Vibration of Multi Degree of Freedom Systems

Equations of motion

A system of three masses $m_1, m_2, m_3$ in a line, connected by springs $k_1$ (wall to $m_1$), $k_2$ and $k_3$ in series, has

$$\begin{bmatrix} m_1 & 0 & 0 \\ 0 & m_2 & 0 \\ 0 & 0 & m_3 \end{bmatrix}\begin{Bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \\ \ddot{x}_3 \end{Bmatrix} + \begin{bmatrix} k_1+k_2 & -k_2 & 0 \\ -k_2 & k_2+k_3 & -k_3 \\ 0 & -k_3 & k_3 \end{bmatrix}\begin{Bmatrix} x_1 \\ x_2 \\ x_3 \end{Bmatrix} = \begin{Bmatrix} F_1 \\ F_2 \\ F_3 \end{Bmatrix}$$

i.e. $[M]\{\ddot{x}\} + [K]\{x\} = \{F\}$.

Influence coefficients

• Stiffness influence coefficient $k_{ij}$: the force at coordinate $i$ required to produce a unit displacement at coordinate $j$, with all other displacements held at zero.
• Flexibility influence coefficient $a_{ij}$: the displacement at coordinate $i$ due to a unit force at coordinate $j$ (other forces zero).

Therefore $\{F\} = [k]\{x\}$ and $\{x\} = [a]\{F\}$.

Relation between them

Substituting $\{x\} = [a]\{F\}$ in $\{F\} = [k]\{x\}$ gives $\{F\} = [k][a]\{F\}$ for every $\{F\}$, so

$$[k][a] = [I] \;\Rightarrow\; [a] = [k]^{-1}$$

The flexibility matrix exists only if the system is restrained (no rigid-body motion). For an unrestrained system $[k]$ is singular.

Maxwell's reciprocity theorem

Statement: The displacement at point $i$ due to a unit load at point $j$ equals the displacement at $j$ due to a unit load at $i$: $a_{ij} = a_{ji}$ (so $[a]$ and $[k]$ are symmetric).

Proof: Apply load $P_i$ first and then $P_j$. The work done (strain energy stored) is

$$U_1 = \tfrac12P_i(a_{ii}P_i) + \tfrac12P_j(a_{jj}P_j) + P_j(a_{ji}P_i)$$

(the last term: $P_j$ moves through the extra deflection at $j$ caused by $P_i$.) Now apply the loads in the reverse order:

$$U_2 = \tfrac12P_j(a_{jj}P_j) + \tfrac12P_i(a_{ii}P_i) + P_i(a_{ij}P_j)$$

The final state is the same and the energy stored in a linear elastic system does not depend on the order of loading, so $U_1 = U_2$, giving $a_{ji} = a_{ij}$.

Data: $m_1 = m_2 = m_3 = 10$ kg, $k = 100\,000$ N/m.

Mass and stiffness matrices

$$[M] = \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix}, \qquad [K] = 10^5\begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{bmatrix}\ \text{N/m}$$

Eigenvalue problem

$$\left([K] - \omega^2[M]\right)\{X\} = \{0\}$$

Let $\lambda = \dfrac{m\omega^2}{k} = \dfrac{10\omega^2}{10^5}$. The characteristic equation $\det\left([K] - \omega^2[M]\right) = 0$ becomes

$$\begin{vmatrix} 2-\lambda & -1 & 0 \\ -1 & 2-\lambda & -1 \\ 0 & -1 & 1-\lambda \end{vmatrix} = 0$$ $$(2-\lambda)\left[(2-\lambda)(1-\lambda) - 1\right] - (1-\lambda) = 0 \;\Rightarrow\; \lambda^3 - 5\lambda^2 + 6\lambda - 1 = 0$$

Solving the cubic (Newton or trial):

$$\lambda_1 = 0.1981,\quad \lambda_2 = 1.5550,\quad \lambda_3 = 3.2470$$

Natural frequencies

$$\omega = \sqrt{\lambda\,\frac{k}{m}} = \sqrt{10000\lambda}$$

Mode shapes

From the first row, $X_2 = (2-\lambda)X_1$; from the second row, $X_3 = (2-\lambda)X_2 - X_1$.

 Mode 1: all in phase, no node
 Mode 2: one node (between m2 and m3)
 Mode 3: two nodes

Orthogonality check (modes 1 and 2)

$$\{X\}_1^T[M]\{X\}_2 = 10\left[(1)(1) + (1.802)(0.445) + (2.247)(-0.802)\right] = 0.000 \approx 0\ \ \checkmark$$

Answer: $\omega_1 = 44.5$, $\omega_2 = 124.7$, $\omega_3 = 180.2$ rad/s; mode shapes as tabulated; orthogonality verified.

Orthogonality

Let $\omega_r, \{\phi\}_r$ and $\omega_s, \{\phi\}_s$ be two natural frequencies with their mode shapes, with $\omega_r \ne \omega_s$. Then

$$[K]\{\phi\}_r = \omega_r^2[M]\{\phi\}_r \quad (1), \qquad [K]\{\phi\}_s = \omega_s^2[M]\{\phi\}_s \quad (2)$$

Pre-multiply (1) by $\{\phi\}_s^T$ and (2) by $\{\phi\}_r^T$:

$$\{\phi\}_s^T[K]\{\phi\}_r = \omega_r^2\{\phi\}_s^T[M]\{\phi\}_r, \qquad \{\phi\}_r^T[K]\{\phi\}_s = \omega_s^2\{\phi\}_r^T[M]\{\phi\}_s$$

Since $[K]$ and $[M]$ are symmetric, the left-hand sides are equal (transpose of a scalar) and so are the mass products. Subtracting:

$$(\omega_r^2 - \omega_s^2)\{\phi\}_r^T[M]\{\phi\}_s = 0 \;\Rightarrow\; \{\phi\}_r^T[M]\{\phi\}_s = 0 \quad (r \ne s)$$

and hence also $\{\phi\}_r^T[K]\{\phi\}_s = 0$ for $r \ne s$.

For $r = s$ the products are the modal mass $M_r = \{\phi\}_r^T[M]\{\phi\}_r$ and modal stiffness $K_r = \{\phi\}_r^T[K]\{\phi\}_r = \omega_r^2M_r$. Normalising so that $M_r = 1$ (mass-normalised) gives $[\Phi]^T[M][\Phi] = [I]$ and $[\Phi]^T[K][\Phi] = \text{diag}(\omega_r^2)$.

Modal analysis for forced response

For $[M]\{\ddot{x}\} + [K]\{x\} = \{F(t)\}$:

1. Find the natural frequencies $\omega_r$ and the mode-shape matrix $[\Phi]$ (columns normalised to unit modal mass).
2. Use the transformation $\{x\} = [\Phi]\{q\}$, where $\{q\}$ are the principal coordinates.
3. Substitute and pre-multiply by $[\Phi]^T$. By orthogonality the equations uncouple:

$$\ddot{q}_r + \omega_r^2q_r = Q_r(t), \qquad Q_r(t) = \{\phi\}_r^T\{F(t)\}$$

4. Solve each single-degree-of-freedom equation, for example with the Duhamel (convolution) integral:

$$q_r(t) = q_r(0)\cos\omega_rt + \frac{\dot{q}_r(0)}{\omega_r}\sin\omega_rt + \frac{1}{\omega_r}\int_0^tQ_r(\tau)\sin\omega_r(t-\tau)\,d\tau$$

5. Transform back: $\{x(t)\} = [\Phi]\{q(t)\} = \sum_r\{\phi\}_rq_r(t)$.

For harmonic force $Q_r\sin\omega t$, the steady-state result is $q_r = \dfrac{Q_r}{\omega_r^2 - \omega^2}\sin\omega t$. Often only the first few modes are needed, which reduces the size of the problem greatly. With proportional damping ($[C] = \alpha[M] + \beta[K]$) the same method works with a damping ratio $\zeta_r$ for each mode.