Linear Spring System - Equations of Motion

[KleptoBear]

This is not exactly a HW problem but a worked out problem that I am trying to understand. Below pictures are from Hutton's "Fundamentals of Finite Element Analysis"

Homework Statement

Derivation of system of equations for given figure.

Homework Equations

3. What confuses me

From the equilibrium condition $f_1+f_2=0$ I could have easily written $f_2=-f_1$ instead. This changes the signs of the diagonals. Yet, when I do this and solve a numerical example that is later done in the book on this same system, I get (expectedly!) same answers in magnitude but opposite in sign. Many books introduce FEM with similar examples and the five or six books that I have checked have the stiffness matrix exactly like above. Some do express the equilibrium condition as $f_2=-f_1$ but then write $\delta=u_1-u_2$ instead.
Is there a general convention that I am missing? I guess this may be a case of staring-at-what-you're-looking-for-but-not-seeing. Any help is appreciated. Thank you

 

[Valkarie]

. A:These types of problems can be tricky to reason through, so I would suggest that you think through the problem conceptually first and then draw out the equations. In this case, we are trying to develop the equations of motion for a 2-node spring system where the displacement of each node is given by $u_1$ and $u_2$. We can represent the displacements of each node in terms of force and stiffness as follows:$$f_1 - k(u_1-u_2) = 0 \\f_2 + k(u_1-u_2) = 0$$where $f_1$ and $f_2$ are the forces applied to the nodes, and $k$ is the stiffness of the spring connecting the nodes. Now, from the equilibrium condition we have that $f_1 + f_2 = 0$, which means that we can write the above equations as $$f_1 - ku_1 + ku_2 = 0 \\-f_1 - ku_1 + ku_2 = 0$$which gives us the following matrix equation:$$\begin{bmatrix}-k & k \\-k & k\end{bmatrix} \begin{bmatrix}u_1 \\ u_2\end{bmatrix}= \begin{bmatrix}f_1 \\ -f_1\end{bmatrix}$$