If you found the logic of how I developed the element stiffness matrix in this lecture hard to follow, or it just simply didn't make much sense to you, here's an alternative approach that get's you to the same place.

Let’s say our bar is pinned at node 2, so no displacement at end 2. When we apply a force $F_1$ to end 1, we observe a displacement of $\delta_{x1}$. The force-displacement relationship at end one is therefore,

Now, since the bar is in equilibrium, we know the reaction force at end 2, $F_2$, is the equal to the negative of $F_1$,

Therefore, we can say,

With equations 1 and 2 put to one side, let’s turn our attention to the other end of the element, node 2. Now imagine node one is pinned and a force is applied at node 2. We can run through the exact same steps to construct equations for $F_1$ and $F_2$ in this scenario too.

The force-displacement relationship at node 2 is,

And again, since the bar is in equilibrium, we can say $F_1 = -F_2$, giving us,

Finally, in the case the neither end is pinned and both ends of the bar undergo axial displacements $\delta_{x1}$ and $\delta_{x2}$, we can obtain the corresponding forces at each end using superposition of equations 1 and 4 for $F_1$ and 2 and 3 for $F_2$,

or, in matrix form,

This gives us the same force-displacement relationship, with the familiar element stiffness matrix emerging, via a slightly different storyline!