Engineering at Alberta Courses » Chapter 9 Examples

Approximate Methods for Multiple Degree of Freedom Systems: Chapter 9 Examples

Example 9-1

At some instant in time, the masses in the system shown below are in positions $x_1$ and $x_2$ and have velocities $\dot{x}_1$ and $\dot{x}_2$ respectively.

Determine the potential and kinetic energies in the system in terms of $x_1$ , $x_2$ , $\dot{x}_1$ and $\dot{x}_2$ from first principles.
Show that the potential and kinetic energies can also be expressed as

$\begin{align*} U =&\ \frac{1}{2}\, \bigl\{q\bigr\}^T \bigl[ K \bigr] \bigl\{ q \bigr\} \\ T =&\ \frac{1}{2}\, \bigl\{\dot{q}\bigr\}^T \bigl[ M \bigr] \bigl\{\dot{q}\bigr\} \end{align*}$

where

$\begin{align*} \bigl[ k \bigr] =&\ \begin{bmatrix} k+k_C & -k_C \\ -k_C & k+k_C \\ \end{bmatrix} \\ % \bigl[ m \bigr] =&\ \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \\ \end{bmatrix} \end{align*}$

and

$\begin{equation*} \bigl\{q\bigr\} = \biggl\{\!\! \begin{array}{c} x_1 \\ x_2 \\ \end{array} \!\!\biggr\}, \qquad % \bigl\{\!\dot{q}\!\bigr\} = \biggl\{\!\! \begin{array}{c} \dot{x}_1 \\ \dot{x}_2 \\ \end{array} \!\!\biggr\}. \end{equation*}$

Example 9-2

For the system shown above, estimate the fundamental natural frequency using Rayleigh’s Quotient. Assume $m_1 = m_2 = m$ and use trial vectors of
$\begin{equation*} \bigl\{\!\Phi\!\bigr\} = \biggl\{\! \begin{array}{r} 1 \\ 1 \\ \end{array} \!\biggr\}, \qquad \bigl\{\!\Phi\!\bigr\} = \biggl\{\! \begin{array}{r} 1 \\ -1 \\ \end{array} \!\biggr\}, \qquad \bigl\{\!\Phi\!\bigr\} = \biggl\{\! \begin{array}{r} 1 \\ 2 \\ \end{array} \!\biggr\}, \qquad \bigl\{\!\Phi\!\bigr\} = \biggl\{\! \begin{array}{r} 1 \\ -2 \\ \end{array} \!\biggr\}, \end{equation*}$

and comment on the results.
It was previously determined that

$\begin{gather*} \bigl[ k \bigr] = \biggl[\! \begin{array}{cc} k+k_C & -k_C \\ -k_C & k+k_C \\ \end{array} \!\biggr] , \qquad % \bigl[ m \bigr] = \begin{bmatrix} m & 0 \\ 0 & m \\ \end{bmatrix}, \\[3mm] % p_1^2 = \frac{k}{m}, \qquad \bigl\{\!\Phi\!\bigr\}^{(1)} = \biggl\{\! \begin{array}{r} 1 \\ 1 \\ \end{array} \!\biggr\}, \qquad % p_2^2 = \frac{k+2k_C}{m}, \qquad \bigl\{\!\Phi\!\bigr\}^{(2)} = \biggl\{\! \begin{array}{r} 1 \\ -1 \\ \end{array} \!\biggr\}. \end{gather*}$
Repeat the calculations in part (1) assuming $m_1 = 2m$ , $m_2=m$ , and $k_C = 3k$ .

Note that the exact solution in this case is
$p_1 \approx 0.809 \sqrt{\frac{k}{m}}, \qquad \bigl\{\Phi\bigr\}^{\textcircled{{\footnotesize{1}}}} ;\approx \biggl\{ \! \begin{array}{c} 1.115 \\ 1 \\ \end{array} \!\biggr\}, \qquad \\$

$p_2 \approx 2.312 \sqrt{\frac{k}{m}}, \qquad \bigl\{\!\Phi\!\bigr\}^{\textcircled{{\footnotesize{2}}}} \approx \biggl\{\! \begin{array}{c} -0.448 \\ 1 \\ \end{array} \!\biggr\}.$

Example 9-3

Use Rayleigh’s method to estimate the fundamental natural frequency for a light simply supported beam supporting three equally spaced identical masses.

Assume trial vectors of

$\begin{equation*} \bigl\{\!\Phi\!\bigr\} = \Biggl\{\! \begin{array}{r} 1 \\ 1 \\ 1 \\ \end{array} \!\Biggr\}, \qquad \bigl\{\!\Phi\!\bigr\} = \Biggl\{\! \begin{array}{r} 1 \\ 2 \\ 1 \\ \end{array} \!\Biggr\}, \qquad \bigl\{\!\Phi\!\bigr\} = \Biggl\{\! \begin{array}{r} 1 \\ 1.5 \\ 1 \\ \end{array} \!\Biggr\}. \end{equation*}$

Note that for this problem
$% \bigl[ a \bigr] =\ % \frac{l^3}{768 EI} % \Biggl[ % \begin{array}{ccc} % 9 & 11 & 7\\ % 11 & 16 & 11\\ % 7 & 11 & 9\\ % \end{array} % \Biggr] \\ \bigl[ k \bigr] =\ \frac{192 EI}{7 l^3} \Biggl[ \begin{array}{rrr} 23 & -22 & 9\\ -22 & 32 & -22\\ 9 & -22 & 23\\ \end{array} \vspace{4mm} \Biggr] \\ \bigl[ m \bigr] =\ m \Biggl[ \begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{array} \Biggr] \\$

Example 9-4

Estimate the fundamental natural frequency for the system shown below using Dunkerley’s Method.

Consider the following cases:

$\begin{gather*} m_1 = m_2 = m_3 = m, \\ k_1 = k_2 = k_3 = k_4 = k. \end{gather*}$

ii)

$\begin{gather*} m_1 = m_3 = m,\qquad m_2 = 2m, \\ k_1 = k_4 = k,\qquad k_2 = k_3 = 2k. \end{gather*}$

iii)

$\begin{gather*} m_1 = 4m,\qquad m_2 = m,\qquad m_3 = 2m, \\ k_1 = k,\qquad k_2 = k_4 = 2k,\qquad k_3 = 5k. \end{gather*}$

Note that the stiffness matrix for this problem is

$\begin{align*} \bigl[ k \bigr] =&\ \Biggl[ \begin{array}{ccc} k_1 + k_2 & -k_2 & 0 \\ -k_2 & k_2 + k_3 & -k_3 \\ 0 & -k_3 & k_3 + k_4 \\ \end{array} \Biggr] \\ \intertext{so that $\bigl[a\bigr] = \bigl[k\bigr]^{-1}$ is} \bigl[ a \bigr] =&\ %\frac{1}{k_1 k_2 k_3 + k_1 k_2 k_4 + k_1 k_3 k_4 + k_2 k_3 k_4} \frac{1}{k_1 (k_2 k_3 + k_2 k_4 + k_3 k_4) + k_2 k_3 k_4} \Biggl[ \begin{array}{ccc} k_2 k_3 + k_2 k_4 + k_3 k_4 & k_2\bigl( k_3 + k_4 \bigr) & k_2 k_3 \\ k_2\bigl( k_3 + k_4 \bigr) & \bigl( k_1 + k_2 \bigr)\bigl( k_3 + k_4 \bigr) & \bigl( k_1 + k_2 \bigr)k_3 \\ k_2 k_3 & \bigl( k_1 + k_2 \bigr)k_3 & k_1 k_2 + k_1 k_3 + k_2 k_3 \\ \end{array} \Biggr] \\ \end{align*}$

Example 9-5

Use Dunkerley’s method to estimate the fundamental natural frequency for a light simply supported beam supporting three equally spaced identical masses.

Note that for this system

$\begin{align*} \bigl[ a \bigr] =&\ \frac{l^3}{768 EI} \Biggl[ \begin{array}{ccc} 9 & 11 & 7\\ 11 & 16 & 11\\ 7 & 11 & 9\\ \end{array} \Biggr], \\ \bigl[ k \bigr] =&\ \frac{192 EI}{7 l^3} \Biggl[ \begin{array}{rrr} 23 & -22 & 9\\ -22 & 32 & -22\\ 9 & -22 & 23\\ \end{array} \Biggr], \\ \bigl[ m \bigr] =&\ m \Biggl[ \begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{array} \Biggr] . \\ \end{align*}$

Example 9-6

For the system shown below:

1. Use matrix iteration to find the lowest and highest natural frequencies and associated mode shapes.
2. Use Rayleigh’s quotient to find an upper bound to the lowest natural frequency. Use a trial vector of
  $\begin{equation*} \left\{ \Phi \right\} = \Biggl\{ \begin{array}{c} 1.0 \\ 1.5 \\ 1.5 \\ \end{array} \Biggr\}. \end{equation*}$
3. Use Dunkerley’s method to find a lower bound for the lowest natural frequency.
4. Use Dunkerley’s method to find an upper bound for the highest natural frequency.

Note that for this problem,

$\begin{align*} \bigl[ k \bigr] =&\ k \Biggl[ \begin{array}{rrr} 3 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \\ \end{array} \Biggr], \\ \bigl[ a \bigr] =&\ \frac{1}{3k} \Biggl[ \begin{array}{rrr} 3 & 3 & 3 \\ 3 & 5 & 4 \\ 3 & 4 & 5 \\ \end{array} \Biggr], \\ \bigl[ m \bigr] =&\ m \Biggl[ \begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{array} \Biggr]. \\ \end{align*}$

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