= Condensed mechanical load vector
{title2=$\widehat{\mathbf F}_r=\mathbf F_r-\mathbf D_{ri}\mathbf D_{ii}^{-1}\mathbf F_i$}
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 $\mathbf D_{ii}$,
$$
\widehat{\mathbf F}_r=\mathbf F_r-\mathbf D_{ri}\mathbf D_{ii}^{-1}\mathbf F_i,
$$
and the retained equation is $\widehat{\mathbf D}_{rr}\mathbf X_r=\widehat{\mathbf F}_r$. 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.<Structural Testing Part I: Mechanical Mobility Measurements by Ole Døssing>[Printed p.11, coupled dynamic models and boundary conditions; pp.14–16, force/response descriptions]{ref}
For an illustrative two-coordinate system at one frequency, let diagonal entries of $\mathbf D$ be175N/m and off-diagonal entries be $-100\,\mathrm{N/m}$. Eliminating coordinate2 gives
$$
\widehat F_1=F_1+\frac{100}{175}F_2,\qquad
\widehat D_{11}=175-\frac{100^2}{175}=\frac{825}{7}\,\mathrm{N/m}.
$$
If $F_1=0$ and $F_2=1\,\mathrm N$, then $\widehat F_1=4/7\,\mathrm N$ 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.
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