\ F\end{Bmatrix}_1^{LB} \\&= \begin{bmatrix} 1 & \frac{1}{k_3} \\ -m_3p^2 &1 - \frac{m_3p^2}{k_3}\end{bmatrix}\begin{Bmatrix} x \\ F\end{Bmatrix}_1^{LB}\end{split}$

Therefore:

$-m_3p^2x_1^{LB} + \bigg(1-\frac{m_3p^2}{k_3}\bigg)F_1^{LB} = 0$

Therefore:

$F_1^{LB} = \frac{m_3p^2}{1 - \frac{m_3p^2}{k_3}}x_1^{LB}$

Let $\alpha = \frac{a}{b}$ :

$x_1^{LB} = \frac{a}{b}x_1$

Therefore:

$\begin{split}x_1^{LB} &= \frac{n_1}{n_1'}x_1^L \\&= \alpha x_1^L\end{split}$

Therefore:

$F_1^{LB} = \frac{m_3p^2\alpha}{1-\frac{m_3p^2}{k_3}}x_1^L$

$\begin{split}+\circlearrowleft \sum M_x &= 0 \\&= F_1^{LB}a + F_1^Rb - F_1^Lb\end{split}$

Therefore:

$F_1^R = -F_1^{LB}\alpha +F_1^L$

$x_1^R = x_1^L$

Therefore:

$\begin{Bmatrix} x \\ F \end{Bmatrix}_1^R = \begin{bmatrix} 1 & 0 \\ -\frac{m_3p^2}{1-\frac{m_3p^2\alpha^2}{k^3}} & 1 \end{bmatrix}\begin{Bmatrix} x \\ F \end{Bmatrix}_1^L$

By analogy, the torsional system is:

$\begin{Bmatrix} \theta \\ T \end{Bmatrix}_1^R = \begin{bmatrix} 1 & 0 \\ -\frac{J_3p^2\alpha^2}{1-\frac{J_3p^2}{k_3}} & 1 \end{bmatrix}\begin{Bmatrix} \theta \\ T \end{Bmatrix}_1^L$

The transfer matrix approach can be extended to handle geared and branched systems. We can derive the results from scratch or use the analog of the translational systems.

We wish to find $\{\mathfrak z\}_1^R$ in terms of ${\mathfrak z}_1^L$ :

$T_1^L = F_{n_1}$

$T_1^R = F_{n_1'}$

$\theta_1^Ln_1 = \theta_1^Rn_1'$

Therefore:

$T_1^R = T_1^L\frac{n_1'}{N_1}$

$\theta_1^R = \frac{n_1}{n_1'}\theta_1^L$

set $\alpha = \frac{n_1}{n_1'}$ :

$\theta_1^R= \alpha\theta_1^L$

$\{\mathfrak z\}_1^R = \begin{bmatrix} \alpha & 0 \\ 0 & \frac{1}{\alpha}\end{bmatrix}\{\mathfrak z\}_1^L$

We can also include the effect of the inertia of the gears:

$\begin{split}+\circlearrowleft \sum M_0 &= J_1\ddot\theta_1^L \\&= -T_1^L + F_{n_1}\end{split}$

$\begin{split}+\circlearrowright \sum M_0' &= J'_1\ddot\theta_1^R \\&= T_1^R - F_{n_1'}\end{split}$

$F = -\frac{J_1'\ddot\theta_1^R}{n_1} + T_1^R$

Therefore:

$J_1\ddot\theta_1^L = -T_1^L - J_1'\ddot\theta_1^R\frac{n_1}{n_1'}+T_1^R\frac{n_1}{n_1'}$

$J_1\ddot\theta_1^L + J_1'\ddot\theta_1^R\frac{n_1}{n_1'} = - T_1^L + T_1^R\frac{n_1}{n_1'}$

Again we assume $\ddot\theta = -p^2\theta$ . Therefore:

$-p^2\bigg(J_1\theta_1^L + J_1'\frac{n_1}{n_1'}\theta_1^R\bigg) = -T_1^L + T_1^R\frac{n_1}{n_1'}$

$-p^2(J_1\theta_1^L + J_1'\alpha^2)\theta_1^L = -T_1^L + \alpha T_1^R$

Therefore:

$T_1^R = \frac{1}{\alpha}T_1^L - p^2\bigg(\frac{J_1}{\alpha} + J_1'\alpha\bigg)\theta_1^L$

Therefore:

$\{\mathfrak z\}_1^R = \begin{bmatrix} \alpha & 0 \\ -p^2\bigg(\frac{J_1}{\alpha} + J_1'\alpha\bigg) & \frac{1}{\alpha}\end{bmatrix}\{\mathfrak z\}_1^L$

$\{\mathfrak z\}_1^L = [z]_{0-1}[p]_0\{\mathfrak z\}_0^L$

$\{\mathfrak z\}_2^R = [p]_2[z]_{1 \to 2}\{\mathfrak z\}_1^R$

To determine the relationship between $\{\mathfrak z\}_1^R$ and $\{\mathfrak z\}_1^L$ , consider the branch (B).

$\{\mathfrak z\}_3^L = [p]_3[z]_{1-3}\{\mathfrak z\}_1^{LB}$

$\begin{split}\begin{Bmatrix}\theta \\ 0 \end{Bmatrix}_3^L &= \begin{bmatrix} 1 & 0 \\ -J_3p^2 & 1 \end{bmatrix}\begin{bmatrix} 1 & \frac{1}{k_3} \\ 0 & 1 \end{bmatrix}\{\mathfrak z\}_1^{LB} \\&= \begin{bmatrix} 1 & \frac{1}{k_3} \\ -J_3p^2 & 1 - \frac{J_3p^2}{k_3}\end{bmatrix}\{\mathfrak z\}_1^{LB}\end{split}$

Therefore:

$-J_3p^2\theta_1^B+ \bigg(1-\frac{J_3p^2}{k_3}\bigg)T_1^{LB} = 0$

$T_1^{LB}=\frac{J_3p^2}{1-\frac{J_3p^2}{k_3}}\theta_1^B$

If we take counterclockwise as positive with increasing numbers as shown above then viewing the gears from the left gives.

$\theta_1^B = \alpha\theta_1$

Therefore:

$T_1^{LB} = \frac{J_3p^2\alpha}{1-\frac{J_3p^2}{k_3}}\theta_1$

Consider a free body diagram of the gears viewed from the left:

$+\circlearrowleft\sum M_0 = -T_1^L+T_1^R-T_1^{LB}+F(n_1+n'_1)$

Therefore:

$T_1^R=T_1^L+T_1{LB}-F(n_1+n'_1)$

$T_1^R = T_1^L-T_1^{LB}\alpha$

Therefore:

$T_1^{LB}=Fn'_1$

Therefore:

$T_1^R=T_1^L-\frac{J_3p^2\alpha^2}{1-\frac{J_3p^2}{k_3}}\theta_1$

$\begin{split}\begin{Bmatrix}\theta \\T \end{Bmatrix}_1^R &= \begin{bmatrix} 1 & 0 \\ -\frac{J_3p^2\alpha^2}{1-\frac{J_3p^2}{k_3}} & 1\end{bmatrix}\begin{Bmatrix} \theta \\ T\end{Bmatrix}_1^L \\&= [H]\{\mathfrak z\}_1^L\end{split}$

Therefore the effect of the branch is embodied in the $H_{21}$ term.

Example 3

$\frac{n_1}{n'_1} = \frac{1}{2}$

Again write down the state vectors and transfers matricies also nondimensionalize by dividing the torques by $\hat k$ .

$\begin{split}\{\mathfrak z\}_3^L &= [Z]_{1-3}\{\mathfrak z\}_1^{LB} \\&= \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}\begin{Bmatrix} \theta \\ \bar T \end{Bmatrix}_1^{LB}\end{split}$

We need this only to calculate the mode shape.

$\{\mathfrak z\}_2^R = [p]_2[Z]_{1-2}[H]_{1L}^{1R}[Z]_{0-1}[p]_0\{\mathfrak z\}_0^L$

$\begin{split}[p]_0 &= \begin{bmatrix} 1 & 0 \\ -\frac{2p^2J}{k} & 1\end{bmatrix}[Z]_{0-1} \\&= \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}\end{split}$

$\begin{split}[H]_{1L}^{1R} &= \begin{bmatrix}1 & 0 \\ -\frac{p^2J}{4k(1-\frac{p^2J}{k})} & 1 \end{bmatrix}[Z]_{1-2} \\&= \begin{bmatrix}1 & 0.5 \\ 0 & 1 \end{bmatrix}\end{split}$

$[p]_2 = \begin{bmatrix}1 & 0 \\ -\frac{p^2J}{k} & 1 \end{bmatrix}$

$\begin{split}x_3^B &= x_1^{LB} + \frac{1}{k_3}F_1^{LB} \\&= \alpha x_1 + \frac{1}{k_3}F_1^{LB} \\&= \alpha x_1 + \frac{\frac{m_3}{k_3}p^2\alpha}{1-\frac{m_3p^2}{k_3}}x_1 \\&= \alpha x_1\bigg[1+\frac{\frac{m_3}{k_3}p^2}{1-\frac{m_3p^2}{k_3}}\bigg] \\&=\alpha x_1\frac{1}{1-\frac{m_3p^2}{k_3}} \\&= \frac{\alpha x_1}{1-\frac{m_3p^2}{k_3}}\end{split}$

For this case $\alpha = \frac{1}{2}$ :

$\theta_3^B = \frac{\frac{1}{2}}{1 - \frac{p^2J}{k}}\theta_1$

Now begin by selecting $p^2\frac{J}{k} = 0.5$ :

$\{\mathfrak z\}_0^R = \begin{bmatrix} 1 & 0 \\ -1 &1\end{bmatrix}\begin{bmatrix} \theta \\ 0\end{bmatrix}_0^L$

$\begin{split}\{\mathfrak z\}_1^L &= \begin{bmatrix} 0 & 1 \\ -1 &1\end{bmatrix}\begin{bmatrix} \theta \\ 0\end{bmatrix}_0^L \\&=\begin{bmatrix} 1 & 1\\0&1\end{bmatrix}\begin{bmatrix}1 & 0 \\ -1& 1\end{bmatrix}\begin{bmatrix}\theta \\ 0 \end{bmatrix}_0^L\end{split}$

$\begin{split}\{\mathfrak z\}_1^R &= \begin{bmatrix} 0 & 1 \\ -1 &0.75\end{bmatrix}\begin{bmatrix} \theta \\0\end{bmatrix}_0^L \\&=\begin{bmatrix} 1 &0\\-0.25&1\end{bmatrix}\begin{bmatrix}0 & 1 \\ -1& 1\end{bmatrix}\begin{bmatrix}\theta \\ 0 \end{bmatrix}_0^L\end{split}$

$\begin{split}\{\mathfrak z\}_2^L &= \begin{bmatrix} -0.5 & 1.375 \\ -1 &0.75\end{bmatrix}\begin{bmatrix} \theta \\0\end{bmatrix}_0^L \\&=\begin{bmatrix} 1 & 0.5\\0&1\end{bmatrix}\begin{bmatrix}0 & 1 \\ -1& 0.75\end{bmatrix}\begin{bmatrix}\theta \\ 0 \end{bmatrix}_0^L\end{split}$

$\begin{split}\{\mathfrak z\}_2^R &= \begin{bmatrix} -0.5 & 1.375 \\ -0.75 & 0.0625\end{bmatrix}\begin{bmatrix} \theta \\ 0 \end{bmatrix} \\&=\begin{bmatrix} 1 & 0 \\ -0.5&1\end{bmatrix}\begin{bmatrix}-0.5 & 1.375 \\ -1 & 0.75 \end{bmatrix}\begin{bmatrix} \theta \\ 0\end{bmatrix}_0^L\end{split}$

Therefore:

$\begin{Bmatrix} \theta_0 \\ \theta_2 \\ \theta_3\end{Bmatrix} = \begin{Bmatrix} 1 \\ -0.5 \\ 0 \end{Bmatrix}$

Now try $p^2\frac{J}{k} = 0.8$ :

$\{\mathfrak z\}_0^R = \begin{bmatrix} 1 & 0 \\ -1.6 &1\end{bmatrix}\begin{bmatrix} \theta \\ 0\end{bmatrix}_0^L$

$\begin{split}\{\mathfrak z\}_1^L &= \begin{bmatrix} -0.6 & 1 \\ -1.6 &1\end{bmatrix}\begin{bmatrix} \theta \\0\end{bmatrix}_1^L \\&=\begin{bmatrix} 1 & 1\\0&1\end{bmatrix}\begin{bmatrix}1 & 0 \\ -1.6& 1\end{bmatrix}\begin{bmatrix}\theta \\ 0 \end{bmatrix}_1^L\end{split}$

$\begin{split}\{\mathfrak z\}_1^R &= \begin{bmatrix} -0.6 & 1 \\ -1 &0\end{bmatrix}\begin{bmatrix} \theta \\0\end{bmatrix}_1^L \\&=\begin{bmatrix} 1 & 0\\-1&1\end{bmatrix}\begin{bmatrix}-0.6 & 1 \\ -1.6& 1\end{bmatrix}\begin{bmatrix}\theta \\ 0 \end{bmatrix}_1^L\end{split}$

$\begin{split}\{\mathfrak z\}_2^L &= \begin{bmatrix} -1.1 & 1 \\ -1 &0\end{bmatrix}\begin{bmatrix} \theta \\0\end{bmatrix}_1^L \\&=\begin{bmatrix} 1 & 0.5\\0&1\end{bmatrix}\begin{bmatrix}-0.6 & 1 \\ -1& 0\end{bmatrix}\begin{bmatrix}\theta \\ 0 \end{bmatrix}_1^L\end{split}$

$\begin{split}\{\mathfrak z\}_2^R &= \begin{bmatrix} -1.1& 1 \\ -0.12 & -0.5\end{bmatrix}\begin{bmatrix} \theta \\ 0 \end{bmatrix} \\&=\begin{bmatrix} 1 & 0 \\ -0.8&1\end{bmatrix}\begin{bmatrix}-1.1 & 1\\ -1 & 0 \end{bmatrix}\begin{bmatrix} \theta \\ 0\end{bmatrix}\end{split}$

$\theta_3 = \frac{\frac{1}{2}}{1-0.8}(1) = -2.5$

$\begin{Bmatrix} \theta_0 \\ \theta_2 \\ \theta_3\end{Bmatrix} = \begin{Bmatrix} 1 \\ -1.1 \\ -2.5 \end{Bmatrix}$

Now try $p^2\frac{J}{k} = 0.85$ :

$\{\mathfrak z\}_1^R = \begin{bmatrix} 1 & 0 \\ -1.7 &1\end{bmatrix}\begin{bmatrix} \theta \\ 0\end{bmatrix}_0^L$

$\begin{split}\{\mathfrak z\}_1^L &= \begin{bmatrix} -0.7 & 1 \\ -1.7 &1\end{bmatrix}\begin{bmatrix} \theta \\0\end{bmatrix} \\&=\begin{bmatrix} 1 & 1\\0&1\end{bmatrix}\begin{bmatrix}1 & 0 \\ -1.7& 1\end{bmatrix}\begin{bmatrix}\theta \\ 0 \end{bmatrix}\end{split}$

$\begin{split}\{\mathfrak z\}_1^R &= \begin{bmatrix} -0.7& 1 \\ -0.7083 &-0.1467\end{bmatrix}\begin{bmatrix} \theta \\0\end{bmatrix}\\&=\begin{bmatrix} 1 & 0\\-1.1467&1\end{bmatrix}\begin{bmatrix}-0.7 & 1 \\ -1.7& 1\end{bmatrix}\begin{bmatrix}\theta \\ 0 \end{bmatrix}\end{split}$

$\begin{split}\{\mathfrak z\}_2^L &= \begin{bmatrix} -1.0542& 0.9267 \\ -0.7083 & -0.1467\end{bmatrix}\begin{bmatrix} \theta \\0\end{bmatrix}_0^L \\&=\begin{bmatrix} 1 & 0.5\\0&1\end{bmatrix}\begin{bmatrix}-0.7 & 1 \\ -0.7083& -0.1467\end{bmatrix}\begin{bmatrix}\theta \\ 0 \end{bmatrix}_0^L\end{split}$

$\begin{split}\{\mathfrak z\}_2^R &= \begin{bmatrix} -1.0542& 0.9267\\ 0.1877 \end{bmatrix}\begin{bmatrix} \theta \\ 0 \end{bmatrix}_0^L \\&=\begin{bmatrix} 1 & 0 \\ -0.85&1\end{bmatrix}\begin{bmatrix}-1.0542 & 0.9267\\ -0.7083 & -0.1467 \end{bmatrix}\begin{bmatrix} \theta \\ 0\end{bmatrix}_0^L\end{split}$

$\theta_3 = \frac{\frac{1}{2}}{1-0.85}\theta_1 = \frac{\frac{1}{2}(-0.7)}{0.15}$

$\begin{Bmatrix} \theta_0 \\ \theta_2 \\ \theta_3\end{Bmatrix} = \begin{Bmatrix} 1 \\ -1.0542 \\ -2.33 \end{Bmatrix}$

Now try $p^2\frac{J}{k} = 2$ :

$\{\mathfrak z\}_1^R = \begin{bmatrix} 1 & 0 \\ -4 &1\end{bmatrix}\begin{bmatrix} \theta \\ 0\end{bmatrix}_1^L$

$\begin{split}\{\mathfrak z\}_1^L &= \begin{bmatrix} -3 & 1 \\ -4 &1\end{bmatrix}\begin{bmatrix} \theta \\0\end{bmatrix}_1^L \\&=\begin{bmatrix} 1 & 1\\0&1\end{bmatrix}\begin{bmatrix}1 & 0 \\ -4& 1\end{bmatrix}\begin{bmatrix}\theta \\ 0 \end{bmatrix}_1^L\end{split}$

$\begin{split}\{\mathfrak z\}_1^R &= \begin{bmatrix} -3& 1 \\ -\frac{11}{2}&\frac{3}{2}\end{bmatrix}\begin{bmatrix} \theta \\0\end{bmatrix}_1^L \\&=\begin{bmatrix} 1 & 0\\\frac{1}{2}&1\end{bmatrix}\begin{bmatrix}-3 & 1 \\ -4& 1\end{bmatrix}\begin{bmatrix}\theta \\ 0 \end{bmatrix}_1^L\end{split}$

$\begin{split}\{\mathfrak z\}_2^L &= \begin{bmatrix} -\frac{23}{4}& \frac{7}{4} \\ -\frac{11}{2} &\frac{3}{2}\end{bmatrix}\begin{bmatrix} \theta \\0\end{bmatrix}_1^L \\&=\begin{bmatrix} 1 & \frac{1}{2}\\0&1\end{bmatrix}\begin{bmatrix}-3 & 1 \\ -\frac{11}{2}& \frac{3}{2}\end{bmatrix}\begin{bmatrix}\theta \\ 0 \end{bmatrix}_1^L\end{split}$

$\begin{split}\{\mathfrak z\}_2^R &= \begin{bmatrix} -\frac{23}{4}& \frac{7}{4}\\ 6 &-2 \end{bmatrix}\begin{bmatrix} \theta \\ 0 \end{bmatrix} \\&=\begin{bmatrix} 1 & 0 \\ -2&1\end{bmatrix}\begin{bmatrix}-\frac{23}{4}& \frac{7}{4}\\ -\frac{11}{2} & \frac{3}{2} \end{bmatrix}\begin{bmatrix} \theta \\ 0\end{bmatrix}_1^L\end{split}$

$\begin{Bmatrix} \theta_0 \\ \theta_2 \\ \theta_3\end{Bmatrix} = \begin{Bmatrix} 1 \\ -\frac{23}{4} \\ -\frac{3}{2} \end{Bmatrix}$

The coefficient $\rightarrow \infty$ as $Jp^2/k \rightarrow 1$ . This is because of the transfer function where the branch occurs. This hold for any number of branches. If the main shaft is excited at the natural frequency of any one of the branches, the amplitude at the branch point is zero and the branch acts as a vibration absorber.

We can extend the transfer matrices to beams and to include a more realistic effect for shafts etc.

Shafts

$\begin{split} T &= \mu I\frac{\partial\theta}{\partial x} \\ &= \frac{\hat{k}L}{I}I\frac{\partial\theta}{\partial x} \end{split}$

$\begin{Bmatrix} z \end{Bmatrix}_1 = \begin{Bmatrix} \theta \\ T \end{Bmatrix}_1$

$\gamma = p\sqrt{\frac{J}{\hat{k}}} \quad \leftarrow \text{ total moment of inertia of rod}$

$\theta = A\cos{\gamma x}{L} + B\sin{\gamma x}{L}$

at $x=0 \qquad \theta = \theta_1$ therefore $A = \theta_1$

at $x=0 \qquad T=T_1$

$T = \hat{k} L \Big[ -\frac{\gamma\theta_1}{L}\sin{\frac{\gamma x}{L}} + \frac{B\gamma}{L}\cos{\frac{\gamma x}{L}} \Big]$

Therefore:

$\begin{split} T_1 &= B\hat{k}\gamma \\ B &= \frac{T_1}{\hat{k}\gamma} \end{split}$

Therefore:

$\theta = \theta_1 \cos{\frac{\gamma x}{L}} + \frac{T_1}{k \gamma}\sin{\frac{\gamma x}{L}}$

$\begin{split} T &= -\gamma \hat{k} \theta_1 \sin{\frac{\gamma x}{L}} + \frac{T_1}{\hat{k}\gamma} \frac{\gamma}{L} \hat{k}L\cos{\frac{\gamma x}{L}} \\ T&= -\gamma \hat{k} \theta_1 \sin{\frac{\gamma x}{L}} + T_1\cos{\frac{\gamma x}{L}} \end{split}$

evaluate these at $x = L$

$\begin{split} \theta_2 &= \theta_1 \cos{\gamma} + \frac{T_1}{\hat{k}\gamma} \sin{\gamma} \\ T_2 &= -\gamma \hat{k} \theta_1 \sin{\gamma} + T_1 \cos{\gamma} \end{split}$

Therefore:

$\begin{pmatrix} \theta \\ T \end{pmatrix}_2 = \begin{bmatrix} \cos{\gamma} && \frac{\sin{\gamma}}{\hat{k}\gamma} \\ -\gamma \hat{k} \sin{\gamma} && \cos{\gamma} \end{bmatrix}$

If the moment of inertia of the shaft is small $\gamma \rightarrow 0$ and the field matrix becomes

$\begin{bmatrix} 1 && \frac{1}{\hat{k}} \\ 0 && 1 \end{bmatrix}$

which is what we had previously.

If the stiffness is very high

$\gamma = p\sqrt{\frac{J}{\hat{k}}} \qquad \gamma \rightarrow 0$

$\begin{split} -\gamma^2k &= -p^2 \frac{J}{\hat{k}}\hat{k} \\ &= -p^2J \end{split}$

and the transfer matrix becomes

$\begin{bmatrix} 1 && 0 \\ -p^2J && 1 \end{bmatrix}$

then the rod is simply a mass.

Beams

The idea of transfer matrices can also be used to handle these complicated continuous system problems. For flexural vibrations the beam is modelled as massless section with a certain stiffness and with discrete masses. We can derive the equations for these various components from elementary strength of materials.

Consider first a section of beam without mass.

The state vector at position $i$ is

$\begin{Bmatrix} z \end{Bmatrix}_i = \begin{Bmatrix} y \\ \phi \\ M \\ V \end{Bmatrix}$

$\begin{split} +\uparrow \Sigma F_y &= 0 \\ &= V_{i+1} - V_i \end{split}$

Therefore $V_{i+1} = V_i$

$\begin{split} + \circlearrowleft \Sigma M_i &= -M_i + M_{i+1} + V_iL_i \\ M_{i+1} &= M_i - v_iL_i \end{split}$

Now must relate the slope and deflection at $i$ to $i+1$

These are the so-called Myosotis Palustris formulae.

Therefore:

$y_{i+1} = y_i + L_i\phi_i + \frac{M_{i+1}L^2_i}{2EI} + \frac{V_{i+1}L_i^3}{3EI}$

but

$\begin{split} M_{i+1} &= M_i - V_iL_i \\ V_{i+1} &= V_i \end{split}$

Therefore:

$\begin{split} y_{i+1} &= y_i + L_i\phi_i + \frac{M_iL^2_i}{2EI} - \frac{V_iL_i^3}{2EI} + \frac{V_{i+1}L_i^3}{3EI} \\ y_{i+1} &= y_i + L_i\phi_i + \frac{M_j}{2}\Big(\frac{L^2}{EI}\Big)_i - \frac{V_i}{6}\Big(\frac{L^3}{EI}\Big)_i \end{split}$

Also the slope

$\begin{split} \phi_{i+1} &= \phi_i + M_{i+1}\Big(\frac{L}{EI}\Big)_i + \frac{V_{i+1}}{2}\Big(\frac{L^2}{EI}\Big)_i \\ \phi_{i+1} &= \phi_i + M_i \Big(\frac{L}{EI}\Big)_i - V_i\Big(\frac{L^2}{EI}\Big)_i + \frac{V_{i+1}}{2}\Big(\frac{L^2}{EI}\Big)_i \\ \phi_{i+1} &= \phi_i + M_i\Big(\frac{L}{EI}\Big)_i - \frac{V_i}{2}\Big(\frac{L^2}{EI}\Big)_i \end{split}$

Therefore:

$\begin{Bmatrix} z \end{Bmatrix}_{i+1} = \begin{bmatrix} 1 && L && \frac{L^2}{2EI} && -\frac{L^3}{6EI} \\ 0 && 1 && \frac{L}{EI} && -\frac{L^2}{2EI} \\ 0 && 0 && 1 && -L \\ 0 && 0 && 0 && 1 \end{bmatrix}_i \begin{Bmatrix} z \end{Bmatrix}_i$

If we consider a point mass

$\uparrow \Sigma F_y = m \Ddot{x}_i = V_i^R - V_i^L$

$\begin{split} V_i^R &= V_i^L + m\Ddot{x}_i \\ &= V_i^L - mp^2x_i \end{split}$

$\begin{split} y_i^L &= y_i^R \\ \phi_i^L &= \phi_i^R \end{split}$

$\begin{Bmatrix} z \end{Bmatrix}_i^R = \begin{bmatrix} 1 && 0 && 0 && 0 \\ 0 && 1 && 0 && 0 \\ 0 && 0 && 1 && 0 \\ -mp^2 && 0 && 0 && 1 \end{bmatrix}_i \begin{Bmatrix} z \end{Bmatrix}_i^L$

If we have a spring

$\uparrow \Sigma F_y = V_i^R -y_ik_i - V_i^L$

Therefore:

$V_i^R = V_i^L + y_ik_i$

$\begin{Bmatrix} z \end{Bmatrix}_i^R = \begin{bmatrix} 1 && 0 && 0 && 0 \\ 0 && 1 && 0 && 0 \\ 0 && 0 && 1 && 0 \\ +k_i && 0 && 0 && 1 \end{bmatrix} \begin{Bmatrix} z \end{Bmatrix}_i^L$

$\begin{split} \hat{m} &= \frac{2m}{2L} \\ &= \frac{m}{L} \end{split}$

$\begin{Bmatrix} z \end{Bmatrix}_0 = \begin{Bmatrix} y_0 \\ \phi_0 \\ 0 \\ 0 \end{Bmatrix} \qquad \begin{Bmatrix} z \end{Bmatrix}_2 = \begin{Bmatrix} 0 \\ 0 \\ M_2 \\ V_2 \end{Bmatrix}$

$\begin{Bmatrix} z \end{Bmatrix}_2^L = \begin{bmatrix} F \end{bmatrix}_{1-2} \begin{bmatrix} P \end{bmatrix}_1 \begin{bmatrix} F \end{bmatrix}_{0-1} \begin{bmatrix} P \end{bmatrix}_0 \begin{Bmatrix} z \end{Bmatrix}_0^L$

$\begin{Bmatrix} 0 \\ 0 \\ M_2 \\ V_2 \end{Bmatrix} = \begin{bmatrix} u_{11} && u_{12} && u_{13} && u_{14} \\ u_{21} && u_{22} && u_{23} && u_{24} \\ u_{31} && u_{32} && u_{33} && u_{34} \\ u_{41} && u_{42} && u_{43} && u_{44} \end{bmatrix} \begin{Bmatrix} y \\ \phi \\ 0 \\ 0 \end{Bmatrix}$

Therefore:

$\begin{split} u_{11} y_0 + u_{12} \phi_0 &= 0 \\ u_{21}y_0 + u_{22} \phi_0 &= 0 \end{split}$

Therefore: $u_{11}u_{22} - u_{12}u_{21} = 0$ is the equation of interest

Note that $\big[F\big] \big[P\big]$ is the same

Therefore:

$\begin{split} \begin{bmatrix} F \end{bmatrix} \begin{bmatrix} P \end{bmatrix} &= \begin{bmatrix} 1 && L && \frac{L^2}{2EI} && -\frac{L^3}{6EI} \\ 0 && 1 && \frac{L}{EI} && -\frac{L^2}{2EI} \\ 0 && 0 && 1 && -L \\ 0 && 0 && 0 && 1 \end{bmatrix} \begin{bmatrix} 1 && 0 && 0 && 0 \\ 0 && 1 && 0 && 0 \\ 0 && 0 && 1 && 0 \\ -mp^2 && 0 && 0 && 1 \end{bmatrix} \\ &= \begin{bmatrix} 1+\frac{mp^2L^3}{6EI} && L && \frac{L^2}{2EI} && -\frac{L^3}{6EI} \\ +\frac{mp^2L^2}{2EI} && 1 && \frac{L}{EI} && -\frac{L^2}{2EI} \\ -mp^2L && 0 && 1 && -L \\ -mp^2 && 0 && 0 && 1 \end{bmatrix} \\ &= A \end{split}$

$\begin{bmatrix} A \end{bmatrix}^2 = \begin{bmatrix} A \end{bmatrix} \begin{bmatrix} A \end{bmatrix} = \begin{bmatrix} U \end{bmatrix}$

$u_{ij} = \Sigma_k a_{ik}a_{kj}$

Therefore:

$\begin{split} u_{11} &= a_{11}a_{11} + a_{12}a_{21} + a_{13}a_{31} + a_{14}a_{41} \\ &= \Big( 1 + \frac{mp^2L^3}{6EI}\Big)^2 + \frac{m^2L^3}{2EI} + \frac{mp^2L^3}{2EI} + \frac{mp^2L^3}{6EI} \\ &= 1 + \frac{9mp^2L^3}{6EI} + \Big( \frac{mp^2L^3}{6EI} \Big)^2 \\ &= 1 + 9\beta + \beta^2 \end{split}$

Where $\beta = \frac{mp^2L^3}{6EI}$

$\begin{split} u_{22} &= a_{21}a_{12} + a_{22}a_{22} + a_{23}a_{32} + a_{24}a_{42} \\ &= \frac{mp^2L^3}{2EI} + 1 + 0 + 0 \\ &= 1 + 3\beta \end{split}$

$\begin{split} u_{12} &= a_{11}a_{12} + a_{12}a_{22} + a_{13}a_{32} + a_{14}a_{42} \\ &= L \Big( 1 + \frac{mp^2L^3}{6EI} \Big) + L \\ &= L(2 + \beta) \end{split}$

$\begin{split} u_{21} &= a_{21}a_{11} + a_{22}a_{21} + a_{23}a_{31} + a_{24}a_{41} \\ &= \frac{mp^2L^2}{2EI}\Big( 1 + \frac{mp^2L^3}{6EI}\Big) + \frac{mp^2L^2}{2EI} + \frac{mp^2L^2}{EI} + \frac{mp^2L^2}{2EI} \\ &= \frac{5}{2}\frac{mp^2L^2}{EI} + \frac{1}{12}\Big( \frac{mp^2L^2}{EI} \Big)^2L \\ &= \frac{6\beta}{L}\Big( \frac{5}{2} + \frac{\beta}{2} \Big) \\ &= \frac{3\beta}{L}(5+\beta) \end{split}$

$\biggr[ 1 + \frac{9mp^2L^3}{6EI} + \Big(\frac{mp^2L^3}{6EI}\Big)^2 \biggr] \biggr[ 1 + \frac{m^2p^2L^3}{2EI} \biggr] - L \biggr[ 2 + \frac{mp^2L^3}{6EI} \biggr] \biggr[ \frac{5}{2} + \frac{1}{12} \frac{mp^2L^3}{EI} \biggr] \frac{mp^2L^2}{EI} = 0$

let $\beta = \frac{mp^2L^3}{6EI}$

$\begin{split} \biggr[ 1 + 9\beta + \beta^2 \biggr] \biggr[ 1 + 3\beta \biggr] -6\beta \biggr[ \Big( 2 + \beta \Big) \Big( \frac{5}{2} + \frac{1}{2} \beta \Big) \biggr] &= 0 \\ \biggr[ 1 + 9\beta + \beta^2 \biggr] \biggr[ 1 + 3\beta \biggr] -3\beta \biggr[ \Big( 2 + \beta \Big) \Big( 5 + \beta \Big) \biggr] &= 0 \\ 1 + 9\beta + \beta^2 + 3\beta + 27\beta^2 + 3\beta^3 -3\beta \big[ 10 + 2\beta + 5\beta + \beta^2 \big] &= 0 \\ 1 + 12\beta +28\beta^2 + 3\beta^3 -30\beta -6\beta^2 - 15\beta^2 -3\beta^3 &= 0 \\ 7\beta^2 - 18\beta + 1 &= 0 \end{split}$

$\begin{split} \beta &= \frac{+18 \pm \sqrt{(18)^2-28}}{14} \\ &= \frac{18 \pm \sqrt{296}}{14} \\ &= \frac{9 \pm \sqrt{74}}{7} \end{split}$

$\beta_{1,2} = 0.05681, \quad 2.5146$

Therefore:

$\begin{split} p_{1,2} &= \sqrt{\beta_{1,2}} \sqrt{6} \sqrt{\frac{EI}{mL^3}} \\ p_{1,2} &= 0.5838 \sqrt{\frac{EI}{mL^3}}, \quad 3.884 \sqrt{\frac{EI}{mL^3}} \end{split}$

Compare with a beam

$\begin{split} p_1 &= 3.52 \sqrt{\frac{EI}{\hat{m} (2L)^4}} \\ &= \frac{3.52}{4} \sqrt{\frac{EI}{\hat{m}L^4}} \\ &= 0.88 \sqrt{\frac{EI}{\hat{m}L^4}} \end{split}$

Consider a Rayleigh approximation

$p^2 \leq \frac{\int_0^L EI \big( \frac{d^2X}{dx^2} \big)^2 dx}{ \int_0^L m(X)^2 dx + \Sigma_{r=1}^k M \big\{ X(r,k)\big\}^2}$

assume $X(x) = Ax^2, \quad X'(x) = 2Ax, \quad X''(x) = 2A$

$\begin{split} p^2 &= \frac{\int_0^2 EI 4A^2 dx}{ m\big\{ AL^2\big\}^2 + m\big\{A4L^2\big\}^2 } \\ p^2 &= \frac{ 4A^2 EI(2L)}{ m\big\{ AL^2\big\}^2 + m\big\{A4L^2\big\}^2 } \\ &= \frac{8EIL}{17mL^4} \\ &= \frac{8EI}{17mL^3} \end{split}$

$p \leq 0.686 \sqrt{\frac{EI}{mL^3}}$

For forced steady state, assume all masses at $X_i\sin\omega t$

picture

Therefore acceleration is $-\omega^2X_i$

$\{\mathfrak z\}_0 = \begin{Bmatrix}0 \\ F \end{Bmatrix}_0$

$\{\mathfrak z\}_1^L = \begin{bmatrix} 1 & \frac{1}{k} \\ 0 & 1 \end{bmatrix}\{\mathfrak z\}_0$

$\{\mathfrak z\}_1^R = \begin{bmatrix} 1 & 0 \\ -m\omega^2 & 1 \end{bmatrix}\{\mathfrak z\}_1^L$

$\begin{split}\begin{Bmatrix} X \\ F_0 \end{Bmatrix} &= \{\mathfrak z\}_1^R \\&= \begin{bmatrix} 1 & 0 \\ -m\omega^2 & 1 \end{bmatrix} \begin{bmatrix} 1 & \frac{1}{k} \\ 0 & 1 \end{bmatrix} \begin{Bmatrix} 0 \\ F \end{Bmatrix} \\&= \begin{bmatrix} 1&\frac{1}{k} \\ -m\omega^2 & -\frac{m\omega^2}{k} + 1\end{bmatrix}\begin{Bmatrix} 0 \\ F \end{Bmatrix} \\&= \begin{Bmatrix}\frac{F}{k} \\ F\bigg(1-\frac{m\omega^2}{k}\bigg) \end{Bmatrix}\end{split}$

Therefore:

$X = \frac{F}{k}$

$F_0 = F\bigg(1-\frac{m\omega^2}{k}\bigg)$

$\begin{split}F &= \frac{F_0}{1 - \frac{m\omega^2}{k}} \\&=Xk\end{split}$

Therefore:

$X = \frac{\frac{F_0}{k}}{1 - \frac{m\omega^2}{k}}$

$\begin{split} Xk &= F \\&= \frac{F_0}{1 - \frac{m\omega^2}{k}} \end{split}$

picture

$\{\mathfrak z\}_0 = \begin{Bmatrix} 0 \\ F \end{Bmatrix}_0$

$\mathfrak z_1^L = \begin{bmatrix} 1 & \frac{1}{2k} \\ 0 & 1 \end{bmatrix}\{\mathfrak z\}_0$

picture

$x_1^L = x_1^R$

$F_1^R -F_1^L = -m_1\omega^2x_1^L$

Therefore:

$\begin{Bmatrix} x \\ F\end{Bmatrix}_1^R = \begin{bmatrix} 1 & 0 \\ -2m\omega^2 & 1 \end{bmatrix} \begin{Bmatrix} x \\ F \end{Bmatrix}_1^L$

$\begin{Bmatrix} x \\ F\end{Bmatrix}_2^L = \begin{bmatrix} 1 & \frac{1}{k} \\ 0 & 1 \end{bmatrix} \begin{Bmatrix} x \\ F \end{Bmatrix}_1^R$

$\begin{split}\begin{Bmatrix} x_2 \\ F_0 \end{Bmatrix} &= \begin{Bmatrix} x \\ F\end{Bmatrix}_2^R \\&= \begin{bmatrix} 1 & 0 \\ -m\omega^2 & 1 \end{bmatrix} \begin{Bmatrix} x \\ F \end{Bmatrix}_2^L\end{split}$

Therefore:

$\begin{split}\begin{Bmatrix} x_2 \\ F_0 \end{Bmatrix}_2^R &= \begin{bmatrix} 1 & 0 \\ -m\omega^2 & 1 \end{bmatrix}\begin{bmatrix} 1 & \frac{1}{k} \\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2m\omega^2 & 1 \end{bmatrix}\begin{bmatrix} 1 & \frac{1}{2k} \\ 0 & 1 \end{bmatrix}\begin{Bmatrix} 0 \\ F \end{Bmatrix} \\&= \begin{bmatrix} 1 & \frac{1}{k}\\ -m\omega^2 & 1 - \frac{m\omega^2}{k}\end{bmatrix}\begin{bmatrix} 1 & \frac{1}{2k} \\ -2m\omega^2 & 1 - \frac{m\omega^2}{k}\end{bmatrix}\begin{Bmatrix} 0 \\ F \end{Bmatrix}\end{split}$

$\begin{Bmatrix} x_2 \\ F_0 \end{Bmatrix} = \begin{bmatrix} 1 - \frac{2m\omega^2}{k} & \frac{1}{2k} + \frac{1}{k}(1-\frac{m\omega^2}{k}) \\ -m\omega^2 - 2m\omega^2(1-\frac{m\omega^2}{k}) & -\frac{m\omega^2}{2k} + (1-\frac{m\omega^2}{k})^2\end{bmatrix}\begin{Bmatrix} 0 \\F \end{Bmatrix}$

Therefore:

$x_2 = F\bigg(\frac{1}{2k} + \frac{1}{k}\bigg(1-\frac{m\omega^2}{k}\bigg)\bigg)$

$F_0 = F\bigg(-\frac{m\omega^2}{2k} + \bigg(1-\frac{m\omega^2}{k}\bigg)^2\bigg)$

$x_2 = \frac{F}{k}\bigg[\frac{3}{2} - \frac{m\omega^2}{k}\bigg]$

$F = \frac{F_0}{-\frac{m\omega^2}{2k} + 1 - \frac{2m\omega^2}{k} + (\frac{m\omega^2}{k})^2}$

Therefore:

$\begin{split}x_2 &=\frac{ \frac{F_0}{k}(\frac{3}{2}- \frac{m\omega^2}{k})}{1 - \frac{5m\omega^2}{2k} + (\frac{m\omega^2}{k})^2} \\&=\frac{ \frac{F_0}{k}(\frac{3}{2}- \frac{m\omega^2}{k})}{(1-\frac{m\omega^2}{2k})(1-\frac{2m\omega^2}{k})}\end{split}$

When $F = 2kx_1$ :

$x_1 = \frac{\frac{F_0}{2k}}{(1-\frac{m\omega^2}{2k})(1-\frac{2m\omega^2}{k})}$

This approach to formulation and solution to vibration problems is generally called the impedance approach and is commonly used in electrical current systems. It can be further expanded to include damping. The approach is identical except that the matrices are complex as this allows the phase relations to be handled. (In the undamped case displacement/forces are either in phase or 180˚ out of phase).

	Deflection (L)	Angle (L)
	$\frac{ML^2}{2EI}$	$\frac{ML}{EI}$
	$\frac{VL^3}{3EI}$	$\frac{VL^2}{2EI}$