Engineering at Alberta Courses » Vibration and Applications of MDOF Systems

Review of single and multi-degree of freedom (mdof) systems: Vibration and Applications of MDOF Systems

For more than a SDOF system the analysis requires a more generalized approach. However, overall it doesn’t matter if there are 2 or 22 DOFs. Consider first a special 2 DOF free vibration system:

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

The simplest approach is to look for Simultaneous Simple Harmonic Motion (SSHM). That is:

$\begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} X_1 \\ X_2 \end{Bmatrix} \sin(pt + \phi)$

Which is a solution (not the general one) if:

$\begin{bmatrix} k+k_c - mp^2 & -k_c \\ -k_c & k+k_c-mp^2 \end{bmatrix} \begin{Bmatrix} X_1 \\ X_2 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix}$

This can be true if and only if (iff), $\text{det}[] = 0$ . Therefore:

$p^4-p^2\Big[\frac{(k+k_c)}{m} + \frac{(k+k_c)}{m}\Big] + \frac{k^2+2k_c k}{m^2} = 0$

There are two soloutions for $p^2 \Rightarrow p_1^2 \ \text{and} \ p_2^2$ .

$\begin{split} p_{1,2}^2 &= \frac{k+k_c}{m} \pm \sqrt{\frac{k^2+2kk_c + k_c^2 - (k^2 + 2kk_c)}{m^2}} \\&= \frac{k+k_c}{m} \pm \frac{k_c}{m} \\&= \frac{k}{m}, \ \frac{k+2k_c}{m} \end{split}$

Therefore:

$p_1 = \sqrt{\frac{k}{m}}, \ p_2 = \sqrt{\frac{k+2k_c}{m}}$

As a result, there are 2 frequencies at which our assumption is true. NOTE: It turns out that the general solution can be determined from these $p_1$ , $p_2$ , and the ratio of the amplitudes between the two masses during each of the two SSHMs. Therefore:

$(k + k_c - mp^2)X_1 -k_cX_2 = 0$

Or:

$-k_cX_1 + (k + k_c - mp^2)X_2 = 0$

Therefore, for either $p_1^2$ or $p_2^2$ :

$\begin{split} \frac{X_2}{X_1} &= \frac{k +k_c -mp^2}{k_c} \\&= \frac{k_c}{k+k_c-mp^2} \end{split}$

If we put $p_1^2$ into either we get:

$\Big(\frac{X_2}{X_1}\Big)_1 = \frac{1}{1} \equiv X_1$

And if we put $p_2^2$ into either we get:

$\Big(\frac{X_2}{X_1}\Big)_2 = \frac{-1}{1} \equiv X_2$

Thus, we define the mode shapes corresponding to each ratio:

$\{u\}_1 = \begin{Bmatrix} 1 \\ X_1 \end{Bmatrix}$

$\{u\}_2 = \begin{Bmatrix} 1 \\ X_2 \end{Bmatrix}$

Therefore, we have a mode shape corresponding to each of the natural frequencies. $\{u\}_1$ and $\{u\}_2$ are “normalized” vectors where the first component is set arbitrarily to unity. The general solution to the free vibration problem is then:

$\begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = c_1 \{u\}_1 \sin(p_1t +\phi_1) + c_2\{u\}_2 \sin(p_2t+\phi_2)$

There are 4 constants $c_1$ , $c_2$ , $\phi_1$ , $\phi_2$ for the 4 initial conditions for $x_1$ , $x_2$ , $\dot{x_1}$ , $\dot{x_2}$ . If the initial conditions are selected then the solution will include both modes in general but only one mode for certain cases. (For example, when $x_1(0) = 1$ , $x_2(0) = 1$ , $\dot{x}_1(0) = 0$ , and $\dot{x}_2(0) = 0$ , $\Rightarrow$ mode 1).

Consider the following initial conditions:

$x_1(0) = 1,\ x_2(0) = 0,\ \dot{x}_1(0) = 0,\ \dot{x}_2(0) = 0$

$x_1(t) = c_1\sin(p_1t+\phi_1) + c_2\sin(p_2t+\phi_2)$

$x_2(t) = c_1\sin(p_1t+\phi_1) - c_2\sin(p_2t+\phi_2)$

At $t = 0$ :

$1 = c_1\sin\phi_1 + c_2\sin\phi_2 \quad (1)$

$0 = c_1\sin\phi_1 - c_2\sin\phi_2 \quad (2)$

$0 = p_1\cos\phi_1 + p_2\cos\phi_2 \quad (3)$

$0 = p_1\cos\phi_1 - p_2\cos\phi_2 \quad (4)$

Therefore:

$\cos\phi_1 = 0 \quad ((3) + (4))$

$\cos\phi_2 = 0 \quad ((3) - (4))$

Therefore:

$\phi_1 = \frac{\pi}{2},\ \phi_2 = \frac{\pi}{2}$

From (1) and (2) (+):

$2c_1\sin\phi_1 = 1$

$c_1 = \frac{1}{2}$

From (1) and (2) (-):

$c_2 = \frac{1}{2}$

Therefore:

$\begin{split} x_1 &= \frac{1}{2}\sin\Big(p_1t + \frac{\pi}{2}\Big) + \frac{1}{2}\sin\Big(p_2t + \frac{\pi}{2}\Big) \\&= \frac{1}{2}\cos p_1t + \frac{1}{2}\cos p_2t \end{split}$

$x_2 = \frac{1}{2}\cos p_1t - \frac{1}{2}\cos p_2t$

Using identities:

$x_1 = \Big[\cos(p_2-p_1)\frac{t}{2} \bullet \cos(p_1+p_2)\frac{t}{2}\Big]$

$x_2 = \Big[\sin(p_2-p_1)\frac{t}{2} \bullet \sin(p_2+p_1)\frac{t}{2}\Big]$

These may be interpreted as a higher frequency oscillation at the average of the two natural frequencies with a variable amplitude, given by a lower frequency given by $\frac{1}{2}$ the difference in natural frequencies.

If $p_1$ and $p_2$ are close to each relative to their magnitude then $(p_2-p_1) << p_1$ , the motion becomes:

Page Comments

Daniel Carratalá Climent says:

August 15, 2023 at 12:14 pm

if sinx(x)=-cos(x+pi/2)
Could it be x1 and x2 negative?

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Daniel Carratalá Climent says:

August 15, 2023 at 12:16 pm

Sorry
sin(x)=-cos(x+pi/2)

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Daniel Carratalá Climent says:

August 15, 2023 at 12:38 pm

Ok.I’m wrong
sin(x+pi/2)=cos(x)
Could it be x1 and x2 were positive
Thanks a lot for the classes.
Best regards

Reply

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Review of single and multi-degree of freedom (mdof) systems: Vibration and Applications of MDOF Systems

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