Engineering at Alberta Courses » Numerical methods for vibrating systems

Advanced Dynamics and Vibrations: Numerical methods for vibrating systems

[Jason’s part goes here]

$m_N \Ddot{x}N = -k(x_N - x{N-1})$

where we can have a different mass as the equation is different. For SHM

$X_{N-1} = \Big( 1 - \frac{p^2m_N}{k}\Big) X_N$

If we choose $m_N = m/2$ then

$X_{N-1} = \Big( 1 - \frac{p^2m}{2k}\Big) X_N$

$\begin{split} \sin{\beta(N-1)} &= \cos{\beta}\sin{\beta N} \\ \sin{\beta N}\cos{\beta} - \cos{\beta N}\sin{\beta} &= \sin{\beta N}\cos{\beta} \end{split}$

Therefore:

$\cos{\beta N}\sin{\beta} = 0$

Note: $\sin{\beta}=0 \text{means} m=0, k \rightarrow \text{as} p^2 = 0$

If $\cos{\beta N} = 0$

$\begin{split} \beta N &= \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2} \\ &= \frac{(2i-1)\pi}{2} \qquad i = 1,2,3, … \\ \frac{\beta}{2} &= \frac{(2i-1)\pi}{4N} \end{split}$

Now using the result above

$\begin{split} \frac{mp^2}{k} &= 4\sin^2\frac{\beta}{2} \\ p &= 2\sqrt{\frac{k}{m}}\sin\frac{\beta}{2} \\ p_i &= 2\sqrt{\frac{k}{m}}\sin\frac{(2i-1)\pi}{4N} \qquad i=1\rightarrow N \end{split}$

Consider $N = 4$

$\begin{split} p_1 &= 2\sqrt{\frac{k}{m}}\sin\frac{\pi}{16} = 0.390\sqrt{\frac{k}{m}} \qquad (11.25^{\circ}) \\ p_2 &= 2\sqrt{\frac{k}{m}}\sin\frac{3\pi}{16} = 1.11\sqrt{\frac{k}{m}} \qquad (33.75^{\circ}) \\ p_3 &= 2\sqrt{\frac{k}{m}}\sin\frac{5\pi}{16} = 1.66\sqrt{\frac{k}{m}} \qquad (56.25^{\circ}) \\ p_4 &= 2\sqrt{\frac{k}{m}}\sin\frac{7\pi}{16} = 1.96\sqrt{\frac{k}{m}} \qquad (78.75^{\circ}) \end{split}$

Note:

$\begin{split} p_5 &= 2\sqrt{\frac{k}{m}}\sin\frac{9\pi}{16} \\ &= 2\sqrt{\frac{k}{m}} \sin\Big(\frac{\pi}{2} + \frac{\pi}{16} \Big) \\ &= 2\sqrt{\frac{k}{m}}\Big(\sin\frac{\pi}{2}\sin\frac{\pi}{16} + \cos\frac{\pi}{2}\cos\frac{\pi}{16} \Big) = p_1 \end{split}$

We can apply this idea to various dynamical systems which are more complicated than the simple spring/mass. Consider an approximation to longitudinal vibration of a bar

Consider a beam of total mass $M$ and cross section $A$ modulus $E$ which we approximate as

$k = \frac{AE}{\ell} \qquad \ell = \frac{L}{N} \qquad m = \frac{M}{N}$

if $N>> 1$ , go to $\textcircled{11}$

if $m_N = m$

$X_{N-1} = X_N \Big( 1 - \frac{p^2m}{k} \Big)$

$\begin{split} \beta\sin\beta (N-1) &= \beta\sin\beta N\big(1-4\sin^2\frac{\beta}{2}\big) \\ \sin\beta(N-1) &= \sin\beta N\big[ 1-(2-2\cos\beta)\big] \\ \sin\beta(N-1) &= \sin\beta N [-1+2\cos\beta] \\ \sin\beta N\cos\beta - \cos\beta N\sin\beta &= 2\cos\beta\sin\beta N - \sin\beta N \\ -\cos\beta\sin\beta N - \cos\beta N\sin\beta &= \sin\beta N \\ -\big[ \cos\beta\sin\beta N + \cos\beta N \sin\beta\big] &= \sin\beta N \\ -\sin\beta(N+1) &= \sin\beta N \end{split}$

$\begin{split} \frac{\sin\beta(N-1)}{\sin\beta N} &= 1-4\sin^2\frac{\beta}{2} \\ 4\sin^2\frac{\beta}{2} &= 1 - \frac{\sin\beta(N-1)}{\sin\beta N} \\ 4\sin^2\frac{\beta}{2} &= \frac{\sin\beta N - \sin\beta (N-1)}{\sin\beta N} \\ \sin\frac{\beta}{2} &= \frac{\cos{\beta(N-\frac{1}{2}}}{\sin\beta N} \\ 2\sin\frac{\beta}{2}\sin\beta N &= \cos\beta(N-\frac{1}{2}) \\ 2\sin\frac{\beta}{2}\sin\beta N &= \cos\beta N\cos\frac{\beta}{2} + \sin\beta N\sin\frac{\beta}{2} \\ \sin\frac{\beta}{2}\sin\beta N - \cos\beta N \cos\frac{\beta}{2} &= 0 \\ \cos\beta(N+\frac{1}{2}) &= 0 \end{split}$

Therefore:

$\begin{split} \beta(N+\frac{1}{2}) &= (2i-1)\frac{\pi}{2} \\ \beta &= \frac{2(2i+1)}{(2N+1)}\frac{\pi}{2} \\ &= \frac{(2i-1)\pi}{(2N+1)} \end{split}$

Therefore:

$p_i = 2\sqrt{\frac{k}{m}}\sin{\frac{(2i-1)}{(2N+1)}\frac{\pi}{2}} \qquad i=1,2,3,4$

N = 4

$\begin{split} p_1 &= 2\sqrt{\frac{k}{m}} \sin{\frac{\pi}{18}} = 0.347\sqrt{\frac{k}{m}} \\ p_2 &= 2\sqrt{\frac{k}{m}} \sin{\frac{3\pi}{18}} = \sqrt{\frac{k}{m}} \\ p_3 &= 2\sqrt{\frac{k}{m}} \sin{\frac{5\pi}{18}} = 1.53\sqrt{\frac{k}{m}} \\ p_4 &= 2\sqrt{\frac{k}{m}} \sin{\frac{7\pi}{18}} = 1.88\sqrt{\frac{k}{m}} \end{split}$

To calculate the mode shapes one can start at the free end e.g.:

$X_{N-1} = \Big(1-\frac{\omega^2m}{2k}\Big)X_N$

Assume $X_N$

$X_{N-1} = \cos\beta X_N$

Now we use the recursion relationship

$X_{n+1} - 2\cos{\beta X_n} + X_{n-1} = 0$

Therefore:

$X_{n-1} = 2\cos{\beta X_n} - X_{n+1}$

For the 4 storey building for the lowest natural frequency is

$\beta = (2i -1)\frac{\pi}{2N} = \frac{\pi}{8} = 22.5^{\circ}$

Therefore:

$X_4 = \underline{1}$

$\begin{split} X_3 &= \cos22.5^{\circ} X_4 \\ &= \underline{0.924} \end{split}$

$\begin{split} X_2 &= (2\cos22.5^{\circ})\cos22.5^{\circ} - 1 \\ &= \underline{0.707} \end{split}$

$\begin{split} X_1 &= 2\cos11.25^{\circ}(0.92) \\ &= 2(\cos22.5^{\circ})^3 - \cos22.5^{\circ} \\ &= \underline{0.383} \end{split}$

Check

$\begin{split} X_2 - 2\cos\beta X_1 &= 0 \\ 0.707 - 2\cos22.5^{\circ}(0.383) &= \underline{0} \end{split}$

$\beta_2 = \frac{3\pi}{8} = 67.5^{\circ}$

$X_4 = 1$

$\begin{split} X_3 &= \cos{\beta_2}X_4 \\ &= 0.383 \end{split}$

$\begin{split} X_2 &= 2\cos{\beta}X_3 - X_4 \\ &= -0.707 \end{split}$

$\begin{split} X_1 &= 2\cos{\beta}X_2 - X_3 \\ &= -0.924 \end{split}$

$\begin{split} X_2 - 2\cos{\beta}X_1 &= -0.707 - 2\cos{67.5^{\circ}}(-0.924) \\ &= \underline{0} \end{split}$

Therefore:

$p_i = 2\sqrt{\frac{k}{m}}\sin\frac{(2i-1)\pi}{4N} \qquad \textcircled{11}$

if $N>>1$ ( $i$ is low) then:

$\sin{\frac{(2i-1)\pi}{4N}} \approx \frac{(2i-1)\pi}{4N}$

for the lower frequencies

$i=1 \qquad \frac{\pi}{4N}$

$\begin{split} p_i &= 2\sqrt{\frac{AEN^2}{ML}}\frac{(2i-1)\pi}{4N} \\ &= \frac{\pi(2i-1)}{2L} \sqrt{\frac{EAL}{M}} \\ &= \frac{\pi(2i-1)}{2L}\sqrt{\frac{E}{\rho}} \end{split}$

where $\rho$ is density

For other boundary conditions, the expression for $\beta$ is determined then the natural frequencies are known. For fixed ends.

$p_n = 2\sqrt{\frac{k}{m}}\sin\frac{n\pi}{2(N+1)} \quad n = 1,2,... N$

$- 2\cos\bigg[\beta(N+\frac{1}{2})\bigg]\sin\frac{\beta}{2} = \frac{K_N}{k}\sin\beta N$

Consider the fixed boundaries:

$\begin{split}\rightarrow\sum F_x &= m\ddot{X}_n \\&= -mp^2X_N \\&= -kX_N - k(X_N-X_{N-1})\end{split}$

$-mp^2X_N = -2kX_n + kX_{N-1}$

$\bigg(2-\frac{p^2m}{k}\bigg)X_N = X_{N-1}$

$2\bigg(1-\frac{p^2m}{2k}\bigg)X_N = X_{N-1}$

$2\cos\beta\sin\beta N = \sin\beta(N-1)$

$2\cos\beta\sin\beta N = \sin\beta N\cos\beta - \cos\beta N\sin\beta$

$\cos\beta\sin\beta N +\cos\beta N \sin\beta = 0$

Therefore:

$\sin\beta(N+1) = 0$

$\beta(N+1) = n\pi \quad n = 1,2,3,...N$

$\beta = \frac{n\pi}{N+1}$

Therefore:

$p_n = 2\sqrt{\frac{k}{m}}\sin\frac{n\pi}{2(N+1)} \quad n = 1,2,3,...N$

Example

$p_i = 2\sqrt{\frac{k}{m}}\sin\frac{n\pi}{2(N+1)}$

$\begin{bmatrix} m & 0 \\ 0 & m\end{bmatrix}\begin{Bmatrix} \ddot x_1 \\ \ddot x_2\end{Bmatrix} + \begin{bmatrix} 2k & -k \\ -k & 2k \end{bmatrix}\begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = 0$