This is an excerpt from Layton Hale’s lecture notes “Design and Analysis of Constraint Systems” from the November 2010 ASPE Conference.

Many engineers make a simple mistake when determining the equivalent stiffness of a spring that is rotated with respect to a coordinate system.  Typically they find $k_x=k \cos{\theta}$ and $k_y=k \sin{\theta}$, which are both incorrect.

When we displace the spring in the x direction $\delta_x$ the compression of the spring along it’s axis is quite a bit less as you can see in the figure below.  From the figures it is quite easy to see:

$\delta_x = \delta / \cos{\theta}$

$\delta_y = \delta / \sin{\theta}$

It follows that the spring stiffness along the x-direction is:

$\begin{matrix}k_{xx}&= &\frac{F_x}{\delta{x}}\\k_{xx}& = &\frac{F\cos{\theta}}{\delta/\cos{\theta}}\\k_{xx}& = &\frac{F}{\delta}\cos^2{\theta}\end{matrix}$

We can then easily solve for the equivalent stiffness matrix as:

$\mathbf{K} = k\cdot \begin{bmatrix} \cos^2{\theta} & \cos{\theta}\cdot\sin{\theta}\\ \cos{\theta}\cdot\sin{\theta} & \sin^2{\theta} \end{bmatrix}$

And now we can compute the forces along the rotated coordinate system as:

$\begin{bmatrix}f_x\\ f_y\end{bmatrix} = \begin{bmatrix}k_{xx} & k_{xy}\\ k_{yx} & k_{yy}\\\end{bmatrix}\cdot\begin{bmatrix}\delta_x\\ \delta_y\end{bmatrix}$

This is a continuation post on simple implementations of tuned-mass-dampers. For the curious please see my previous posts about TMDs here …

Say we have the 6-DOF model with frequency response as shown below.  The FRF’s show the response of a leadscrew actuator to an acceleration disturbance at it’s base (continue reading about the leadscrew actuator here…).

We can use a TMD to dampen the first few modes. As a first pass damper design, let’s set the mass ratio $\alpha=\frac{m_{damper}}{m_{sys}}$ to be 10%.  We then have $m_{damper}=0.2$ kg.  Also, the first interesting system mode occurs around $\omega_n=300$hz, a general rule of thumb for determining the damper’s resonance is $\frac{\omega_{damper}}{\omega_{sys}}=\frac{1}{1-\alpha}$.  We then set $\omega_d = 278$hz.  It is trivial to determine the damper stiffness $k_{damper}=m_{damper}{\omega_d}^2$.  We can start with a damping value of $c=\frac{1}{2}m_{damper}\omega_d$.

So we create a 6-DOF tuned-mass-damper by appending six 1-DOF TMD’s, and if so desired, each 1-DOF TMD can target different modes as they are independent of eachother.

$tmd_{6d} = \left[\begin{matrix}tmd_{1dx}\\tmd_{1dy}\\tmd_{1dz}\\tmd_{1drx}\\tmd_{1dry}\\tmd_{1drz}\end{matrix}\right]$

The 6-DOF TMD is then feedback to the dynamic system and the resultant FRF’s are shown below.  The dashed lines are the undamped and solid lines are the damped FRF’s.  Notice the original resonance is quite suppressed, this will dramatically reduce the POI motion from base disturbances.

For further reading on tuned mass dampers can by Lei Zuo and Samir Nayfeh be found here.

I proposed using an improper transfer function to model a simple tuned-mass damper (read more here) with the transfer function:

$\frac{y}{u} = \frac{mcs^{3}+mks^{2}}{ms^{2}+cs+k}$

To convert this to state-space representation we need to use the more generalized descriptor state-space notation which introduces the “E” term:

$E\dot{x}(t)=Ax(t)+Bu(t)$

$y(t)=Cx(t)+Du(t)$

where $E,A \in \mathbb{R}^{n\times n}$$B \in \mathbb{R}^{n\times m}$$C \in \mathbb{R}^{p\times n}$ and $D \in \mathbb{R}^{p\times m}$.

We derive $C$ as:

$y = C \times \left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right]$

$y = \left[\begin{matrix}mc,mk,0,0\end{matrix}\right]\times \left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right]$

We then have the following states:

$\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right]=\left[\begin{matrix}us^3/d\\us^2/d\\us/d\\u/d\end{matrix}\right]$

$\left[\begin{matrix}\dot{x}_1\\\dot{x}_2\\\dot{x}_3\\\dot{x}_4\end{matrix}\right]=\left[\begin{matrix}us^4/d\\us^3/d\\us^2/d\\us/d\end{matrix}\right]$

where $d = ms^2+cs+k$

we can see that

$\begin{matrix}u&=&d\times x_4&\\u&=&ms^2x_4+csx_4+kx_4&\\u&=&m\dot{x}_3+cx_3+kx_4&\\\end{matrix}$

On to fill out the rest of the matrices as:

$\begin{matrix}E&\dot{X}&=&A&X&+&B&u\\ \left[\begin{matrix}0&1&0&0\\0&0&1&0\\0&0&m&0\\0&0&0&1\end{matrix}\right]&\left[\begin{matrix}\dot{x}_1\\\dot{x}_2\\\dot{x}_3\\\dot{x}_4\end{matrix}\right]&=&\left[\begin{matrix} 1&0&0&0\\0&1&0&0\\0&0&-c&-k\\0&0&0&1 \end{matrix}\right]&\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right]&+&\left[\begin{matrix}0\\0\\1\\0\end{matrix}\right]&u\end{matrix}$

$\begin{matrix}Y&=&C&X&+&D&u&\\Y&=&\begin{matrix}\left[mc,mk,0,0\right]\end{matrix}&\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right]&&\end{matrix}$

where $D=0$.

Just a quick reminder about converting relative motion to absolute motion.  Say we already have the transfer function for input acceleration to relative motion:

$sys_{rel} = \frac{\delta}{\ddot{x}_1} = \frac{x_2-x_1}{\ddot{x}_1}$

The absolute motion of the second mass can be found as

$\frac{x_{2}}{\ddot{x}_{1}} = \frac{\delta}{\ddot{x}_{1}} + \frac{1}{s^{2}}$

The derivation is as follows:

$sys_{rel} = \frac{\delta}{\ddot{x}_{1}}= \frac{x_{2}-x_{1}}{\ddot{x}_{1}}=\frac{x_2}{\ddot{x}_1}-\frac{x_1}{\ddot{x}_1}$

$\frac{x_2}{\ddot{x}_1} = \frac{\delta}{\ddot{x}_{1}}+\frac{x_1}{\ddot{x}_1}=\frac{\delta}{\ddot{x}_{1}}+\frac{1}{s^2}$