Elements in stiffness matrix I do not know if I am at the right address here, but I'll just ask.
Is the following correct?
Every element in the stiffness matrix represents the displacement of every element, when exerting an force on one of the elements (i.e. element (2,1) of the stiffness matrix gives the force on mass 2, for a displacement on mass 1).
Can I say this?
For example:
$$Mx'' + Kx = F$$
gives,
$$\begin{align}
M_{11} x_1'' + M_{12} x_2'' + K_{11} x_1 + K_{12} x_2 &= F_1
\\
M_{21} x_1'' + M_{22} x_2'' + K_{21} x_1 + K_{22} x_2 &= F_2
\end{align}$$
So then, $K_{21}$ gives the force on mass 2, for an input displacement on mass 1 right?
Thanks for helping :)
 A: Stiffness coefficient $\kappa_{ij}$ is static force $F_i$ required per displacement $x_j$ while keeping all other displacements $x_k$ zero. Note that condition. This usually implies that there are balancing Forces $F_k\neq 0$ required at other nodes. That is, if you multiply $K$ with the displacement vector $(0,\ldots,0,x_j,0,\ldots,0)$, you get several nonzero entries $(F_1,\ldots,F_n)$ in the resulting force vector, one of which is $F_i$.
If you want displacement per force, you would have to invert $K$. But that is not possible in general because there are usually some combinations of displacements that do not represent actual distortion or dilatations, but combinations of rigid shifts or rotations that can be performed by a rigid body as well. This means that $K$ must map some displacement vectors to a zero force vector, and therefore the linear system $f = K\,x$ is not invertible in general. To overcome this, you have to add boundary conditions that fix the node positions $x$ when there are no forces applied.
