Participation Factor And Effective Mass

The modal mass for mode $i$ is defined as

$${\widehat{m}}_{i} = {\boldsymbol{\varphi}_{i}}^{T}\mathbf{M}\boldsymbol{\varphi}_{i}$$

The direction information can be extracted using the participation factor. The participation factor for mode $i$ in the $j$ direction is given by

$$\Gamma_{ij} = \frac{{\boldsymbol{\varphi}_{i}}^{T}\mathbf{M}\mathbf{r}_{j}}{{\widehat{m}}_{i}}$$

where the $\mathbf{r}_{j}$ vector is a rigid body vector in the $j$ direction. The effective mass is similar but defined as

$${\widetilde{m}}_{ij} = \frac{\left( {\boldsymbol{\varphi}_{i}}^{T}\mathbf{M}\mathbf{r}_{j} \right)^{2}}{{\widehat{m}}_{i}} = {\widehat{m}}_{i}{\Gamma_{ij}}^{2}$$

The rigid body vectors are defined as

$$r_{x} = \begin{Bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ \vdots \\ \end{Bmatrix},r_{y} = \begin{Bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ \vdots \\ \end{Bmatrix},...$$

So a rigid body vector for a rotation of $\theta$ about global z can be defined as

$$r_{\theta} = \begin{Bmatrix} \cos\theta \\ \sin\theta \\ 0 \\ 0 \\ 0 \\ 0 \\ \cos\theta \\ \sin\theta \\ \vdots \\ \end{Bmatrix}$$

The effective mass in a rotated axis system can be calculated from the participation factors and effective masses.

$$\begin{aligned}\Gamma_{i\theta} &= \frac{{\boldsymbol{\varphi}_{i}}^{T}\mathbf{M}\mathbf{r}_{\theta}}{{\widehat{m}}_{i}} \\ &= \frac{{\boldsymbol{\varphi}_{i}}^{T}\mathbf{M}\mathbf{r}_{x}}{{\widehat{m}}_{i}}\cos\theta + \frac{{\boldsymbol{\varphi}_{i}}^{T}\mathbf{M}\mathbf{r}_{y}}{{\widehat{m}}_{i}}\sin\theta \\ &= \Gamma_{ix}\cos\theta + \Gamma_{iy}\sin\theta\end{aligned}$$

So

$$\begin{aligned}{\widetilde{m}}_{i\theta} &= \widehat{m}{\Gamma_{i\theta}}^{2} \\ &= \widehat{m}\left( \Gamma_{ix}\cos\theta + \Gamma_{iy}\sin\theta \right)^{2}\end{aligned}$$

The sum of the effective mass in any given direction over all the modes is the total unrestrained mass. Staring with the definition of effective mass

$${\widetilde{m}}_{ij} = \frac{\left( {\boldsymbol{\varphi}_{i}}^{T}\mathbf{M}\mathbf{r}_{j} \right)^{2}}{{\widehat{m}}_{i}}$$

The rigid body vector can be written as

$$\mathbf{r}_{j} = \mathbf{\Phi}\mathbf{a}_{j}$$

So the term in the numerator of the effective mass becomes ${\boldsymbol{\varphi}_{i}}^{T}\mathbf{M}\boldsymbol{\Phi}\mathbf{a}_{j}$ so

$${\widetilde{m}}_{ij} = \frac{\left( {\boldsymbol{\varphi}_{i}}^{T}\mathbf{M}\boldsymbol{\Phi}\mathbf{a}_{j} \right)^{2}}{{\widehat{m}}_{i}} = \frac{\left( {\widehat{m}}_{i}\mathbf{a}i_{j} \right)^{2}}{{\widehat{m}}_{i}} = {\widehat{m}}_{i}{\mathbf{a}_{ij}}^{2}$$

Also the total mass ${\mathbf{r}_{j}}^{T}\mathbf{M}\mathbf{r}_{j} = {\mathbf{a}_{j}}^{T}\boldsymbol{\Phi}^{T}\mathbf{M}\boldsymbol{\Phi}\mathbf{a}_{j}$ and $\boldsymbol{\Phi}^{T}\mathbf{M}\boldsymbol{\Phi} = diag\left( {\widetilde{m}}_{i} \right)$ so

$${\mathbf{a}_{j}}^{T}\mathbf{\Phi}^{T}\mathbf{M}\boldsymbol{\Phi}\mathbf{a}_{j} = \sum_{i}^{}{\widetilde{m}}_{i}{a_{ij}}^{2}$$

So the sum of the effective masses over all the modes is the total unrestrained mass.