Mechanical dynamic condensation

Mechanical dynamic condensation eliminates selected coordinates from a frequency-dependent mechanical response while retaining their effect on the remaining coordinates. It is exact algebraic elimination at the chosen transform variable when the eliminated block is invertible. It is not equivalent to clamping those coordinates. Starting from the partitioned mechanical dynamic-stiffness matrix,solve the internal equation as ${\mathbf{X}}_i={\mathbf{D}}_{ii}^{-1}({\mathbf{F}}_i-{\mathbf{D}}_{ir}{\mathbf{X}}_r)$. Substitution givesThe effective matrix is a Schur complement; the right side is the condensed mechanical load vector. Both matter when the internal coordinate is externally forced. The coupled dynamic equations and measurement boundary conditions are the starting model, rather than independent scalar stiffness reciprocals.^[ref]

For the two-mass example at $\omega =5rad/s$, take $D_{11}=D_{22}=175N/m$ and $D_{12}=D_{21}=-100N/m$. With $F_1=0$ and $F_2=1\mathrm{N}$, eliminating coordinate2 gives effective stiffness $117.8571N/m$ and effective load $100/175\mathrm{N}$. Hence $X_1=100/20625\mathrm{m}\approx 4.848\mathrm{mm}$. Omitting the load correction would incorrectly predict no response at coordinate1.

For constant original mass matrix, damping and mechanical stiffness matrix, the condensed matrix is generally a rational function of frequency, not simply a smaller constant mass–damping–stiffness triple. A low-frequency approximation requires its own justification. The elimination can also become singular at an internal-block frequency even when the complete system matrix remains invertible. For example, $\left(\begin{array}{cc}0& -k_c\\ -k_c& 0\end{array}\right)$ has invertible full matrix for $k_c\ne 0$, but its lower-right scalar block cannot be inverted. One must then retain that coordinate or choose a different elimination.

The condensed mechanical load vector is the effective forcing on retained coordinates after mechanical dynamic condensation eliminates internal coordinates. It includes the effect of forces applied to the eliminated coordinates; it is not simply the retained subset of the original load vector. With partitioned mechanical dynamic-stiffness matrix $\mathbf{D}$ and invertible internal block ${\mathbf{D}}_{ii}$,and the retained equation is ${\hat{\mathbf{D}}}_{rr}{\mathbf{X}}_r={\hat{\mathbf{F}}}_r$. This follows by solving the internal block equation and substituting it into the retained equation, using the same Schur complement that constructs the reduced dynamic stiffness. Coupled force/response models require their boundary conditions to be stated.^[ref]

For an illustrative two-coordinate system at one frequency, let diagonal entries of $\mathbf{D}$ be175N/m and off-diagonal entries be $-100N/m$. Eliminating coordinate2 givesIf $F_1=0$ and $F_2=1\mathrm{N}$, then ${\hat{F}}_1=4/7\mathrm{N}$ and $X_1=100/20625\mathrm{m}$. Discarding the eliminated-coordinate force would incorrectly remove this response.

Conversely, choose simultaneous real complex amplitudes $F_1=1\mathrm{N}$ and $F_2=-1.75\mathrm{N}$. The condensed load vanishes and $X_1=0$, even though neither applied force is zero. The recovered internal displacement is $X_2=-0.010\mathrm{m}$, which satisfies both original equations. This is a frequency-specific response cancellation, rather than a vanishing total physical force: $F_1+F_2=-0.75\mathrm{N}$.

Because the elimination uses ${\mathbf{D}}_{ii}^{-1}(s)$, the effective load can depend on frequency and phase even when the original force amplitudes are simple. An arbitrary change of load or frequency therefore requires recomputing that transformation; a reduced matrix alone does not preserve every forced-response problem.