Orthogonality of modes and modal analysis

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.